Exam 2024_retake.txt — openai/gpt-oss-120b

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Quiz: Reguläre und kontextfreie Sprachen

**Teilaufgaben (a-e):** Für diese Fragen ist eine Begründung nicht gefordert. Sie erhalten die Punkte auf eine Teilaufgabe genau dann, wenn Sie alle Antwortmöglichkeiten korrekt angekreuzt haben.
**Es ist immer mindestens eine Antwortmöglichkeit richtig.**

In dieser Aufgabe verwenden wir durchgehend das Alphabet $\Sigma := \{a, b\}$.

## subproblem a (2.0 points)
Welche der folgenden Sprachen sind Residualsprachen von $L((bba)^*b)$?

<options>
<option id="1">$\{b\}$</option>
<option id="2">$L(ba(bba)^*b | \varepsilon)$</option>
<option id="3">$\emptyset$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

We need to determine which of the given languages are residual (i.e., left quotients) of L = L((bba)^* b). Language defined by regex (bba)^* b. Over alphabet {a,b}. Let's compute language: strings consisting of zero or more repetitions of "bba" followed by a final "b". So strings: b, (bba)b = bbab, (bba)(bba)b = bba bba b = bbabbab, etc. Formally: (bba)^k b for k >=0. So set = { (bba)^k b | k≥0 }.

Residual languages (left quotients) are: for any w ∈ Σ*, w^{-1} L = { x | wx ∈ L }. For each w there is a residual; possibly duplicates. We need to check which of the given languages are among them.

List options:

1. {b}. Is {b} a residual? Need w such that {x | wx ∈ L} = {b}. That means exactly x = b works. So for some prefix w, the only continuation that yields a word in L is b. Let's test: suppose w = (bba)^k . Then we need (bba)^k x ∈ L iff x = b? Let's compute: L consists of strings (bba)^m b. If we have w = (bba)^k, then w x = (bba)^k x need to be of form (bba)^m b. So x must equal (bba)^{m-k} b, with m>=k. So set of x is { (bba)^{t} b | t>=0 }. That's not just {b} (except maybe when k maximal?). If k is such that any extra (bba) not allowed? If we pick w = (bba)^k b? Let's test w = (bba)^k b. Then w x = (bba)^k b x. To be in L, the word must end with final b and previous pattern (bba)^*. But after that b, any extension x must produce a string maybe if x empty? Actually after a word already in L, extending with any non-empty string cannot stay in L because L strings always end with b and have pattern ...b at end but extra characters would break pattern. The only way w x ∈ L is if x = ε, because adding more characters would produce longer strings not in L (since the pattern is exactly (bba)^* b). So residual for w = (bba)^k b is { ε }. Not {b}.

What about w = (bba)^k a? Not possible as prefix? Let's test w = (bba)^k b a? Not sure. But we need residual {b}. Could there be w such that only continuation b yields a word in L? Let's think: If w is such that after w we need to complete to a word that ends with b and possibly some preceding (bba) repetitions. The set of possible completions is either empty, {ε}, or set of strings of form (bba)^{t} b. The smallest non-empty set is {ε}. {b} is not of that form unless we allow t = 0 and preceded by something? Actually (bba)^{t} b with t=0 gives b. So residual could be {b} if w is such that completion must be exactly b. Means after w, we need exactly b and cannot have extra (bba). That would be if w = (bba)^k a? Let's test: Suppose w ends with an 'a' that is the second a in 'bba' sequence? The pattern is ...b b a ... So after 'a', the next required symbols to finish pattern maybe b? Actually pattern (bba)^* b: segments of "b b a". So after a, you are at end of a segment. The next needed is either start of next segment with 'b' (if more repeats) or final b if no more repeats. So completions are either just 'b' (to finish) or 'b b a ... b' (more). So residual after w where w ends with a last character being a (i.e., you have consumed a complete "bba" and are ready to either stop or continue), the set of completions is {b} ∪ { (b b a)^{t} b | t >=1 }. That's {b} plus longer strings. So not just {b}.

Thus {b} is not a residual? Wait the residual after w = (bba)^k b b a? Let's think: w = (bba)^k b b a yields after an 'a', same as above.

Thus only possible residual sets are: ∅, {ε}, and { (bba)^{t} b | t >=0 } = L itself. Also maybe {b}? Let's confirm: Consider w = (bba)^k b b. That is after two b's of a segment but before a. The completions: need an 'a' then maybe continue. So set = { a (bba)^{t} b | t >=0 }. That's not {b}.

Thus {b} not a residual. Option 1 = 0.

Option 2: $L(ba(bba)^*b \mid \varepsilon)$. This language: either ε or strings of form ba (bba)^* b. Let's examine if this is a residual. Possibly after prefix 'b'? Let's compute w = 'b'. Then we need completions x such that 'b' x ∈ L. L strings are (bba)^k b. Starting with 'b'. For k=0, word is 'b' alone; for k≥1, first three chars are b b a. So if w='b', then x must be either ε (to get 'b') or start with 'b a ...'? Wait if we have w='b', then we need to complete to (bba)^k b. If k=0, then x = ε works. If k≥1, then remaining pattern is (b b a)^{k-1} (b b a) b? Actually overall pattern: (b b a)^k b. After consuming first 'b', left to produce (b b a)^{k-1} b a ? let's compute: The full string: (b b a)^k b = b (b a)?? Not simple. Better: For k=1, string = b b a b. Starting with 'b', remaining "b a b". That's "b a b". For k=2, string = b b a b b a b -> after first b, remaining "b a b b a b". So completions after leading b are strings that start with b a ... and end with b. That set is exactly: either ε (k=0) or b a (b b a)^* b? Let's test: For k=1: remaining = b a b = "b a b". That's b a ( (b b a)^0 ) b -> correct. For k=2: remaining = b a b b a b = b a (b b a) b. Yes pattern b a (b b a)^{k-1} b. So completions set = { ε } ∪ { b a (b b a)^{t} b | t ≥0 }. That's exactly language described in option 2: ε ∪ ba (bba)^* b. Yes matches.

Thus residual after w='b' is exactly that language. So option 2 is a residual => 1.

Option 3: ∅. Is empty set a residual? There exists w such that no continuation yields a word in L. For any w not a prefix of any word in L, the residual is ∅. E.g., w = 'a' (since L strings start with b). 'a' cannot be prefix of any word in L (all strings start with b). So w='a' leads to empty residual. So ∅ is a residual. Hence option 3 =1.

Thus answers:

<option id="1">0</option>
<option id="2">1</option>
<option id="3">1</option>

<option id="1">0</option>
<option id="2">1</option>
<option id="3">1</option>

Welche der folgenden Sprachen sind Residualsprachen von $L((bba)^*b)$?

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Quiz: Reguläre und kontextfreie Sprachen

**Teilaufgaben (a-e):** Für diese Fragen ist eine Begründung nicht gefordert. Sie erhalten die Punkte auf eine Teilaufgabe genau dann, wenn Sie alle Antwortmöglichkeiten korrekt angekreuzt haben.
**Es ist immer mindestens eine Antwortmöglichkeit richtig.**

In dieser Aufgabe verwenden wir durchgehend das Alphabet $\Sigma := \{a, b\}$.

## subproblem a (2.0 points)
Welche der folgenden Sprachen sind Residualsprachen von $L((bba)^*b)$?

<options>
<option id="1">$\{b\}$</option>
<option id="2">$L(ba(bba)^*b | \varepsilon)$</option>
<option id="3">$\emptyset$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">0</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem b (2.0 points)
Sei $L \subseteq \Sigma^*$ eine Sprache mit $LL = \Sigma^*$. Welche Aussagen sind wahr?

<options>
<option id="1">$L$ ist unendlich</option>
<option id="2">$\varepsilon \in L$</option>
<option id="3">$LLL = \Sigma^*$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

We need to answer subproblem b. We have language L over {a,b} with property LL = Σ* (the concatenation of L with itself yields all strings). Determine which statements are true: (1) L is infinite. (2) ε ∈ L. (3) LLL = Σ*.

We must think: L subset of Σ* such that L·L = Σ*.

First consider possibilities. Σ* = all strings over a,b, including ε. If LL = Σ*, then for any string w ∈ Σ*, there exist x,y ∈ L such that w = xy. Does that imply L is infinite? Likely yes because Σ* is infinite. If L were finite, the concatenation of two finite sets yields a finite set. Finite×finite -> finite. So LL would be finite, cannot be Σ* (which is infinite). Therefore L must be infinite. So statement 1 true (1).

Statement 2: ε ∈ L? If ε ∉ L, could LL still be Σ*? Let's see. Choose any w in Σ*. We need x,y in L with xy = w. If ε not in L, then any factorization must have both non-empty. Is it possible to cover all strings without ε? For example, take L = Σ+ (all non-empty strings). Then L·L = Σ+ Σ+ = all strings of length at least 2? Actually concatenation of two non-empty strings yields strings of length at least 2, but also could produce any length ≥2. Strings of length 0 or 1 cannot be formed. So LL would not be Σ*. So need ε perhaps in L to get length 0 strings? To get ε as a product, need x = ε and y = ε, but if ε ∈ L then ε∈LL. Without ε, can't get ε in LL. But we need LL = Σ*, which includes ε. Therefore ε must be in LL; need some x,y ∈ L such that xy = ε. Only possibility is both ε. So ε must be in L. Hence statement 2 true.

Statement 3: LLL = Σ*? Since LL already equals Σ*, then LLL = (LL)L = Σ* L = Σ* (since Σ* concatenated with any subset yields Σ*). Indeed Σ*·L = Σ* (because for any w ∈ Σ*, you can write w = uv where u ∈ Σ*, v ∈ L? Wait concatenation Σ*·L = {xy | x∈Σ*, y∈L}. This set is Σ* because you can always take x = w and y = ε if ε∈L; but if ε∈L, yes. But is ε definitely in L? We argued yes, so Σ*·L = Σ*. Thus LLL = Σ*. So statement 3 true. So all three true => 1.

Thus output all 1.

<option id="1">1</option> 
<option id="2">1</option> 
<option id="3">1</option>

Sei $L \subseteq \Sigma^*$ eine Sprache mit $LL = \Sigma^*$. Welche Aussagen sind wahr?

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Quiz: Reguläre und kontextfreie Sprachen

**Teilaufgaben (a-e):** Für diese Fragen ist eine Begründung nicht gefordert. Sie erhalten die Punkte auf eine Teilaufgabe genau dann, wenn Sie alle Antwortmöglichkeiten korrekt angekreuzt haben.
**Es ist immer mindestens eine Antwortmöglichkeit richtig.**

In dieser Aufgabe verwenden wir durchgehend das Alphabet $\Sigma := \{a, b\}$.

## subproblem a (2.0 points)
Welche der folgenden Sprachen sind Residualsprachen von $L((bba)^*b)$?

<options>
<option id="1">$\{b\}$</option>
<option id="2">$L(ba(bba)^*b | \varepsilon)$</option>
<option id="3">$\emptyset$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">0</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem b (2.0 points)
Sei $L \subseteq \Sigma^*$ eine Sprache mit $LL = \Sigma^*$. Welche Aussagen sind wahr?

<options>
<option id="1">$L$ ist unendlich</option>
<option id="2">$\varepsilon \in L$</option>
<option id="3">$LLL = \Sigma^*$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option> 
<option id="2">1</option> 
<option id="3">1</option>

user:
## subproblem c (2.0 points)
Sei $L \subseteq \Sigma^*$ regulär. Welche der folgenden Sprachen sind regulär?

<options>
<option id="1">$\{u^R v \mid u, v \in L\}$</option>
<option id="2">$\{u^R u \mid u \in L\}$</option>
<option id="3">$\{uv \mid u, v \in L \land |u| = |v|\}$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

We need to analyze each subproblem.

Subproblem a: Determine which languages are residual languages (left quotients) of L = L((bba)^* b). So L = language of regex (bba)^* b over alphabet {a,b}. Let's compute L set: strings that consist of zero or more repetitions of "bba" followed by a final "b". So strings: b, (bba)b = bbab, (bba)(bba)b = bbabba b? Wait compute: (bba)^2 b = bba bba b = b b a b b a b => "bbab bab"? Actually "bba bba b" = "bbab bab"? Let's just think: pattern is (bba)^k b for k>=0. So strings length: each "bba" contributes 3 chars, plus final "b". So lengths: 1 (k=0), 4 (k=1), 7 (k=2), etc. The strings are: for k=0: "b". k=1: "bba"+"b" = "bbab". k=2: "bba"+"bba"+"b" = "bbabbab". So basically "b" followed by k copies of "ba"? Not exactly. Let's write pattern: (bba)^k = (bb a)^k . Then plus final b. So final strings end with "b". They have pattern: b (ba b?) Wait the base is "b" then each extra iteration adds "ba b"? Let's compute: Suppose we have k=0: "b". For k=1: "bba"+"b" = "bba b". That's "bbab". For k=2: "bba bba b" = "bba bba b". That's "bbab bab". So strings can be seen as "b (ba b)^{k}"? Let's check: "b (ba b)" = "b"+"bab" = "bbab". That's consistent for k=1. For k=2: "b (ba b)^2" = "b"+"bab"+"bab" = "bbab bab". Indeed matches. So L = { b (bab)^k | k >= 0 }. Good.

Now residual languages: For any word w, residual w^{-1} L = { x | wx ∈ L }. The set of all such residual languages (quotients) of L.

We need to see which of the given languages equal some residual.

Options:

1) {b}. Is there w such that w^{-1}L = {b} ?

Check if the set {b} can be a quotient. To have only b as possible continuation. So need w such that concatenating w with b is in L, but any other continuation not in L.

Let w be something that forces remaining string must be exactly "b". Since L strings all end with b, and have certain structure. Could w be prefix that leaves exactly one possible suffix "b". Let's see L strings start with b. If we choose w = (bba)^k (maybe) . Then w^{-1} L = suffixes such that w + suffix ∈ L, i.e., suffix must be the remaining part to complete pattern. If w is a prefix of some string in L. For w = b (the first character). Then w^{-1}L = { ε? Actually after consuming b, remaining strings that make full string in L are (bab)^k . Since L = b (bab)^*. So b^{-1} L = (bab)^*. That's includes ε, "bab", "babbab"... Not {b}.

If w = "bb"? Not prefix? Let's see if "bb" is possible prefix? Strings start "b". The second character could be either 'b' (if k>=1) else directly final b? But after first b, next char could be 'b' if there is a 'b' from "bab" pattern? Actually after first b, the string may be just end (if k=0) i.e., "b". There's no second character. So "bb" is not a prefix of any string in L. Then w^{-1}L = empty set, because there is no string x with w x ∈ L. So option 3 says ∅ is residual, indeed empty language is residual for any non-prefix w. So option 3 likely true.

Option 2: L(ba(bba)^*b | ε) . Let's parse: Language given: L( ba (bba)^* b | ε ). So it's either ε or strings that start with ba then (bba)^* b. That is { ε } ∪ { ba (bba)^* b }. Let's see if this equals some residual.

Consider w = "b". Then w^{-1} L = (bab)^*. Not the same. Not include ε? (bab)^* includes ε, but also includes strings like "bab". Does (bab)^* equal the given set? Let's examine given set: strings of form ba (bba)^* b . Let's compute sample: ba + (bba)^0 b = ba b = "bab". For (bba)^1 b: ba bba b = "bab bab"? Actually "ba"+"bba"+"b" = "bab bab". That's "babbab". So set is { "bab", "babbab", "babbabbab", ...} plus ε. This matches (bab)^*? Let's compute (bab)^* = sequences of "bab". So strings: ε, "bab", "babbab", "babbabbab", ... Indeed same. So option 2 is exactly (bab)^*, which equals b^{-1}L as we found. So option 2 is a residual.

Option 1 {b}: Is there w where residual is exactly {b}? Let's try w = ??? Suppose w = "b a"? Wait if w = "ba"? Then w^{-1}L = set of x such that "ba"+"x" ∈ L. We need strings where prefix is "ba". In L, strings after first "b" possibly "a"? No, after 'b' next could be 'a' only in pattern "bab" (since after b we need 'a' then 'b'). So "ba" is prefix of strings where after that we have "b" then maybe more pattern. So "ba"+x ∈ L => "bab"+rest must be in L. L = b(bab)^*. So "ba"+x must be equal to b(bab)^*. But L strings start with b. If we already have "ba", the first char is b, second a, fine. So remaining part must be suffix of some L after "ba". Let's see strings in L: b (bab)^k . After initial "b", the remainder is (bab)^k . Starting with "a"? The first character of (bab)^k is 'a' if k>0 (since "bab" starts with 'b'? wait 'bab' starts with b. Actually "bab": b a b. Starting with b. Wait "bab" starts with b, not a. So after initial b, we have (bab)^k which start with b if k>0. Thus L strings after first b always have next char b (if k>0) or empty (if k=0). So "ba" never occurs as a prefix (except maybe not). Indeed "ba" is not a prefix of any string in L because after b, either end or b. So "ba" can't be prefix. Therefore residual for w = "ba" is empty.

Thus {b} not a residual. So answer: option1 0, option2 1, option3 1. This matches assistant's first answer (option1 0, option2 1, option3 1). Good.

Now subproblem b: L subset of Σ* with LL = Σ*. Need to evaluate statements.

Given any language L such that concatenation of L with itself yields all strings over {a,b}. What can we infer?

First, note that if LL = Σ*, then particularly ε (empty string) must be in Σ* obviously. Is there some x such that x ∈ LL? Means there exist u,v ∈ L with uv = ε, implying u = ε and v = ε, therefore ε must be in L. So statement 2 (ε ∈ L) must be true.

Is L infinite? Could L be finite and still LL = Σ*? Let's think. If L is finite, its concatenation yields a finite set of strings (the set of concatenations of two elements from a finite set) which would also be finite. Since Σ* is infinite, L must be infinite. So statement1 true.

Statement3: Does LL L = Σ*? If LL = Σ*, then (LL)L = Σ*L = Σ* (since Σ* concatenated with anything equals Σ*). More formally, Σ* L = Σ* (since for any string w, there exists w = uv with u ∈ Σ* and v ∈ L? Actually concatenation of Σ* with L yields Σ*, because Σ* already includes all strings, and concatenating any string from L after any string yields all strings again? Let's verify: Σ* L = { xy | x ∈ Σ*, y ∈ L }. Since Σ* includes all strings, for any string w ∈ Σ*, we can choose x = w and y = ε (if ε ∈ L). We have already shown ε ∈ L, so each w = w·ε is in Σ*L. Hence Σ* L = Σ*. Therefore LLL = (LL)L = Σ* L = Σ*. So statement3 true.

Thus all three statements true: answer 1.

Thus assistant answered all 1? He gave 1 for all 3, which matches.

Now subproblem c: L regular. Determine which of the following languages are regular.

1) { u^R v | u,v ∈ L }. That's set of strings that are reverse of some u in L concatenated with any v from L. Since regular languages are closed under reversal and concatenation, we have L^R is regular (reversal). L^R concatenated with L: L^R L. Since L^R regular, L regular, concatenation of regular languages is regular. So language is regular. So option1 true.

2) { u^R u | u ∈ L }. That's set of palindrome formed by reverse of u plus u. This is like L intersected with its reverse? Wait it's not intersection. It's { u^R u }. This language is essentially L concatenated with its reversal but reversed order reversed? Actually we can express as { u^R u | u ∈ L } = { x y | x = u^R, y = u, u∈L }. That's equal to L^R ⋂?? Not exactly. This is a kind of "copy" language where we need matching halves, which may be non-regular in general even if L is regular. Example: If L = a^* b, then elements u are a^n b. Then u^R = b a^n. So strings are b a^n a^n b = b a^{2n} b. That's regular (since a^{2n} is a^*). But need counterexample where resulting language is non-regular. Take L = a^* (i.e., all a's). Then u = a^n, u^R = a^n (same). So strings = a^{2n}. That's regular (a^*). Hmm.

But we can use L = { a^n b^n | n ≥ 0 }? That is not regular. Not allowed. L must be regular. We need to find a regular L such that { u^R u } is not regular. Classic result: The set { w w^R | w ∈ Σ* } (palindromes of even length) is not regular. That's { u u^R } not { u^R u }. But reversing yields same set (since palindrome). Actually { u u^R } = { u^R u } (since all strings, reverse order same as palindrome). Not exactly; { u u^R } yields palindromes where first half is original. { u^R u } yields strings where second half is reversal of first? Wait if u = "ab", u^R = "ba", then u^R u = "baab". That's also palindrome? "baab" reversed = "baab". Yes it's palindrome. So both forms produce even-length palindromes. So language = set of even-length palindromes over Σ such that first half reversed appears. That's exactly the set of all even-length palindromes = { w w^R | w∈ Σ* }. This language is known not regular (requires matching). Therefore if we set L = Σ*, which is regular, then { u^R u | u ∈ Σ* } = set of even palindromes, not regular. Since L = Σ* is regular, the constructed language is not regular. Hence statement2 is false (not guaranteed regular).

Thus option2 not regular in general => answer 0.

3) { uv | u, v ∈ L and |u| = |v| }. This is set of concatenations of equal length strings from L. Is this regular? Not necessarily. Example L = Σ*. Then set = all strings where can split into two halves of equal length (any string of even length). That language is regular (even length). But need a counterexample where L regular but language not regular. For L = a^* b a^* (i.e., any string with exactly one b). Let's think: L = a^* b a^*. This is regular. Consider strings u,v each have exactly one b. Condition |u| = |v|. The concatenation uv will have exactly two b's. Also length constraint that the distance between b's is something? Could this be non-regular? Perhaps we can encode equality of lengths again.

Consider L = { a^n b | n ≥ 0 }. This is regular? Yes: a* b. Actually a* b is regular. For u = a^m b, v = a^n b, with |u| = |v| => m+1 = n+1 => m=n. So uv = a^n b a^n b. Language = { a^n b a^n b | n≥0 }. That's not regular (requires same n). This is like { a^n b a^n b } which is not regular (similar to a^n b a^n). Since can use pumping lemma. So language not regular. Thus option3 not guaranteed regular. So answer 0.

Thus final answers: option1 1, option2 0, option3 0.

Provide format.

<option id="1">1</option>
<option id="2">0</option>
<option id="3">0</option>

Sei $L \subseteq \Sigma^*$ regulär. Welche der folgenden Sprachen sind regulär?

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Quiz: Reguläre und kontextfreie Sprachen

**Teilaufgaben (a-e):** Für diese Fragen ist eine Begründung nicht gefordert. Sie erhalten die Punkte auf eine Teilaufgabe genau dann, wenn Sie alle Antwortmöglichkeiten korrekt angekreuzt haben.
**Es ist immer mindestens eine Antwortmöglichkeit richtig.**

In dieser Aufgabe verwenden wir durchgehend das Alphabet $\Sigma := \{a, b\}$.

## subproblem a (2.0 points)
Welche der folgenden Sprachen sind Residualsprachen von $L((bba)^*b)$?

<options>
<option id="1">$\{b\}$</option>
<option id="2">$L(ba(bba)^*b | \varepsilon)$</option>
<option id="3">$\emptyset$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">0</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem b (2.0 points)
Sei $L \subseteq \Sigma^*$ eine Sprache mit $LL = \Sigma^*$. Welche Aussagen sind wahr?

<options>
<option id="1">$L$ ist unendlich</option>
<option id="2">$\varepsilon \in L$</option>
<option id="3">$LLL = \Sigma^*$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option> 
<option id="2">1</option> 
<option id="3">1</option>

user:
## subproblem c (2.0 points)
Sei $L \subseteq \Sigma^*$ regulär. Welche der folgenden Sprachen sind regulär?

<options>
<option id="1">$\{u^R v \mid u, v \in L\}$</option>
<option id="2">$\{u^R u \mid u \in L\}$</option>
<option id="3">$\{uv \mid u, v \in L \land |u| = |v|\}$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">0</option>
<option id="3">0</option>

user:
## subproblem d (2.0 points)
Welche der folgenden kontextfreien Grammatiken sind mehrdeutig?

<options>
<option id="1">$S \rightarrow SS \mid a$</option>
<option id="2">$S \rightarrow Sa \mid Sb \mid \varepsilon$</option>
<option id="3">$S \rightarrow TT\quad T \rightarrow aT \mid bT \mid a \mid b$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

We need to decide which grammars are ambiguous. Grammar 1: S->SS | a. This generates all strings of a's with any binary tree decomposition; ambiguous because multiple parse trees for same string with more than one a? For string aaaa (4 a's) there are many ways to split. So ambiguous.

Grammar 2: S->Sa | Sb | ε. This generates all strings over {a,b} (any order) but right-recursive. It's unambiguous because each string has unique leftmost derivation? Actually S -> S a ... expands leftmost S repeatedly; but you can choose between Sa and Sb at each step; the sequence of choices determines the string uniquely, and no other parse tree yields same string because the recursion is right-linear. So unambiguous.

Grammar 3: S->TT, T->aT | bT | a | b. This generates strings that are concatenation of two nonempty strings over {a,b}. Since T generates any nonempty string. Is it ambiguous? Possibly yes because a string can be split into two parts in multiple ways, leading to different parse trees. For example string "ab": could be T -> a (first) and T -> b (second). That's unique. For longer strings, multiple splits yield ambiguity. Eg "aba": could split as "a"+"ba" or "ab"+"a". Both are possible. Thus ambiguous. So options 1 and 3 ambiguous.

Thus answer: 1=1,2=0,3=1.

<option id="1">1</option>
<option id="2">0</option>
<option id="3">1</option>

Welche der folgenden kontextfreien Grammatiken sind mehrdeutig?

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Quiz: Reguläre und kontextfreie Sprachen

**Teilaufgaben (a-e):** Für diese Fragen ist eine Begründung nicht gefordert. Sie erhalten die Punkte auf eine Teilaufgabe genau dann, wenn Sie alle Antwortmöglichkeiten korrekt angekreuzt haben.
**Es ist immer mindestens eine Antwortmöglichkeit richtig.**

In dieser Aufgabe verwenden wir durchgehend das Alphabet $\Sigma := \{a, b\}$.

## subproblem a (2.0 points)
Welche der folgenden Sprachen sind Residualsprachen von $L((bba)^*b)$?

<options>
<option id="1">$\{b\}$</option>
<option id="2">$L(ba(bba)^*b | \varepsilon)$</option>
<option id="3">$\emptyset$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">0</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem b (2.0 points)
Sei $L \subseteq \Sigma^*$ eine Sprache mit $LL = \Sigma^*$. Welche Aussagen sind wahr?

<options>
<option id="1">$L$ ist unendlich</option>
<option id="2">$\varepsilon \in L$</option>
<option id="3">$LLL = \Sigma^*$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option> 
<option id="2">1</option> 
<option id="3">1</option>

user:
## subproblem c (2.0 points)
Sei $L \subseteq \Sigma^*$ regulär. Welche der folgenden Sprachen sind regulär?

<options>
<option id="1">$\{u^R v \mid u, v \in L\}$</option>
<option id="2">$\{u^R u \mid u \in L\}$</option>
<option id="3">$\{uv \mid u, v \in L \land |u| = |v|\}$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">0</option>
<option id="3">0</option>

user:
## subproblem d (2.0 points)
Welche der folgenden kontextfreien Grammatiken sind mehrdeutig?

<options>
<option id="1">$S \rightarrow SS \mid a$</option>
<option id="2">$S \rightarrow Sa \mid Sb \mid \varepsilon$</option>
<option id="3">$S \rightarrow TT\quad T \rightarrow aT \mid bT \mid a \mid b$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">0</option>
<option id="3">1</option>

user:
## subproblem e (2.0 points)
Sei $A \subseteq \Sigma^*$ regulär und $B \subseteq \Sigma^*$ kontextfrei. Welche der folgenden Sprachen sind kontextfrei?

<options>
<option id="1">$A \cup \overline{B}$</option>
<option id="2">$\overline{A} \cup B$</option>
<option id="3">$AB$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

We must evaluate each subproblem.

We need to produce correct yes/no (1 true, 0 false) for each option.

First, we must review previous answers possibly wrong; but we now focus on subproblem e only.

We need to check each language: given regular A, context-free B.

Option 1: A ∪ complement of B. Is it necessarily context-free? B is CFL. Complement of a CFL is not necessarily CFL (CFL not closed under complement). However union of regular with complement of CFL: regular union anything? CFL class is closed under union with regular? Actually CFLs are closed under union with regular languages (since regular is CFL). But complement of B may not be CFL. So A ∪ complement B might not be CFL. Could be CFL in some cases but not guaranteed. So the statement "which of the following languages are context-free?" Likely they ask which are guaranteed to be context-free given any regular A and any CFL B. So we need to answer which are always CFL (i.e., for all such A,B).

Option 1: Not guaranteed. So answer 0.

Option 2: complement of A union B. Complement of regular is regular, and regular is CFL. Union of two CFLs is CFL (CFL closed under union). So complement A (regular) is regular, thus CFL. Union with B (CFL) yields CFL. So yes, guaranteed context-free. So answer 1.

Option 3: concatenation AB. Regular concatenated with CFL? CFL closed under concatenation with regular on either side? Yes, CFLs are closed under concatenation with regular languages (since we can build a PDA that first processes regular part then non-deterministically switches to CFL). Actually CFLs are closed under concatenation with regular languages on either side (regular*CFL and CFL*regular). So AB is CFL. So answer 1.

Thus final: option1 0, option2 1, option3 1.

Check nuance: In Option 1, A ∪ complement of B could be CFL? If B is CFL, complement may be not CFL; union with regular doesn't guarantee. So 0.

Thus produce answer.

<option id="1">0</option>
<option id="2">1</option>
<option id="3">1</option>

Sei $A \subseteq \Sigma^*$ regulär und $B \subseteq \Sigma^*$ kontextfrei. Welche der folgenden Sprachen sind kontextfrei?

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Quiz: Reguläre und kontextfreie Sprachen

**Teilaufgaben (a-e):** Für diese Fragen ist eine Begründung nicht gefordert. Sie erhalten die Punkte auf eine Teilaufgabe genau dann, wenn Sie alle Antwortmöglichkeiten korrekt angekreuzt haben.
**Es ist immer mindestens eine Antwortmöglichkeit richtig.**

In dieser Aufgabe verwenden wir durchgehend das Alphabet $\Sigma := \{a, b\}$.

## subproblem a (2.0 points)
Welche der folgenden Sprachen sind Residualsprachen von $L((bba)^*b)$?

<options>
<option id="1">$\{b\}$</option>
<option id="2">$L(ba(bba)^*b | \varepsilon)$</option>
<option id="3">$\emptyset$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">0</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem b (2.0 points)
Sei $L \subseteq \Sigma^*$ eine Sprache mit $LL = \Sigma^*$. Welche Aussagen sind wahr?

<options>
<option id="1">$L$ ist unendlich</option>
<option id="2">$\varepsilon \in L$</option>
<option id="3">$LLL = \Sigma^*$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option> 
<option id="2">1</option> 
<option id="3">1</option>

user:
## subproblem c (2.0 points)
Sei $L \subseteq \Sigma^*$ regulär. Welche der folgenden Sprachen sind regulär?

<options>
<option id="1">$\{u^R v \mid u, v \in L\}$</option>
<option id="2">$\{u^R u \mid u \in L\}$</option>
<option id="3">$\{uv \mid u, v \in L \land |u| = |v|\}$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">0</option>
<option id="3">0</option>

user:
## subproblem d (2.0 points)
Welche der folgenden kontextfreien Grammatiken sind mehrdeutig?

<options>
<option id="1">$S \rightarrow SS \mid a$</option>
<option id="2">$S \rightarrow Sa \mid Sb \mid \varepsilon$</option>
<option id="3">$S \rightarrow TT\quad T \rightarrow aT \mid bT \mid a \mid b$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">0</option>
<option id="3">1</option>

user:
## subproblem e (2.0 points)
Sei $A \subseteq \Sigma^*$ regulär und $B \subseteq \Sigma^*$ kontextfrei. Welche der folgenden Sprachen sind kontextfrei?

<options>
<option id="1">$A \cup \overline{B}$</option>
<option id="2">$\overline{A} \cup B$</option>
<option id="3">$AB$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">0</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem f (3.0 points)
Geben Sie reguläre Ausdrücke $X$, $Y$, $Z$ über $\Sigma$ an, sodass die folgenden Gleichungen erfüllt sind.
$X \equiv a^*b^*aX \mid ab$
$Y \equiv (b\mid a)a^*Y \mid bbY \mid \epsilon$
$Z \equiv a^*(a \mid b)Z$

<answer-template id="1">
$X=?$
$Y=?$
$Z=?$
</answer-template>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to produce regular expressions X, Y, Z satisfying given equations (meaning the language defined by the regex equals the language defined recursively? They gave equations with equivalence (≡) using regex operations and recursion. We need to find regex that solves those equations.

Interpretation: We need to find expressions for X, Y, Z over Σ = {a,b} such that the languages satisfy the equations. Essentially we need closed-form regex solutions.

Equation1: X ≡ a^* b^* a X | ab

Interpret: Language X = (a^* b^* a X) ∪ {ab}. This is a linear equation L = R L ∪ S. The solution: L = R* S where R = a^* b^* a, S = ab. So X = (a^* b^* a)* ab.

Check: If we take X = (a^* b^* a)* ab, then patterns: either directly ab (0 iterations of R), or any number of repetitions of a^* b^* a then ab. That matches equation: any string w in X must be either ab or of form a^* b^* a w' where w' in X. Indeed that recursion corresponds to left factoring: any w in X either ab or a^* b^* a concatenated with a string in X. So solution X = (a^* b^* a)* ab.

Could also simplify: a^* b^* a repeated zero or more times then ab. Fine.

Equation2: Y ≡ (b | a) a^* Y | bb Y | ε

Interpret: Y = ((b|a) a^* ) Y ∪ (bb) Y ∪ {ε}. That's Y = R Y ∪ ε where R = (b|a) a^* ∪ bb = (a|b) a^* ∪ bb. But note (a|b) a^* includes strings starting with a or b followed by any number of a's; bb is just two bs, which also is covered? (b) a^* includes b followed by a*. That includes "b", "ba", "baa"... not "bb"? Actually a* includes only a's, not b, so "bb" is not in (b) a^*. So we need union with "bb". So overall R = (a|b) a^* ∪ bb.

Equation: Y = R Y ∪ ε gives solution Y = R*.

Thus Y = ((a|b) a^* ∪ bb)* . Could simplify: ( (a|b) a* ∪ bb )* . Could also rewrite (a|b) a* = a a* ∪ b a* . That's a+? Actually a a* = a+, b a* = b a*. So overall R = a+ ∪ b a* ∪ bb. But note b a* includes 'b' and 'ba', 'baa', etc. So union with 'bb' might be redundant? 'bb' is not in b a* because a* only a's, so 'bb' separate. So final.

Thus Y = ((a|b) a* ∪ bb)* . Could also write (a a* ∪ b a* ∪ bb)* = (a+ ∪ b a* ∪ bb)*.

Equation3: Z ≡ a^*(a | b) Z

Interpret: Z = a^* (a|b) Z . Note no base case; maybe Z can be empty string? Not included. But solving: This is a linear equation L = R L where R = a^* (a|b). The only solution is L = ∅ or maybe infinite? Typically L = R* ∅ = ∅. However there is also possibility of infinite language if we allow infinite recursion? For regular languages, equation X = R X has solution X = ∅. But also any language that is subset of R*? Actually if L = R L, then L ⊆ R^+ L? Wait. This is typical left-linear equation: L = R L ∪ S; if S empty, then minimal solution is ∅, but also there are others like R* S = ∅. But also L = {ε}? test: If Z = {ε}, then RHS = a^* (a|b) {ε} = a^* (a|b) which includes strings like "a", "b", "aa"... not empty. So not equal.

Thus only solution is empty set. However sometimes they consider greatest fixpoint maybe Σ*? Let's test if Z = Σ* satisfies? RHS = a^* (a|b) Σ* = a^* (a|b) Σ* = all strings that have at least one a or b after some a's? That is all non-empty strings? Σ* includes ε, but RHS doesn't include ε. So not equal.

Thus solution Z = ∅. But maybe they expect Z = a^* (a|b) Z which leads to infinite languages: any string that can be written as a^* (a|b) then again a^* (a|b) etc, i.e., any non-empty string composed of a and b where each a-block can be any number of a's, then a single a or b? Actually any non-empty string can be factorized as a^* (a|b) repeated. That's just any non-empty string over {a,b}. So language is Σ+ . Let's see: definition: Σ+ = Σ Σ* . But is Σ+ equal to a^* (a|b) Σ+? Let's consider closure: Let L = Σ+ . Then RHS = a^* (a|b) L = a^* (a|b) Σ+ . Any string w in Σ+ can be written as a^* (a|b) x where x is also non-empty? For example w="b": we need a^* (a|b) prefix, a* could be empty, (a|b) = 'b', and then x∈ Σ+. But then we need remainder x to be non-empty, but we have no remainder. This fails. So Σ+ not a solution.

What about L = a^* (a|b)* = a* (a|b)* = Σ*? Actually a* (a|b)* = Σ* (since (a|b)* includes a and b). So L = Σ*. RHS = a* (a|b) Σ* = a* (a|b) Σ* = Σ+ (non-empty). Not equal.

Thus only empty set.

But maybe they allow solution with Z = ∅ because there is no base case. So answer Z = ∅. In regex, ∅ can be expressed as a regular expression that matches nothing, e.g., a b? Actually we can use ∅ directly; but perhaps we should write something like a a? Make explicit: \emptyset or maybe a? But typical regex syntax: ∅ or {}. We'll output ∅.

Thus final answer: Provide X = (a^* b^* a)* ab ; Y = ((a|b) a* ∪ bb)* ; Z = ∅.

Let's format per answer-template: <answer id="1">... lines.

We need to follow template: lines with $X=?$ etc.

Thus:

<answer id="1">
$X=(a^*b^*a)^*ab$
$Y=((a|b)a^*|bb)^*$
$Z=\emptyset$
</answer>

Make sure syntax matches. Use LaTeX in $$? they show inside $...$. Use as given.

Let's output.

<answer id="1">
$X=(a^*b^*a)^*ab$  
$Y=((a|b)a^* \;|\; bb)^*$  
$Z=\emptyset$
</answer>

Geben Sie reguläre Ausdrücke $X$, $Y$, $Z$ über $\Sigma$ an, sodass die folgenden Gleichungen erfüllt sind.
$X \equiv a^*b^*aX \mid ab$
$Y \equiv (b\mid a)a^*Y \mid bbY \mid \epsilon$
$Z \equiv a^*(a \mid b)Z$

$X=(a^*b^*a)^*ab$  
$Y=((a|b)a^* \;|\; bb)^*$  
$Z=\emptyset$

1: mismatch
1: could not parse value ((a|b)a^* \;|\; bb)^* for type re

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Quiz: Reguläre und kontextfreie Sprachen

**Teilaufgaben (a-e):** Für diese Fragen ist eine Begründung nicht gefordert. Sie erhalten die Punkte auf eine Teilaufgabe genau dann, wenn Sie alle Antwortmöglichkeiten korrekt angekreuzt haben.
**Es ist immer mindestens eine Antwortmöglichkeit richtig.**

In dieser Aufgabe verwenden wir durchgehend das Alphabet $\Sigma := \{a, b\}$.

## subproblem a (2.0 points)
Welche der folgenden Sprachen sind Residualsprachen von $L((bba)^*b)$?

<options>
<option id="1">$\{b\}$</option>
<option id="2">$L(ba(bba)^*b | \varepsilon)$</option>
<option id="3">$\emptyset$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">0</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem b (2.0 points)
Sei $L \subseteq \Sigma^*$ eine Sprache mit $LL = \Sigma^*$. Welche Aussagen sind wahr?

<options>
<option id="1">$L$ ist unendlich</option>
<option id="2">$\varepsilon \in L$</option>
<option id="3">$LLL = \Sigma^*$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option> 
<option id="2">1</option> 
<option id="3">1</option>

user:
## subproblem c (2.0 points)
Sei $L \subseteq \Sigma^*$ regulär. Welche der folgenden Sprachen sind regulär?

<options>
<option id="1">$\{u^R v \mid u, v \in L\}$</option>
<option id="2">$\{u^R u \mid u \in L\}$</option>
<option id="3">$\{uv \mid u, v \in L \land |u| = |v|\}$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">0</option>
<option id="3">0</option>

user:
## subproblem d (2.0 points)
Welche der folgenden kontextfreien Grammatiken sind mehrdeutig?

<options>
<option id="1">$S \rightarrow SS \mid a$</option>
<option id="2">$S \rightarrow Sa \mid Sb \mid \varepsilon$</option>
<option id="3">$S \rightarrow TT\quad T \rightarrow aT \mid bT \mid a \mid b$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">0</option>
<option id="3">1</option>

user:
## subproblem e (2.0 points)
Sei $A \subseteq \Sigma^*$ regulär und $B \subseteq \Sigma^*$ kontextfrei. Welche der folgenden Sprachen sind kontextfrei?

<options>
<option id="1">$A \cup \overline{B}$</option>
<option id="2">$\overline{A} \cup B$</option>
<option id="3">$AB$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">0</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem f (3.0 points)
Geben Sie reguläre Ausdrücke $X$, $Y$, $Z$ über $\Sigma$ an, sodass die folgenden Gleichungen erfüllt sind.
$X \equiv a^*b^*aX \mid ab$
$Y \equiv (b\mid a)a^*Y \mid bbY \mid \epsilon$
$Z \equiv a^*(a \mid b)Z$

<answer-template id="1">
$X=?$
$Y=?$
$Z=?$
</answer-template>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
$X=(a^*b^*a)^*ab$  
$Y=((a|b)a^* \;|\; bb)^*$  
$Z=\emptyset$
</answer>

user:
## subproblem g (3.0 points)
Ein NFA $(Q,\Sigma,\delta,q_0,F)$ heißt *reversibel*, falls für alle Zustände $q_1, q_2 \in Q$ und für alle Zeichen $x \in \Sigma$ gilt: Falls $q_2 \in \delta(q_1,x)$, so $q_1 \in \delta(q_2,x)$. Mit anderen Worten: Falls man von $q_1$ mit $x$ in $q_2$ übergehen kann, so kann man auch von $q_2$ mit $x$ in $q_1$ übergehen.

Gibt es einen reversiblen NFA $N$ über $\Sigma$ mit $L(N) = L((aa)^*(bb)^*)$? Falls ja, geben Sie einen solchen NFA an; falls nein, beweisen Sie es.

<options>
<option id="1">Wahr, reversibler NFA</option>
<option id="2">Falsch, Beweis</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<answer id="3">
...
</answer>

We need to decide if there exists a reversible NFA for language L = (aa)*(bb)* over alphabet {a,b}. Language consists of strings of any number of "aa" blocks followed by any number of "bb" blocks. So strings of even number of a's (maybe zero) then even number of b's (maybe zero). No mixing. For example "", "aa", "bbbb", "aabb", "aaaa", "aabbbb" etc. Actually (aa)^* gives even number of a's (including zero). Similarly (bb)^* even number of b's. Concatenated gives even a's then even b's.

We need reversible NFA definition: transitions must be symmetric: for each x, edges are undirected (bidirectional). So essentially underlying graph for each symbol is an undirected relation: if there is an edge labeled x from q1 to q2, there must also be edge from q2 to q1. This also allows self-loops.

Question: does such NFA exist? Let's think. Reversible NFA essentially models an undirected labeled graph. The language recognized must be such that there exists a set of start states (single start) and final states where there is a path spelling the word from start to a final. Reversibility imposes that edges are symmetric.

Can we construct such NFA for (aa)*(bb)*? Let's attempt.

We need to enforce that a's come in pairs before switching to b's, also b's in pairs.

Idea: have two states for parity of a's? Since we need exact even a's, we could have self-loop a from state q0 to q1 and back? Let's design states: q0 start and accept (allow zero a). Also need to ensure we can move to b part only after even number of a's. Use two-state chain for a's: q0 -(a)-> q1, q1 -(a)-> q0. Because edges must be symmetric, we need also a from q1 to q0 which already is there (since we have a from q0 to q1, we need also a from q1 to q0). That's okay: both directions exist same label a. So we can have a-cycle of length 2 labeled a: q0--a-->q1, q1--a-->q0. This ensures that reading single a moves between states; reading two a's returns to start. Accepting after any even number of a's: q0 is accept (since after 0 a's also accept). While still reading a's we can stay in q0 after even count.

Now need to allow b's after we stop a's. After finishing a's we might be in q0. Then we need to accept b's in pairs and stay within perhaps a separate set of states, while not allowing a again. Since edges must be symmetric, we can have similar two-state for b: q0 -(b)-> q2, q2-(b)-> q0 (and symmetric). However this also allows b at any time (including interleaved with a). For example, from start q0 reading a gives q1; then reading b from q1? No transition labeled b from q1 defined; but we defined b only from q0 and q2. So from q1 cannot read b (since we only have b transition from q0 to q2 and q2 to q0). After reading a (one a) we are in q1, cannot read b, good. After two a's we are back to q0 and can start reading b's. That seems okay.

Now after reading b's, we need to stay within q0 and q2 toggling; can we allow further a's? From q0 after reading b we go to q2, could we read a from q2? No a transition defined from q2 (only from q0 and q1). So after entering b-phase, you cannot go back to a. Also you cannot read a when in q2, which ensures no mixing.

Thus we have reversible NFA with states {q0,q1,q2}. Transitions: a: q0<->q1 (both directions). b: q0<->q2 (both directions). Start state q0. Accepting states: q0 (since after even a's and any even b's you end at q0). Also maybe q2? Let's see: after one b you are at q2, which is odd count of b's, not acceptable (language requires even b count). So final states should be q0 only. But also empty string accepted, fine.

Now does this accept strings like "aa bb"? Let's trace: start q0, a to q1, a to q0, b to q2, b to q0 -> accept. Works. Does it accept "bbbb"? start q0, b to q2, b to q0, b to q2, b to q0 accept. Even number of b's. Accepts "aabb". Works.

Does it accept "ab"? start q0, a to q1. From q1 cannot b, so not accepted. Good.

Thus language recognized is exactly strings where a's appear only in pairs and after any b's you cannot return to a and b also appear in pairs (since you need even number to end in q0). Is there any string not in language but accepted? Example: "aa a"? That's odd a's leads to q1 not accept. Not accepted. "bb a"? start q0 b to q2 b to q0 a to q1; ends at q1 not accept => not accepted. So seems matches language.

Thus a reversible NFA exists. So answer: option 1 true. Provide NFA details.

We must output 0/1 for both options and then answer id=3 with NFA.

Option labeling: option id="1": 1 if true exists, option id="2": 0. Provide NFA in answer.

Define NFA as tuple: Q={q0,q1,q2}, start q0, F={q0}, delta: δ(q0,a)={q1}, δ(q1,a)={q0}, δ(q0,b)={q2}, δ(q2,b)={q0}. No other transitions.

We should format answer likely as description.

Now produce final.

<option id="1">1</option>
<option id="2">0</option>
<answer id="3">
**Reversibler NFA \(N\) für \(L = (aa)^*(bb)^*\)**  

- **Zustandsmenge:** \(Q = \{q_0, q_1, q_2\}\)  
- **Startzustand:** \(q_0\)  
- **Endzustände:** \(F = \{q_0\}\)  

- **Übergangsfunktion \(\delta\):**  

\[
\begin{aligned}
\delta(q_0, a) &= \{q_1\}, & \delta(q_1, a) &= \{q_0\}, \\
\delta(q_0, b) &= \{q_2\}, & \delta(q_2, b) &= \{q_0\}, \\
\delta(q_1, b) &= \emptyset, & \delta(q_2, a) &= \emptyset.
\end{aligned}
\]

Alle übrigen Übergänge sind leer (\(\emptyset\)).  

**Warum ist der NFA reversibel?**  
Für jedes Symbol \(x \in \{a,b\}\) gilt:
- Wenn \(q_2 \in \delta(q_1,x)\) (z.B. \(q_1 \xrightarrow{a} q_0\)), dann ist zugleich \(q_1 \in \delta(q_2,x)\) (z.B. \(q_0 \xrightarrow{a} q_1\)).  
- Analog für die \(b\)-Übergänge zwischen \(q_0\) und \(q_2\).  

Damit erfüllt \(N\) die Definition eines reversiblen NFA.

**Sprachakzeptanz:**  
- Durch abwechselnde \(a\)-Übergänge zwischen \(q_0\) und \(q_1\) wird nur eine gerade Anzahl von \(a\)’s akzeptiert (Endzustand muss \(q_0\) sein).  
- Nachdem ein \(b\) gelesen wurde, wechselt man zu \(q_2\); weitere \(b\)’s pendeln zwischen \(q_2\) und \(q_0\), also ebenfalls nur gerade Anzahl von \(b\)’s führen zum Endzustand \(q_0\).  
- Sobald man sich in \(q_2\) befindet, gibt es keinen \(a\)-Übergang, sodass nach Beginn des \(b\)-Blocks kein weiteres \(a\) mehr gelesen werden kann.  

Somit akzeptiert \(N\) exakt die Sprache \((aa)^*(bb)^*\).  
</answer>

Ein NFA $(Q,\Sigma,\delta,q_0,F)$ heißt reversibel, falls für alle Zustände $q_1, q_2 \in Q$ und für alle Zeichen $x \in \Sigma$ gilt: Falls $q_2 \in \delta(q_1,x)$, so $q_1 \in \delta(q_2,x)$. Mit anderen Worten: Falls man von $q_1$ mit $x$ in $q_2$ übergehen kann, so kann man auch von $q_2$ mit $x$ in $q_1$ übergehen.

Gibt es einen reversiblen NFA $N$ über $\Sigma$ mit $L(N) = L((aa)^*(bb)^*)$? Falls ja, geben Sie einen solchen NFA an; falls nein, beweisen Sie es.

wrong answer for choice (0)

Reversibler NFA $N$ für $L = (aa)^*(bb)^*$  

- Zustandsmenge: $Q = \{q_0, q_1, q_2\}$  
- Startzustand: $q_0$  
- Endzustände: $F = \{q_0\}$  

- Übergangsfunktion $\delta$:  

$$\begin{aligned}
\delta(q_0, a) &= \{q_1\}, & \delta(q_1, a) &= \{q_0\}, \\
\delta(q_0, b) &= \{q_2\}, & \delta(q_2, b) &= \{q_0\}, \\
\delta(q_1, b) &= \emptyset, & \delta(q_2, a) &= \emptyset.
\end{aligned}$$

Alle übrigen Übergänge sind leer ($\emptyset$).  

Warum ist der NFA reversibel?  
Für jedes Symbol $x \in \{a,b\}$ gilt:
- Wenn $q_2 \in \delta(q_1,x)$ (z.B. $q_1 \xrightarrow{a} q_0$), dann ist zugleich $q_1 \in \delta(q_2,x)$ (z.B. $q_0 \xrightarrow{a} q_1$).  
- Analog für die $b$-Übergänge zwischen $q_0$ und $q_2$.  

Damit erfüllt $N$ die Definition eines reversiblen NFA.

Sprachakzeptanz:  
- Durch abwechselnde $a$-Übergänge zwischen $q_0$ und $q_1$ wird nur eine gerade Anzahl von $a$’s akzeptiert (Endzustand muss $q_0$ sein).  
- Nachdem ein $b$ gelesen wurde, wechselt man zu $q_2$; weitere $b$’s pendeln zwischen $q_2$ und $q_0$, also ebenfalls nur gerade Anzahl von $b$’s führen zum Endzustand $q_0$.  
- Sobald man sich in $q_2$ befindet, gibt es keinen $a$-Übergang, sodass nach Beginn des $b$-Blocks kein weiteres $a$ mehr gelesen werden kann.  

Somit akzeptiert $N$ exakt die Sprache $(aa)^*(bb)^*$.

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: DFA-Algorithmen


## subproblem a (6.0 points)
Wir betrachten den folgenden NFA $N$ über dem Alphabet $\Sigma=\{a,b\}$:

Zustände: 0, 1, 2, 3
Startzustand: 0
Finalzustände: 0, 2, 3
Transitionen:
    0, b -> 0
    0, b -> 1
    1, b -> 1
    1, b -> 2
    2, a -> 3
    2, b -> 3
    3, a -> 1
    3, a -> 2

Sei $M$ der DFA, den man erhält, wenn man $N$ mithilfe des Verfahrens aus der Vorlesung (Potenzmengenkonstruktion) zu einem DFA konvertiert. Tragen Sie $M$ in untere Schablone ein, indem Sie (1) die Transitionen beschriften, (2) die Finalzustände markieren, und (3) in der Tabelle rechts für jeden Zustand von $M$ die korrespondierende Menge an Zuständen aus $N$ eintragen.

Zustände: q0, q1, q2, q3, q4, q5, q6, q7
Startzustand: q0
Finalzustände: ?
Transitionen:
    q0 -> q1
    q0 -> q4
    q1 -> q2
    q1 -> q4
    q2 -> q3
    q2 -> q5
    q3 -> q3
    q3 -> q7
    q7 -> q7
    q7 -> q7
    q5 -> q4
    q5 -> q6
    q6 -> q5
    q6 -> q7
    q4 -> q4
    q4 -> q4
Zustandstabelle:
    q0:
    q1:
    q2:
    q3:
    q4:
    q5:
    q6:
    q7:


<answer-template id="1">
Finalzustände: ?
Transitionen:
    ???
Zustandstabelle:
    q0: ?
    q1: ?
    q2: ?
    q3: ?
    q4: ?
    q5: ?
    q6: ?
    q7: ?

</answer-template>

<example>
Finalzustände: q0, q1, q2
Transitionen:
    q0, b -> q1
    q0, a -> q4
    ...
Zustandstabelle:
    q0: 0, 1
    q1: \emptyset
    q2: 0, 1, 2
    ...
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to compute DFA via subset construction from given NFA. Let's parse NFA.

States: 0,1,2,3. Start: 0. Final: 0,2,3. Alphabet {a,b}. Transitions:
- From 0 on b -> 0
- From 0 on b -> 1 (so nondet)
- From 1 on b -> 1
- From 1 on b -> 2
- From 2 on a -> 3
- From 2 on b -> 3
- From 3 on a -> 1
- From 3 on a -> 2

No epsilon transitions.

Now subset construction: start state = epsilon-closure of {0} = {0}. So DFA state q0 corresponds to subset {0}. In given template they label q0...q7.

We must compute reachable subsets via a and b.

Let us compute.

Define function δ_N(S, a) etc.

First, for each subset we compute transitions.

Start subset S0 = {0}. Need δ on a and b.

From 0: there are no transitions on a. So δ({0}, a) = ∅.

On b: from 0 we have to 0 and 1. So result = {0,1}. No further epsilon closure needed (none). So δ({0}, b) = {0,1}.

Thus we have DFA transitions from q0:
- on a -> empty set (dead state)
- on b -> subset {0,1}

So there is a dead state representing empty set. We'll likely label it q7? We'll see mapping.

Now subset {0,1}: compute transitions.

From 0 on a: none. From 1 on a: none. So a => ∅.

On b: from 0: 0,1. From 1:1,2. Union = {0,1,2}. So result {0,1,2}.

Thus {0,1} on b -> {0,1,2}.

Now subset {0,1,2}. Compute.

On a: from 0 none, from1 none, from2 a->3. So result {3}.

On b: from0 ->0,1 ; from1 ->1,2 ; from2 ->3. Union {0,1,2,3}. So result {0,1,2,3}.

Now subset {3}. Compute.

On a: from3 ->1,2. So result {1,2}.

On b: from3 no b transition (none). So ∅.

Subset {0,1,2,3}. Compute.

On a: from0 none, from1 none, from2 ->3, from3 ->1,2 . So union {1,2,3}. also maybe from2 a->3 already included; from3 a->1,2 gives {1,2,3}. So result {1,2,3}.

On b: from0 ->0,1 ; from1 ->1,2 ; from2 ->3 ; from3 none. Union {0,1,2,3}. So result {0,1,2,3} (self-loop).

Subset {1,2}. Compute.

On a: from1 none, from2 ->3. So {3}.

On b: from1 ->1,2 ; from2 ->3 . Union {1,2,3}.

Subset {1,2,3}. Compute.

On a: from1 none, from2 ->3, from3 ->1,2 . Union {1,2,3}. (since 3 from 2, plus 1,2 from 3) -> {1,2,3}.

On b: from1 ->1,2 ; from2 ->3 ; from3 none . Union {1,2,3}. So both a,b stay in {1,2,3} (self-loop).

Now we have subsets discovered: ∅, {0}, {0,1}, {0,1,2}, {3}, {0,1,2,3}, {1,2}, {1,2,3}. That's up to 8 subsets (including empty). The template has q0..q7 eight states, so they correspond.

Now we must map each q_i to subset according to transition structure already given in the prompt (they gave transitions between q's but not labeled with a/b). Let's examine given transition list:

Transitionen:
    q0 -> q1
    q0 -> q4
    q1 -> q2
    q1 -> q4
    q2 -> q3
    q2 -> q5
    q3 -> q3
    q3 -> q7
    q7 -> q7
    q7 -> q7
    q5 -> q4
    q5 -> q6
    q6 -> q5
    q6 -> q7
    q4 -> q4
    q4 -> q4

There are 2 outgoing per state? Usually each state has two transitions (a and b). So each arrow corresponds to either a or b, not specified which is which. We need to label them accordingly.

Let's analyze structure to map.

From our subset DFA, start state {0} (q0). Its a transition goes to ∅, b to {0,1}. So q0 should have two outgoing edges: one to empty subset (maybe q7?) and one to subset {0,1} (maybe q1). In given list, q0 -> q1 and q0 -> q4. So which is empty? Let's see other transitions: q7 appears to have self loops (both transitions go to q7). So q7 looks like dead state (empty set). Indeed they have q7 -> q7 twice, consistent with empty state loops to itself on both symbols. So q7 is empty set.

Thus q0's transition to empty set must be q0 -> q7, but they have q0->q4 and q0->q1. So maybe q4 is empty set? Let's examine q4 transitions: they have q4 -> q4 (twice). That's also a dead state. So both q4 and q7 have self loops. So which one is actually dead? There may be two dead states? But subset construction can have just one dead (empty set). However they maybe have duplicate but unreachable state? Let's analyze.

Our reachable subsets list includes ∅ plus others. We'll assign each q to subset.

Given transitions list, identify loops:

- q3 -> q3 (once) and q3 -> q7 (second). So q3 has a self loop on one symbol (maybe a) and goes to q7 on other symbol.
- q7 has both self loops.
- q4 has both self loops.

Thus we have multiple dead-like states.

Consider subsets: which subsets have self loops on both symbols? {1,2,3} self loops on both a and b (as computed). That's not dead but stable. Could correspond to q7? But q7 also self loops, but also maybe there are other loops.

But q7 is reachable? Let's check our DFA: from start {0}, we go by a to ∅, then ∅ loops to itself on both. So empty set is reachable (q0-> a). So q7 reachable.

Also maybe {1,2,3} reachable? Let's see path: from start {0} b -> {0,1}, b -> {0,1,2}, a -> {3}, a -> {1,2}. b -> {1,2,3}. That's reachable. So subset {1,2,3} is reachable.

Now we have multiple subsets with self loops. {1,2,3} loops to itself on both symbols (as we saw). That matches q3 maybe? But q3 also has one self-loop and other transition to q7. Wait q3 has self-loop and transition to q7. {1,2,3} would have both self loops, no transition elsewhere. So q3 not {1,2,3}.

What subset has a self-loop on one symbol and transition to empty on other? Let's compute: subset {3} on a goes to {1,2} (not empty), on b goes to ∅. So {3} has transition to empty on b, and to {1,2} on a. Not self-loop. So not.

Subset {0,1,2,3} loops to itself on b, but on a goes to {1,2,3}. So one self-loop on b, other to {1,2,3}. That matches q3 maybe: self-loop (b) and transition to q7 (empty?) wait but {0,1,2,3} goes to {1,2,3} not empty. So not.

Subset {0,1,2} on a goes to {3}, on b to {0,1,2,3}. So one to {3}, one to larger set.

Subset {0,1} on a goes to ∅, on b to {0,1,2}.

Subset {1,2} on a -> {3}, b -> {1,2,3}.

Subset {1,2,3} loops to itself.

Empty set loops to itself.

Thus we need mapping where q0 start corresponds to {0}. q0 has two outgoing: q1 and q4. So either q1 is empty and q4 is {0,1} or vice versa.

Let's inspect q1's outgoing: q1 -> q2 and q1 -> q4. So from q1 two targets: q2 and q4. Which subset fits? If q1 corresponds to {0,1}, its a transition goes to ∅, b to {0,1,2}. So need two distinct targets: one empty (could be q2) and one maybe {0,1,2} (q4). But q1 currently goes to q2 and q4. So maybe q2 = empty, q4 = {0,1,2}. But earlier we thought q0 also goes to empty maybe q4. But we need consistency.

Let's examine q2's outgoing: q2 -> q3 and q2 -> q5.

If q2 is empty, its outgoing both self loops to empty; but they point to q3 and q5, not self. So empty state cannot be q2. Therefore empty must be q4 or q7 (since those have self loops). So q2 is not empty.

Thus q0's transition to empty must be q0 -> q4 (since q4 loops to itself). So q4 is empty set.

Thus q0: on a (maybe) -> q4 (empty) ; on b -> q1 (subset {0,1}). Need to decide labeling later.

Now q1 -> q2 and q1 -> q4. Since q4 is empty, q1 must have one transition to empty (a) and other to subset {0,1,2} (b). So q2 must be {0,1,2}.

Thus assign: q2 = {0,1,2}.

Now q2 -> q3 and q2 -> q5. For subset {0,1,2}: we computed a -> {3}, b -> {0,1,2,3}. So q3 should be {3}, q5 should be {0,1,2,3}.

Now q3 -> q3 and q3 -> q7. Subset {3}: a-> {1,2}, b-> empty. So one self-loop maybe on a? No, {3} on a goes to {1,2} not self. So which subset has self-loop? None. So maybe mapping reversed: For subset {3}, one transition goes to {1,2} (some q) and other to empty (q4). But q3 -> q3 self-loop suggests {3} self loops on something, which is not correct. Let's reconsider: maybe our identification of symbols swapped? That doesn't change self-loop property.

Perhaps our mapping earlier wrong: maybe q3 corresponds to {1,2,3} which does have self-loop on both symbols. Then q3 -> q3 (self loop on a) and q3 -> q7 (other symbol) contradictory, as {1,2,3} loops on both, not to q7. So not.

What subset has self-loop on one symbol and transition to q7 (empty) on other? let's see subsets:

- {0,1,2,3}: on b self-loop to itself, on a goes to {1,2,3}. Not empty.
- {0,1,2}: neither self-loop.
- {3}: on b empty (q4), on a to {1,2} (some q).
- {1,2}: on b to {1,2,3}, on a to {3}.
- {1,2,3}: loops both.

Thus none have self-loop and empty transition. So maybe q7 is not empty but corresponds to {1,2,3}? But q7 also self-loops both, matches {1,2,3}. While q4 maybe empty? But q4 also self-loops both. So duplicate dead states; one may be empty, other {1,2,3} (both loop). That's possible: both have loops on both symbols, consistent with their description.

Check q7 -> q7 twice (both symbols). If q7 = {1,2,3}, that matches loops. If q4 = empty set, also loops.

Now q3 -> q3 and q3 -> q7. Which subset could have self-loop on one symbol to itself, and other symbol to {1,2,3}. That matches subset {0,1,2,3}: on b self-loop, on a goes to {1,2,3}. Yes! So q3 = {0,1,2,3} and q7 = {1,2,3}.

Thus mapping:

q0 = {0}
q1 = {0,1}
q2 = {0,1,2}
q3 = {0,1,2,3}
q4 = ∅ (empty)
q5 = {3}
q6 = {1,2}
q7 = {1,2,3}

Check remaining transitions:

From q5 -> q4 and q5 -> q6. Subset {3} on a -> {1,2} (which is q6), on b -> ∅ (q4). So matches.

From q6 -> q5 and q6 -> q7. Subset {1,2} on a -> {3} (q5), on b -> {1,2,3} (q7). Works.

From q4 -> q4 (both) as empty loops – OK.

Now q3 transitions earlier: q3 -> q3 and q3 -> q7. Subset {0,1,2,3} on b -> itself (q3), on a -> {1,2,3} (q7). Good.

Thus mapping consistent.

Now final states: DFA final states are any subset containing at least one NFA final state (0,2,3). So which subsets contain 0 or 2 or 3?

List subsets:

q0 {0} contains 0 => final.
q1 {0,1} contains 0 => final.
q2 {0,1,2} contains 0 and 2 => final.
q3 {0,1,2,3} contains final => final.
q4 ∅ => not final.
q5 {3} contains 3 => final.
q6 {1,2} contains 2 => final.
q7 {1,2,3} contains 2 or 3 => final.

Thus all except q4 are final. So final states: q0,q1,q2,q3,q5,q6,q7 (maybe also q4? no).

Now we need to produce answer in template: list final states, then list transitions with symbols a,b; then state table mapping each q to subset.

We must fill transitions with correct labeling a or b.

From our mapping:

From q0 ({0}):
- a -> q4 (∅)
- b -> q1 ({0,1})

From q1 ({0,1}):
- a -> q4 (∅) (since both 0 and1 have no a)
- b -> q2 ({0,1,2})

From q2 ({0,1,2}):
- a -> q5 ({3})
- b -> q3 ({0,1,2,3})

From q3 ({0,1,2,3}):
- a -> q7 ({1,2,3})
- b -> q3 (self)

From q4 (∅):
- a -> q4
- b -> q4

From q5 ({3}):
- a -> q6 ({1,2})
- b -> q4 (∅)

From q6 ({1,2}):
- a -> q5 ({3})
- b -> q7 ({1,2,3})

From q7 ({1,2,3}):
- a -> q7
- b -> q7

Now list according to template format: they gave transitions without symbols; we must include them.

Thus output:

Finalzustände: q0, q1, q2, q3, q5, q6, q7

Transitionen:
    q0, a -> q4
    q0, b -> q1
    q1, a -> q4
    q1, b -> q2
    q2, a -> q5
    q2, b -> q3
    q3, a -> q7
    q3, b -> q3
    q4, a -> q4
    q4, b -> q4
    q5, a -> q6
    q5, b -> q4
    q6, a -> q5
    q6, b -> q7
    q7, a -> q7
    q7, b -> q7

State table:

q0: 0
q1: 0,1
q2: 0,1,2
q3: 0,1,2,3
q4: (empty) maybe \emptyset
q5: 3
q6: 1,2
q7: 1,2,3

Provide in template.

Now produce answer.

<answer id="1">
Finalzustände: q0, q1, q2, q3, q5, q6, q7
Transitionen:
    q0, a -> q4
    q0, b -> q1
    q1, a -> q4
    q1, b -> q2
    q2, a -> q5
    q2, b -> q3
    q3, a -> q7
    q3, b -> q3
    q4, a -> q4
    q4, b -> q4
    q5, a -> q6
    q5, b -> q4
    q6, a -> q5
    q6, b -> q7
    q7, a -> q7
    q7, b -> q7
Zustandstabelle:
    q0: 0
    q1: 0, 1
    q2: 0, 1, 2
    q3: 0, 1, 2, 3
    q4: \emptyset
    q5: 3
    q6: 1, 2
    q7: 1, 2, 3
</answer>

Wir betrachten den folgenden NFA $N$ über dem Alphabet $\Sigma=\{a,b\}$:
Zustände: 0, 1, 2, 3
Startzustand: 0
Finalzustände: 0, 2, 3
Transitionen:
    0, b -> 0
    0, b -> 1
    1, b -> 1
    1, b -> 2
    2, a -> 3
    2, b -> 3
    3, a -> 1
    3, a -> 2
Sei $M$ der DFA, den man erhält, wenn man $N$ mithilfe des Verfahrens aus der Vorlesung (Potenzmengenkonstruktion) zu einem DFA konvertiert. Tragen Sie $M$ in untere Schablone ein, indem Sie (1) die Transitionen beschriften, (2) die Finalzustände markieren, und (3) in der Tabelle rechts für jeden Zustand von $M$ die korrespondierende Menge an Zuständen aus $N$ eintragen.
Zustände: q0, q1, q2, q3, q4, q5, q6, q7
Startzustand: q0
Finalzustände: ?
Transitionen:
    q0 -> q1
    q0 -> q4
    q1 -> q2
    q1 -> q4
    q2 -> q3
    q2 -> q5
    q3 -> q3
    q3 -> q7
    q7 -> q7
    q7 -> q7
    q5 -> q4
    q5 -> q6
    q6 -> q5
    q6 -> q7
    q4 -> q4
    q4 -> q4
Zustandstabelle:
    q0:
    q1:
    q2:
    q3:
    q4:
    q5:
    q6:
    q7:

Finalzustände: q0, q1, q2, q3, q5, q6, q7
Transitionen:
    q0, a -> q4
    q0, b -> q1
    q1, a -> q4
    q1, b -> q2
    q2, a -> q5
    q2, b -> q3
    q3, a -> q7
    q3, b -> q3
    q4, a -> q4
    q4, b -> q4
    q5, a -> q6
    q5, b -> q4
    q6, a -> q5
    q6, b -> q7
    q7, a -> q7
    q7, b -> q7
Zustandstabelle:
    q0: 0
    q1: 0, 1
    q2: 0, 1, 2
    q3: 0, 1, 2, 3
    q4: \emptyset
    q5: 3
    q6: 1, 2
    q7: 1, 2, 3

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: DFA-Algorithmen


## subproblem a (6.0 points)
Wir betrachten den folgenden NFA $N$ über dem Alphabet $\Sigma=\{a,b\}$:

Zustände: 0, 1, 2, 3
Startzustand: 0
Finalzustände: 0, 2, 3
Transitionen:
    0, b -> 0
    0, b -> 1
    1, b -> 1
    1, b -> 2
    2, a -> 3
    2, b -> 3
    3, a -> 1
    3, a -> 2

Sei $M$ der DFA, den man erhält, wenn man $N$ mithilfe des Verfahrens aus der Vorlesung (Potenzmengenkonstruktion) zu einem DFA konvertiert. Tragen Sie $M$ in untere Schablone ein, indem Sie (1) die Transitionen beschriften, (2) die Finalzustände markieren, und (3) in der Tabelle rechts für jeden Zustand von $M$ die korrespondierende Menge an Zuständen aus $N$ eintragen.

Zustände: q0, q1, q2, q3, q4, q5, q6, q7
Startzustand: q0
Finalzustände: ?
Transitionen:
    q0 -> q1
    q0 -> q4
    q1 -> q2
    q1 -> q4
    q2 -> q3
    q2 -> q5
    q3 -> q3
    q3 -> q7
    q7 -> q7
    q7 -> q7
    q5 -> q4
    q5 -> q6
    q6 -> q5
    q6 -> q7
    q4 -> q4
    q4 -> q4
Zustandstabelle:
    q0:
    q1:
    q2:
    q3:
    q4:
    q5:
    q6:
    q7:


<answer-template id="1">
Finalzustände: ?
Transitionen:
    ???
Zustandstabelle:
    q0: ?
    q1: ?
    q2: ?
    q3: ?
    q4: ?
    q5: ?
    q6: ?
    q7: ?

</answer-template>

<example>
Finalzustände: q0, q1, q2
Transitionen:
    q0, b -> q1
    q0, a -> q4
    ...
Zustandstabelle:
    q0: 0, 1
    q1: \emptyset
    q2: 0, 1, 2
    ...
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
Finalzustände: q0, q1, q2, q3, q5, q6, q7
Transitionen:
    q0, a -> q4
    q0, b -> q1
    q1, a -> q4
    q1, b -> q2
    q2, a -> q5
    q2, b -> q3
    q3, a -> q7
    q3, b -> q3
    q4, a -> q4
    q4, b -> q4
    q5, a -> q6
    q5, b -> q4
    q6, a -> q5
    q6, b -> q7
    q7, a -> q7
    q7, b -> q7
Zustandstabelle:
    q0: 0
    q1: 0, 1
    q2: 0, 1, 2
    q3: 0, 1, 2, 3
    q4: \emptyset
    q5: 3
    q6: 1, 2
    q7: 1, 2, 3
</answer>

user:
## subproblem b (1.0 points)
Ist M minimal? Begründen Sie Ihre Antwort kurz.

<options>
<option id="1">Ja</option>
<option id="2">Nein</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<answer id="3">
...
</answer>

We need to evaluate minimality of DFA M derived from NFA.

But my previous answer likely wrong. Need to compute correctly.

Let's construct NFA N transitions given:

States {0,1,2,3}
Start 0
Final {0,2,3}
Transitions:
0, b -> 0
0, b -> 1  (so two b transitions from 0 to 0 and 1)
1, b -> 1
1, b -> 2
2, a -> 3
2, b -> 3
3, a -> 1
3, a -> 2

No other transitions (implicitly empty for other symbols).

Now perform powerset construction. Alphabet {a,b}. Start state of DFA = epsilon-closure of {0}. No epsilon transitions, so {0}.

Compute moves:

From set S, on a: union over states in S of δ(state,a). For each state, a transitions:

- 0: none.
- 1: none.
- 2: a -> 3.
- 3: a -> 1,2.

Similarly for b: δ.

Let's compute reachable subsets.

Start q0 = {0}. On a: none => empty set ∅. On b: from 0 b->0 and 1 => {0,1}. So we have ∅ and {0,1}.

Define q4 = ∅ maybe.

Now process {0,1} (call q1). From {0,1}:

- a: none (0 no a, 1 no a) => ∅ (q4).
- b: from 0 ->0,1 ; from1 ->1,2 => {0,1,2}. call q2.

Now process {0,1,2} (q2). a: from 2 ->3, from 3? Wait 3 not in set. So a gives {3}. call q5 perhaps.

b: from 0 ->0,1 ; from1 ->1,2 ; from2 ->3 => {0,1,2,3}. call q3 maybe.

Process {3} (q5). a: from 3 ->1,2 => {1,2}. call q6.

b: from 3? No b transition from 3 (only a). So b -> ∅ (q4).

Process {0,1,2,3} (q3). a: from 2 ->3 ; from3 ->1,2 ; others none => {1,2,3}. call q7? Wait we also may have 0? No a from 0,1 none. So {1,2,3}. call q6? Actually q6 already {1,2}. This is {1,2,3} new. call q6 maybe but we need distinct.

Let's keep systematic:

List subsets:

S0 = {0}
S1 = ∅
S2 = {0,1}
S3 = {0,1,2}
S4 = {0,1,2,3}
S5 = {3}
S6 = {1,2}
S7 = {1,2,3}
S8 = {1}? maybe later.

We need to compute transitions for each.

We already have for S0:
 a -> S1
 b -> S2

S1 (∅): a -> ∅, b -> ∅ (dead state). That's S1 itself.

S2 = {0,1}:
 a -> ∅ (S1)
 b -> {0,1,2} = S3

S3 = {0,1,2}:
 a -> {3}=S5
 b -> {0,1,2,3}=S4

S4 = {0,1,2,3}:
 a -> from 2->3, from3->1,2 => {1,2,3}=S7
 b -> from 0->0,1 ; 1->1,2 ; 2->3 ; 3 no b => {0,1,2,3}=S4 (self)

S5 = {3}:
 a -> {1,2}=S6
 b -> ∅ = S1

S6 = {1,2}:
 a -> from 2->3 => {3}=S5 (also from 1 none) So a -> S5
 b -> from 1->1,2 ; from2->3 => {1,2,3}=S7

S7 = {1,2,3}:
 a -> from 2->3 ; from3->1,2 => {1,2,3}=S7 (self)
 b -> from1->1,2 ; from2->3 ; from3 none => {1,2,3}=S7 (self)

Now any other subsets like {1}? Not reachable.

Thus DFA states = S0..S7 (8 states) as given q0..q7.

Now final states are those containing any NFA final state (0,2,3). So check each subset:

S0 {0} -> contains 0 => final.
S1 ∅ -> not final.
S2 {0,1} -> contains 0 => final.
S3 {0,1,2} -> contains 0,2 => final.
S4 {0,1,2,3} -> contains 0,2,3 => final.
S5 {3} -> contains 3 => final.
S6 {1,2} -> contains 2 => final.
S7 {1,2,3} -> contains 2,3 => final.

Thus all states except the dead state (∅) are final. So final set = all except q1 (if q1 corresponds to ∅). In given transition list they used q4 as dead? Actually from earlier answer they put q4 as dead (self-loop both a,b). In my mapping: S1=∅ likely q4 in their list (since they have q4 transition to itself both letters). Indeed q4 is dead state. So final states: q0,q1,q2,q3,q5,q6,q7 (i.e., all except q4). The provided answer earlier had that set? I gave q0,q1,q2,q3,q5,q6,q. That matches.

Now minimality: Since all states except dead are final, many may be indistinguishable? For DFA minimal, if all final states are equivalent with respect to language, they might be merged if reachable and equivalent. Need to see if any distinguishable final states. Since all final accept the same set of suffixes? Let's test.

Given that from any final state, reading any string results in acceptance? Not necessarily because there is dead state reachable only from some final states via a? Actually from q0 (S0={0}) reading a leads to dead, which then leads to dead for all further letters, which is non-accepting. So q0 distinguishes from other finals: after 'a' you go to dead => reject, but maybe from q1 (S2={0,1}) reading a also leads to dead. So also rejects. Let's examine.

We need to see whether any two final states have different futures.

Compute transitions for each final state (excluding dead):

- q0 (S0): a->dead, b->q1 (S2)
- q1 (S2): a->dead, b->q2 (S3)
- q2 (S3): a->q5 (S5), b->q3 (S4)
- q3 (S4): a->q7 (S7), b->q3 (self)
- q5 (S5): a->q6 (S6), b->dead
- q6 (S6): a->q5, b->q7
- q7 (S7): a->q7, b->q7

Now we can see distinctions: For example, from q0 reading b then a: q0--b->q1, then a->dead (reject). From q5 reading b: dead (reject). But many patterns.

We need to check equivalence relation: Two states are equivalent if for all strings they both accept or both reject.

Given q7 is a sink accepting state (any further input stays in q7 which is accepting). So from q7 any string leads to accepting. For other states, some inputs may lead to dead (reject). Hence q7 is distinguishable from others. Similarly q5: there exists string 'a' leads to q6 (final), but also leads to accepting; but 'b' leads to dead (reject). For q0, string 'a' leads to dead (reject). So both have at least one rejecting string, but need compare all strings.

Are any two states indistinguishable? Let's test q0 and q1: Both have a->dead. Their b-transitions go to q1 and q2 respectively. Since q1 and q2 may be distinguishable, then maybe q0 and q1 are distinguishable. Need partition:

Initial partition: Final set F = {q0,q1,q2,q3,q5,q6,q7}, Non-final {q4}.

Now refine based on transitions to partitions.

Create table of transitions (on a,b) to which partition.

Currently: All finals in same block.

Check each final state's a- and b- destinations:

- q0: a->q4 (non-final), b->q1 (final)
- q1: a->q4 (non-final), b->q2 (final)
- q2: a->q5 (final), b->q3 (final)
- q3: a->q7 (final), b->q3 (final)
- q5: a->q6 (final), b->q4 (non-final)
- q6: a->q5 (final), b->q7 (final)
- q7: a->q7 (final), b->q7 (final)

Now we need to split block based on (a-target block, b-target block). Since currently two blocks: F and NF. So map (a->NF?, b->F?). For each:

q0: (NF, F)
q1: (NF, F) same as q0! So q0 and q1 are currently indistinguishable at this step.

q2: (F, F)
q3: (F, F)
q5: (F, NF)
q6: (F, F)
q7: (F, F)

Thus we split F into sub-blocks:

Block A: {q0,q1}
Block B: {q2,q3,q6,q7}
Block C: {q5}

Now iterate: Check within each block if further split.

Block A: q0 and q1 have transitions: a->q4 (NF) both same; b-> q1 (A) vs q2 (B). Since destinations differ (one to same block A, other to B), they are distinguishable. So split into {q0} and {q1}.

Block B: members q2,q3,q6,q7.

Compute their transitions:

- q2: a->q5 (C), b->q3 (B)
- q3: a->q7 (B), b->q3 (B)
- q6: a->q5 (C), b->q7 (B)
- q7: a->q7 (B), b->q7 (B)

Now see equivalence: Need pairwise check.

Consider q2 vs q6: both a->C same, b-> q3 vs q7 (both in B). Since q3 and q7 are currently in same block B, their b-targets are same block. So q2 and q6 may be equivalent for now.

q3 vs q7: both a->B (q7 vs q7) (both B), b->B also. So q3 and q7 potentially equivalent.

Now q2 vs q3: a-target C vs B => different, so distinguish.

q2 vs q7: a-target C vs B => distinguish.

q6 vs q3: a-target C vs B => distinguish.

Thus block B splits into {q2,q6} and {q3,q7}.

Now we have blocks: {q0}, {q1}, {q2,q6}, {q3,q7}, {q5}, {q4}. Need further refinement.

Check block {q2,q6}: 

- q2: a->q5 (C), b->q3 (B2)
- q6: a->q5 (C), b->q7 (B3)

Now their b-targets go to different blocks: q3 in {q3,q7} vs q7 also same block {q3,q7}. Actually q3 and q7 are in same block currently. So both go to same block. So still equivalent. Need check if later q3 and q7 split? They are currently in same block; we will see.

Check block {q3,q7}: 

- q3: a->q7 (same block), b->q3 (same block)
- q7: a->q7 (same), b->q7 (same). Both destinations within same block (could be themselves). However note that q3's b-loop goes to q3 (in block), q7's b-loop goes to q7 (also block). Since block contains both, they are indistinguishable for now.

Now consider block {q5}: its transitions a->q6 (block {q2,q6}), b->q4 (NF). Unique.

Now iterate: Need to see if any further splitting occurs due to differences in that block partitions.

Check block {q2,q6}: Both have a->q5 (C) same, b-> {q3,q7} same block. So stable.

Block {q3,q7}: a->q7 (within block) both, b-> respective states within block. Since both destinations stay within block, they equivalent.

Thus final partition: states: q0, q1, q2/q6, q3/q7, q5, q4. So are 6 equivalence classes (or 7? Let's count: q0, q1, {q2,q6}, {q3,q7}, q5, q4 = 6). Actually that's 6 distinct after merging: q0, q1, (q2,q6) merged, (q3,q7) merged, q5, q4.

Thus DFA is NOT minimal because we can merge q2 with q6, and q3 with q7. Also maybe q0 and q1 are distinct. So answer: No.

Provide option selections: Option 1 (Ja) = 0, Option 2 (Nein) = 1.

Also include brief justification.

Thus output:

<option id="1">0</option>
<option id="2">1</option>
<answer id="3">
The DFA can be reduced: states q₂ and q₆ are equivalent, as are q₃ and q₇; after merging them the automaton has fewer states, so the original DFA M is not minimal.
</answer>

<option id="1">0</option>
<option id="2">1</option>
<answer id="3">
The DFA is not minimal.  In the reachable part of M the following states are equivalent:
- q₂ and q₆ have identical future behaviour,
- q₃ and q₇ have identical future behaviour.

Merging each pair yields a smaller DFA, therefore the original DFA M is not minimal.
</answer>

Ist M minimal? Begründen Sie Ihre Antwort kurz.

The DFA is not minimal.  In the reachable part of M the following states are equivalent:
- q₂ and q₆ have identical future behaviour,
- q₃ and q₇ have identical future behaviour.

Merging each pair yields a smaller DFA, therefore the original DFA M is not minimal.

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: DFA-Algorithmen


## subproblem a (6.0 points)
Wir betrachten den folgenden NFA $N$ über dem Alphabet $\Sigma=\{a,b\}$:

Zustände: 0, 1, 2, 3
Startzustand: 0
Finalzustände: 0, 2, 3
Transitionen:
    0, b -> 0
    0, b -> 1
    1, b -> 1
    1, b -> 2
    2, a -> 3
    2, b -> 3
    3, a -> 1
    3, a -> 2

Sei $M$ der DFA, den man erhält, wenn man $N$ mithilfe des Verfahrens aus der Vorlesung (Potenzmengenkonstruktion) zu einem DFA konvertiert. Tragen Sie $M$ in untere Schablone ein, indem Sie (1) die Transitionen beschriften, (2) die Finalzustände markieren, und (3) in der Tabelle rechts für jeden Zustand von $M$ die korrespondierende Menge an Zuständen aus $N$ eintragen.

Zustände: q0, q1, q2, q3, q4, q5, q6, q7
Startzustand: q0
Finalzustände: ?
Transitionen:
    q0 -> q1
    q0 -> q4
    q1 -> q2
    q1 -> q4
    q2 -> q3
    q2 -> q5
    q3 -> q3
    q3 -> q7
    q7 -> q7
    q7 -> q7
    q5 -> q4
    q5 -> q6
    q6 -> q5
    q6 -> q7
    q4 -> q4
    q4 -> q4
Zustandstabelle:
    q0:
    q1:
    q2:
    q3:
    q4:
    q5:
    q6:
    q7:


<answer-template id="1">
Finalzustände: ?
Transitionen:
    ???
Zustandstabelle:
    q0: ?
    q1: ?
    q2: ?
    q3: ?
    q4: ?
    q5: ?
    q6: ?
    q7: ?

</answer-template>

<example>
Finalzustände: q0, q1, q2
Transitionen:
    q0, b -> q1
    q0, a -> q4
    ...
Zustandstabelle:
    q0: 0, 1
    q1: \emptyset
    q2: 0, 1, 2
    ...
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
Finalzustände: q0, q1, q2, q3, q5, q6, q7
Transitionen:
    q0, a -> q4
    q0, b -> q1
    q1, a -> q4
    q1, b -> q2
    q2, a -> q5
    q2, b -> q3
    q3, a -> q7
    q3, b -> q3
    q4, a -> q4
    q4, b -> q4
    q5, a -> q6
    q5, b -> q4
    q6, a -> q5
    q6, b -> q7
    q7, a -> q7
    q7, b -> q7
Zustandstabelle:
    q0: 0
    q1: 0, 1
    q2: 0, 1, 2
    q3: 0, 1, 2, 3
    q4: \emptyset
    q5: 3
    q6: 1, 2
    q7: 1, 2, 3
</answer>

user:
## subproblem b (1.0 points)
Ist M minimal? Begründen Sie Ihre Antwort kurz.

<options>
<option id="1">Ja</option>
<option id="2">Nein</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<answer id="3">
...
</answer>

assistant:<option id="1">0</option>
<option id="2">1</option>
<answer id="3">
The DFA is not minimal.  In the reachable part of M the following states are equivalent:
- q₂ and q₆ have identical future behaviour,
- q₃ and q₇ have identical future behaviour.

Merging each pair yields a smaller DFA, therefore the original DFA M is not minimal.

user:
## subproblem c (4.0 points)
Im Folgenden sind zwei DFAs $M_1$ und $M_2$ angegeben. Berechnen Sie nach dem Algorithmus aus der Vorlesung (Produktkonstruktion) einen DFA $M$ mit $L(M)=L(M_1)\cap L(M_2)$ und tragen Sie das Ergebnis unten ein. Konstruieren Sie nur die erreichbaren Zustände.

M_1:
Zustände: 0, 1, 2
Startzustand: 0
Finalzustände: 1, 2
Transitionen:
    0, a -> 1
    0, b -> 0
    1, a -> 2
    1, b -> 2
    2, a -> 2
    2, b -> 2


M_2:
Zustände: 0, 1, 2
Startzustand: 0
Finalzustände: 1, 2
Transitionen:
    0, a -> 2
    0, b -> 0
    1, a -> 2
    1, b -> 0
    2, a -> 1
    2, b -> 2


<answer-template id="1">
Zustände: ?
Startzustand: ?
Finalzustände: ?
Transitionen:
???

</answer-template>

<example>
Zustände: (0, 0), (0, 1), (1, 1)
Startzustand: (0, 0)
Finalzustände: (0, 0), (0, 1)
Transitionen:
(0, 0), a -> (0, 0)
(0, 0), b -> (0, 1)
(0, 1), a -> (1, 1)
(0, 1), b -> (0, 0)
(1, 1), a -> (1, 1)
(1, 1), b -> (0, 1)
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to compute DFA for intersection of M1 and M2. Only reachable states.

Given M1, M2 both with same alphabet {a,b}. Let's compute product states (p,q). Start (0,0). Determine transitions.

M1 transitions:
- From 0: a->1, b->0
- From 1: a->2, b->2
- From 2: a->2, b->2

M2 transitions:
- From 0: a->2, b->0
- From 1: a->2, b->0
- From 2: a->1, b->2

Now product: For each pair (p,q) reachable, define transition on a: (δ1(p,a), δ2(q,a)). On b: (δ1(p,b), δ2(q,b)).

Start (0,0). Compute its transitions.

- (0,0) on a: (1,2)
- on b: (0,0)

Add (1,2) reachable.

Now compute transitions for (1,2).

- a: (δ1(1,a)=2, δ2(2,a)=1) => (2,1)
- b: (δ1(1,b)=2, δ2(2,b)=2) => (2,2)

Add (2,1) and (2,2).

Now compute (2,1):

- a: (δ1(2,a)=2, δ2(1,a)=2) => (2,2) already.
- b: (δ1(2,b)=2, δ2(1,b)=0) => (2,0)

Add (2,0).

Compute (2,2):

- a: (2,1) already.
- b: (2,2) self-loop.

Compute (2,0):

- a: (2,2) already.
- b: (2,0) self-loop (since δ1(2,b)=2, δ2(0,b)=0) => (2,0) self-loop.

Thus reachable set: (0,0), (1,2), (2,1), (2,2), (2,0). Are there any others from these? No.

Now final states: Intersection: both components must be final in respective DFAs. M1 final states: {1,2}. M2 final states: {1,2}. So a product state (p,q) is final if p∈{1,2} and q∈{1,2}. Determine among reachable:

- (0,0): p=0 not final => not final.
- (1,2): p=1 (final), q=2 (final) => final.
- (2,1): both final => final.
- (2,2): both final => final.
- (2,0): q=0 not final => not final.

Thus final states = {(1,2), (2,1), (2,2)}.

Now we need to list states maybe as tuples like (0,0) etc. Provide transitions list.

Let's enumerate all reachable states listing transitions:

State (0,0):
- a -> (1,2)
- b -> (0,0)

State (1,2):
- a -> (2,1)
- b -> (2,2)

State (2,1):
- a -> (2,2)
- b -> (2,0)

State (2,2):
- a -> (2,1)
- b -> (2,2)

State (2,0):
- a -> (2,2)
- b -> (2,0)

That's all.

Now produce answer in required template:

<answer id="1">
Zustände: (0,0), (1,2), (2,1), (2,2), (2,0)
Startzustand: (0,0)
Finalzustände: (1,2), (2,1), (2,2)
Transitionen:
(0,0), a -> (1,2)
(0,0), b -> (0,0)
(1,2), a -> (2,1)
(1,2), b -> (2,2)
(2,1), a -> (2,2)
(2,1), b -> (2,0)
(2,2), a -> (2,1)
(2,2), b -> (2,2)
(2,0), a -> (2,2)
(2,0), b -> (2,0)
</answer>

Make sure formatting matches exactly.

<answer id="1">
Zustände: (0,0), (1,2), (2,1), (2,2), (2,0)
Startzustand: (0,0)
Finalzustände: (1,2), (2,1), (2,2)
Transitionen:
(0,0), a -> (1,2)
(0,0), b -> (0,0)
(1,2), a -> (2,1)
(1,2), b -> (2,2)
(2,1), a -> (2,2)
(2,1), b -> (2,0)
(2,2), a -> (2,1)
(2,2), b -> (2,2)
(2,0), a -> (2,2)
(2,0), b -> (2,0)
</answer>

Im Folgenden sind zwei DFAs $M_1$ und $M_2$ angegeben. Berechnen Sie nach dem Algorithmus aus der Vorlesung (Produktkonstruktion) einen DFA $M$ mit $L(M)=L(M_1)\cap L(M_2)$ und tragen Sie das Ergebnis unten ein. Konstruieren Sie nur die erreichbaren Zustände.
M_1:
Zustände: 0, 1, 2
Startzustand: 0
Finalzustände: 1, 2
Transitionen:
    0, a -> 1
    0, b -> 0
    1, a -> 2
    1, b -> 2
    2, a -> 2
    2, b -> 2
M_2:
Zustände: 0, 1, 2
Startzustand: 0
Finalzustände: 1, 2
Transitionen:
    0, a -> 2
    0, b -> 0
    1, a -> 2
    1, b -> 0
    2, a -> 1
    2, b -> 2

Zustände: (0,0), (1,2), (2,1), (2,2), (2,0)
Startzustand: (0,0)
Finalzustände: (1,2), (2,1), (2,2)
Transitionen:
(0,0), a -> (1,2)
(0,0), b -> (0,0)
(1,2), a -> (2,1)
(1,2), b -> (2,2)
(2,1), a -> (2,2)
(2,1), b -> (2,0)
(2,2), a -> (2,1)
(2,2), b -> (2,2)
(2,0), a -> (2,2)
(2,0), b -> (2,0)

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: DFA-Algorithmen


## subproblem a (6.0 points)
Wir betrachten den folgenden NFA $N$ über dem Alphabet $\Sigma=\{a,b\}$:

Zustände: 0, 1, 2, 3
Startzustand: 0
Finalzustände: 0, 2, 3
Transitionen:
    0, b -> 0
    0, b -> 1
    1, b -> 1
    1, b -> 2
    2, a -> 3
    2, b -> 3
    3, a -> 1
    3, a -> 2

Sei $M$ der DFA, den man erhält, wenn man $N$ mithilfe des Verfahrens aus der Vorlesung (Potenzmengenkonstruktion) zu einem DFA konvertiert. Tragen Sie $M$ in untere Schablone ein, indem Sie (1) die Transitionen beschriften, (2) die Finalzustände markieren, und (3) in der Tabelle rechts für jeden Zustand von $M$ die korrespondierende Menge an Zuständen aus $N$ eintragen.

Zustände: q0, q1, q2, q3, q4, q5, q6, q7
Startzustand: q0
Finalzustände: ?
Transitionen:
    q0 -> q1
    q0 -> q4
    q1 -> q2
    q1 -> q4
    q2 -> q3
    q2 -> q5
    q3 -> q3
    q3 -> q7
    q7 -> q7
    q7 -> q7
    q5 -> q4
    q5 -> q6
    q6 -> q5
    q6 -> q7
    q4 -> q4
    q4 -> q4
Zustandstabelle:
    q0:
    q1:
    q2:
    q3:
    q4:
    q5:
    q6:
    q7:


<answer-template id="1">
Finalzustände: ?
Transitionen:
    ???
Zustandstabelle:
    q0: ?
    q1: ?
    q2: ?
    q3: ?
    q4: ?
    q5: ?
    q6: ?
    q7: ?

</answer-template>

<example>
Finalzustände: q0, q1, q2
Transitionen:
    q0, b -> q1
    q0, a -> q4
    ...
Zustandstabelle:
    q0: 0, 1
    q1: \emptyset
    q2: 0, 1, 2
    ...
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
Finalzustände: q0, q1, q2, q3, q5, q6, q7
Transitionen:
    q0, a -> q4
    q0, b -> q1
    q1, a -> q4
    q1, b -> q2
    q2, a -> q5
    q2, b -> q3
    q3, a -> q7
    q3, b -> q3
    q4, a -> q4
    q4, b -> q4
    q5, a -> q6
    q5, b -> q4
    q6, a -> q5
    q6, b -> q7
    q7, a -> q7
    q7, b -> q7
Zustandstabelle:
    q0: 0
    q1: 0, 1
    q2: 0, 1, 2
    q3: 0, 1, 2, 3
    q4: \emptyset
    q5: 3
    q6: 1, 2
    q7: 1, 2, 3
</answer>

user:
## subproblem b (1.0 points)
Ist M minimal? Begründen Sie Ihre Antwort kurz.

<options>
<option id="1">Ja</option>
<option id="2">Nein</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<answer id="3">
...
</answer>

assistant:<option id="1">0</option>
<option id="2">1</option>
<answer id="3">
The DFA is not minimal.  In the reachable part of M the following states are equivalent:
- q₂ and q₆ have identical future behaviour,
- q₃ and q₇ have identical future behaviour.

Merging each pair yields a smaller DFA, therefore the original DFA M is not minimal.

user:
## subproblem c (4.0 points)
Im Folgenden sind zwei DFAs $M_1$ und $M_2$ angegeben. Berechnen Sie nach dem Algorithmus aus der Vorlesung (Produktkonstruktion) einen DFA $M$ mit $L(M)=L(M_1)\cap L(M_2)$ und tragen Sie das Ergebnis unten ein. Konstruieren Sie nur die erreichbaren Zustände.

M_1:
Zustände: 0, 1, 2
Startzustand: 0
Finalzustände: 1, 2
Transitionen:
    0, a -> 1
    0, b -> 0
    1, a -> 2
    1, b -> 2
    2, a -> 2
    2, b -> 2


M_2:
Zustände: 0, 1, 2
Startzustand: 0
Finalzustände: 1, 2
Transitionen:
    0, a -> 2
    0, b -> 0
    1, a -> 2
    1, b -> 0
    2, a -> 1
    2, b -> 2


<answer-template id="1">
Zustände: ?
Startzustand: ?
Finalzustände: ?
Transitionen:
???

</answer-template>

<example>
Zustände: (0, 0), (0, 1), (1, 1)
Startzustand: (0, 0)
Finalzustände: (0, 0), (0, 1)
Transitionen:
(0, 0), a -> (0, 0)
(0, 0), b -> (0, 1)
(0, 1), a -> (1, 1)
(0, 1), b -> (0, 0)
(1, 1), a -> (1, 1)
(1, 1), b -> (0, 1)
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
Zustände: (0,0), (1,2), (2,1), (2,2), (2,0)
Startzustand: (0,0)
Finalzustände: (1,2), (2,1), (2,2)
Transitionen:
(0,0), a -> (1,2)
(0,0), b -> (0,0)
(1,2), a -> (2,1)
(1,2), b -> (2,2)
(2,1), a -> (2,2)
(2,1), b -> (2,0)
(2,2), a -> (2,1)
(2,2), b -> (2,2)
(2,0), a -> (2,2)
(2,0), b -> (2,0)
</answer>

user:
## subproblem d (2.0 points)
Geben Sie jeweils ein kürzestes Wort in $L(M_1)\cap L(M_2)$, $L(M_1)\setminus L(M_2)$ und $L(M_2)\setminus L(M_1)$ an, oder $\emptyset$, falls kein solches Wort existiert.

*Hinweis:* Verwenden Sie Ihr Ergebnis aus c).

<answer-template id="1">
$L(M_1)\cap L(M_2)$: $?$
$L(M_1)\setminus L(M_2)$: $?$
$L(M_2)\setminus L(M_1)$: $?$
</answer-template>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to compute shortest words for intersection, M1\ M2, M2\M1.

First, recall M1 and M2 definitions.

M1:
states 0,1,2; start 0; finals 1,2.
Transitions:
0 a ->1
0 b ->0
1 a ->2
1 b ->2
2 a ->2
2 b ->2

So M1 accepts any word that leads to state 1 or 2. Let's analyze language: from start 0, reading a gets to 1 (accept). reading b stays 0 (non-accept). Once in 1, any next symbol (a or b) goes to 2 (accept). From 2 stays accept forever. So language: all strings over {a,b} that contain at least one 'a' (because need to move from 0 to 1 via a). Actually also strings where the first a may be later: any string with at least one a. Because once an a occurs, we go to 1 then accept. After that any further symbols stay accept. So L(M1) = { w | w contains at least one a }.

Check: what about a at position later after some b's: start 0, b loop stays 0, then a to 1 accept. yes.

Thus L(M1) = Σ* a Σ* (with at least one a). So complement is strings with no a: only b*.

M2:

states 0,1,2; start 0; finals 1,2.
Transitions:
0 a ->2
0 b ->0
1 a ->2
1 b ->0
2 a ->1
2 b ->2

Let's analyze.

Start 0. b loops to 0 (non-accept). a from 0 goes to 2 (accept). From 2, a ->1 (accept), b ->2 (accept). From 1, a ->2 (accept), b ->0 (non-accept). So accept states 1 and 2.

We need to characterize language.

We can think of automaton as: states represent parity of a's? Let's compute reachable behavior.

We can consider reading a toggles between 2 and 1? Actually from 0 on a ->2. From 2 on a ->1. From 1 on a ->2. So a toggles between 2 and 1. b from 2 stays 2, from 1 goes to 0, from 0 stays 0. So b resets to 0 only if currently in state 1; otherwise stays 2 or 0.

Thus accept if after reading whole word, we are in 1 or 2. So the only non-accept state is 0.

Thus language L(M2) = all strings that do NOT end in state 0. Since start 0 is non-accept, we need to avoid being in 0 at end.

When does automaton end in 0? The only transition to 0 is from 1 on b. Also from 0 on b stays 0. So to be in 0 at end, either we never left 0 (only b's), or we were in 1 then read a b. Also from 2 cannot go to 0 directly.

Thus words that keep us in 0: sequences of b's only (b*). Also any word that after some prefix leads to 1 then a b ends in 0, and further symbols after that might move out again? From 0 again only b's keep 0; a moves to 2.

So generally language is complement of set of words that end in 0.

Let's try to compute language description.

Let's consider automaton as deterministic; we can compute reachable states for each word.

We can attempt to find shortest word in intersection L(M1)∩L(M2): need word containing at least one a (M1) and not ending in state 0 (M2). Let's test small words.

Length 1:
a: M1? contains a -> accept. M2: start 0, a->2 (accept). So a is in intersection. So shortest word is "a". Good.

Now L(M1) \ L(M2): words accepted by M1 but not by M2, i.e., contain at least one a, but end in state 0 in M2.

We need shortest such word.

Check "b": does not contain a, not accepted M1. So not.

Length 1 "a": accepted by both, not.

Length 2: possibilities:

"aa": M1 accept (contains a). M2: start 0 a->2, then a -> from 2 a->1 (accept). So accept, not.

"ab": M1 accept. M2: 0 a->2, then b from 2 ->2 (accept). So accept.

"ba": M1 accept. M2: 0 b->0, then a ->2 (accept). So accept.

"bb": M1 reject.

Thus none length2.

Length3: need contain a, end in 0. Let's try patterns.

We need that after reading whole word, M2 state is 0. Need transition to 0 at final step from state 1 via b, or stay in 0 whole time (i.e., word of only b's, but that lacks a, so not). So we need to end with ...b where prior state is 1.

Thus we need a word that leads to state 1 before final b. How to reach state 1? Starting from 0, we need an odd number of a's? Actually route: 0 a ->2 (state2). From 2, a ->1 (state1). Also from 2, b stays 2. From 1, a->2, b->0. So state1 reachable after an odd number (>0) of a's? Starting with a yields state2; then another a yields state1. So after two a's (aa) we are in 1. Or after pattern that leads to 2 then a ->1.

Thus need a word that after some prefix ends in state1, then final b to go to 0.

Shortest: "aab". Let's test.

M1: contains a, accept.

M2: 0 a->2, a->1, b->0 => ends in 0 (reject). So it's in M1\ M2. Length 3.

Check if any length 2 works: need to end in 0, so final step must be b from state1, but to be in state1 after 1 symbol? Starting 0, after first symbol must be state1. Is there a one-symbol transition to state1? No. So need at least two symbols before final, so length at least 2? Actually to be in state1 after first symbol we need something that goes to 1 directly, none. So length >=2. For length2, need prefix of length1 be state1, impossible. So minimal length is 3; "aab" works. Could there be "aba"? ends a not b. Not. "baa"? 0 b->0, a ->2, a->1, ends in 1 not 0. Need final b. So "baa b"? that's length4. So shortest is "aab". Good.

Now L(M2) \ L(M1): words accepted by M2 but not by M1: must not contain any a (since M1 requires a). So must be strings over only b, but also accepted by M2 (i.e., not end in state0). However only b's keep M2 in 0 always, because from start 0 b loops to 0; from 0 always stay 0 with b. So any all-b word ends in 0 (reject). Thus there is no word accepted by M2 without a. Therefore L(M2)\L(M1) = empty set. So answer is ∅.

Thus we provide shortest words: intersection "a", M1\M2 "aab", M2\M1 "∅".

Provide answer using template.

<answer id="1">
$L(M_1)\cap L(M_2)$: $a$  
$L(M_1)\setminus L(M_2)$: $aab$  
$L(M_2)\setminus L(M_1)$: $\emptyset$
</answer>

Geben Sie jeweils ein kürzestes Wort in $L(M_1)\cap L(M_2)$, $L(M_1)\setminus L(M_2)$ und $L(M_2)\setminus L(M_1)$ an, oder $\emptyset$, falls kein solches Wort existiert.

Hinweis: Verwenden Sie Ihr Ergebnis aus c).

$L(M_1)\cap L(M_2)$: $a$  
$L(M_1)\setminus L(M_2)$: $aab$  
$L(M_2)\setminus L(M_1)$: $\emptyset$

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: DFA-Algorithmen


## subproblem a (6.0 points)
Wir betrachten den folgenden NFA $N$ über dem Alphabet $\Sigma=\{a,b\}$:

Zustände: 0, 1, 2, 3
Startzustand: 0
Finalzustände: 0, 2, 3
Transitionen:
    0, b -> 0
    0, b -> 1
    1, b -> 1
    1, b -> 2
    2, a -> 3
    2, b -> 3
    3, a -> 1
    3, a -> 2

Sei $M$ der DFA, den man erhält, wenn man $N$ mithilfe des Verfahrens aus der Vorlesung (Potenzmengenkonstruktion) zu einem DFA konvertiert. Tragen Sie $M$ in untere Schablone ein, indem Sie (1) die Transitionen beschriften, (2) die Finalzustände markieren, und (3) in der Tabelle rechts für jeden Zustand von $M$ die korrespondierende Menge an Zuständen aus $N$ eintragen.

Zustände: q0, q1, q2, q3, q4, q5, q6, q7
Startzustand: q0
Finalzustände: ?
Transitionen:
    q0 -> q1
    q0 -> q4
    q1 -> q2
    q1 -> q4
    q2 -> q3
    q2 -> q5
    q3 -> q3
    q3 -> q7
    q7 -> q7
    q7 -> q7
    q5 -> q4
    q5 -> q6
    q6 -> q5
    q6 -> q7
    q4 -> q4
    q4 -> q4
Zustandstabelle:
    q0:
    q1:
    q2:
    q3:
    q4:
    q5:
    q6:
    q7:


<answer-template id="1">
Finalzustände: ?
Transitionen:
    ???
Zustandstabelle:
    q0: ?
    q1: ?
    q2: ?
    q3: ?
    q4: ?
    q5: ?
    q6: ?
    q7: ?

</answer-template>

<example>
Finalzustände: q0, q1, q2
Transitionen:
    q0, b -> q1
    q0, a -> q4
    ...
Zustandstabelle:
    q0: 0, 1
    q1: \emptyset
    q2: 0, 1, 2
    ...
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
Finalzustände: q0, q1, q2, q3, q5, q6, q7
Transitionen:
    q0, a -> q4
    q0, b -> q1
    q1, a -> q4
    q1, b -> q2
    q2, a -> q5
    q2, b -> q3
    q3, a -> q7
    q3, b -> q3
    q4, a -> q4
    q4, b -> q4
    q5, a -> q6
    q5, b -> q4
    q6, a -> q5
    q6, b -> q7
    q7, a -> q7
    q7, b -> q7
Zustandstabelle:
    q0: 0
    q1: 0, 1
    q2: 0, 1, 2
    q3: 0, 1, 2, 3
    q4: \emptyset
    q5: 3
    q6: 1, 2
    q7: 1, 2, 3
</answer>

user:
## subproblem b (1.0 points)
Ist M minimal? Begründen Sie Ihre Antwort kurz.

<options>
<option id="1">Ja</option>
<option id="2">Nein</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<answer id="3">
...
</answer>

assistant:<option id="1">0</option>
<option id="2">1</option>
<answer id="3">
The DFA is not minimal.  In the reachable part of M the following states are equivalent:
- q₂ and q₆ have identical future behaviour,
- q₃ and q₇ have identical future behaviour.

Merging each pair yields a smaller DFA, therefore the original DFA M is not minimal.

user:
## subproblem c (4.0 points)
Im Folgenden sind zwei DFAs $M_1$ und $M_2$ angegeben. Berechnen Sie nach dem Algorithmus aus der Vorlesung (Produktkonstruktion) einen DFA $M$ mit $L(M)=L(M_1)\cap L(M_2)$ und tragen Sie das Ergebnis unten ein. Konstruieren Sie nur die erreichbaren Zustände.

M_1:
Zustände: 0, 1, 2
Startzustand: 0
Finalzustände: 1, 2
Transitionen:
    0, a -> 1
    0, b -> 0
    1, a -> 2
    1, b -> 2
    2, a -> 2
    2, b -> 2


M_2:
Zustände: 0, 1, 2
Startzustand: 0
Finalzustände: 1, 2
Transitionen:
    0, a -> 2
    0, b -> 0
    1, a -> 2
    1, b -> 0
    2, a -> 1
    2, b -> 2


<answer-template id="1">
Zustände: ?
Startzustand: ?
Finalzustände: ?
Transitionen:
???

</answer-template>

<example>
Zustände: (0, 0), (0, 1), (1, 1)
Startzustand: (0, 0)
Finalzustände: (0, 0), (0, 1)
Transitionen:
(0, 0), a -> (0, 0)
(0, 0), b -> (0, 1)
(0, 1), a -> (1, 1)
(0, 1), b -> (0, 0)
(1, 1), a -> (1, 1)
(1, 1), b -> (0, 1)
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
Zustände: (0,0), (1,2), (2,1), (2,2), (2,0)
Startzustand: (0,0)
Finalzustände: (1,2), (2,1), (2,2)
Transitionen:
(0,0), a -> (1,2)
(0,0), b -> (0,0)
(1,2), a -> (2,1)
(1,2), b -> (2,2)
(2,1), a -> (2,2)
(2,1), b -> (2,0)
(2,2), a -> (2,1)
(2,2), b -> (2,2)
(2,0), a -> (2,2)
(2,0), b -> (2,0)
</answer>

user:
## subproblem d (2.0 points)
Geben Sie jeweils ein kürzestes Wort in $L(M_1)\cap L(M_2)$, $L(M_1)\setminus L(M_2)$ und $L(M_2)\setminus L(M_1)$ an, oder $\emptyset$, falls kein solches Wort existiert.

*Hinweis:* Verwenden Sie Ihr Ergebnis aus c).

<answer-template id="1">
$L(M_1)\cap L(M_2)$: $?$
$L(M_1)\setminus L(M_2)$: $?$
$L(M_2)\setminus L(M_1)$: $?$
</answer-template>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
$L(M_1)\cap L(M_2)$: $a$  
$L(M_1)\setminus L(M_2)$: $aab$  
$L(M_2)\setminus L(M_1)$: $\emptyset$
</answer>

user:
## subproblem e (5.0 points)
Sei $L\subseteq\{a,b,c\}^*$ eine Sprache. Wir definieren die Sprache $R_{c}(L)\subseteq\{a,b\}^*$ als die Menge der Wörter, die man erhält, wenn man aus jedem Wort von $L$ jedes Zeichen $c$ löscht.
Zum Beispiel gilt $R_{c}(\{ba, bca, cc, ca\})=\{ba,\varepsilon,a\}$.

Sei $M$ ein DFA mit $n$ Zuständen. Wählen Sie eine Funktion $f:\N\rightarrow\N$, sodass es einen DFA $M'$ für $R_{c}(L(M))$ mit höchstens $f(n)$ Zuständen gibt **und** beschreiben Sie einen Algorithmus, um $M'$ zu berechnen uns argumentieren Sie, wieso $M'$ höchstens $f(n)$ Zustände hat.

<answer-template id="1">
Ihre Wahl: $f(n):=?$
Begründung:
???
</answer-template>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to answer subproblem e. Provide a function f(n) and describe algorithm to construct DFA M' for R_c(L(M)). Language over {a,b,c}; we delete all c's from words. So we need to simulate original DFA but ignore c's. Classic construction: replace each transition on c with epsilon (i.e., stay in same state). Equivalent to compute DFA where c-transitions become self-loops? Actually deleting c means that when reading c, we do not change state (since c is removed). But careful: In original DFA M, reading c may change state, but after deletion we won't have c in the word, meaning we want to accept a word w over {a,b}* if there exists a word w' over {a,b,c}* that after deleting c's yields w and w' is accepted by M. That is: Language of M' = { w ∈ {a,b}* | ∃ u ∈ {a,b,c}* such that u ∈ L(M) and R_c(u)=w }. That's a projection/homomorphism that deletes c. This is like applying a homomorphism h: Σ* → Σ'*, where h(c)=ε, h(a)=a, h(b)=b. The preimage under homomorphism of a regular language is regular; we can construct NFA by adding epsilon transitions for c, then determinize. Standard method: build NFA N' that simulates M but when reading c we take ε-move (i.e., stay without consuming input). Actually need to allow arbitrary number of c's interleaved; we can replace each c-transition of M by an ε-transition (i.e., allow moving without consuming input), also add ε-loop at each state for c? Wait: The original M reads c and changes state; but when we delete c from the word, we need to allow skipping c's, i.e., we can simulate M while reading w but also may guess that there were some c's inserted in between, potentially causing state changes. That corresponds to allowing ε-transitions that simulate the effect of reading any sequence of c's. So we construct ε-NFA N where for each c-transition p --c--> q in M, we add an ε-transition from p to q (since we can ignore c). Also we need to allow any number of c's, thus also we could have a path of multiple c's, which correspond to sequences of ε-transitions. Additionally we should also keep original a,b transitions same as M (consuming input). Accepting states same as M. Then we need to eliminate ε by ε-closure and determinize (subset construction). The resulting DFA has at most 2^n states, but we can potentially do better: we can compute ε-closure of each state (including reachable via any number of c-transitions). This is a set of states reachable by only c-transitions (including zero). Precompute closure for each state; then define new DFA M' with same set of states as original M (n) but where transition on a is from state p to closure of δ(p,a) (i.e., take a transition, then take closure). Similarly for b. Accepting states are those whose closure includes some accepting state of M. So we can stay with at most n states. Indeed we can construct a DFA with same number of states n (or maybe less). The function f(n) = n. Or could be n (or n+1). Let's argue.

Algorithm: compute for each state q the set C(q) = { p ∈ Q | q ⇒_c* p } i.e., reachable via zero or more c-transitions (including q). This can be done with graph reachability (DFS) on c-transitions only. Then define M' = (Q, {a,b}, δ', q0, F'), where δ'(q, x) = C(δ(q, x)) for x∈{a,b}. If δ(q,x) undefined, treat as dead sink (could add a dead state). Accepting set F' = { q | C(q) ∩ F ≠ ∅ }. This yields at most n states (maybe need extra dead sink if missing transitions). So f(n)= n+1 to include dead sink. Provide algorithm and justification.

Answer accordingly using template.

<answer id="1">
Ihre Wahl: $f(n)=n+1$

Begründung:<br>
Gegeben ein DFA  
\(M=(Q,\{a,b,c\},\delta,q_{0},F)\) mit \(|Q|=n\).

1. **c‑Closure berechnen**  
   Für jeden Zustand \(q\in Q\) bestimmen wir die Menge  
   \[
   C(q)=\{p\in Q \mid q \xrightarrow{c^{*}} p\},
   \]
   also alle Zustände, die von \(q\) über **beliebig viele** c‑Übergänge (auch 0) erreichbar sind.  
   Da die c‑Übergänge allein einen gerichteten Graphen bilden, lässt sich \(C(q)\) für alle \(q\) in Zeit \(O(n^{2})\) mittels einer Tiefensuche bzw. Floyd‑Warshall ermitteln.

2. **DFA ohne c‑Eingaben konstruieren**  
   Wir definieren einen neuen Automaten  
   \[
   M'=(Q\cup\{q_{\text{sink}}\},\{a,b\},\delta',q_{0},F')
   \]
   mit den gleichen Zuständen wie \(M\) (plus einem einzigen Fehlzustand \(q_{\text{sink}}\), falls ein Übergang nicht definiert ist).

   * **Übergangsfunktion**  
     Für \(x\in\{a,b\}\) und \(q\in Q\) setzen wir
     \[
     \delta'(q,x)=
     \begin{cases}
        \text{ein beliebiger Zustand } p\in C(\delta(q,x)) & \text{wenn }\delta(q,x)\text{ definiert ist},\\[4pt]
        q_{\text{sink}} & \text{sonst}.
     \end{cases}
     \]
     Das bedeutet: wir führen zunächst den normalen \(x\)-Übergang von \(M\) aus und „überspringen“ danach beliebig viele c‑Übergänge, indem wir sofort in die entsprechende c‑Closure gehen.

   * **Endzustände**  
     Ein Zustand \(q\) ist Endzustand von \(M'\), genau dann wenn die c‑Closure von \(q\) einen Endzustand von \(M\) enthält:
     \[
     F'=\{\,q\in Q \mid C(q)\cap F\neq\emptyset\,\}\;\cup\;\{q_{\text{sink}}\;:\;q_{\text{sink}}\in F\}.
     \]

3. **Korrektheit**  
   Für ein Wort \(w\in\{a,b\}^{*}\) lässt sich ein Wort \(\hat w\in\{a,b,c\}^{*}\) mit \(R_{c}(\hat w)=w\) exakt dann erzeugen, wenn man in \(M\) von \(q_{0}\) aus die Buchstaben von \(w\) liest und zwischendurch beliebig viele c‑Übergänge ausführt.  Durch die Definition von \(\delta'\) und \(F'\) akzeptiert \(M'\) genau die Wörter, für die ein solches \(\hat w\) existiert; also
   \[
   L(M') = R_{c}\bigl(L(M)\bigr).
   \]

4. **Zustandsanzahl**  
   Der konstruierte Automat verwendet höchstens die ursprünglichen \(n\) Zustände von \(M\) plus ggf. einen zusätzlichen Fehlzustand, also höchstens \(n+1\) Zustände.  Damit erfüllt er die geforderte Schranke \(f(n)=n+1\).

Der beschriebene Algorithmus läuft in polynomialer Zeit (O(n²) für die Closure‑Berechnung und O(n) für das Anlegen der neuen Übergänge) und liefert einen DFA \(M'\) mit \(\le f(n)=n+1\) Zuständen, der exakt die Sprache \(R_{c}(L(M))\) akzeptiert.
</answer>

Sei $L\subseteq\{a,b,c\}^*$ eine Sprache. Wir definieren die Sprache $R_{c}(L)\subseteq\{a,b\}^*$ als die Menge der Wörter, die man erhält, wenn man aus jedem Wort von $L$ jedes Zeichen $c$ löscht.
Zum Beispiel gilt $R_{c}(\{ba, bca, cc, ca\})=\{ba,\varepsilon,a\}$.

Sei $M$ ein DFA mit $n$ Zuständen. Wählen Sie eine Funktion $f:\N\rightarrow\N$, sodass es einen DFA $M'$ für $R_{c}(L(M))$ mit höchstens $f(n)$ Zuständen gibt und beschreiben Sie einen Algorithmus, um $M'$ zu berechnen uns argumentieren Sie, wieso $M'$ höchstens $f(n)$ Zustände hat.

Ihre Wahl: $f(n)=n+1$

Begründung:<br>
Gegeben ein DFA  
$M=(Q,\{a,b,c\},\delta,q_{0},F)$ mit $|Q|=n$.

1. c‑Closure berechnen  
   Für jeden Zustand $q\in Q$ bestimmen wir die Menge  
   $$C(q)=\{p\in Q \mid q \xrightarrow{c^{*}} p\},$$
   also alle Zustände, die von $q$ über beliebig viele c‑Übergänge (auch 0) erreichbar sind.  
   Da die c‑Übergänge allein einen gerichteten Graphen bilden, lässt sich $C(q)$ für alle $q$ in Zeit $O(n^{2})$ mittels einer Tiefensuche bzw. Floyd‑Warshall ermitteln.

2. DFA ohne c‑Eingaben konstruieren  
   Wir definieren einen neuen Automaten  
   $$M'=(Q\cup\{q_{\text{sink}}\},\{a,b\},\delta',q_{0},F')$$
   mit den gleichen Zuständen wie $M$ (plus einem einzigen Fehlzustand $q_{\text{sink}}$, falls ein Übergang nicht definiert ist).

   * Übergangsfunktion  
     Für $x\in\{a,b\}$ und $q\in Q$ setzen wir
     $$\delta'(q,x)=
     \begin{cases}
        \text{ein beliebiger Zustand } p\in C(\delta(q,x)) & \text{wenn }\delta(q,x)\text{ definiert ist},\\[4pt]
        q_{\text{sink}} & \text{sonst}.
     \end{cases}$$
     Das bedeutet: wir führen zunächst den normalen $x$-Übergang von $M$ aus und „überspringen“ danach beliebig viele c‑Übergänge, indem wir sofort in die entsprechende c‑Closure gehen.

   * Endzustände  
     Ein Zustand $q$ ist Endzustand von $M'$, genau dann wenn die c‑Closure von $q$ einen Endzustand von $M$ enthält:
     $$F'=\{\,q\in Q \mid C(q)\cap F\neq\emptyset\,\}\;\cup\;\{q_{\text{sink}}\;:\;q_{\text{sink}}\in F\}.$$

3. Korrektheit  
   Für ein Wort $w\in\{a,b\}^{*}$ lässt sich ein Wort $\hat w\in\{a,b,c\}^{*}$ mit $R_{c}(\hat w)=w$ exakt dann erzeugen, wenn man in $M$ von $q_{0}$ aus die Buchstaben von $w$ liest und zwischendurch beliebig viele c‑Übergänge ausführt.  Durch die Definition von $\delta'$ und $F'$ akzeptiert $M'$ genau die Wörter, für die ein solches $\hat w$ existiert; also
   $$L(M') = R_{c}\bigl(L(M)\bigr).$$

4. Zustandsanzahl  
   Der konstruierte Automat verwendet höchstens die ursprünglichen $n$ Zustände von $M$ plus ggf. einen zusätzlichen Fehlzustand, also höchstens $n+1$ Zustände.  Damit erfüllt er die geforderte Schranke $f(n)=n+1$.

Der beschriebene Algorithmus läuft in polynomialer Zeit (O(n²) für die Closure‑Berechnung und O(n) für das Anlegen der neuen Übergänge) und liefert einen DFA $M'$ mit $\le f(n)=n+1$ Zuständen, der exakt die Sprache $R_{c}(L(M))$ akzeptiert.

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Rekursive Funktionen

Sei $\Sigma:=\{0,1\}$. Für eine Sprache $L\subseteq\Sigma^*$ bezeichnet $N(L)$ die Sprache, die man erhält, wenn man den Wörtern aus $L$ **höchstens eine** Null an einer beliebigen Stelle hinzufügt. Formal definieren wir
\[N(L):= \{vxw \mid v,w\in \Sigma^*\wedge x\in \{0,\varepsilon\}\wedge vw\in L\}\]

## subproblem a (1.0 points)
Listen Sie alle Wörter der Sprache $N(\{01, 0\})$ explizit auf:

<answer-template id="1">
$N(\{01, 0\})=\{?\}$
</answer-template>

<example>
$N(\{01, 0\})=\{\varepsilon, 0, 01\}$
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to list all words of N({01,0}). Definition: N(L) = { v x w | v,w ∈ Σ* and x ∈ {0, ε} and v w ∈ L }. So we take any decomposition of a word in L into v w (concatenation). Then we optionally insert a 0 at any position between v and w (i.e., between them). Actually x is either 0 or ε, placed between v and w. So we can either not insert (x=ε) giving original word, or insert a 0 at any split point.

We need to consider all possible splits of each word in L.

L = { "01", "0" }.

First word "01". All splits v|w where v,w ∈ Σ* and v w = "01". Splits: (v="", w="01"), (v="0", w="1"), (v="01", w=""). For each split, we can insert ε (i.e., keep original) or insert 0.

- Split ("", "01"): inserting ε gives ""+""+ "01" = "01". inserting 0 gives "0"+"01" = "001".
- Split ("0","1"): inserting ε gives "0"+"1" = "01". inserting 0 gives "0"+"0"+"1" = "001"? Wait "0"+"0"+"1" = "001". Same as before.
- Split ("01",""): inserting ε gives "01". inserting 0 gives "01"+"0" = "010".

Thus from "01" we get words: "01", "001", "010". Note duplicates: "01" appears multiple times but set.

Now word "0". Splits: (v="", w="0") and (v="0", w="").

- Split ("","0"): ε insertion gives "0". inserting 0 gives "0"+"0" = "00".
- Split ("0",""): ε insertion gives "0". inserting 0 gives "0"+"0" = "00". So we get "0" and "00".

Collect all: {"01","001","010","0","00"}.

Check if any other words like "": empty? Could we get empty? No because original L words are non-empty, and we only insert 0 optionally, not delete. So no empty.

Thus answer: N({01,0}) = {0,00,01,010,001}. Order not important.

Provide in template: $N(\{01,0\})=\{0,00,01,010,001\}$.

Make sure format exactly.

<answer id="1">
$N(\{01, 0\})=\{0,00,01,010,001\}$
</answer>

Listen Sie alle Wörter der Sprache $N(\{01, 0\})$ explizit auf:

$N(\{01, 0\})=\{0,00,01,010,001\}$

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Rekursive Funktionen

Sei $\Sigma:=\{0,1\}$. Für eine Sprache $L\subseteq\Sigma^*$ bezeichnet $N(L)$ die Sprache, die man erhält, wenn man den Wörtern aus $L$ **höchstens eine** Null an einer beliebigen Stelle hinzufügt. Formal definieren wir
\[N(L):= \{vxw \mid v,w\in \Sigma^*\wedge x\in \{0,\varepsilon\}\wedge vw\in L\}\]

## subproblem a (1.0 points)
Listen Sie alle Wörter der Sprache $N(\{01, 0\})$ explizit auf:

<answer-template id="1">
$N(\{01, 0\})=\{?\}$
</answer-template>

<example>
$N(\{01, 0\})=\{\varepsilon, 0, 01\}$
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:

user:
## subproblem b (6.0 points)
Geben Sie eine rekursive Funktion $\Null$ an, sodass $\Null(r)$ ein regulärer Ausdruck mit $L(\Null(r)) = N(L(r))$ ist, für jeden regulären Ausdruck $r$. Im Folgenden sind $r,r_1,r_2$ reguläre Ausdrücke, und $a\in \Sigma$.

<answer-template id="1">
$\Null(\emptyset) := ?$
$\Null(\epsilon) := ?$
$\Null(a) := ?$
$\Null(r_1 \mid r_2) := ?$
$\Null(r_1 r_2) := ?$
$\Null(r^*) := ?$
</answer-template>

<example>
$\Null(\emptyset) := \epsilon$
...
$\Null(r_1 \mid r_2) := \Null(r_1) \Null(r_2)$
...
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to answer two subproblems. Subproblem a: list all words of N({01,0}). Using definition: N(L): add at most one 0 at any position (including none) to strings in L. For each word w in L, we can either keep as is, or insert a single 0 somewhere (including at beginning, between any two symbols, or at end). Also we can choose not to insert (since "höchstens eine Null"). So compute.

L = { "01", "0" }.

For "01": length 2. Possible insert positions: before first char, between chars, after last char. Insert a "0". Also no insertion. So:

- No insertion: "01"
- Insert at position 0: "001"
- Insert between: "001"? Wait between chars: original "0 1". Insert between yields "001"? Actually first char is '0', then insert '0', then '1' => "001". That's same as before? Before first char also yields "001". Wait before first char: "0"+"01" => "001". So both produce same word. Insert after last: "010". So distinct results: "01", "001", "010". Also note insertion of ε (no insertion) already counted.

Now for "0": length 1. Positions: before, after. Insert a 0:

- No insertion: "0".
- Insert before: "00".
- Insert after: "00". Both same, just "00". So distinct words: "0", "00".

Combine sets: {"01","001","010","0","00"}.

Check duplicates: "001" may be duplicate with something else? Not from other word. So final set: {0, 00, 01, 001, 010}. Should we also include epsilon? No, because no word in L is epsilon, and can't insert a 0 into empty string because we only add 0 if we have a word from L, and we have at most one 0 added. Could we take vw in L and choose x=ε (no 0) -> yields same word. So epsilon not included. So answer.

Subproblem b: define recursive function Null (maybe they want to add a 0 optionally). We need to produce a regular expression that denotes N(L(r)). N(L) is at most one extra 0 inserted anywhere.

We need Null(r) such that its language is N(L(r)). So define function Null recursively based on structure.

We need to produce regular expression that matches any string that is either in L(r), or can be obtained by inserting exactly one 0 somewhere in a string of L(r). That's like closure over insertion of a single 0.

One way: N(L) = L ∪ ( (0)·L ∪ Σ*0Σ*? Wait needs exactly one insertion of 0 at any position in a word of L. Equivalent to: all strings w such that either w∈L, or there exist u,v∈Σ* such that u0v ∈ L? Actually we insert a 0 into a word from L to get w. So w = v x w' where x∈{0, ε} and vw'∈L . That's given: w = v x w' where x is either 0 or ε and vw'∈L . So either w = vw' (no insertion) => original word, or w = v 0 w' where v,w' any split of a word from L. So N(L) = L ∪ {v0w' | ∃ u∈L, u = v w'}. That's basically L ∪ Σ*0Σ* intersect words that can be formed by inserting 0 inside an L word.

But we need a regular expression built from r. A classic construction: N(L(r)) = L(r) ∪ ( ( ( Σ* )0( Σ* ) ) ∩ L(r)·? Not simple.

But maybe they expect a definition of Null that expands regex by allowing extra optional 0 at each concatenation point. Could be define Null recursively as:

- Null(∅) = ∅ (or maybe ε? Let's test. N(∅) = ∅ because no words to add 0 to. So Null(∅) should denote ∅. So Null(∅) = ∅.

- Null(ε) = ε ∪ 0 (since from empty string you can insert at most one 0. With original ε, no insertion => ε. With insertion of a 0 at any position (only position) => 0. So Null(ε) = ε | 0.

- Null(a) where a∈Σ: L = {a}. Adding at most one 0 anywhere gives strings: a, 0a, a0. Also if we insert before a (makes 0a), after a (a0). So Null(a) = a | 0a | a0 . That's a|(0a)|(a0). Could be expressed as (0? a 0?)? Actually with at most one 0, we cannot have both sides 0, just one. So expression: a | 0a | a0.

- Null(r1 | r2) = Null(r1) | Null(r2). Because union just union languages, and N distributes over union.

- Null(r1 r2): We need language of concatenation of L(r1) L(r2) plus one extra 0 possibly inserted anywhere within the concatenated word. Inserting a 0 could be: before the whole concatenation, after the whole concatenation, or somewhere inside r1, or at the boundary between r1 and r2 (i.e., at the split). So N(L1·L2) = N(L1)·L2 ∪ L1·N(L2) ∪ L1·L2 (but the latter already covered by both? Actually N includes original words, so union includes that). But careful: inserting a 0 in L1 part yields strings v0w·x where v w ∈ L1. That's Null(r1)·r2. Similarly, insert inside r2 yields r1·Null(r2). Also inserting at the concatenation boundary, i.e., after some prefix from L1 and before start of L2, corresponds to adding a 0 between them: that's also covered by Null(r1)·r2 where Null adds the 0 at the end of L1? Wait Null(r1) allows insertion anywhere in L1, including at end. So a 0 inserted exactly at end of L1 then concatenated with L2 yields same as insertion at boundary. Similarly insertion at beginning of L2 is covered by r1·Null(r2) where Null inserts at beginning. So union of both covers all possibilities, plus original concatenation (included in both). So define:

Null(r1 r2) = Null(r1) r2 | r1 Null(r2). But we should avoid duplication; we can define as (Null(r1) r2) | (r1 Null(r2)). That works.

- Null(r*) : The language L(r*) = (L(r))*. Adding at most one 0 anywhere inside a string from L(r)*. This could be: either the 0 is inserted in one of the factors (including before first factor or after last). Equivalent to: (Null(r))*? Not exactly. Actually we need to allow strings that are concatenations of zero or more copies of L(r), possibly with a single extra 0 inserted somewhere. This can be expressed as (Null(r))*? Hmm.

Consider w ∈ N(L(r*)). This means w is either in L(r*) (no extra 0) or w = v0u where v u ∈ L(r*) (i.e., there is some split of a string from L(r*) such that 0 inserted). Since L(r*) is closed under concatenation, we might write: N(L(r*)) = L(r*) ∪ ( Σ*0Σ* intersect L(r*)? Actually any split of a word from L(r*) yields prefix and suffix each in L(r*). So we can think of N(L(r*)) = L(r*) ∪ (L(r*)·0·L(r*)). But careful: Inserting a 0 anywhere means we take a word s ∈ L(r*) and split it into s = uv . Then w = u 0 v. Both u and v are themselves in L(r*) (since L(r*) is prefix-closed? Not exactly; any prefix of a word in L(r*) may not be in L(r*). Example: r = 'a', L(r*) = a*, prefix 'aa' is in a*, yes because any prefix of a* is also a* (including empty). Indeed any prefix of a word from L(r*) is also in L(r*), because a* is closed under prefix? For any k ≤ n, the prefix of a^n is a^k ∈ a*. For arbitrary L(r) not necessarily closed under prefix. Example r = 'ab', L(r*) = (ab)*. The word 'ab ab' = 'abab'. Split after first 'a' gives prefix 'a' which is not in (ab)*. So prefix may not be in the star language. So can't just use L(r*)·0·L(r*).

Better approach: N(L*) = (N(L))*? Not exactly. Let's consider constructing Null(r*) as (Null(r))* | (Null(r)*)? Actually we need at most one additional 0 anywhere across concatenated sequence of factors This could be seen as either the 0 is inside one of the factors, handled by Null(r) in that factor, while other factors are plain r. So expression: (Null(r) r*) | (r* Null(r))? Wait we need exactly one factor possibly containing the inserted 0; others are normal. So we can write: (Null(r) r*) ∪ (r+ Null(r) r*)? But also the 0 could be before first factor or after last (i.e., in a Null(r) with epsilon factor?). Let's treat Null on a star as: (Null(r) r*) ∪ (r* Null(r)) ??? However Null(r) already includes original r and with an extra 0 inside r. So concatenating Null(r) with any number of r (including zero) before and after gives strings where the 0 is inside that specific factor, and there could also be extra 0 before the first factor (i.e., Null(ε) inserted as a separate factor?). Maybe easier: N(L*) = (N(L))* ∪ (N(L))*0(N(L))*? Wait we already allowed at most one extra 0 overall, but (N(L))* may contain multiple extra 0s across factors because each factor could have a 0. So can't use that.

Instead we define Null for star recursively: Null(r*) = (Null(r) r*) ∪ (r* Null(r)) ∪ ε? But that might double count.

Let's think: Any word w ∈ N(L(r)*) can be represented as: take a decomposition of w into a sequence of factors, each either in L(r) (unchanged) except at most one factor may be in N(L(r))\L(r) i.e., containing the inserted 0. So expression: (L(r))* ∪ (L(r))* (N(L(r))\L(r)) (L(r))* . Since N(L(r)) includes L(r) as well, the union simplifies to (L(r))* ∪ (L(r))* N(L(r)) (L(r))* . But N(L(r)) includes L(r), so the second term already includes the first (by taking factor that has no 0). However, the second term allows one factor with possibly extra 0; if that factor is just L(r) (no extra 0), then overall string is just (L(r))* which is covered. So we can simply define Null(r*) = (r*) | (r*) Null(r) (r*). But Null(r) already includes L(r). So expression could be: (r*) ∪ (r*) Null(r) (r*). However this may generate strings with two extra 0s if Null(r) itself includes an extra 0 (which it does) and the surrounding r* may be empty, that's okay (only one 0). Good.

Thus define Null(r*) = r* | (r*) Null(r) (r*). But we can also write as (r*) Null(r) (r*) because Null(r) includes ε? Wait Null(r) includes original strings and those with inserted 0, but not ε unless r can produce ε. If Null(r) contains ε, then (r*) Null(r) (r*) already includes pure r* (by taking ε from Null(r)). So we can just define Null(r*) = (r*) Null(r) (r*) . Equivalent.

Simpler: Null(r*) := (r*)·Null(r)·(r*).

Now we need to provide definitions in answer template.

We need to decide values for base cases:

- Null(∅) = ∅ because N(empty)=∅.

- Null(ε) = ε | 0 (since can insert 0 before/after? There is only one position, yield 0). So expression: ε ∪ 0.

But need to use notation; they used `|` for union and concatenation plain. So we can write: `ε | 0`.

- Null(a) = a | 0a | a0.

- Null(r1 | r2) = Null(r1) | Null(r2).

- Null(r1 r2) = Null(r1) r2 | r1 Null(r2). Could also add union with r1 r2 but that's covered.

- Null(r*) = (r*) Null(r) (r*). (Maybe also union with r* but covered). Provide as: `(r*) Null(r) (r*)`.

Make sure syntax consistent: parentheses around subexpressions.

Thus answer.

Now produce final answer format with <answer id="1"> ... </answer> for subproblem a, and another for subproblem b? Wait subproblem a also uses id=1. Both have same id 1; but they are separate sections. The instructions earlier: each answer template has its own id. They both use id="1". We need produce two separate <answer id="1"> sections presumably.

We'll output both in order: first answer for a, then answer for b.

Let's craft:

<answer id="1">
$N(\{01, 0\})=\{0,00,01,001,010\}$
</answer>

Second answer:

<answer id="1">
$\Null(\emptyset) := \emptyset$\\
$\Null(\epsilon) := \epsilon \mid 0$\\
$\Null(a) := a \mid 0a \mid a0$\\
$\Null(r_1 \mid r_2) := \Null(r_1) \mid \Null(r_2)$\\
$\Null(r_1 r_2) := \Null(r_1) \, r_2 \mid r_1 \, \Null(r_2)$\\
$\Null(r^{*}) := (r^{*}) \, \Null(r) \, (r^{*})$
</answer>

Make sure line breaks with \\.

Now final.

<answer id="1">
$\Null(\emptyset) := \emptyset$
$\Null(\epsilon) := \epsilon \mid 0$
$\Null(a) := a \mid 0a \mid a0$
$\Null(r_1 \mid r_2) := \Null(r_1) \mid \Null(r_2)$
$\Null(r_1 r_2) := \Null(r_1)\, r_2 \mid r_1\, \Null(r_2)$
$\Null(r^*) := (r^{*})\, \Null(r)\, (r^{*})$
</answer>

Geben Sie eine rekursive Funktion $\Null$ an, sodass $\Null(r)$ ein regulärer Ausdruck mit $L(\Null(r)) = N(L(r))$ ist, für jeden regulären Ausdruck $r$. Im Folgenden sind $r,r_1,r_2$ reguläre Ausdrücke, und $a\in \Sigma$.

$\Null(\emptyset) := \emptyset$
$\Null(\epsilon) := \epsilon \mid 0$
$\Null(a) := a \mid 0a \mid a0$
$\Null(r_1 \mid r_2) := \Null(r_1) \mid \Null(r_2)$
$\Null(r_1 r_2) := \Null(r_1)\, r_2 \mid r_1\, \Null(r_2)$
$\Null(r^*) := (r^{*})\, \Null(r)\, (r^{*})$

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Rekursive Funktionen

Sei $\Sigma:=\{0,1\}$. Für eine Sprache $L\subseteq\Sigma^*$ bezeichnet $N(L)$ die Sprache, die man erhält, wenn man den Wörtern aus $L$ **höchstens eine** Null an einer beliebigen Stelle hinzufügt. Formal definieren wir
\[N(L):= \{vxw \mid v,w\in \Sigma^*\wedge x\in \{0,\varepsilon\}\wedge vw\in L\}\]

## subproblem a (1.0 points)
Listen Sie alle Wörter der Sprache $N(\{01, 0\})$ explizit auf:

<answer-template id="1">
$N(\{01, 0\})=\{?\}$
</answer-template>

<example>
$N(\{01, 0\})=\{\varepsilon, 0, 01\}$
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:

user:
## subproblem b (6.0 points)
Geben Sie eine rekursive Funktion $\Null$ an, sodass $\Null(r)$ ein regulärer Ausdruck mit $L(\Null(r)) = N(L(r))$ ist, für jeden regulären Ausdruck $r$. Im Folgenden sind $r,r_1,r_2$ reguläre Ausdrücke, und $a\in \Sigma$.

<answer-template id="1">
$\Null(\emptyset) := ?$
$\Null(\epsilon) := ?$
$\Null(a) := ?$
$\Null(r_1 \mid r_2) := ?$
$\Null(r_1 r_2) := ?$
$\Null(r^*) := ?$
</answer-template>

<example>
$\Null(\emptyset) := \epsilon$
...
$\Null(r_1 \mid r_2) := \Null(r_1) \Null(r_2)$
...
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
$N(\{01, 0\})=\{0,00,01,001,010\}$
</answer>

<answer id="1">
$\Null(\emptyset) := \emptyset$\\
$\Null(\epsilon) := \epsilon \mid 0$\\
$\Null(a) := a \mid 0a \mid a0$\\
$\Null(r_1 \mid r_2) := \Null(r_1) \mid \Null(r_2)$\\
$\Null(r_1 r_2) := \Null(r_1)\, r_2 \mid r_1\, \Null(r_2)$\\
$\Null(r^{*}) := (r^{*})\, \Null(r)\, (r^{*})$
</answer>

user:
## subproblem c (6.0 points)
Gegeben sei die folgende rekursive Funktion über dem Alphabet $\Sigma:=\{0,1\}$:
$\Cut(\emptyset) := \emptyset$
$\Cut(0) := \emptyset$
$\Cut(r_1 \mid r_2) := \Cut(r_1) \mid \Cut(r_2)$
$\Cut(r^*) := r^*\Cut(r)$
$\Cut(\epsilon) := \emptyset$
$\Cut(1) := \epsilon$
$\Cut(r_1 r_2) := \Cut(r_1)\mid r_1\Cut(r_2)$

Per struktureller Induktion über $r$ lässt sich zeigen, dass für alle $u,v\in\Sigma^*$ mit $u1v\in L(r)$ gilt, dass $u\in L(\Cut(r))$.

Vervollständigen Sie den Beweis für den Fall $r=r_1r_2$. Ergänzen Sie die Induktionshypothese(n) und führen Sie dann den Beweis. Kennzeichnen Sie Verwendungen der Induktionshypothesen im Beweis deutlich.

*Zu zeigen:* Für alle $u1v\in L(r_1r_2)$ mit $u,v\in\Sigma^*$ gilt $u\in L(\Cut(r_1r_2))$.

<answer-template id="1">
Induktionshypothese(n) für den Fall $r=r_1r_2$:
???

Beweis des Falls $r=r_1r_2$:
???
</answer-template>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to provide inductive proof for r = r1 r2. Provide induction hypotheses: for r1 and r2 individually, the property holds. Then prove for concatenation.

We need to output in <answer id="1"> with the template.

We should write something like:

Induktionshypothese(n) für den Fall r = r1 r2:
- For all u1v ∈ L(r1) with u,v ∈ Σ* holds u ∈ L(Cut(r1)).
- For all u1v ∈ L(r2) with u,v ∈ Σ* holds u ∈ L(Cut(r2)).

Now proof: suppose u1v ∈ L(r1 r2). Then there exists split x,y such that u1v = xy, x ∈ L(r1), y ∈ L(r2). There are cases where the 1 is in x or in y. Use IH accordingly. Then show u ∈ L(Cut(r1r2)) = Cut(r1) | r1 Cut(r2). Show that either u comes from first part (if 1 in x) then by IH u ∈ L(Cut(r1)) ⊆ L(Cut(r1) | r1 Cut(r2)). Else 1 in y, then x is prefix; write x = u' maybe? Actually u1v = x y, with y containing 1. Let y = u1v' where u corresponds to prefix of concatenated string? Let's reason: If 1 is in x, then x = u1v' and y arbitrary; then by IH for r1 we get u ∈ L(Cut(r1)) and then u ∈ L(Cut(r1) | r1 Cut(r2)). If 1 is in y, then x is entirely before the 1; so x ∈ L(r1) and y = u1v' with u after x. Then by IH for r2 we get u ∈ L(Cut(r2)). Then r1 Cut(r2) includes strings of form w w' where w∈L(r1) and w'∈L(Cut(r2)). So prefix x concatenated with u (the part from Cut(r2)) yields u in L(r1 Cut(r2))? Wait definition: L(Cut(r1 r2)) = L(Cut(r1)) ∪ L(r1 L(Cut(r2))) i.e., strings either from Cut(r1) or strings formed by concatenating a string from L(r1) with a string from L(Cut(r2)). The u we need to show belongs to L(Cut(r1 r2)). In case 1 in y, we have u is suffix after x? Actually overall string is x y = x u 1 v' (since y contains the 1). Then total string u1v (global) = x u' ?? Let's denote original global u (the part before the 1) is maybe x u'? Wait we have original decomposition: u 1 v ∈ L(r1 r2). Want to show u ∈ L(Cut(r1 r2)). The location of the 1 inside concatenated string determines where u splits. If 1 lies inside r1, then u is some prefix of string from r1, and by IH get u in Cut(r1). If 1 lies inside r2, then u can be written as x u', where x is entire string from r1, and u' is prefix of part of r2 before the 1. Then by IH for r2, u' ∈ L(Cut(r2)). Then x u' ∈ L(r1) L(Cut(r2)) ⊆ L(Cut(r1 r2)). So done.

We must mark uses of IH clearly, maybe with **IH**.

Now produce answer.

<answer id="1">
**Induktionshypothese(n) für den Fall \(r = r_{1}r_{2}\):**  

1. *(IH 1)* Für alle \(u_{1}v\in L(r_{1})\) mit \(u_{1},v\in\Sigma^{*}\) gilt  
   \(\displaystyle u_{1}\in L(\Cut(r_{1})).\)

2. *(IH 2)* Für alle \(u_{2}v\in L(r_{2})\) mit \(u_{2},v\in\Sigma^{*}\) gilt  
   \(\displaystyle u_{2}\in L(\Cut(r_{2})).\)

---

**Beweis des Falls \(r = r_{1}r_{2}\):**  

Sei \(u1v\in L(r_{1}r_{2})\) mit \(u,v\in\Sigma^{*}\).  
Nach der Definition der Konkatenation gibt es Wörter \(x\in L(r_{1})\) und \(y\in L(r_{2})\) derart, dass  

\[
u1v = xy .
\]

Der einzelne Buchstabe \(1\) befindet sich entweder in \(x\) **oder** in \(y\). Wir behandeln beide Möglichkeiten getrennt.

---

**Fall 1 – die 1 liegt in \(x\).**  

Dann lässt sich \(x\) schreiben als \(x = u_{1}1v_{1}\) mit \(u_{1},v_{1}\in\Sigma^{*}\) und \(y\) beliebig.  
Da \(x\in L(r_{1})\) und \(u_{1}1v_{1}\in L(r_{1})\), dürfen wir die Induktionshypothese **(IH 1)** anwenden und erhalten  

\[
u_{1}\in L(\Cut(r_{1})).
\]

Weil nach Definition  

\[
\Cut(r_{1}r_{2}) = \Cut(r_{1}) \mid r_{1}\Cut(r_{2}),
\]

ist jedes Wort aus \(L(\Cut(r_{1}))\) zugleich ein Wort aus \(L(\Cut(r_{1}r_{2}))\). Also  

\[
u = u_{1}\in L(\Cut(r_{1}r_{2})).
\]

---

**Fall 2 – die 1 liegt in \(y\).**  

Dann hat \(y\) die Form \(y = u_{2}1v_{2}\) mit \(u_{2},v_{2}\in\Sigma^{*}\) und \(x\) enthält kein 1.  
Wir haben also  

\[
x\in L(r_{1}) \quad\text{und}\quad u_{2}1v_{2}\in L(r_{2}).
\]

Aus \(u_{2}1v_{2}\in L(r_{2})\) und **(IH 2)** folgt  

\[
u_{2}\in L(\Cut(r_{2})).
\]

Damit ist das Wort \(xu_{2}\) ein Element von  

\[
L(r_{1})\,L(\Cut(r_{2})) \subseteq L(\Cut(r_{1}) \mid r_{1}\Cut(r_{2}))
      = L(\Cut(r_{1}r_{2})).
\]

Da das ursprüngliche Präfix vor dem Symbol \(1\) genau \(u = xu_{2}\) ist, erhalten wir  

\[
u\in L(\Cut(r_{1}r_{2})).
\]

---

In beiden Fällen folgt aus der Annahme \(u1v\in L(r_{1}r_{2})\), dass \(u\in L(\Cut(r_{1}r_{2}))\). Damit ist die zu zeigende Aussage für den Fall \(r = r_{1}r_{2}\) bewiesen. ∎
</answer>

Gegeben sei die folgende rekursive Funktion über dem Alphabet $\Sigma:=\{0,1\}$:
$\Cut(\emptyset) := \emptyset$
$\Cut(0) := \emptyset$
$\Cut(r_1 \mid r_2) := \Cut(r_1) \mid \Cut(r_2)$
$\Cut(r^*) := r^*\Cut(r)$
$\Cut(\epsilon) := \emptyset$
$\Cut(1) := \epsilon$
$\Cut(r_1 r_2) := \Cut(r_1)\mid r_1\Cut(r_2)$

Per struktureller Induktion über $r$ lässt sich zeigen, dass für alle $u,v\in\Sigma^*$ mit $u1v\in L(r)$ gilt, dass $u\in L(\Cut(r))$.

Vervollständigen Sie den Beweis für den Fall $r=r_1r_2$. Ergänzen Sie die Induktionshypothese(n) und führen Sie dann den Beweis. Kennzeichnen Sie Verwendungen der Induktionshypothesen im Beweis deutlich.

Zu zeigen: Für alle $u1v\in L(r_1r_2)$ mit $u,v\in\Sigma^*$ gilt $u\in L(\Cut(r_1r_2))$.

**Induktionshypothese(n) für den Fall \(r = r_{1}r_{2}\):**  

1. *(IH 1)* Für alle \(u_{1}v\in L(r_{1})\) mit \(u_{1},v\in\Sigma^{*}\) gilt  
   \(\displaystyle u_{1}\in L(\Cut(r_{1})).\)

2. *(IH 2)* Für alle \(u_{2}v\in L(r_{2})\) mit \(u_{2},v\in\Sigma^{*}\) gilt  
   \(\displaystyle u_{2}\in L(\Cut(r_{2})).\)

---

**Beweis des Falls \(r = r_{1}r_{2}\):**  

Sei \(u1v\in L(r_{1}r_{2})\) mit \(u,v\in\Sigma^{*}\).  
Nach der Definition der Konkatenation gibt es Wörter \(x\in L(r_{1})\) und \(y\in L(r_{2})\) derart, dass  

\[
u1v = xy .
\]

Der einzelne Buchstabe \(1\) befindet sich entweder in \(x\) **oder** in \(y\). Wir behandeln beide Möglichkeiten getrennt.

---

**Fall 1 – die 1 liegt in \(x\).**  

Dann lässt sich \(x\) schreiben als \(x = u_{1}1v_{1}\) mit \(u_{1},v_{1}\in\Sigma^{*}\) und \(y\) beliebig.  
Da \(x\in L(r_{1})\) und \(u_{1}1v_{1}\in L(r_{1})\), dürfen wir die Induktionshypothese **(IH 1)** anwenden und erhalten  

\[
u_{1}\in L(\Cut(r_{1})).
\]

Weil nach Definition  

\[
\Cut(r_{1}r_{2}) = \Cut(r_{1}) \mid r_{1}\Cut(r_{2}),
\]

ist jedes Wort aus \(L(\Cut(r_{1}))\) zugleich ein Wort aus \(L(\Cut(r_{1}r_{2}))\). Also  

\[
u = u_{1}\in L(\Cut(r_{1}r_{2})).
\]

---

**Fall 2 – die 1 liegt in \(y\).**  

Dann hat \(y\) die Form \(y = u_{2}1v_{2}\) mit \(u_{2},v_{2}\in\Sigma^{*}\) und \(x\) enthält kein 1.  
Wir haben also  

\[
x\in L(r_{1}) \quad\text{und}\quad u_{2}1v_{2}\in L(r_{2}).
\]

Aus \(u_{2}1v_{2}\in L(r_{2})\) und **(IH 2)** folgt  

\[
u_{2}\in L(\Cut(r_{2})).
\]

Damit ist das Wort \(xu_{2}\) ein Element von  

\[
L(r_{1})\,L(\Cut(r_{2})) \subseteq L(\Cut(r_{1}) \mid r_{1}\Cut(r_{2}))
      = L(\Cut(r_{1}r_{2})).
\]

Da das ursprüngliche Präfix vor dem Symbol \(1\) genau \(u = xu_{2}\) ist, erhalten wir  

\[
u\in L(\Cut(r_{1}r_{2})).
\]

---

In beiden Fällen folgt aus der Annahme \(u1v\in L(r_{1}r_{2})\), dass \(u\in L(\Cut(r_{1}r_{2}))\). Damit ist die zu zeigende Aussage für den Fall \(r = r_{1}r_{2}\) bewiesen. ∎

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: CFG-Algorithmen

In den Teilaufgaben (a)-(d) geht es darum, eine kontextfreie Grammatik (CFG) in Chomsky-Normalform (CNF) zu konvertieren. Wir führen jeden Schritt einzeln aus, und jeweils auf einer anderen Grammatik – Sie können also die Aufgabenteile unabhängig voneinander bearbeiten.

Eine CFG $G=(V,\Sigma,P,S)$ ist in CNF, wenn jede Produktion $(X\rightarrow\alpha)\in P$, mit $X\in V$ und $\alpha\in(\Sigma\cup V)^*$, folgende Bedingungen erfüllt:
(1) $\alpha\in\Sigma\cup V^*$; Terminale dürfen nur in Produktionen der Länge 1 erzeugt werden.
(2) $|\alpha|\le2$; jede Produktion hat höchstens Länge $2$.
(3) $\alpha\ne\varepsilon$; es gibt keine $\varepsilon$-Produktionen.
(4) $\alpha\notin V$; es gibt keine Kettenproduktionen.

**Achtung:** Die Teilaufgaben fragen jeweils spezifisch nach dem Ergebnis, das sich durch die Ausführung des Algorithmus aus der Vorlesung ergibt, nicht nach einer beliebigen äquivalenten CFG, die den Bedingungen genügt. Details, wie etwa die Namen der Variablen oder die Reihenfolge, in der Produktionen betrachtet werden, können Sie frei wählen.
Wir nennen $A \rightarrow \epsilon$ eine *$\epsilon$-Produktion*.
\begin{Lemma}
Zu jeder CFG $G = (V,\Sigma,P,S)$ kann man eine CFG $G'$ konstruieren, die keine $\epsilon$-Produktionen enthält, so dass gilt
$L(G') = L(G)\setminus\{\epsilon\}$
\end{Lemma}
Wir erweitern $P$ induktiv zu eine Obermenge $\hat{P}$:
\begin{enumerate}
\item Jede Produktion aus $P$ ist in $\hat{P}$
\item Sind \alert{$B \rightarrow \epsilon$} und \alert{$A \rightarrow \alpha B \beta$} in $\hat{P}$, so füge auch \alert{$A \rightarrow \alpha\beta$} hinzu.
\end{enumerate}
Offensichtlich gilt $L(\hat{G}) = L(G)$: Jede neue Produktion kann von 2 alten Productionensimuliert werden.

Wir definieren $G'$ als $\hat{G}$ ohne die $\epsilon$-Produktionen. Denn diese sind nun überflüssig.

Wir nennen $A \rightarrow B$ eine *Kettenproduktion*.
\begin{Lemma}
Zu jeder CFG $G = (V,\Sigma,P,S)$ kann man eine CFG $G'$ konstruieren, die keine Kettenproduktionen enthält, so dass gilt $L(G') = L(G)$.
\end{Lemma}
\begin{Beweis}
Wir erweitern $P$ induktiv zu eine Obermenge $\hat{P}$:
\begin{enumerate}
\item Jede Produktion aus $P$ ist in $\hat{P}$
\item Sind \alert{$A \rightarrow B$} und \alert{$B \rightarrow \alpha$} in $\hat{P}$ mit $\alpha \neq A$, so füge auch \alert{$A \rightarrow \alpha$} hinzu.
\end{enumerate}
Das Ergebnis $G'$ ist $\hat{G}$ ohne die (nun überflüssigen) Kettenproduktionen.

\paragraph{Konstruktion einer Chomsky-Normalform}

Eingabe: Eine kontextfreie Grammatik $G = (V,\Sigma,P,S)$
\begin{enumerate}
\item Füge für jedes $a\in\Sigma$, das in einer rechten Seite der Länge $\geq2$ vorkommt, ein neues Nichtterminal $A_a$ zu $V$ hinzu, ersetze $a$ in allen rechten Seiten der Länge $\geq 2$ durch $A_a$, und füge $A_a\rightarrow a$ zu $P$ hinzu.
\item Ersetze jede Produktion der Form
\[A\rightarrow B_1B_2 \cdots B_k (k \geq 3)\]
durch
\[A \rightarrow B_1C_2,\ C_2 \rightarrow B_2C_3, ..., C_{k-1} \rightarrow B_{k-1}B_k \]
wobei $C_2, ..., C_{k-1}$ neue Nichtterminale sind.
\item Eliminiere alle $\epsilon$-Produktionen.
\item Eliminiere alle Kettenproduktionen.
\end{enumerate}

## subproblem a (2.0 points)
*Entfernen von Terminalen in langen Produktionen.* Die CFG $G_a$ ist gegeben durch folgende Produktionen:
$S \rightarrow aX \mid Xa \mid ab$
$X \rightarrow SS \mid S \mid \varepsilon \mid bXX$
Führen Sie den ersten Schritt des Algorithmus zur Überführung in CNF aus und geben Sie die Produktionen einer CFG $G'_a$ an, so dass $L(G_a) = L(G'_a)$ gilt und $G'_a$ Bedingung (1) erfüllt.

<example>
$S \rightarrow XX \mid aa \mid \varepsilon$
$X \rightarrow SX \mid X$
...
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to perform first step of algorithm: replace terminals in RHS of length >=2 with new nonterminals. Also add productions for those new nonterminals: For each terminal a that appears in RHS of length >=2, create new nonterminal A_a (or any name) with production A_a -> a. Then replace all occurrences of a in RHS of length >=2 with A_a.

We have grammar G_a:

Productions:

1. S -> aX
2. S -> Xa
3. S -> ab
4. X -> SS
5. X -> S
6. X -> ε
7. X -> bXX

Terminals: a, b.

We need to consider RHS of length >=2:

- aX length 2 includes terminal a
- Xa length 2 includes terminal a
- ab length 2 includes terminals a and b
- SS length 2 no terminals
- bXX length 3 includes terminal b

Thus we need new nonterminals for a and b: let's call A_a and A_b.

Add productions:

A_a -> a
A_b -> b

Now replace terminals in those RHS:

- S -> aX becomes S -> A_a X
- S -> Xa becomes S -> X A_a
- S -> ab becomes S -> A_a A_b
- X -> bXX becomes X -> A_b X X (since length>=2, replace b with A_b). Note RHS length 3 => after replacement becomes A_b X X.

Other productions unchanged (since they have no terminals or length 1 terminals? there's X -> S (length 1 with nonterminal), X -> ε (epsilon stays for later steps), X -> SS unchanged).

Thus G'_a productions:

S -> A_a X | X A_a | A_a A_b
X -> S S | S | ε | A_b X X
A_a -> a
A_b -> b

Make sure condition (1) satisfied: terminals only appear alone on RHS length 1 (A_a -> a, A_b -> b). All other productions have only nonterminals.

Thus answer. Provide in answer tag.

<answer id="1">
$S \rightarrow A_aX \mid XA_a \mid A_aA_b$
$X \rightarrow SS \mid S \mid \varepsilon \mid A_bXX$
$A_a \rightarrow a$
$A_b \rightarrow b$
</answer>

Entfernen von Terminalen in langen Produktionen. Die CFG $G_a$ ist gegeben durch folgende Produktionen:
$S \rightarrow aX \mid Xa \mid ab$
$X \rightarrow SS \mid S \mid \varepsilon \mid bXX$
Führen Sie den ersten Schritt des Algorithmus zur Überführung in CNF aus und geben Sie die Produktionen einer CFG $G'_a$ an, so dass $L(G_a) = L(G'_a)$ gilt und $G'_a$ Bedingung (1) erfüllt.

$S \rightarrow A_aX \mid XA_a \mid A_aA_b$
$X \rightarrow SS \mid S \mid \varepsilon \mid A_bXX$
$A_a \rightarrow a$
$A_b \rightarrow b$

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: CFG-Algorithmen

In den Teilaufgaben (a)-(d) geht es darum, eine kontextfreie Grammatik (CFG) in Chomsky-Normalform (CNF) zu konvertieren. Wir führen jeden Schritt einzeln aus, und jeweils auf einer anderen Grammatik – Sie können also die Aufgabenteile unabhängig voneinander bearbeiten.

Eine CFG $G=(V,\Sigma,P,S)$ ist in CNF, wenn jede Produktion $(X\rightarrow\alpha)\in P$, mit $X\in V$ und $\alpha\in(\Sigma\cup V)^*$, folgende Bedingungen erfüllt:
(1) $\alpha\in\Sigma\cup V^*$; Terminale dürfen nur in Produktionen der Länge 1 erzeugt werden.
(2) $|\alpha|\le2$; jede Produktion hat höchstens Länge $2$.
(3) $\alpha\ne\varepsilon$; es gibt keine $\varepsilon$-Produktionen.
(4) $\alpha\notin V$; es gibt keine Kettenproduktionen.

**Achtung:** Die Teilaufgaben fragen jeweils spezifisch nach dem Ergebnis, das sich durch die Ausführung des Algorithmus aus der Vorlesung ergibt, nicht nach einer beliebigen äquivalenten CFG, die den Bedingungen genügt. Details, wie etwa die Namen der Variablen oder die Reihenfolge, in der Produktionen betrachtet werden, können Sie frei wählen.
Wir nennen $A \rightarrow \epsilon$ eine *$\epsilon$-Produktion*.
\begin{Lemma}
Zu jeder CFG $G = (V,\Sigma,P,S)$ kann man eine CFG $G'$ konstruieren, die keine $\epsilon$-Produktionen enthält, so dass gilt
$L(G') = L(G)\setminus\{\epsilon\}$
\end{Lemma}
Wir erweitern $P$ induktiv zu eine Obermenge $\hat{P}$:
\begin{enumerate}
\item Jede Produktion aus $P$ ist in $\hat{P}$
\item Sind \alert{$B \rightarrow \epsilon$} und \alert{$A \rightarrow \alpha B \beta$} in $\hat{P}$, so füge auch \alert{$A \rightarrow \alpha\beta$} hinzu.
\end{enumerate}
Offensichtlich gilt $L(\hat{G}) = L(G)$: Jede neue Produktion kann von 2 alten Productionensimuliert werden.

Wir definieren $G'$ als $\hat{G}$ ohne die $\epsilon$-Produktionen. Denn diese sind nun überflüssig.

Wir nennen $A \rightarrow B$ eine *Kettenproduktion*.
\begin{Lemma}
Zu jeder CFG $G = (V,\Sigma,P,S)$ kann man eine CFG $G'$ konstruieren, die keine Kettenproduktionen enthält, so dass gilt $L(G') = L(G)$.
\end{Lemma}
\begin{Beweis}
Wir erweitern $P$ induktiv zu eine Obermenge $\hat{P}$:
\begin{enumerate}
\item Jede Produktion aus $P$ ist in $\hat{P}$
\item Sind \alert{$A \rightarrow B$} und \alert{$B \rightarrow \alpha$} in $\hat{P}$ mit $\alpha \neq A$, so füge auch \alert{$A \rightarrow \alpha$} hinzu.
\end{enumerate}
Das Ergebnis $G'$ ist $\hat{G}$ ohne die (nun überflüssigen) Kettenproduktionen.

\paragraph{Konstruktion einer Chomsky-Normalform}

Eingabe: Eine kontextfreie Grammatik $G = (V,\Sigma,P,S)$
\begin{enumerate}
\item Füge für jedes $a\in\Sigma$, das in einer rechten Seite der Länge $\geq2$ vorkommt, ein neues Nichtterminal $A_a$ zu $V$ hinzu, ersetze $a$ in allen rechten Seiten der Länge $\geq 2$ durch $A_a$, und füge $A_a\rightarrow a$ zu $P$ hinzu.
\item Ersetze jede Produktion der Form
\[A\rightarrow B_1B_2 \cdots B_k (k \geq 3)\]
durch
\[A \rightarrow B_1C_2,\ C_2 \rightarrow B_2C_3, ..., C_{k-1} \rightarrow B_{k-1}B_k \]
wobei $C_2, ..., C_{k-1}$ neue Nichtterminale sind.
\item Eliminiere alle $\epsilon$-Produktionen.
\item Eliminiere alle Kettenproduktionen.
\end{enumerate}

## subproblem a (2.0 points)
*Entfernen von Terminalen in langen Produktionen.* Die CFG $G_a$ ist gegeben durch folgende Produktionen:
$S \rightarrow aX \mid Xa \mid ab$
$X \rightarrow SS \mid S \mid \varepsilon \mid bXX$
Führen Sie den ersten Schritt des Algorithmus zur Überführung in CNF aus und geben Sie die Produktionen einer CFG $G'_a$ an, so dass $L(G_a) = L(G'_a)$ gilt und $G'_a$ Bedingung (1) erfüllt.

<example>
$S \rightarrow XX \mid aa \mid \varepsilon$
$X \rightarrow SX \mid X$
...
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
S \;\rightarrow\; A_a\,X \;\mid\; X\,A_a \;\mid\; A_a\,A_b  
\\[4pt]
X \;\rightarrow\; S\,S \;\mid\; S \;\mid\; \varepsilon \;\mid\; A_b\,X\,X  
\\[4pt]
A_a \;\rightarrow\; a  
\\[4pt]
A_b \;\rightarrow\; b
</answer>

user:
## subproblem b (2.0 points)
*Entfernen langer Produktionen.* Die CFG $G_b$ ist gegeben durch die folgenden Produktionen:
$S \rightarrow A \mid BAA \mid \varepsilon$
$A \rightarrow SABB \mid a$
$B \rightarrow AS \mid b$
Führen Sie den zweiten Schritt des Algorithmus zur Überführung in CNF aus und geben Sie die Produktionen einer CFG $G'_b$ an, so dass $L(G_b) = L(G'_b)$ und $G'_b$ Bedingungen (1) und (2) erfüllt.

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to apply second step: replace productions with right side length >=3 (nonterminals only, as after step1 terminals in long productions already replaced). The grammar G_b provided: productions:

S -> A | BAA | ε
A -> SABB | a
B -> AS | b

First step not applied yet? The instruction says: "Führen Sie den zweiten Schritt des Algorithmus zur Überführung in CNF aus und geben Sie die Produktionen einer CFG G'_b an, so dass L(G_b) = L(G'_b) und G'_b Bedingungen (1) und (2) erfüllt."

We need to ensure condition (1) (terminals only in length-1 productions) - that would require step1 first. But they specifically ask for second step after we have removed terminals from long productions. However the grammar as given includes terminals a, b in productions with length >1: e.g., A->SABB (contains terminals a, b). Actually SABB: S A B B? Wait capital letters are nonterminals, lowercase are terminals. S A B B: S nonterminal, A nonterminal, B nonterminal, B nonterminal. That's okay, all nonterminals, no terminals. But "a" and "b" appear as terminals in productions A->a, B->b, etc. That's fine as they are length 1.

However S -> BAA is three symbols B A A all nonterminals, okay.

Thus condition (1) already satisfied. So we can directly apply step 2 to split long productions of length >=3.

Identify productions with RHS length >=3:

- S -> B A A (length 3)
- A -> S A B B (length 4)

Also maybe others? No.

We need to replace each with binary productions using new nonterminals.

Procedure: For each production A -> X1 X2 ... Xk (k>=3), replace with A -> X1 C2, C2 -> X2 C3, ..., C_{k-1} -> X_{k-1} X_k, where C_i are new nonterminals.

We need to introduce new nonterminals, perhaps denote them as e.g., C1, C2, etc, but ensure uniqueness per each set.

Let's do for S -> B A A.

k=3. Introduce new nonterminal, say C1 (or maybe C_{S,1})? We'll call it X1 maybe.

We need: S -> B C1, C1 -> A A? Wait formula B1 C2, C2 -> B2 C3,... C_{k-1} -> B_{k-1} B_k.

For k=3: we need one new nonterminal C2 (since C_{k-1}=C2?), Actually we start with C2. So we create new nonterminal, call it C1 (or C_2). Let's just call it X1. Then productions:

S -> B X1
X1 -> A A

Ok.

Now for A -> S A B B (k=4). Need new nonterminals: C2, C3? Starting with C2 for first split.

We need:

A -> S C2
C2 -> A C3
C3 -> B B

Thus we introduced two new nonterminals.

We must also keep other productions unchanged: S -> A, S -> ε, A -> a, B -> A S, B -> b.

Now condition (2): all productions RHS length <=2. We have:

S -> A (length 1)
S -> B X1 (2)
S -> ε (allowed? CNF condition (3) later would eliminate epsilon, but step2 still allowed; we just need condition (2) satisfied: length <=2, epsilon length 0 fine)
A -> S C2 (2)
A -> a (1)
B -> A S (2)
B -> b (1)
X1 -> A A (2)
C2 -> A C3 (2)
C3 -> B B (2)

All good.

Thus answer should list productions.

We need to output in <answer id="1"> ... </answer>. Use same notation.

Let's produce.

<answer id="1">
$S \;\rightarrow\; A \;\mid\; B\,X_1 \;\mid\; \varepsilon$
$X_1 \;\rightarrow\; A\,A$
$A \;\rightarrow\; S\,C_2 \;\mid\; a$
$C_2 \;\rightarrow\; A\,C_3$
$C_3 \;\rightarrow\; B\,B$
$B \;\rightarrow\; A\,S \;\mid\; b$
</answer>

Entfernen langer Produktionen. Die CFG $G_b$ ist gegeben durch die folgenden Produktionen:
$S \rightarrow A \mid BAA \mid \varepsilon$
$A \rightarrow SABB \mid a$
$B \rightarrow AS \mid b$
Führen Sie den zweiten Schritt des Algorithmus zur Überführung in CNF aus und geben Sie die Produktionen einer CFG $G'_b$ an, so dass $L(G_b) = L(G'_b)$ und $G'_b$ Bedingungen (1) und (2) erfüllt.

$S \;\rightarrow\; A \;\mid\; B\,X_1 \;\mid\; \varepsilon$
$X_1 \;\rightarrow\; A\,A$
$A \;\rightarrow\; S\,C_2 \;\mid\; a$
$C_2 \;\rightarrow\; A\,C_3$
$C_3 \;\rightarrow\; B\,B$
$B \;\rightarrow\; A\,S \;\mid\; b$

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: CFG-Algorithmen

In den Teilaufgaben (a)-(d) geht es darum, eine kontextfreie Grammatik (CFG) in Chomsky-Normalform (CNF) zu konvertieren. Wir führen jeden Schritt einzeln aus, und jeweils auf einer anderen Grammatik – Sie können also die Aufgabenteile unabhängig voneinander bearbeiten.

Eine CFG $G=(V,\Sigma,P,S)$ ist in CNF, wenn jede Produktion $(X\rightarrow\alpha)\in P$, mit $X\in V$ und $\alpha\in(\Sigma\cup V)^*$, folgende Bedingungen erfüllt:
(1) $\alpha\in\Sigma\cup V^*$; Terminale dürfen nur in Produktionen der Länge 1 erzeugt werden.
(2) $|\alpha|\le2$; jede Produktion hat höchstens Länge $2$.
(3) $\alpha\ne\varepsilon$; es gibt keine $\varepsilon$-Produktionen.
(4) $\alpha\notin V$; es gibt keine Kettenproduktionen.

**Achtung:** Die Teilaufgaben fragen jeweils spezifisch nach dem Ergebnis, das sich durch die Ausführung des Algorithmus aus der Vorlesung ergibt, nicht nach einer beliebigen äquivalenten CFG, die den Bedingungen genügt. Details, wie etwa die Namen der Variablen oder die Reihenfolge, in der Produktionen betrachtet werden, können Sie frei wählen.
Wir nennen $A \rightarrow \epsilon$ eine *$\epsilon$-Produktion*.
\begin{Lemma}
Zu jeder CFG $G = (V,\Sigma,P,S)$ kann man eine CFG $G'$ konstruieren, die keine $\epsilon$-Produktionen enthält, so dass gilt
$L(G') = L(G)\setminus\{\epsilon\}$
\end{Lemma}
Wir erweitern $P$ induktiv zu eine Obermenge $\hat{P}$:
\begin{enumerate}
\item Jede Produktion aus $P$ ist in $\hat{P}$
\item Sind \alert{$B \rightarrow \epsilon$} und \alert{$A \rightarrow \alpha B \beta$} in $\hat{P}$, so füge auch \alert{$A \rightarrow \alpha\beta$} hinzu.
\end{enumerate}
Offensichtlich gilt $L(\hat{G}) = L(G)$: Jede neue Produktion kann von 2 alten Productionensimuliert werden.

Wir definieren $G'$ als $\hat{G}$ ohne die $\epsilon$-Produktionen. Denn diese sind nun überflüssig.

Wir nennen $A \rightarrow B$ eine *Kettenproduktion*.
\begin{Lemma}
Zu jeder CFG $G = (V,\Sigma,P,S)$ kann man eine CFG $G'$ konstruieren, die keine Kettenproduktionen enthält, so dass gilt $L(G') = L(G)$.
\end{Lemma}
\begin{Beweis}
Wir erweitern $P$ induktiv zu eine Obermenge $\hat{P}$:
\begin{enumerate}
\item Jede Produktion aus $P$ ist in $\hat{P}$
\item Sind \alert{$A \rightarrow B$} und \alert{$B \rightarrow \alpha$} in $\hat{P}$ mit $\alpha \neq A$, so füge auch \alert{$A \rightarrow \alpha$} hinzu.
\end{enumerate}
Das Ergebnis $G'$ ist $\hat{G}$ ohne die (nun überflüssigen) Kettenproduktionen.

\paragraph{Konstruktion einer Chomsky-Normalform}

Eingabe: Eine kontextfreie Grammatik $G = (V,\Sigma,P,S)$
\begin{enumerate}
\item Füge für jedes $a\in\Sigma$, das in einer rechten Seite der Länge $\geq2$ vorkommt, ein neues Nichtterminal $A_a$ zu $V$ hinzu, ersetze $a$ in allen rechten Seiten der Länge $\geq 2$ durch $A_a$, und füge $A_a\rightarrow a$ zu $P$ hinzu.
\item Ersetze jede Produktion der Form
\[A\rightarrow B_1B_2 \cdots B_k (k \geq 3)\]
durch
\[A \rightarrow B_1C_2,\ C_2 \rightarrow B_2C_3, ..., C_{k-1} \rightarrow B_{k-1}B_k \]
wobei $C_2, ..., C_{k-1}$ neue Nichtterminale sind.
\item Eliminiere alle $\epsilon$-Produktionen.
\item Eliminiere alle Kettenproduktionen.
\end{enumerate}

## subproblem a (2.0 points)
*Entfernen von Terminalen in langen Produktionen.* Die CFG $G_a$ ist gegeben durch folgende Produktionen:
$S \rightarrow aX \mid Xa \mid ab$
$X \rightarrow SS \mid S \mid \varepsilon \mid bXX$
Führen Sie den ersten Schritt des Algorithmus zur Überführung in CNF aus und geben Sie die Produktionen einer CFG $G'_a$ an, so dass $L(G_a) = L(G'_a)$ gilt und $G'_a$ Bedingung (1) erfüllt.

<example>
$S \rightarrow XX \mid aa \mid \varepsilon$
$X \rightarrow SX \mid X$
...
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
S \;\rightarrow\; A_a\,X \;\mid\; X\,A_a \;\mid\; A_a\,A_b  
\\[4pt]
X \;\rightarrow\; S\,S \;\mid\; S \;\mid\; \varepsilon \;\mid\; A_b\,X\,X  
\\[4pt]
A_a \;\rightarrow\; a  
\\[4pt]
A_b \;\rightarrow\; b
</answer>

user:
## subproblem b (2.0 points)
*Entfernen langer Produktionen.* Die CFG $G_b$ ist gegeben durch die folgenden Produktionen:
$S \rightarrow A \mid BAA \mid \varepsilon$
$A \rightarrow SABB \mid a$
$B \rightarrow AS \mid b$
Führen Sie den zweiten Schritt des Algorithmus zur Überführung in CNF aus und geben Sie die Produktionen einer CFG $G'_b$ an, so dass $L(G_b) = L(G'_b)$ und $G'_b$ Bedingungen (1) und (2) erfüllt.

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
S \;\rightarrow\; A \;\mid\; B\,X_1 \;\mid\; \varepsilon \\[4pt]
X_1 \;\rightarrow\; A\,A \\[4pt]
A \;\rightarrow\; S\,C_2 \;\mid\; a \\[4pt]
C_2 \;\rightarrow\; A\,C_3 \\[4pt]
C_3 \;\rightarrow\; B\,B \\[4pt]
B \;\rightarrow\; A\,S \;\mid\; b
</answer>

user:
## subproblem c (3.0 points)
*Entfernen von $\varepsilon$-Produktionen.* Die CFG $G_c$ ist gegeben durch folgende Produktionen:
$S \rightarrow AB \mid DE$
$A \rightarrow AB \mid a \mid \varepsilon$
$B \rightarrow b \mid BB \mid D$
$C \rightarrow A \mid c$
$D \rightarrow BC \mid BS$
$E \rightarrow SS \mid CA$
Führen Sie den dritten Schritt des Algorithmus zur Überführung in CNF aus. Geben Sie die Produktionen einer CFG $G'_c$ an, so dass $L(G_c) = L(G'_c)$ und $G'_c$ Bedingungen (1), (2) und (3) erfüllt.

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to remove epsilon-productions from the given grammar, after earlier steps (but we only have to do step 3). The grammar already maybe has step 1 (terminal removal) and step 2 (binary productions) applied? The problem says "Entfernen von ε-Produktionen". So we assume grammar is already in CNF except possibly epsilon-productions. But we must also ensure condition (1) (terminals only in length-1 productions) and (2) length <=2. Probably the given grammar already respects those except epsilon. Let's check:

Given productions:

S -> AB | DE (both length 2, all nonterminals) OK.

A -> AB | a | ε (contains terminal a length 1, OK. ε to be removed.)

B -> b | BB | D (b is terminal length1 OK; BB length2, both nonterminals; D nonterminal single, violates condition (4) (no unit productions) but step 4 is later (remove chain). For step 3 we only care about epsilon, not chain. But condition (1) satisfied? D alone is a nonterminal producing D, that is length 1 nonterminal, which is allowed in (1)?? Actually condition (1) says α ∈ Σ ∪ V*. Means either a single terminal or a (possibly empty) string of nonterminals? Wait typical CNF requires not unit productions, but condition (4) says α not in V, i.e., no unit productions. So B -> D is a unit production, not allowed yet but step 4 later. For step 3 we only need to ensure (1), (2), (3). Condition (4) can be ignored for now.

C -> A | c (unit production A, also terminal c length1 OK)

D -> BC | BS (both length2 nonterminals OK)

E -> SS | CA (both length2 nonterminals OK)

Thus only epsilon productions: A -> ε. Also maybe others produce epsilon via chain? No.

Algorithm for eliminating epsilon productions: Compute nullable nonterminals (that can derive ε). Initially include A (since A -> ε). Also any nonterminal that has a production where all symbols are nullable becomes nullable.

Let's compute:

- A is nullable.
- Check other productions:
S -> AB, both A nullable? A nullable, B? Not yet nullable. So AB not all nullable (B not). DE: D? unknown, E? unknown.

B -> b (not), BB (both B), D (nonterminal). So B not nullable unless B -> ε via something.

C -> A | c. Since A nullable, C can derive ε? Wait C -> A, and A can go to ε, so C is nullable (through unit chain). So C becomes nullable.

D -> BC | BS. B not nullable, C is nullable, so BC not all nullable. BS: B not, S? unknown. So D not nullable.

E -> SS | CA. SS: both S? unknown. CA: C nullable, A nullable, so CA all nullable => E becomes nullable! Because C and A nullable, thus CA can produce ε, so E nullable.

Now with E nullable, we need to consider other productions where E appears:

S -> DE: D nonnullable, E nullable -> D (i.e., we can produce D alone). So S gets new production S -> D.

Also S -> AB: A nullable, B not => B alone: S -> B.

Similarly other productions:

B -> D (unit), D not nullable.

D -> BC: B not nullable, C nullable -> B alone: D -> B.

D -> BS: B not nullable, S maybe nullable? Not yet. S not known to be nullable (we haven't proven S nullable). We'll compute after.

E -> SS: if S nullable then we could get ε, but S not nullable currently.

E -> CA: both nullable already gave ε (which we already have). Also we need to create productions where we drop some nullable symbols.

For each production, generate all combos where we delete some nullable symbols, except we must not delete all symbols (unless we are eliminating epsilon from start maybe). But algorithm typically: for each production A->α, for each subset of nullable nonterminals in α (non-empty), produce A-> α' where those are removed. Then after processing all, remove original ε productions (except maybe for start symbol). Also need to keep start S perhaps allow ε if original language includes ε (but original grammar does not have S->ε). So we will remove A->ε, not add new start epsilon.

Now compute all new productions.

List all original productions (excluding A->ε after removal) and generate.

I'll denote nullable set N = {A, C, E}. Also maybe after adding new productions S may become nullable? Let's see: S new productions could be S -> B, S -> D, plus original S->AB, S->DE. Could S become nullable via these? B not nullable, D not nullable. So S not nullable.

Now for each original production:

1. S -> AB
nullable symbols: A (nullable) B (non). Subsets: remove A -> produce S -> B. Already accounted. Also could remove both A and B? B not nullable, can't remove. So only S->B.

2. S -> DE
nullable: D not, E yes. Remove E -> S -> D. Already.

3. A -> AB
nullable: A nullable, B not. Remove A => A -> B. Also can we keep original? yes (but we will keep after epsilon elimination, but A still exists as variable). Since we will remove epsilon production A->ε, but keep other productions.

So add A -> B.

Also A -> AB (original) remains.

4. A -> a (terminal) stays.

5. B -> b remains.

6. B -> BB: nullable symbols? B is not currently nullable (but could become nullable via something? Not yet). B not in N, so no change.

7. B -> D (unit) stays (though later chain removal).

8. C -> A: A nullable. For production C -> A, we can drop A to get ε. That would make C -> ε new production, but algorithm for removing epsilon will add ε productions for variables that can derive ε (C already nullable). But final step we remove all ε productions except possibly start? The algorithm includes adding new productions that result from removing nullable symbols; this may create ε productions for other variables (C). Indeed we need to add C -> ε (since C->A and A nullable). But we will later remove ε productions (only explicit ones). Actually we must ensure resulting grammar has no ε productions at all. So if we add C->ε we need to later remove it? The standard algorithm: After generating new productions, we delete all ε productions (including the original ones). If the start symbol is nullable, we keep S->ε. Here start is not nullable, so we will delete all ε productions, including the newly added C->ε and maybe others. But we must also consider that we might have produced new productions where a variable may derive ε via multiple steps that are not direct epsilon. That is okay as long as no direct ε productions exist in final grammar. However, we need to ensure language equivalence after removal: Since C is nullable, but we will not have ε productions, C can still derive ε via chain? Actually without ε production C cannot derive ε unless there is a chain to something that does produce ε. Since we removed all ε productions, we need to make sure that any string that originally could be derived with C deriving ε can still be derived maybe via alternative productions that we added (like we removed C from RHS). So direct C->ε is not needed. In standard algorithm, we add productions that omit nullable symbols, but we do not add new ε productions except possibly for start symbol. So for production C->A where A nullable, we add C->ε? Wait standard algorithm: For each production A→α containing nullable symbols, for each subset of nullable occurrences (excluding the empty set) produce a new production A→α' where those symbols are omitted. If after omission α' becomes empty, then we add A→ε (if A is not start). Then later we delete original ε productions. So indeed we will generate C->ε. Then later we delete ε productions (including the one we just added) unless start. That would lose the ability for C to derive ε; but we need C to be nullable in the resulting grammar, i.e., capable of being omitted in contexts. That is achieved because we have added productions that omit C when it appears on RHS of other productions (like D->BC we added D->B, D->C, etc). Since we already added those, we may not need C->ε. However algorithm typical final grammar may still have ε productions for variables that were nullable but not start? Let's recall the lemma: "Given CFG G, construct G' without ε-productions such that L(G') = L(G) \ {ε}". So final grammar has no ε productions at all. So we must ensure *no* ε productions remain. Thus we will delete C->ε.

Hence we generate C->ε but later remove it. So final set will not include C->ε.

9. C -> c (terminal) stays.

10. D -> BC
nullable: B not, C nullable => produce D -> B (remove C). Also if both nullable? B not, so no ε.

11. D -> BS
nullable: B not, S not => unchanged.

12. E -> SS: none nullable (S not). unchanged.

13. E -> CA
nullable: C nullable, A nullable. Subsets:
- remove C => E -> A
- remove A => E -> C
- remove both => E -> ε (but will be removed later). So we add E->A and E->C. (E already nullable due to CA; we will later remove ε productions.)

Now also consider that after adding new productions, some variables might become nullable (like D maybe? D->B; B not nullable). No new nullable.

Now we need to remove all ε productions (including original A->ε and newly added C->ε, E->ε). The start symbol S is not nullable, so we just delete them.

Thus final production set G'_c (after step 3) should be the original productions plus the new ones, without any ε productions.

Let's compile list:

Original (except A->ε):

S -> AB | DE
A -> AB | a
B -> b | BB | D
C -> A | c
D -> BC | BS
E -> SS | CA

Add new ones:

From S -> AB: S -> B
From S -> DE: S -> D

From A -> AB: A -> B

From C -> A: (we considered C->ε, but we delete). No other.

From D -> BC: D -> B

From D -> BS (no change)

From D -> BC we also could have D -> C? Wait removal of B? B not nullable, but C nullable we already did D->B. Also if we considered removal of B (not nullable) can't.

From E -> CA: added E -> A, E -> C

From maybe other productions with nullable symbols?

Check B -> D (unit) D not nullable.

Check B -> BB: none nullable.

Check E -> SS: none nullable.

Check D -> BS: no.

Check maybe AB again? Already.

Thus final set:

S → AB | DE | B | D
A → AB | a | B
B → b | BB | D
C → A | c
D → BC | BS | B
E → SS | CA | A | C

We should also consider possible productions from removing nullable symbols from RHS containing multiple nullable symbols more than one at a time. For example, if there is a production with two nullable symbols, we need to generate version removing each individually and also both. Already handled for E -> CA (both nullable). For D->BC we handled removal of C only. For S->DE we removed E only. For S->AB removed A only. For A->AB removed A only. For D->BC removal of B not allowed (not nullable). For others none.

Now ensure condition (1) still holds: only terminals appear alone. Indeed we have terminals: a appears alone in A -> a, b appears alone in B -> b, c appears alone in C -> c. No other terminals appear in longer RHS. So condition (1) satisfied.

Condition (2): RHS length <=2. Let's check each production:

S -> AB (2), DE (2), B (1), D (1) OK.

A -> AB (2), a (1), B (1) OK.

B -> b (1), BB (2), D (1) OK.

C -> A (1), c (1) OK.

D -> BC (2), BS (2), B (1) OK.

E -> SS (2), CA (2), A (1), C (1) OK.

All fine.

Condition (3): no ε productions. None present.

Thus final answer list.

We need to output using <answer id="1"> ... </answer>.

Let's format each production line with proper LaTeX maybe. But example shows just productions separated by line breaks. We'll follow similar.

<answer id="1">
$S \rightarrow AB \mid DE \mid B \mid D$
$A \rightarrow AB \mid a \mid B$
$B \rightarrow b \mid BB \mid D$
$C \rightarrow A \mid c $
$D \rightarrow BC \mid BS \mid B $
$E \rightarrow SS \mid CA \mid A \mid C$
</answer>

Entfernen von $\varepsilon$-Produktionen. Die CFG $G_c$ ist gegeben durch folgende Produktionen:
$S \rightarrow AB \mid DE$
$A \rightarrow AB \mid a \mid \varepsilon$
$B \rightarrow b \mid BB \mid D$
$C \rightarrow A \mid c$
$D \rightarrow BC \mid BS$
$E \rightarrow SS \mid CA$
Führen Sie den dritten Schritt des Algorithmus zur Überführung in CNF aus. Geben Sie die Produktionen einer CFG $G'_c$ an, so dass $L(G_c) = L(G'_c)$ und $G'_c$ Bedingungen (1), (2) und (3) erfüllt.

$S \rightarrow AB \mid DE \mid B \mid D$
$A \rightarrow AB \mid a \mid B$
$B \rightarrow b \mid BB \mid D$
$C \rightarrow A \mid c$
$D \rightarrow BC \mid BS \mid B$
$E \rightarrow SS \mid CA \mid A \mid C$

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: CFG-Algorithmen

In den Teilaufgaben (a)-(d) geht es darum, eine kontextfreie Grammatik (CFG) in Chomsky-Normalform (CNF) zu konvertieren. Wir führen jeden Schritt einzeln aus, und jeweils auf einer anderen Grammatik – Sie können also die Aufgabenteile unabhängig voneinander bearbeiten.

Eine CFG $G=(V,\Sigma,P,S)$ ist in CNF, wenn jede Produktion $(X\rightarrow\alpha)\in P$, mit $X\in V$ und $\alpha\in(\Sigma\cup V)^*$, folgende Bedingungen erfüllt:
(1) $\alpha\in\Sigma\cup V^*$; Terminale dürfen nur in Produktionen der Länge 1 erzeugt werden.
(2) $|\alpha|\le2$; jede Produktion hat höchstens Länge $2$.
(3) $\alpha\ne\varepsilon$; es gibt keine $\varepsilon$-Produktionen.
(4) $\alpha\notin V$; es gibt keine Kettenproduktionen.

**Achtung:** Die Teilaufgaben fragen jeweils spezifisch nach dem Ergebnis, das sich durch die Ausführung des Algorithmus aus der Vorlesung ergibt, nicht nach einer beliebigen äquivalenten CFG, die den Bedingungen genügt. Details, wie etwa die Namen der Variablen oder die Reihenfolge, in der Produktionen betrachtet werden, können Sie frei wählen.
Wir nennen $A \rightarrow \epsilon$ eine *$\epsilon$-Produktion*.
\begin{Lemma}
Zu jeder CFG $G = (V,\Sigma,P,S)$ kann man eine CFG $G'$ konstruieren, die keine $\epsilon$-Produktionen enthält, so dass gilt
$L(G') = L(G)\setminus\{\epsilon\}$
\end{Lemma}
Wir erweitern $P$ induktiv zu eine Obermenge $\hat{P}$:
\begin{enumerate}
\item Jede Produktion aus $P$ ist in $\hat{P}$
\item Sind \alert{$B \rightarrow \epsilon$} und \alert{$A \rightarrow \alpha B \beta$} in $\hat{P}$, so füge auch \alert{$A \rightarrow \alpha\beta$} hinzu.
\end{enumerate}
Offensichtlich gilt $L(\hat{G}) = L(G)$: Jede neue Produktion kann von 2 alten Productionensimuliert werden.

Wir definieren $G'$ als $\hat{G}$ ohne die $\epsilon$-Produktionen. Denn diese sind nun überflüssig.

Wir nennen $A \rightarrow B$ eine *Kettenproduktion*.
\begin{Lemma}
Zu jeder CFG $G = (V,\Sigma,P,S)$ kann man eine CFG $G'$ konstruieren, die keine Kettenproduktionen enthält, so dass gilt $L(G') = L(G)$.
\end{Lemma}
\begin{Beweis}
Wir erweitern $P$ induktiv zu eine Obermenge $\hat{P}$:
\begin{enumerate}
\item Jede Produktion aus $P$ ist in $\hat{P}$
\item Sind \alert{$A \rightarrow B$} und \alert{$B \rightarrow \alpha$} in $\hat{P}$ mit $\alpha \neq A$, so füge auch \alert{$A \rightarrow \alpha$} hinzu.
\end{enumerate}
Das Ergebnis $G'$ ist $\hat{G}$ ohne die (nun überflüssigen) Kettenproduktionen.

\paragraph{Konstruktion einer Chomsky-Normalform}

Eingabe: Eine kontextfreie Grammatik $G = (V,\Sigma,P,S)$
\begin{enumerate}
\item Füge für jedes $a\in\Sigma$, das in einer rechten Seite der Länge $\geq2$ vorkommt, ein neues Nichtterminal $A_a$ zu $V$ hinzu, ersetze $a$ in allen rechten Seiten der Länge $\geq 2$ durch $A_a$, und füge $A_a\rightarrow a$ zu $P$ hinzu.
\item Ersetze jede Produktion der Form
\[A\rightarrow B_1B_2 \cdots B_k (k \geq 3)\]
durch
\[A \rightarrow B_1C_2,\ C_2 \rightarrow B_2C_3, ..., C_{k-1} \rightarrow B_{k-1}B_k \]
wobei $C_2, ..., C_{k-1}$ neue Nichtterminale sind.
\item Eliminiere alle $\epsilon$-Produktionen.
\item Eliminiere alle Kettenproduktionen.
\end{enumerate}

## subproblem a (2.0 points)
*Entfernen von Terminalen in langen Produktionen.* Die CFG $G_a$ ist gegeben durch folgende Produktionen:
$S \rightarrow aX \mid Xa \mid ab$
$X \rightarrow SS \mid S \mid \varepsilon \mid bXX$
Führen Sie den ersten Schritt des Algorithmus zur Überführung in CNF aus und geben Sie die Produktionen einer CFG $G'_a$ an, so dass $L(G_a) = L(G'_a)$ gilt und $G'_a$ Bedingung (1) erfüllt.

<example>
$S \rightarrow XX \mid aa \mid \varepsilon$
$X \rightarrow SX \mid X$
...
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
S \;\rightarrow\; A_a\,X \;\mid\; X\,A_a \;\mid\; A_a\,A_b  
\\[4pt]
X \;\rightarrow\; S\,S \;\mid\; S \;\mid\; \varepsilon \;\mid\; A_b\,X\,X  
\\[4pt]
A_a \;\rightarrow\; a  
\\[4pt]
A_b \;\rightarrow\; b
</answer>

user:
## subproblem b (2.0 points)
*Entfernen langer Produktionen.* Die CFG $G_b$ ist gegeben durch die folgenden Produktionen:
$S \rightarrow A \mid BAA \mid \varepsilon$
$A \rightarrow SABB \mid a$
$B \rightarrow AS \mid b$
Führen Sie den zweiten Schritt des Algorithmus zur Überführung in CNF aus und geben Sie die Produktionen einer CFG $G'_b$ an, so dass $L(G_b) = L(G'_b)$ und $G'_b$ Bedingungen (1) und (2) erfüllt.

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
S \;\rightarrow\; A \;\mid\; B\,X_1 \;\mid\; \varepsilon \\[4pt]
X_1 \;\rightarrow\; A\,A \\[4pt]
A \;\rightarrow\; S\,C_2 \;\mid\; a \\[4pt]
C_2 \;\rightarrow\; A\,C_3 \\[4pt]
C_3 \;\rightarrow\; B\,B \\[4pt]
B \;\rightarrow\; A\,S \;\mid\; b
</answer>

user:
## subproblem c (3.0 points)
*Entfernen von $\varepsilon$-Produktionen.* Die CFG $G_c$ ist gegeben durch folgende Produktionen:
$S \rightarrow AB \mid DE$
$A \rightarrow AB \mid a \mid \varepsilon$
$B \rightarrow b \mid BB \mid D$
$C \rightarrow A \mid c$
$D \rightarrow BC \mid BS$
$E \rightarrow SS \mid CA$
Führen Sie den dritten Schritt des Algorithmus zur Überführung in CNF aus. Geben Sie die Produktionen einer CFG $G'_c$ an, so dass $L(G_c) = L(G'_c)$ und $G'_c$ Bedingungen (1), (2) und (3) erfüllt.

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
S \rightarrow AB \mid DE \mid B \mid D \\
A \rightarrow AB \mid a \mid B \\
B \rightarrow b \mid BB \mid D \\
C \rightarrow A \mid c \\
D \rightarrow BC \mid BS \mid B \\
E \rightarrow SS \mid CA \mid A \mid C
</answer>

user:
## subproblem d (3.0 points)
*Entfernen von Kettenproduktionen.* Die CFG $G_d$ ist gegeben durch die Produktionen:
$S \rightarrow a \mid BA \mid B$
$A \rightarrow a \mid AA$
$B \rightarrow b \mid SC \mid A$
$C \rightarrow B \mid SS$
Führen Sie den vierten Schritt des Algorithmus zur Überführung in CNF aus und geben Sie die Produktionen einer CFG $G'_d$ in CNF an, so dass $L(G_d) = L(G'_d)$ gilt.

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to perform step 4: eliminate chain productions after earlier steps? The algorithm order: step1 remove terminals in long productions, step2 break long productions, step3 eliminate epsilon, step4 eliminate chain productions. For subproblem d, we must apply step4 to given grammar. But we must also ensure conditions (1)-(4) of CNF: no epsilon, no chain, length ≤2, terminals only alone. However earlier steps (1-3) may not have been applied? The problem says "Führen Sie den vierten Schritt des Algorithmus ...". So we assume previous steps already applied? Usually we need to eliminate chain productions from a grammar that already satisfies (1) and (2) and (3). But the given grammar likely has terminals in length>1 productions (like BA?), actually BA length 2 but both nonterminals, okay. Terminals alone ok. There are chain productions: S -> B, B -> A, A -> AA? That's not chain because RHS length 2 nonterminals. Chain is A -> B (single nonterminal). So we need to eliminate them.

First, let's list productions:

1. S → a | BA | B
2. A → a | AA
3. B → b | SC | A
4. C → B | SS

We need to eliminate epsilon? There are none (no ε). Terminals appear alone (a, b). No productions length >2? B → SC (length 2) okay. C → SS length 2. All good. So only chain productions: S → B, B → A, C → B. Also possibly? A → a (terminal), not chain. S → a terminal. B → b terminal. So we need to remove those.

Standard method: compute for each nonterminal A the set of nonterminals reachable via chain (including itself). Then add productions A -> α for each production B -> α where B is in reachable set and α is not a single nonterminal. Then remove all chain productions.

Let's compute:

Nonterminals: S, A, B, C.

Chain edges:
S -> B
B -> A
C -> B

Also maybe there are indirect: S -> B -> A, so S can reach A.
C -> B -> A, so C can reach A.
Also S can reach B and A.
Also B can reach A only.
C can reach B and A.

Also each can reach itself.

Now collect non-chain productions (i.e., RHS not a single nonterminal):

- S → a (terminal)
- S → BA (two nonterminals)
- A → a (terminal)
- A → AA (two nonterminals)
- B → b (terminal)
- B → SC (two nonterminals)
- C → SS (two nonterminals)

Now for each nonterminal X, we add productions X → α for each α from any Y in reachable set of X where Y has a non-chain production.

Let's compute reachable sets:

R(S) = {S, B, A}
R(B) = {B, A}
R(C) = {C, B, A}
R(A) = {A}

Now generate:

For S:
Include productions from S itself: a, BA.
From B: b, SC.
From A: a, AA.
Thus we add to S:
S → b
S → SC
S → AA
(and also S → a already there)
S → BA already.

For B:
Include from B: b, SC
From A: a, AA
Thus add:
B → a
B → AA

For C:
Include from C: SS
From B: b, SC
From A: a, AA
Thus add:
C → b
C → SC
C → a
C → AA

Also maybe we need to keep original non-chain productions. Also we may keep chain productions for now but later we will delete all chain productions S→B, B→A, C→B.

Now after adding these, we must ensure we haven't introduced new chain productions (none, because RHS are terminals or two symbols). However note that some added productions may have length 2 with one being terminal? No, all added RHS are either terminal (single) or two nonterminals (AA, SC). That's fine.

Now we can remove chain productions:

Remove: S → B, B → A, C → B. Also any other chain? No.

Now final grammar G'_d consists of:

S productions:
- S → a
- S → BA
- S → b
- S → SC
- S → AA

A productions:
- A → a
- A → AA

B productions:
- B → b
- B → SC
- B → a
- B → AA

C productions:
- C → SS
- C → b
- C → SC
- C → a
- C → AA

Check length condition: All RHS are either single terminal (a,b) or two nonterminals. However note SC: S and C are nonterminals, ok. AA, BA, SS are two nonterminals. Good. No epsilon, no chain. So CNF satisfied.

We might also need to ensure terminals appear only alone; they do.

Thus answer.

We need to output in specified format:

<answer id="1">
...
</answer>

Let's format with each production line.

Probably they'd like each line like "S → a | BA | b | SC | AA". But any format is okay as long as clear.

Let's produce.

<answer id="1">
$S \;\rightarrow\; a \mid BA \mid b \mid SC \mid AA $
$A \;\rightarrow\; a \mid AA $
$B \;\rightarrow\; b \mid SC \mid a \mid AA $
$C \;\rightarrow\; SS \mid b \mid SC \mid a \mid AA$
</answer>

Entfernen von Kettenproduktionen. Die CFG $G_d$ ist gegeben durch die Produktionen:
$S \rightarrow a \mid BA \mid B$
$A \rightarrow a \mid AA$
$B \rightarrow b \mid SC \mid A$
$C \rightarrow B \mid SS$
Führen Sie den vierten Schritt des Algorithmus zur Überführung in CNF aus und geben Sie die Produktionen einer CFG $G'_d$ in CNF an, so dass $L(G_d) = L(G'_d)$ gilt.

$S \;\rightarrow\; a \mid BA \mid b \mid SC \mid AA$
$A \;\rightarrow\; a \mid AA$
$B \;\rightarrow\; b \mid SC \mid a \mid AA$
$C \;\rightarrow\; SS \mid b \mid SC \mid a \mid AA$

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: CFG-Algorithmen

In den Teilaufgaben (a)-(d) geht es darum, eine kontextfreie Grammatik (CFG) in Chomsky-Normalform (CNF) zu konvertieren. Wir führen jeden Schritt einzeln aus, und jeweils auf einer anderen Grammatik – Sie können also die Aufgabenteile unabhängig voneinander bearbeiten.

Eine CFG $G=(V,\Sigma,P,S)$ ist in CNF, wenn jede Produktion $(X\rightarrow\alpha)\in P$, mit $X\in V$ und $\alpha\in(\Sigma\cup V)^*$, folgende Bedingungen erfüllt:
(1) $\alpha\in\Sigma\cup V^*$; Terminale dürfen nur in Produktionen der Länge 1 erzeugt werden.
(2) $|\alpha|\le2$; jede Produktion hat höchstens Länge $2$.
(3) $\alpha\ne\varepsilon$; es gibt keine $\varepsilon$-Produktionen.
(4) $\alpha\notin V$; es gibt keine Kettenproduktionen.

**Achtung:** Die Teilaufgaben fragen jeweils spezifisch nach dem Ergebnis, das sich durch die Ausführung des Algorithmus aus der Vorlesung ergibt, nicht nach einer beliebigen äquivalenten CFG, die den Bedingungen genügt. Details, wie etwa die Namen der Variablen oder die Reihenfolge, in der Produktionen betrachtet werden, können Sie frei wählen.
Wir nennen $A \rightarrow \epsilon$ eine *$\epsilon$-Produktion*.
\begin{Lemma}
Zu jeder CFG $G = (V,\Sigma,P,S)$ kann man eine CFG $G'$ konstruieren, die keine $\epsilon$-Produktionen enthält, so dass gilt
$L(G') = L(G)\setminus\{\epsilon\}$
\end{Lemma}
Wir erweitern $P$ induktiv zu eine Obermenge $\hat{P}$:
\begin{enumerate}
\item Jede Produktion aus $P$ ist in $\hat{P}$
\item Sind \alert{$B \rightarrow \epsilon$} und \alert{$A \rightarrow \alpha B \beta$} in $\hat{P}$, so füge auch \alert{$A \rightarrow \alpha\beta$} hinzu.
\end{enumerate}
Offensichtlich gilt $L(\hat{G}) = L(G)$: Jede neue Produktion kann von 2 alten Productionensimuliert werden.

Wir definieren $G'$ als $\hat{G}$ ohne die $\epsilon$-Produktionen. Denn diese sind nun überflüssig.

Wir nennen $A \rightarrow B$ eine *Kettenproduktion*.
\begin{Lemma}
Zu jeder CFG $G = (V,\Sigma,P,S)$ kann man eine CFG $G'$ konstruieren, die keine Kettenproduktionen enthält, so dass gilt $L(G') = L(G)$.
\end{Lemma}
\begin{Beweis}
Wir erweitern $P$ induktiv zu eine Obermenge $\hat{P}$:
\begin{enumerate}
\item Jede Produktion aus $P$ ist in $\hat{P}$
\item Sind \alert{$A \rightarrow B$} und \alert{$B \rightarrow \alpha$} in $\hat{P}$ mit $\alpha \neq A$, so füge auch \alert{$A \rightarrow \alpha$} hinzu.
\end{enumerate}
Das Ergebnis $G'$ ist $\hat{G}$ ohne die (nun überflüssigen) Kettenproduktionen.

\paragraph{Konstruktion einer Chomsky-Normalform}

Eingabe: Eine kontextfreie Grammatik $G = (V,\Sigma,P,S)$
\begin{enumerate}
\item Füge für jedes $a\in\Sigma$, das in einer rechten Seite der Länge $\geq2$ vorkommt, ein neues Nichtterminal $A_a$ zu $V$ hinzu, ersetze $a$ in allen rechten Seiten der Länge $\geq 2$ durch $A_a$, und füge $A_a\rightarrow a$ zu $P$ hinzu.
\item Ersetze jede Produktion der Form
\[A\rightarrow B_1B_2 \cdots B_k (k \geq 3)\]
durch
\[A \rightarrow B_1C_2,\ C_2 \rightarrow B_2C_3, ..., C_{k-1} \rightarrow B_{k-1}B_k \]
wobei $C_2, ..., C_{k-1}$ neue Nichtterminale sind.
\item Eliminiere alle $\epsilon$-Produktionen.
\item Eliminiere alle Kettenproduktionen.
\end{enumerate}

## subproblem a (2.0 points)
*Entfernen von Terminalen in langen Produktionen.* Die CFG $G_a$ ist gegeben durch folgende Produktionen:
$S \rightarrow aX \mid Xa \mid ab$
$X \rightarrow SS \mid S \mid \varepsilon \mid bXX$
Führen Sie den ersten Schritt des Algorithmus zur Überführung in CNF aus und geben Sie die Produktionen einer CFG $G'_a$ an, so dass $L(G_a) = L(G'_a)$ gilt und $G'_a$ Bedingung (1) erfüllt.

<example>
$S \rightarrow XX \mid aa \mid \varepsilon$
$X \rightarrow SX \mid X$
...
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
S \;\rightarrow\; A_a\,X \;\mid\; X\,A_a \;\mid\; A_a\,A_b  
\\[4pt]
X \;\rightarrow\; S\,S \;\mid\; S \;\mid\; \varepsilon \;\mid\; A_b\,X\,X  
\\[4pt]
A_a \;\rightarrow\; a  
\\[4pt]
A_b \;\rightarrow\; b
</answer>

user:
## subproblem b (2.0 points)
*Entfernen langer Produktionen.* Die CFG $G_b$ ist gegeben durch die folgenden Produktionen:
$S \rightarrow A \mid BAA \mid \varepsilon$
$A \rightarrow SABB \mid a$
$B \rightarrow AS \mid b$
Führen Sie den zweiten Schritt des Algorithmus zur Überführung in CNF aus und geben Sie die Produktionen einer CFG $G'_b$ an, so dass $L(G_b) = L(G'_b)$ und $G'_b$ Bedingungen (1) und (2) erfüllt.

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
S \;\rightarrow\; A \;\mid\; B\,X_1 \;\mid\; \varepsilon \\[4pt]
X_1 \;\rightarrow\; A\,A \\[4pt]
A \;\rightarrow\; S\,C_2 \;\mid\; a \\[4pt]
C_2 \;\rightarrow\; A\,C_3 \\[4pt]
C_3 \;\rightarrow\; B\,B \\[4pt]
B \;\rightarrow\; A\,S \;\mid\; b
</answer>

user:
## subproblem c (3.0 points)
*Entfernen von $\varepsilon$-Produktionen.* Die CFG $G_c$ ist gegeben durch folgende Produktionen:
$S \rightarrow AB \mid DE$
$A \rightarrow AB \mid a \mid \varepsilon$
$B \rightarrow b \mid BB \mid D$
$C \rightarrow A \mid c$
$D \rightarrow BC \mid BS$
$E \rightarrow SS \mid CA$
Führen Sie den dritten Schritt des Algorithmus zur Überführung in CNF aus. Geben Sie die Produktionen einer CFG $G'_c$ an, so dass $L(G_c) = L(G'_c)$ und $G'_c$ Bedingungen (1), (2) und (3) erfüllt.

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
S \rightarrow AB \mid DE \mid B \mid D \\
A \rightarrow AB \mid a \mid B \\
B \rightarrow b \mid BB \mid D \\
C \rightarrow A \mid c \\
D \rightarrow BC \mid BS \mid B \\
E \rightarrow SS \mid CA \mid A \mid C
</answer>

user:
## subproblem d (3.0 points)
*Entfernen von Kettenproduktionen.* Die CFG $G_d$ ist gegeben durch die Produktionen:
$S \rightarrow a \mid BA \mid B$
$A \rightarrow a \mid AA$
$B \rightarrow b \mid SC \mid A$
$C \rightarrow B \mid SS$
Führen Sie den vierten Schritt des Algorithmus zur Überführung in CNF aus und geben Sie die Produktionen einer CFG $G'_d$ in CNF an, so dass $L(G_d) = L(G'_d)$ gilt.

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
S \;\rightarrow\; a \mid BA \mid b \mid SC \mid AA \\
A \;\rightarrow\; a \mid AA \\
B \;\rightarrow\; b \mid SC \mid a \mid AA \\
C \;\rightarrow\; SS \mid b \mid SC \mid a \mid AA
</answer>

user:
## subproblem e (3.0 points)
Die CFG $G$ ist gegeben durch die Produktionen:
$S \rightarrow AB\mid C$
$A \to aA \mid AS$
$B \to bS$
$C \to aCb \mid \varepsilon$
Geben Sie die erzeugenden, erreichbaren und nützlichen Nichtterminale von $G$ an.

<answer-template id="1">
Erzeugend: $?$
Erreichbar: $?$
Nützlich: $?$
</answer-template>

<example>
Erzeugend: $S$
Erreichbar: $S, A, B$
Nützlich: $S$
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to compute generating (erzeugend), reachable (erreichbar), and useful (nützlich = both). Grammar:

Nonterminals: S, A, B, C.

Productions:
1. S -> AB | C
2. A -> aA | AS
3. B -> bS
4. C -> a C b | ε

Terminals: a, b.

We need to find which nonterminals can derive some terminal string (generating). Also reachable from start S (via productions). Useful = both generating and reachable.

Compute generating set:

Base: any nonterminal with production that yields only terminals (or ε). C -> ε : C is generating. Also does C have production a C b which includes C, not terminal-only yet. So C generating (since it can derive ε). Also check others.

Consider B: B -> b S. S is not yet known generating. So not yet.

A: A -> aA (needs A) or A -> A S (needs A and S). No terminal-only. Not generating yet.

S: S -> AB, need A and B both generating. S -> C, C generating, so S is generating (since C can produce ε, so S can produce ε). So S generating.

Now B: B -> b S, and S generating, so B is generating (produces b followed by something generated by S, e.g., ε, etc.). So B generating.

Now A: A -> aA not generating (needs A). A -> A S needs A and S. Since S generating, but A still not. There is no production with only terminals. So A is not generating (non-generating). Could there be a derivation that eliminates A? No. So A non-generating.

Thus generating set = {S, B, C}. Possibly also others? Check if any production yields only terminals via combination: B -> bS, S->C, C->ε yields b ε = b, so B can produce b. So yes.

Now reachable from start S: Start S reachable trivially.

From S productions: S -> AB gives A and B reachable. S -> C gives C reachable.

Thus reachable set = {S, A, B, C}. All four reachable.

Now useful = intersection: {S, B, C}. Since A reachable but not generating, not useful.

Thus answer:

Erzeugend: $S, B, C$

Erreichbar: $S, A, B, C$

Nützlich: $S, B, C$

We must follow template: <answer id="1"> with lines as in template. Use LaTeX formatting? Use same placeholder $...$.

Thus produce.

<answer id="1">
Erzeugend: $S, B, C$  
Erreichbar: $S, A, B, C$  
Nützlich: $S, B, C$
</answer>

Die CFG $G$ ist gegeben durch die Produktionen:
$S \rightarrow AB\mid C$
$A \to aA \mid AS$
$B \to bS$
$C \to aCb \mid \varepsilon$
Geben Sie die erzeugenden, erreichbaren und nützlichen Nichtterminale von $G$ an.

Erzeugend: $S, B, C$  
Erreichbar: $S, A, B, C$  
Nützlich: $S, B, C$

2: mismatch

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Verlorene Buchsten

Sei $\Sigma:=\{a,b\}$. Für eine Sprache $L\subseteq\Sigma^*$ bezeichnet $\Del(L)$ die Sprache,
die man erhält, wenn man den Wörtern aus $L$ **genau einmal** das Wort $ab$ an einer beliebigen Stelle entfernt. Formal definieren wir
$\Del(L):= \{vw \mid v,w\in \Sigma^* \wedge vabw\in L\}$

## subproblem a (1.0 points)
Listen Sie alle Wörter der Sprache $\Del(\{abbaba,baa,ab,a,b\})$ explizit auf:

<answer-template id="1">
$\Del(\{abbaba,baa,ab,a,b\}) = \{?\}$
</answer-template>

<example>
$\Del(\{abbaba,baa,ab,a,b\}) = \{\varepsilon, ab, bbab\}$
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to compute Del of given set. Definition: Del(L) = { v w | v,w ∈ Σ* and v a b w ∈ L }. That is, we take each word in L, find a occurrence of substring "ab" and delete it exactly once, i.e., remove the letters a and b that appear consecutively as "ab". The result is the concatenation of the prefix before that "ab" (v) and suffix after (w). So for each word in L, we consider all possible ways to locate a substring "ab" and delete it, producing possibly multiple results. Then union over all words.

Given L = { abbaba, baa, ab, a, b }. Let's compute.

Alphabet Σ = {a,b}.

Word1: "abbaba". Let's list positions of "ab". The string: a b b a b a. Indices 1-6. Substrings length2:
1-2: "ab"? positions 1 a,2 b => "ab". Yes at start.
2-3: "bb" no.
3-4: "ba" no.
4-5: "ab"? positions 4 a,5 b => "ab". Yes at positions 4-5.
5-6: "ba" no.

So two occurrences: at 1-2 and at 4-5. Deleting each yields:

- Delete first "ab" (positions1-2): remove a,b => remaining characters: positions 3-6: "b a b a"? Actually original: a b b a b a. Remove first two => left "b a b a"? That's "baba". So result "baba".

- Delete second "ab" (positions4-5): remove a at pos4 and b at pos5 => remaining: positions1-3: "a b b" = "abb", and position6: "a". So "abb a" => "abba". So result "abba".

Thus from "abbaba" we get "baba" and "abba".

Word2: "baa". String: b a a. Substrings length2:
1-2: "ba" not "ab".
2-3: "aa" not.
Thus no "ab". So cannot delete exactly one "ab". So no result from this word.

Word3: "ab". Only one occurrence at positions1-2. Delete yields empty string ε. So result ε.

Word4: "a". No "ab". No result.

Word5: "b". No "ab". No result.

Thus Del set = { "baba", "abba", ε }. Are there any duplicates? No.

Thus answer: $\Del(\{abbaba,baa,ab,a,b\}) = \{\varepsilon, ab, bbab\}$? Wait example shows that but it's just example not correct. Our computed set: {ε, baba, abba}. Let's double-check "baba" and "abba". Are there any other possibilities? Could we also consider overlapping? No, "ab" is length2, cannot overlap with itself except if pattern "aba" but not here.

Thus answer.

We must follow answer-template: $\Del(\{abbaba,baa,ab,a,b\}) = \{?\}$ where ? is list of words separated by commas. Use LaTeX? The template expects the whole expression inside. So we output:

<answer id="1">
$\Del(\{abbaba,baa,ab,a,b\}) = \{\varepsilon,\,baba,\,abba\}$
</answer>

Make sure ordering maybe any. Use \varepsilon for empty string. Provide exactly.

<answer id="1">
$\Del(\{abbaba,baa,ab,a,b\}) = \{\varepsilon,baba,abba\}$
</answer>

Listen Sie alle Wörter der Sprache $\Del(\{abbaba,baa,ab,a,b\})$ explizit auf:

$\Del(\{abbaba,baa,ab,a,b\}) = \{\varepsilon,baba,abba\}$

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Verlorene Buchsten

Sei $\Sigma:=\{a,b\}$. Für eine Sprache $L\subseteq\Sigma^*$ bezeichnet $\Del(L)$ die Sprache,
die man erhält, wenn man den Wörtern aus $L$ **genau einmal** das Wort $ab$ an einer beliebigen Stelle entfernt. Formal definieren wir
$\Del(L):= \{vw \mid v,w\in \Sigma^* \wedge vabw\in L\}$

## subproblem a (1.0 points)
Listen Sie alle Wörter der Sprache $\Del(\{abbaba,baa,ab,a,b\})$ explizit auf:

<answer-template id="1">
$\Del(\{abbaba,baa,ab,a,b\}) = \{?\}$
</answer-template>

<example>
$\Del(\{abbaba,baa,ab,a,b\}) = \{\varepsilon, ab, bbab\}$
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
$\Del(\{abbaba,baa,ab,a,b\}) = \{\varepsilon,\;baba,\;abba\}$
</answer>

user:
## subproblem b (2.0 points)
Wir definieren $\DelTwo(L):=\{uvw \mid u,v,w\in \Sigma^* \wedge u ab v ab w\in L \}$. Zeigen Sie oder widerlegen Sie mit einem Gegenbeispiel: Für jede **endliche** Sprache $L\subseteq\Sigma^*$ gilt
$\Del(\Del(L))=\DelTwo(L)$.

<options>
<option id="1">Wahr</option>
<option id="2">Falsch</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<answer id="3">
...
</answer>

We need to decide statement: For any finite language L, Del(Del(L)) = DelTwo(L). Del removes exactly one occurrence of "ab". DelTwo removes two occurrences (not necessarily consecutive) but with one v in between (i.e., remove two "ab" strings, possibly with something between). Let's analyze.

Definition: Del(L) = { v w | ∃v,w: v a b w ∈ L }. Here remove exactly one "ab". So if a word in L contains at least one "ab", we can delete one occurrence, yielding new word.

Then Del(Del(L)): apply again, i.e., from a word that is result of deleting one "ab" from some original word, delete another "ab". This yields words that are obtained by deleting two (maybe overlapping?) occurrences of "ab" from original L? But careful: after first deletion, the resulting word may have new "ab" substrings formed by concatenating parts around removed "ab". Could yield different results than directly removing two distinct "ab" substrings from original? Let's examine.

DelTwo(L) = { u v w | ∃u,v,w: u a b v a b w ∈ L }. So you remove two "ab" strings that are separated by some substring v (which can be empty?). Yes v ∈ Σ*; could be empty meaning two "ab" consecutive? Actually if v = ε, then you remove "ab" "ab" as consecutive substring "abab". So DelTwo removes two occurrences that are non-overlapping and maintaining original order.

Now Del(Del(L)) could possibly delete overlapping occurrences? For instance, original word contains "aab". If you delete the first "ab" (starting at position 2?), but there is only one "ab". Overlap cannot happen because "ab" is length 2, overlapping would require "aba" pattern maybe? Let's consider a word "a b a b": "abab". Removing first "ab" yields "ab". Then second removal yields ε. This matches removing two "ab" as in DelTwo where v = ε, we would get ε as well. So maybe they are equal for finite languages? Need to find counterexample.

Potential issue: Del(Del(L)) could remove the same "ab" twice via new "ab" formed after first removal that wasn't present in original, leading to removal of something not originally two distinct "ab". Example: original word "aab". It has substring "ab" at positions 2-3. Remove it -> left "a". No further "ab". So not a problem.

Consider "abb". Contains "ab" at positions 1-2. Remove it -> "b". No further. So can't get two deletions.

Consider "aab b"? hmm.

Maybe overlapping case: original word "aab". Actually "aab" after removing "ab" (positions 2-3) gives "a". Not two.

Consider "ababa". It has two overlapping occurrences? The substrings "ab" at positions 1-2 and 3-4. They are separate, not overlapping. Remove first -> yields "aba". Which still contains "ab" at positions 1-2. So you can delete again, giving "a". This corresponds to original word "ababa": remove two "ab" strings (positions 1-2 and after removal, the second originally at 3-4 becomes now at 2-3) which corresponds to removing two ab's maybe not captured by DelTwo? DelTwo would need to find u,v,w such that uabvabw is original. For "ababa": can we write as u a b v a b w? Let's see use u = ε, then need "ab v ab w" = "ababa". Choose v = ε? "ab"+"ab" = "abab", w = "a"? That gives "ababa". So u=ε, v=ε, w="a": gives uabvabw = "ab"+"ab"+"a" = "ababa". Thus DelTwo yields "a". So matches.

Maybe need example where after first deletion a new "ab" is formed from letters that were not adjacent originally, meaning the two deletions correspond to overlapping original segments that are not two distinct "ab" substrings. Example: original word "a a b". No. Let's think: letters "a", then some stuff, then "b" ... after deletion of middle "ab", the preceding "a" and following "b" become adjacent, forming new "ab". In original these 'a' and 'b' were separated by the removed "ab". So they were not part of a single "ab" substring. So Del(Del(L)) could delete a composite formed by concatenating ending 'a' of first part and starting 'b' of latter part. That may not correspond to two separate "ab" substrings in original (they are overlapping across the deleted part). The definition of DelTwo requires that both "ab" substrings exist in the original as separate substrings, not overlapping across the middle region. So we can find counterexample.

Take original word "a a b b": i.e., "aabb". Does it contain "ab"? Yes at positions 2-3? Actually letters: 1:a,2:a,3:b,4:b. Substring "ab" at positions 2-3. Remove it -> we get "ab" (first a and last b). That "ab" is new but originally letters were a (position1) and b (position4) not adjacent. So second removal yields ε. So Del(Del) yields ε from original "aabb". Does DelTwo produce ε? Need original word to have two "ab" substrings. Does "aabb" have two disjoint "ab"? Only one "ab". So DelTwo of language containing "aabb" can't produce ε because need uabvabw pattern, which requires two occurrences. So ε ∈ Del(Del({aabb})) but ε ∉ DelTwo({aabb}). Therefore equality fails. So statement false. Provide counterexample L = {aabb}. Finite language.

Thus answer: option 1 false (0), option 2 true (1). Provide explanation in answer.

<option id="1">0</option>
<option id="2">1</option>
<answer id="3">
Die Beh ist **falsch**.  

Ein Gegenbeispiel ist die endliche Sprache  

\[
L=\{\,aabb\,\}.
\]

*   **Berechnung von \(\Del(\Del(L))\):**  
    - Das Wort \(aabb\) enthält das Teilwort \(ab\) (Position 2‑3).  
    - Entfernen wir dieses \(ab\), erhalten wir das Wort \(ab\).  
    - Das resultierende Wort \(ab\) enthält erneut das Teilwort \(ab\); nach erneutem Entfernen erhalten wir \(\varepsilon\).  

    Also \(\varepsilon \in \Del(\Del(L))\).

*   **Berechnung von \(\DelTwo(L)\):**  
    - Für \(\DelTwo(L)\) muss das Ausgangswort die Form \(u\,ab\,v\,ab\,w\) haben, also mindestens **zwei** vorkommende Teilwörter \(ab\).  
    - Das Wort \(aabb\) enthält nur ein einziges \(ab\) und lässt sich nicht als \(u\,ab\,v\,ab\,w\) schreiben.  

    Deshalb \(\varepsilon \notin \DelTwo(L)\).

Da \(\varepsilon\) in \(\Del(\Del(L))\) liegt, aber nicht in \(\DelTwo(L)\), gilt die Gleichheit nicht für alle endlichen Sprachen \(L\).

\[
\boxed{\Del(\Del(L))\neq\DelTwo(L)\text{ im Allgemeinen}}
\]

Damit ist die Aussage **falsch**.
</answer>

Wir definieren $\DelTwo(L):=\{uvw \mid u,v,w\in \Sigma^* \wedge u ab v ab w\in L \}$. Zeigen Sie oder widerlegen Sie mit einem Gegenbeispiel: Für jede endliche Sprache $L\subseteq\Sigma^*$ gilt
$\Del(\Del(L))=\DelTwo(L)$.

Die Beh ist falsch.  

Ein Gegenbeispiel ist die endliche Sprache  

$$L=\{\,aabb\,\}.$$

*   Berechnung von $\Del(\Del(L))$:  
    - Das Wort $aabb$ enthält das Teilwort $ab$ (Position 2‑3).  
    - Entfernen wir dieses $ab$, erhalten wir das Wort $ab$.  
    - Das resultierende Wort $ab$ enthält erneut das Teilwort $ab$; nach erneutem Entfernen erhalten wir $\varepsilon$.  

    Also $\varepsilon \in \Del(\Del(L))$.

*   Berechnung von $\DelTwo(L)$:  
    - Für $\DelTwo(L)$ muss das Ausgangswort die Form $u\,ab\,v\,ab\,w$ haben, also mindestens zwei vorkommende Teilwörter $ab$.  
    - Das Wort $aabb$ enthält nur ein einziges $ab$ und lässt sich nicht als $u\,ab\,v\,ab\,w$ schreiben.  

    Deshalb $\varepsilon \notin \DelTwo(L)$.

Da $\varepsilon$ in $\Del(\Del(L))$ liegt, aber nicht in $\DelTwo(L)$, gilt die Gleichheit nicht für alle endlichen Sprachen $L$.

$$\boxed{\Del(\Del(L))\neq\DelTwo(L)\text{ im Allgemeinen}}$$

Damit ist die Aussage falsch.

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Verlorene Buchsten

Sei $\Sigma:=\{a,b\}$. Für eine Sprache $L\subseteq\Sigma^*$ bezeichnet $\Del(L)$ die Sprache,
die man erhält, wenn man den Wörtern aus $L$ **genau einmal** das Wort $ab$ an einer beliebigen Stelle entfernt. Formal definieren wir
$\Del(L):= \{vw \mid v,w\in \Sigma^* \wedge vabw\in L\}$

## subproblem a (1.0 points)
Listen Sie alle Wörter der Sprache $\Del(\{abbaba,baa,ab,a,b\})$ explizit auf:

<answer-template id="1">
$\Del(\{abbaba,baa,ab,a,b\}) = \{?\}$
</answer-template>

<example>
$\Del(\{abbaba,baa,ab,a,b\}) = \{\varepsilon, ab, bbab\}$
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
$\Del(\{abbaba,baa,ab,a,b\}) = \{\varepsilon,\;baba,\;abba\}$
</answer>

user:
## subproblem b (2.0 points)
Wir definieren $\DelTwo(L):=\{uvw \mid u,v,w\in \Sigma^* \wedge u ab v ab w\in L \}$. Zeigen Sie oder widerlegen Sie mit einem Gegenbeispiel: Für jede **endliche** Sprache $L\subseteq\Sigma^*$ gilt
$\Del(\Del(L))=\DelTwo(L)$.

<options>
<option id="1">Wahr</option>
<option id="2">Falsch</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<answer id="3">
...
</answer>

assistant:<option id="1">0</option>
<option id="2">1</option>
<answer id="3">
Die Beh ist **falsch**.  

Ein Gegenbeispiel ist die endliche Sprache  

\[
L=\{\,aabb\,\}.
\]

*   **Berechnung von \(\Del(\Del(L))\):**  
    - Das Wort \(aabb\) enthält das Teilwort \(ab\) (Position 2‑3).  
    - Entfernen wir dieses \(ab\), erhalten wir das Wort \(ab\).  
    - Das resultierende Wort \(ab\) enthält erneut das Teilwort \(ab\); nach erneutem Entfernen erhalten wir \(\varepsilon\).  

    Also \(\varepsilon \in \Del(\Del(L))\).

*   **Berechnung von \(\DelTwo(L)\):**  
    - Für \(\DelTwo(L)\) muss das Ausgangswort die Form \(u\,ab\,v\,ab\,w\) haben, also mindestens **zwei** vorkommende Teilwörter \(ab\).  
    - Das Wort \(aabb\) enthält nur ein einziges \(ab\) und lässt sich nicht als \(u\,ab\,v\,ab\,w\) schreiben.  

    Deshalb \(\varepsilon \notin \DelTwo(L)\).

Da \(\varepsilon\) in \(\Del(\Del(L))\) liegt, aber nicht in \(\DelTwo(L)\), gilt die Gleichheit nicht für alle endlichen Sprachen \(L\).

\[
\boxed{\Del(\Del(L))\neq\DelTwo(L)\text{ im Allgemeinen}}
\]

Damit ist die Aussage **falsch**.

user:
## subproblem c (7.0 points)
Sei $G=(V,\Sigma,P,S)$ eine kontextfreie Grammatik in Chomsky-Normalform. Wir möchten eine kontextfreie Grammatik $G'=(V',\Sigma,P',S')$ konstruieren, sodass $L(G')=\Del(L(G))$ gilt. Hierzu setzen wir
$V':= V\cup \{X_{ab}\mid X\in V\}\cup \{X_a\mid X\in V\}\cup \{X_b\mid X\in V\}$
wobei $X_{ab},X_a,X_b\notin V$ für alle $X\in V$. Definieren Sie nun $P'$ und $S'$, sodass
1. $L_{G'}(X)=L_G(X)$
2. $L_{G'}(X_{ab})=\Del(L_G(X))$
3. $L_{G'}(X_{a})=\{w\in\Sigma^*\mid wa\in L_G(X)\}$
4. $L_{G'}(X_{b})=\{w\in\Sigma^*\mid bw\in L_G(X)\}$
für alle $X\in V$ gilt. Geben Sie die Produktionen $P'$ präzise an; vermeiden Sie unstrukturierten Fließtext.

*Hinweis:* $G$ ist in CNF, hat also keine $\varepsilon$-Produktionen. $G'$ muss nicht in CNF sein.

<answer-template id="1">
Startsymbol $S':=?$
Produktionen $P'$:
???
</answer-template>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to provide productions for G' given G in CNF. We need to define start symbol S' and productions P' that satisfy the four conditions.

We have original nonterminals X in V. For each X we add X_ab, X_a, X_b.

Goal: L_{G'}(X) = L_G(X) (so keep original productions for X). L_{G'}(X_ab) = Del(L_G(X)): remove exactly one occurrence of "ab". L_{G'}(X_a) = { w | w a ∈ L_G(X) } i.e., strings that can be extended by a to be in L_G(X). Similarly X_b.

We need to define productions for these new nonterminals using original productions.

Given G is CNF: productions are of form X → Y Z or X → a (terminal) (no epsilon). No unit productions.

We need to construct P' that includes:

- All original productions for X unchanged.

- For X_ab: we need to generate strings that are derived from X with exactly one "ab" removed. That means we need to simulate removal of an "ab" somewhere in the derivation. Approach: For each production X → Y Z, we need to consider possibilities where the "ab" is removed crossing boundaries or inside Y or Z. Also for terminal productions X → a or X → b, we need to handle removal if the terminal is part of "ab". Since "ab" is two terminals, removal can happen when we have a terminal a followed by b consecutively in the yield. In CNF, terminals appear only at leaves, so adjacency can happen when Y yields a and Z yields b, via X → Y Z where Y → a and Z → b. So we need to capture that.

Standard construction: define X_ab → Y_ab Z | Y Z_ab | Y_a Z_b (when Y yields a and Z yields b). Similarly for X_a: strings that when appended a become in L_G(X). So X_a can derive from Y_a Z, Y Z_a, and also if X → a then X_a → ε? Actually if X → a, then strings w such that w a ∈ L_G(X). If X → a, then w must be ε because ε a = a ∈ L_G(X). So X_a can produce ε. Similarly for X_b: if X → b then X_b → ε.

Also need to handle X_a when X → Y Z: we need strings w such that w a ∈ L_G(X). That means either w a is derived from Y Z with a at end. Cases: w a = (string from Y) (string from Z) and the final a is either from Y's suffix or Z's suffix. Since concatenation, the final a could be the last terminal of Y Z. So we need productions: X_a → Y_a Z | Y Z_a. Similarly X_b: X_b → Y_b Z | Y Z_b.

For X_ab: we need to remove exactly one "ab". Cases: removal occurs entirely inside Y, inside Z, or across boundary where Y yields a and Z yields b. So productions: X_ab → Y_ab Z | Y Z_ab | Y_a Z_b. Also need base cases: if X → a then X_ab cannot remove because need both a and b; but maybe if X → a then cannot produce anything? Actually removal of "ab" from a single terminal a is impossible, so no production. If X → b similarly none.

If X → a, then X_a → ε (as above). If X → b, then X_b → ε.

Also need productions for X_ab when X → a b? Not possible directly because CNF doesn't have X → a b (two terminals). So only via binary nonterminals.

Thus P' includes:

- For each original production X → Y Z: include same in P'.

- For each original production X → a: include same.

- For each X ∈ V: add productions:

  X_ab → Y_ab Z | Y Z_ab | Y_a Z_b   for each X → Y Z.

  X_a → Y_a Z | Y Z_a   for each X → Y Z.

  X_b → Y_b Z | Y Z_b   for each X → Y Z.

- For terminal productions:

  If X → a then add X_a → ε.

  If X → b then add X_b → ε.

- For X_ab base: if X → a and X → b? Not needed.

Also need productions for X_a and X_b when X → a? Already added ε.

Similarly for X_ab when X → a? No.

Thus start symbol S' should be S_ab? Actually we need L(G') = Del(L(G)). That means we need to generate strings that are original strings with exactly one "ab" removed. So start symbol should be S_ab. So S' = S_{ab}.

Thus answer: start symbol S' = S_{ab}. Provide productions list.

We need to present in answer template: "Startsymbol $S':=?$" and "Produktionen $P'$:" then list.

We'll write in formal notation.

Let's craft answer.

<answer id="1">
Startsymbol $S':=S_{ab}$  

Produktionen $P'$:

**1.  Original‑Produktionen (beibehalten)**  
\[
\forall X\in V,\;X\rightarrow YZ\in P\quad\Longrightarrow\quad X\rightarrow YZ\in P' 
\]
\[
\forall X\in V,\;X\rightarrow a\in P\quad\Longrightarrow\quad X\rightarrow a\in P' 
\]

**2.  Hilfs‑Nichtterminal $X_{ab}$** – Entfernt genau ein Vorkommen von $ab$  
\[
\begin{aligned}
X_{ab}&\rightarrow Y_{ab}\,Z \\
X_{ab}&\rightarrow Y\,Z_{ab} \\
X_{ab}&\rightarrow Y_{a}\,Z_{b}
\end{aligned}
\qquad\text{für jede Produktion }X\rightarrow YZ\in P
\]

**3.  Hilfs‑Nichtterminal $X_{a}$** – Wörter, die durch Anhängen von $a$ zu $L_G(X)$ führen  
\[
\begin{aligned}
X_{a}&\rightarrow Y_{a}\,Z \\
X_{a}&\rightarrow Y\,Z_{a}
\end{aligned}
\qquad\text{für jede Produktion }X\rightarrow YZ\in P
\]

Zusatz für terminale Produktionen:  
\[
\begin{aligned}
X\rightarrow a &\in P \;\Longrightarrow\; X_{a}\rightarrow \varepsilon \\
X\rightarrow b &\in P \;\Longrightarrow\; X_{b}\rightarrow \varepsilon
\end{aligned}
\]

**4.  Hilfs‑Nichtterminal $X_{b}$** – Wörter, die durch Voranstellen von $b$ zu $L_G(X)$ führen  
\[
\begin{aligned}
X_{b}&\rightarrow Y_{b}\,Z \\
X_{b}&\rightarrow Y\,Z_{b}
\end{aligned}
\qquad\text{für jede Produktion }X\rightarrow YZ\in P
\]

Zusatz für terminale Produktionen:  
\[
\begin{aligned}
X\rightarrow a &\in P \;\Longrightarrow\; X_{b}\text{ hat keine neue Produktion} \\
X\rightarrow b &\in P \;\Longrightarrow\; X_{b}\rightarrow \varepsilon
\end{aligned}
\]

Damit gelten für alle $X\in V$:

\[
\begin{aligned}
L_{G'}(X)      &= L_G(X)\\
L_{G'}(X_{ab}) &= \Del\bigl(L_G(X)\bigr)\\
L_{G'}(X_{a})  &= \{w\mid wa\in L_G(X)\}\\
L_{G'}(X_{b})  &= \{w\mid bw\in L_G(X)\}
\end{aligned}
\]

und das Startsymbol $S_{ab}$ erzeugt genau $\Del\!\bigl(L(G)\bigr)$.  
</answer>

Sei $G=(V,\Sigma,P,S)$ eine kontextfreie Grammatik in Chomsky-Normalform. Wir möchten eine kontextfreie Grammatik $G'=(V',\Sigma,P',S')$ konstruieren, sodass $L(G')=\Del(L(G))$ gilt. Hierzu setzen wir
$V':= V\cup \{X_{ab}\mid X\in V\}\cup \{X_a\mid X\in V\}\cup \{X_b\mid X\in V\}$
wobei $X_{ab},X_a,X_b\notin V$ für alle $X\in V$. Definieren Sie nun $P'$ und $S'$, sodass
1. $L_{G'}(X)=L_G(X)$
2. $L_{G'}(X_{ab})=\Del(L_G(X))$
3. $L_{G'}(X_{a})=\{w\in\Sigma^*\mid wa\in L_G(X)\}$
4. $L_{G'}(X_{b})=\{w\in\Sigma^*\mid bw\in L_G(X)\}$
für alle $X\in V$ gilt. Geben Sie die Produktionen $P'$ präzise an; vermeiden Sie unstrukturierten Fließtext.

Hinweis: $G$ ist in CNF, hat also keine $\varepsilon$-Produktionen. $G'$ muss nicht in CNF sein.

Startsymbol $S':=S_{ab}$  

Produktionen $P'$:

1.  Original‑Produktionen (beibehalten)  
$$\forall X\in V,\;X\rightarrow YZ\in P\quad\Longrightarrow\quad X\rightarrow YZ\in P'$$
$$\forall X\in V,\;X\rightarrow a\in P\quad\Longrightarrow\quad X\rightarrow a\in P'$$

2.  Hilfs‑Nichtterminal $X_{ab}$ – Entfernt genau ein Vorkommen von $ab$  
$$\begin{aligned}
X_{ab}&\rightarrow Y_{ab}\,Z \\
X_{ab}&\rightarrow Y\,Z_{ab} \\
X_{ab}&\rightarrow Y_{a}\,Z_{b}
\end{aligned}
\qquad\text{für jede Produktion }X\rightarrow YZ\in P$$

3.  Hilfs‑Nichtterminal $X_{a}$ – Wörter, die durch Anhängen von $a$ zu $L_G(X)$ führen  
$$\begin{aligned}
X_{a}&\rightarrow Y_{a}\,Z \\
X_{a}&\rightarrow Y\,Z_{a}
\end{aligned}
\qquad\text{für jede Produktion }X\rightarrow YZ\in P$$

Zusatz für terminale Produktionen:  
$$\begin{aligned}
X\rightarrow a &\in P \;\Longrightarrow\; X_{a}\rightarrow \varepsilon \\
X\rightarrow b &\in P \;\Longrightarrow\; X_{b}\rightarrow \varepsilon
\end{aligned}$$

4.  Hilfs‑Nichtterminal $X_{b}$ – Wörter, die durch Voranstellen von $b$ zu $L_G(X)$ führen  
$$\begin{aligned}
X_{b}&\rightarrow Y_{b}\,Z \\
X_{b}&\rightarrow Y\,Z_{b}
\end{aligned}
\qquad\text{für jede Produktion }X\rightarrow YZ\in P$$

Zusatz für terminale Produktionen:  
$$\begin{aligned}
X\rightarrow a &\in P \;\Longrightarrow\; X_{b}\text{ hat keine neue Produktion} \\
X\rightarrow b &\in P \;\Longrightarrow\; X_{b}\rightarrow \varepsilon
\end{aligned}$$

Damit gelten für alle $X\in V$:

$$\begin{aligned}
L_{G'}(X)      &= L_G(X)\\
L_{G'}(X_{ab}) &= \Del\bigl(L_G(X)\bigr)\\
L_{G'}(X_{a})  &= \{w\mid wa\in L_G(X)\}\\
L_{G'}(X_{b})  &= \{w\mid bw\in L_G(X)\}
\end{aligned}$$

und das Startsymbol $S_{ab}$ erzeugt genau $\Del\!\bigl(L(G)\bigr)$.

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Verlorene Buchsten

Sei $\Sigma:=\{a,b\}$. Für eine Sprache $L\subseteq\Sigma^*$ bezeichnet $\Del(L)$ die Sprache,
die man erhält, wenn man den Wörtern aus $L$ **genau einmal** das Wort $ab$ an einer beliebigen Stelle entfernt. Formal definieren wir
$\Del(L):= \{vw \mid v,w\in \Sigma^* \wedge vabw\in L\}$

## subproblem a (1.0 points)
Listen Sie alle Wörter der Sprache $\Del(\{abbaba,baa,ab,a,b\})$ explizit auf:

<answer-template id="1">
$\Del(\{abbaba,baa,ab,a,b\}) = \{?\}$
</answer-template>

<example>
$\Del(\{abbaba,baa,ab,a,b\}) = \{\varepsilon, ab, bbab\}$
</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
$\Del(\{abbaba,baa,ab,a,b\}) = \{\varepsilon,\;baba,\;abba\}$
</answer>

user:
## subproblem b (2.0 points)
Wir definieren $\DelTwo(L):=\{uvw \mid u,v,w\in \Sigma^* \wedge u ab v ab w\in L \}$. Zeigen Sie oder widerlegen Sie mit einem Gegenbeispiel: Für jede **endliche** Sprache $L\subseteq\Sigma^*$ gilt
$\Del(\Del(L))=\DelTwo(L)$.

<options>
<option id="1">Wahr</option>
<option id="2">Falsch</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<answer id="3">
...
</answer>

assistant:<option id="1">0</option>
<option id="2">1</option>
<answer id="3">
Die Beh ist **falsch**.  

Ein Gegenbeispiel ist die endliche Sprache  

\[
L=\{\,aabb\,\}.
\]

*   **Berechnung von \(\Del(\Del(L))\):**  
    - Das Wort \(aabb\) enthält das Teilwort \(ab\) (Position 2‑3).  
    - Entfernen wir dieses \(ab\), erhalten wir das Wort \(ab\).  
    - Das resultierende Wort \(ab\) enthält erneut das Teilwort \(ab\); nach erneutem Entfernen erhalten wir \(\varepsilon\).  

    Also \(\varepsilon \in \Del(\Del(L))\).

*   **Berechnung von \(\DelTwo(L)\):**  
    - Für \(\DelTwo(L)\) muss das Ausgangswort die Form \(u\,ab\,v\,ab\,w\) haben, also mindestens **zwei** vorkommende Teilwörter \(ab\).  
    - Das Wort \(aabb\) enthält nur ein einziges \(ab\) und lässt sich nicht als \(u\,ab\,v\,ab\,w\) schreiben.  

    Deshalb \(\varepsilon \notin \DelTwo(L)\).

Da \(\varepsilon\) in \(\Del(\Del(L))\) liegt, aber nicht in \(\DelTwo(L)\), gilt die Gleichheit nicht für alle endlichen Sprachen \(L\).

\[
\boxed{\Del(\Del(L))\neq\DelTwo(L)\text{ im Allgemeinen}}
\]

Damit ist die Aussage **falsch**.

user:
## subproblem c (7.0 points)
Sei $G=(V,\Sigma,P,S)$ eine kontextfreie Grammatik in Chomsky-Normalform. Wir möchten eine kontextfreie Grammatik $G'=(V',\Sigma,P',S')$ konstruieren, sodass $L(G')=\Del(L(G))$ gilt. Hierzu setzen wir
$V':= V\cup \{X_{ab}\mid X\in V\}\cup \{X_a\mid X\in V\}\cup \{X_b\mid X\in V\}$
wobei $X_{ab},X_a,X_b\notin V$ für alle $X\in V$. Definieren Sie nun $P'$ und $S'$, sodass
1. $L_{G'}(X)=L_G(X)$
2. $L_{G'}(X_{ab})=\Del(L_G(X))$
3. $L_{G'}(X_{a})=\{w\in\Sigma^*\mid wa\in L_G(X)\}$
4. $L_{G'}(X_{b})=\{w\in\Sigma^*\mid bw\in L_G(X)\}$
für alle $X\in V$ gilt. Geben Sie die Produktionen $P'$ präzise an; vermeiden Sie unstrukturierten Fließtext.

*Hinweis:* $G$ ist in CNF, hat also keine $\varepsilon$-Produktionen. $G'$ muss nicht in CNF sein.

<answer-template id="1">
Startsymbol $S':=?$
Produktionen $P'$:
???
</answer-template>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

assistant:<answer id="1">
Startsymbol $S':=S_{ab}$  

Produktionen $P'$:

**1.  Original‑Produktionen (beibehalten)**  
\[
\forall X\in V,\;X\rightarrow YZ\in P\quad\Longrightarrow\quad X\rightarrow YZ\in P' 
\]
\[
\forall X\in V,\;X\rightarrow a\in P\quad\Longrightarrow\quad X\rightarrow a\in P' 
\]

**2.  Hilfs‑Nichtterminal $X_{ab}$** – Entfernt genau ein Vorkommen von $ab$  
\[
\begin{aligned}
X_{ab}&\rightarrow Y_{ab}\,Z \\
X_{ab}&\rightarrow Y\,Z_{ab} \\
X_{ab}&\rightarrow Y_{a}\,Z_{b}
\end{aligned}
\qquad\text{für jede Produktion }X\rightarrow YZ\in P
\]

**3.  Hilfs‑Nichtterminal $X_{a}$** – Wörter, die durch Anhängen von $a$ zu $L_G(X)$ führen  
\[
\begin{aligned}
X_{a}&\rightarrow Y_{a}\,Z \\
X_{a}&\rightarrow Y\,Z_{a}
\end{aligned}
\qquad\text{für jede Produktion }X\rightarrow YZ\in P
\]

Zusatz für terminale Produktionen:  
\[
\begin{aligned}
X\rightarrow a &\in P \;\Longrightarrow\; X_{a}\rightarrow \varepsilon \\
X\rightarrow b &\in P \;\Longrightarrow\; X_{b}\rightarrow \varepsilon
\end{aligned}
\]

**4.  Hilfs‑Nichtterminal $X_{b}$** – Wörter, die durch Voranstellen von $b$ zu $L_G(X)$ führen  
\[
\begin{aligned}
X_{b}&\rightarrow Y_{b}\,Z \\
X_{b}&\rightarrow Y\,Z_{b}
\end{aligned}
\qquad\text{für jede Produktion }X\rightarrow YZ\in P
\]

Zusatz für terminale Produktionen:  
\[
\begin{aligned}
X\rightarrow a &\in P \;\Longrightarrow\; X_{b}\text{ hat keine neue Produktion} \\
X\rightarrow b &\in P \;\Longrightarrow\; X_{b}\rightarrow \varepsilon
\end{aligned}
\]

Damit gelten für alle $X\in V$:

\[
\begin{aligned}
L_{G'}(X)      &= L_G(X)\\
L_{G'}(X_{ab}) &= \Del\bigl(L_G(X)\bigr)\\
L_{G'}(X_{a})  &= \{w\mid wa\in L_G(X)\}\\
L_{G'}(X_{b})  &= \{w\mid bw\in L_G(X)\}
\end{aligned}
\]

und das Startsymbol $S_{ab}$ erzeugt genau $\Del\!\bigl(L(G)\bigr)$.  
</answer>

user:
## subproblem d (2.0 points)
Wenden Sie Ihre Konstruktion auf das Beispiel $G:=(\{S,A\},\{a,b\},P,S)$ an, wobei $P$ gegeben ist durch:
$S\to SA\mid a\qquad A\to a\mid b$
*Hinweis:* Falls ein Nichtterminal nicht auf der linken Seite einer Produktion vorkommt, lassen Sie das entsprechende Feld leer.

<answer-template id="1">
Startsymbol $S':=?$
Produktionen $P'$:
$S\rightarrow ?$
$S_{ab}\rightarrow ?$
$S_a\rightarrow ?$
$S_b\rightarrow ?$
$A\rightarrow ?$
$A_{ab}\rightarrow ?$
$A_a\rightarrow ?$
$A_b\rightarrow ?$

</answer-template>

<example>
Startsymbol $S':=S$
Produktionen $P'$:
$S\rightarrow SS\mid AA\mid a"$
$S_{ab}\rightarrow S_{ab}A\mid\varepsilon$
...

</example>

### answer format
Give your answer in the following format:

<answer id="1">
...
</answer>

We need to apply the construction from part c to the specific grammar G.

Grammar G: V={S,A}, Σ={a,b}, productions:
1. S → S A
2. S → a
3. A → a
4. A → b

In CNF? Actually S→SA is binary, S→a is terminal, A→a, A→b are terminal. So yes.

We need to construct G' with V' = V ∪ {X_{ab},X_a,X_b | X∈V}. So we have S, A, plus S_ab, S_a, S_b, A_ab, A_a, A_b.

We need to define P' as per the specification earlier. In part c we gave a generic set of productions: keep original productions for X, also productions for X_ab, X_a, X_b derived from each binary production X→YZ, plus epsilon productions for terminals.

But we must output explicit productions for this particular G.

Let's recall the definitions:

Goal: For each X,
- L_{G'}(X) = L_G(X) (so keep original productions)
- L_{G'}(X_{ab}) = Del(L_G(X)) (remove exactly one occurrence of "ab")
- L_{G'}(X_a) = { w | w a ∈ L_G(X) } (words that become in X after appending a)
- L_{G'}(X_b) = { w | b w ∈ L_G(X) } (words that become in X after prefixing b)

Construction we gave: For each binary production X → Y Z,
- X_ab → Y_ab Z
- X_ab → Y Z_ab
- X_ab → Y_a Z_b   (since we can delete the 'ab' across the boundary: the a could be last symbol of string from Y, b first from Z)
Similarly for X_a:
- X_a → Y_a Z
- X_a → Y Z_a
(and also if X → a then X_a → ε because removing the trailing a yields empty string)
Similarly for X_b:
- X_b → Y_b Z
- X_b → Y Z_b
(and if X → b then X_b → ε)

We also need to handle terminal productions: for X → a (or b) we keep them for X; and also define X_a or X_b epsilon if appropriate.

Now apply to our grammar.

Binary productions:
- S → S A
That's the only binary production.

Terminal productions:
- S → a
- A → a
- A → b

Now define P'.

First keep original productions:

S → S A
S → a
A → a
A → b

Now define S_ab productions using binary production S → S A:

- S_ab → S_ab A
- S_ab → S A_ab
- S_ab → S_a A_b   (since a from left side, b from right side)

Similarly define A_ab: but A has no binary productions (no A → Y Z). So no such productions derived from binary ones. But we might need productions for A_ab from terminal productions? Deleting "ab" inside a string generated by A (which are just "a" or "b") can't happen because they are length 1. So A_ab should generate empty set? Could we have epsilon? Actually Del(L_G(A)) for L_G(A) = {a, b}. Removing "ab" from "a" or "b" yields nothing because they don't contain "ab". So L_{G'}(A_ab) = ∅. We can achieve that by having no productions for A_ab (or only non-generating). So leave empty.

Now define S_a productions:

From binary S → S A:

- S_a → S_a A
- S_a → S A_a

Additionally from terminal production S → a: we need S_a → ε (since if X→a, then X_a can derive ε). So add S_a → ε.

Now define S_b productions:

From binary S → S A:

- S_b → S_b A
- S_b → S A_b

No terminal S→b, so no epsilon rule for S_b.

Now define A_a productions:

A has no binary productions. From terminal productions:
- A → a gives A_a → ε (since removing trailing a)
- A → b does not give anything for A_a.

So A_a → ε.

Similarly A_b productions:
- From A → b, we get A_b → ε.
- A → a does not affect.

Thus A_b → ε.

Now we also need productions for A_ab? none.

Potential extra productions for X_ab when X → a or b? No, because deletion of "ab" across terminal? Not needed.

Now we need to define start symbol S' = S_ab (as per generic construction we set S' = S_{ab}) to generate Del(L(G)). So start symbol S' = S_ab.

Thus final answer list:

Startsymbol S' = S_ab

Productions:
List each.

Let's format as required: they gave template with placeholders:

$S\rightarrow ?$
$S_{ab}\rightarrow ?$
$S_a\rightarrow ?$
$S_b\rightarrow ?$
$A\rightarrow ?$
$A_{ab}\rightarrow ?$
$A_a\rightarrow ?$
$A_b\rightarrow ?$

We need to fill each with right-hand sides (possibly multiple alternatives separated by |). Use standard notation.

S → SA | a
S_ab → S_ab A | S A_ab | S_a A_b
S_a → S_a A | S A_a | ε
S_b → S_b A | S A_b
A → a | b
A_ab → (nothing) – leave empty field
A_a → ε
A_b → ε

Make sure to follow format: each line ends with the production RHS (could be empty). For empty maybe just leave blank after arrow.

Thus final answer.

<answer id="1">
Startsymbol $S':=S_{ab}$  
Produktionen $P'$:
$S\rightarrow SA\mid a$
$S_{ab}\rightarrow S_{ab}A\mid SA_{ab}\mid S_{a}A_{b}$
$S_a\rightarrow S_{a}A\mid SA_{a}\mid \varepsilon$
$S_b\rightarrow S_{b}A\mid SA_{b}$
$A\rightarrow a\mid b$
$A_{ab}\rightarrow $
$A_a\rightarrow \varepsilon$
$A_b\rightarrow \varepsilon$
</answer>

Wenden Sie Ihre Konstruktion auf das Beispiel $G:=(\{S,A\},\{a,b\},P,S)$ an, wobei $P$ gegeben ist durch:
$S\to SA\mid a\qquad A\to a\mid b$
Hinweis: Falls ein Nichtterminal nicht auf der linken Seite einer Produktion vorkommt, lassen Sie das entsprechende Feld leer.

Startsymbol $S':=S_{ab}$  
Produktionen $P'$:
$S\rightarrow SA\mid a$
$S_{ab}\rightarrow S_{ab}A\mid SA_{ab}\mid S_{a}A_{b}$
$S_a\rightarrow S_{a}A\mid SA_{a}\mid \varepsilon$
$S_b\rightarrow S_{b}A\mid SA_{b}$
$A\rightarrow a\mid b$
$A_{ab}\rightarrow$
$A_a\rightarrow \varepsilon$
$A_b\rightarrow \varepsilon$

2 mistakes

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Quiz: Berechenbarkeit und Komplexität

**Teilaufgaben (a-e):** Für diese Fragen ist eine Begründung nicht gefordert. Sie erhalten die Punkte auf eine Teilaufgabe genau dann, wenn Sie alle Antwortmöglichkeiten korrekt angekreuzt haben.
**Es ist immer mindestens eine Antwortmöglichkeit richtig.**
In dieser Aufgabe verwenden wir durchgehend das Alphabet $\Sigma := \{0, 1\}$.

## subproblem a (2.0 points)
Sei $L$ eine kontextfreie Sprache. Welche Aussagen sind wahr?

<options>
<option id="1">$L$ ist entscheidbar</option>
<option id="2">$\overline{L}$ ist entscheidbar</option>
<option id="3">$L$ ist semi-entscheidbar</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

We need to answer which statements are true for a context-free language L.

We know: CFLs are decidable (membership can be decided in O(n^3) etc). So L is decidable. So statement 1 true.

Complement of a CFL is not necessarily context-free but decidable because CFLs are decidable and complements of decidable languages are decidable. Since L is decidable, its complement is also decidable. So statement 2 true.

Semi-decidable (recursively enumerable) is weaker: any decidable language is also semi-decidable (both RE and co-RE). So statement 3 true.

Thus all three true => 1.

Return 1 for each.

<option id="1">1</option>
<option id="2">1</option>
<option id="3">1</option>

Sei $L$ eine kontextfreie Sprache. Welche Aussagen sind wahr?

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Quiz: Berechenbarkeit und Komplexität

**Teilaufgaben (a-e):** Für diese Fragen ist eine Begründung nicht gefordert. Sie erhalten die Punkte auf eine Teilaufgabe genau dann, wenn Sie alle Antwortmöglichkeiten korrekt angekreuzt haben.
**Es ist immer mindestens eine Antwortmöglichkeit richtig.**
In dieser Aufgabe verwenden wir durchgehend das Alphabet $\Sigma := \{0, 1\}$.

## subproblem a (2.0 points)
Sei $L$ eine kontextfreie Sprache. Welche Aussagen sind wahr?

<options>
<option id="1">$L$ ist entscheidbar</option>
<option id="2">$\overline{L}$ ist entscheidbar</option>
<option id="3">$L$ ist semi-entscheidbar</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem b (2.0 points)
Seien $L_1,L_2\subseteq\Sigma^*$ entscheidbare Sprachen. Welche der folgenden Sprachen sind entscheidbar?

<options>
<option id="1">$L_1\cap L_2$</option>
<option id="2">$L_1\setminus L_2$</option>
<option id="3">$L_1L_2$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

We need evaluate each statement about decidability.

Given L1, L2 decidable (recursive). Then their intersection is decidable (by running both deciders, accept if both accept). So true.

Set difference L1\L2 = {x | x∈L1 and x∉L2}. Since complement of L2 is decidable (decidable languages closed under complement), and intersection of decidable languages is decidable, so L1\L2 is decidable. So true.

Concatenation L1L2 = {xy | x∈L1, y∈L2}. Is concatenation of two decidable languages always decidable? Yes: we can decide membership by enumerating all possible splits of input w into w = uv and run deciders for L1 on u and L2 on v. Since there are finitely many splits (|w|+1), we can decide by checking each split until one works, else reject. Since each check is decidable and finite, overall decidable. So true.

Thus all three are decidable => 1 each.

Return format.

<option id="1">1</option>
<option id="2">1</option>
<option id="3">1</option>

Seien $L_1,L_2\subseteq\Sigma^*$ entscheidbare Sprachen. Welche der folgenden Sprachen sind entscheidbar?

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Quiz: Berechenbarkeit und Komplexität

**Teilaufgaben (a-e):** Für diese Fragen ist eine Begründung nicht gefordert. Sie erhalten die Punkte auf eine Teilaufgabe genau dann, wenn Sie alle Antwortmöglichkeiten korrekt angekreuzt haben.
**Es ist immer mindestens eine Antwortmöglichkeit richtig.**
In dieser Aufgabe verwenden wir durchgehend das Alphabet $\Sigma := \{0, 1\}$.

## subproblem a (2.0 points)
Sei $L$ eine kontextfreie Sprache. Welche Aussagen sind wahr?

<options>
<option id="1">$L$ ist entscheidbar</option>
<option id="2">$\overline{L}$ ist entscheidbar</option>
<option id="3">$L$ ist semi-entscheidbar</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem b (2.0 points)
Seien $L_1,L_2\subseteq\Sigma^*$ entscheidbare Sprachen. Welche der folgenden Sprachen sind entscheidbar?

<options>
<option id="1">$L_1\cap L_2$</option>
<option id="2">$L_1\setminus L_2$</option>
<option id="3">$L_1L_2$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem c (2.0 points)
Sei $L\subseteq\{0\}^*$ unentscheidbar. Welche der folgenden Funktionen $f:\N\rightarrow\N$ sind berechenbar?

<options>
<option id="1">$f(n):=|L\cap \Sigma^n|$</option>
<option id="2">$f(n):=\min\{k\in\N: 0^k\in L\}$</option>
<option id="3">$f(n):=1$, falls $0^n\in L$, sonst $f(n):=0$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

We must evaluate each. Subproblem a: L context-free language. Are CF languages decidable? Yes, emptiness, membership are decidable (CYK). So L is decidable. Its complement of a CFL is not necessarily CFL, but decidability: complement of a decidable language is decidable. Since L is decidable, complement is also decidable. Semi-decidable: decidable implies semi-decidable, yes. So all true => 1 for each. However need check nuance: Are all CFL decidable? Yes, membership (given string) is decidable, but the language as a set of strings is decidable? The decision problem "given string w, is w in L?" is decidable, so language is decidable set. So correct.

Subproblem b: L1, L2 decidable languages. Intersection and set difference and concatenation? Intersection of decidable languages is decidable. Difference also decidable (L1 \ L2 = L1 ∩ complement L2). complement of decidable is decidable, so yes. Concatenation L1L2 = { xy | x in L1, y in L2 }. Given an input w, we can decide membership by trying all splits w = uv and checking u in L1 and v in L2. Since there are finitely many splits (|w|+1) and each check decider halts, we can decide. So concatenation is decidable. So all true.

Subproblem c: L subset of {0}* (unary strings) is undecidable. We need to evaluate functions.

Option1: f(n) = |L ∩ Σ^n| . Wait Σ = {0,1}. Σ^n strings length n over {0,1}. Intersection with L (only strings of form 0^k). So among all binary strings length n, only one is 0^n. So |L ∩ Σ^n| is either 1 if 0^n ∈ L, else 0. So same as characteristic function. Computing this would decide membership in L, which is undecidable. So not computable. Option1 false.

Option2: f(n) = min { k ∈ ℕ : 0^k ∈ L }. That's the smallest length of string in L, independent of n? Actually f does not depend on n, returns same value for all n (if defined). But the set may be empty (if L empty). Assuming L nonempty maybe. Computing this requires searching for smallest k such that 0^k ∈ L, which is essentially solving membership for each k until success. Since L undecidable, can't in finite time guarantee termination? But we could simulate decider for each k, but decider doesn't exist. However we could run enumerator for L (since L is semi-decidable? Not given). It might be not even semi-decidable. So cannot guarantee compute. Moreover f may be undefined if L empty. So not computable. Option2 false.

Option3: f(n) =1 if 0^n ∈ L else 0. That's characteristic function again, same as option1. So not computable. So all false => 0 for each.

Thus answer.

Check subtlety: Could any of those be computable despite undecidability? Since L is undecidable, its characteristic function is not computable. Option1 & 3 equal characteristic. Option2 requires solving global property, which also not computable. So all 0.

Thus final answer.

<option id="1">0</option>
<option id="2">0</option>
<option id="3">0</option>

Sei $L\subseteq\{0\}^*$ unentscheidbar. Welche der folgenden Funktionen $f:\N\rightarrow\N$ sind berechenbar?

wrong answer for choice (1)

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Quiz: Berechenbarkeit und Komplexität

**Teilaufgaben (a-e):** Für diese Fragen ist eine Begründung nicht gefordert. Sie erhalten die Punkte auf eine Teilaufgabe genau dann, wenn Sie alle Antwortmöglichkeiten korrekt angekreuzt haben.
**Es ist immer mindestens eine Antwortmöglichkeit richtig.**
In dieser Aufgabe verwenden wir durchgehend das Alphabet $\Sigma := \{0, 1\}$.

## subproblem a (2.0 points)
Sei $L$ eine kontextfreie Sprache. Welche Aussagen sind wahr?

<options>
<option id="1">$L$ ist entscheidbar</option>
<option id="2">$\overline{L}$ ist entscheidbar</option>
<option id="3">$L$ ist semi-entscheidbar</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem b (2.0 points)
Seien $L_1,L_2\subseteq\Sigma^*$ entscheidbare Sprachen. Welche der folgenden Sprachen sind entscheidbar?

<options>
<option id="1">$L_1\cap L_2$</option>
<option id="2">$L_1\setminus L_2$</option>
<option id="3">$L_1L_2$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem c (2.0 points)
Sei $L\subseteq\{0\}^*$ unentscheidbar. Welche der folgenden Funktionen $f:\N\rightarrow\N$ sind berechenbar?

<options>
<option id="1">$f(n):=|L\cap \Sigma^n|$</option>
<option id="2">$f(n):=\min\{k\in\N: 0^k\in L\}$</option>
<option id="3">$f(n):=1$, falls $0^n\in L$, sonst $f(n):=0$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">0</option>
<option id="2">0</option>
<option id="3">0</option>

user:
## subproblem d (2.0 points)
Seien $A,B\subseteq\Sigma^*$ Sprachen mit $A\le_p\SAT\le_p B$. Welche Aussagen sind wahr?

<options>
<option id="1">$A\in\NP$</option>
<option id="2">$B\in\NP$</option>
<option id="3">$B$ ist NP-hart</option>
<option id="4">$A\le_p B$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>
<option id="4">0 or 1</option>

We need to answer subproblem d. Let's analyze.

We have languages A, B subsets of Σ* with A ≤_p SAT ≤_p B. Means A polynomial-time many-one reducible to SAT, and SAT polynomial-time many-one reducible to B. So SAT is NP-complete. Given that reductions exist.

We need to evaluate statements:

1. A ∈ NP
2. B ∈ NP
3. B is NP-hard
4. A ≤_p B

We know A ≤_p SAT. SAT is NP-complete, meaning SAT is in NP and NP-hard. However, from A ≤_p SAT we cannot infer A ∈ NP; we only have that A is polynomial-time reducible to an NP language. That does not guarantee A ∈ NP (it could be harder, e.g., any language reduces to SAT via trivial reduction? Actually any language reduces to SAT if you map yes-instances to a fixed satisfiable formula and no-instances to unsatisfiable formula; but that reduction may be not computable if you need to decide membership of A. Many-one reduction must be computable. For arbitrary language A, you cannot compute reduction without solving A. So existence of reduction from A to SAT implies A is at most as hard as SAT under ≤_p. Since SAT is NP-complete, this means A ∈ NP? Wait, ≤_p is many-one reduction that is computable in polynomial time. If we have a reduction f such that x ∈ A iff f(x) ∈ SAT. Then if we could decide SAT, we could decide A via f and SAT oracle; but does it guarantee A ∈ NP? Not necessarily; we need a nondeterministic polynomial verifier for A. We could use reduction to SAT and then NP algorithm for SAT. So we can decide A in NP time: given input x, compute f(x) (poly time), then nondeterministically guess assignment for SAT of size poly(|x|) and check. That yields NP algorithm for A. So yes, A ∈ NP. Because composition of polynomial-time computable reduction with NP machine yields NP machine. So statement 1 is true.

2. B ∈ NP? We have SAT ≤_p B. So B is at least as hard as SAT. But we don't know if B is in NP; could be even harder (e.g., PSPACE-complete). Reduction from SAT to B doesn't give membership in NP. So cannot guarantee B ∈ NP. So statement 2 is not necessarily true (could be false). So answer 0.

3. B is NP-hard. Since SAT is NP-complete, SAT ≤_p B implies B is NP-hard (by definition: a language is NP-hard if every language in NP reduces to it; because SAT is NPC, there is a poly reduction from any L∈NP to SAT, then compose with SAT ≤_p B gives reduction from L to B). So yes, B is NP-hard. So statement 3 true (1).

4. A ≤_p B? Since A ≤_p SAT and SAT ≤_p B, we can compose reductions: A ≤_p SAT ≤_p B => A ≤_p B. So true (1).

Thus final answer:

<option id="1">1</option>
<option id="2">0</option>
<option id="3">1</option>
<option id="4">1</option>

<option id="1">1</option>
<option id="2">0</option>
<option id="3">1</option>
<option id="4">1</option>

Seien $A,B\subseteq\Sigma^*$ Sprachen mit $A\le_p\SAT\le_p B$. Welche Aussagen sind wahr?

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Quiz: Berechenbarkeit und Komplexität

**Teilaufgaben (a-e):** Für diese Fragen ist eine Begründung nicht gefordert. Sie erhalten die Punkte auf eine Teilaufgabe genau dann, wenn Sie alle Antwortmöglichkeiten korrekt angekreuzt haben.
**Es ist immer mindestens eine Antwortmöglichkeit richtig.**
In dieser Aufgabe verwenden wir durchgehend das Alphabet $\Sigma := \{0, 1\}$.

## subproblem a (2.0 points)
Sei $L$ eine kontextfreie Sprache. Welche Aussagen sind wahr?

<options>
<option id="1">$L$ ist entscheidbar</option>
<option id="2">$\overline{L}$ ist entscheidbar</option>
<option id="3">$L$ ist semi-entscheidbar</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem b (2.0 points)
Seien $L_1,L_2\subseteq\Sigma^*$ entscheidbare Sprachen. Welche der folgenden Sprachen sind entscheidbar?

<options>
<option id="1">$L_1\cap L_2$</option>
<option id="2">$L_1\setminus L_2$</option>
<option id="3">$L_1L_2$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem c (2.0 points)
Sei $L\subseteq\{0\}^*$ unentscheidbar. Welche der folgenden Funktionen $f:\N\rightarrow\N$ sind berechenbar?

<options>
<option id="1">$f(n):=|L\cap \Sigma^n|$</option>
<option id="2">$f(n):=\min\{k\in\N: 0^k\in L\}$</option>
<option id="3">$f(n):=1$, falls $0^n\in L$, sonst $f(n):=0$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">0</option>
<option id="2">0</option>
<option id="3">0</option>

user:
## subproblem d (2.0 points)
Seien $A,B\subseteq\Sigma^*$ Sprachen mit $A\le_p\SAT\le_p B$. Welche Aussagen sind wahr?

<options>
<option id="1">$A\in\NP$</option>
<option id="2">$B\in\NP$</option>
<option id="3">$B$ ist NP-hart</option>
<option id="4">$A\le_p B$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>
<option id="4">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">0</option>
<option id="3">1</option>
<option id="4">1</option>

user:
## subproblem e (2.0 points)
Sei $L\subseteq \Sigma^*\{\#\}\Sigma^*$ eine Sprache, sodass $L\in\P$; $L$ enthält also Wörter der Form $u\#v$, mit $u,v\in\Sigma^*$. Welche der folgenden Sprachen sind in $\NP$?

<options>
<option id="1">$L$</option>
<option id="2">$\{u\mid\exists v\in\Sigma^*:u\#v\in L\}$</option>
<option id="3">$\{u\mid\exists v\in\Sigma^*:u\#v\in L\wedge |u|=|v|\}$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

We need to evaluate each subproblem.

We must output answer IDs with 0 or 1.

We need to verify correctness.

Subproblem a: "Sei L eine kontextfreie Sprache. Welche Aussagen sind wahr? Options: 1: L ist entscheidbar. 2: complement of L ist entscheidbar. 3: L ist semi-entscheidbar."

In theory: All context-free languages are decidable (they are recognizable by pushdown automata and emptiness? Actually membership problem for CFG is decidable in polynomial time). So yes, L is decidable. So option 1 true.

Is complement of a CFL necessarily decidable? Complement of a CFL is also decidable because complement of a decidable language is decidable. Since L is decidable, its complement is also decidable. So option 2 true.

Option 3: L is semi-decidable (recursively enumerable). Since decidable implies semi-decidable, true. So all three true. So answer (1,1,1). The assistant gave (1,1,1) earlier. That is correct.

Subproblem b: L1, L2 decidable. Which of these languages are decidable? Intersection of decidable languages is decidable. Set difference decidable. Concatenation of decidable languages? Not necessarily decidable? Actually given two decidable languages, L1L2 = { xy | x∈L1, y∈L2 }. Decidability: membership given w, we need to split w into u v and test both. Since we can try all possible splits O(|w|) possibilities; each test usesider for L1 and L2. Since finite number, we can accept if any split works. That yields a decider (in O(n) * T). So concatenation of decidable languages is decidable. So all 3 true. So answer (1,1,1). The assistant gave (1,1,1).

Subproblem c: L subset of {0}* (only strings of zeros) unentscheidbar (undecidable). Which functions are computable?

Option1: f(n) = |L ∩ Σ^n|. But Σ = {0,1}. However L is only zeros, so L ∩ Σ^n = elements of length n in L, but only possible element is 0^n if that belongs to L. So cardinality is either 0 or 1. So f(n) = 1 if 0^n ∈ L else 0. That's basically the characteristic function of L on unary strings. Since L is undecidable, this function is not computable (if it were, we could decide membership). So f not computable. So option1 false.

Option2: f(n) = min { k ∈ N : 0^k ∈ L }. This is the smallest length of string in L, independent of n. It's either undefined if L empty? But function defined maybe returns minimal k (maybe infinite). Since L is undecidable, the minimal element may be not computable (you could search through strings until you find one in L; if L is nonempty, you'll eventually find it; that algorithm halts, but if L empty it runs forever). Since function must be total, unclear. Likely not computable. Indeed if L nonempty and is undecidable, you cannot compute the minimal element because that would let you decide membership by comparing n to min? Maybe you could? Suppose you compute minimal k, then to decide if 0^n ∈ L, you check if n = k? Not enough; there could be other elements. So maybe it's still uncomputable. Actually function min {k: 0^k in L} could be computed by enumerating all strings in L (semi-decidable?), but L may not be RE? L is just any language subset of {0}*; it's undecidable but could be RE? Not guaranteed. Since complement may also be not RE. So we cannot necessarily enumerate L. So function likely not computable.

Option3: f(n)=1 if 0^n ∈ L else 0. Same as option1. Not computable.

Thus all false? Option1 false, 2 false, 3 false. The assistant responded 0 0 0. That seems correct.

Subproblem d: Languages A,B with A ≤_p SAT ≤_p B. That means A polynomial-time many-one reducible to SAT, and SAT reducible to B. Which statements are true?

Option1: A ∈ NP. If A ≤_p SAT, does that imply A ∈ NP? Not necessarily; we only know A reduces to SAT, which is NP-complete. If A is reducible to SAT, that only gives that A is at most as hard as SAT (i.e., A is in P? Actually if A ≤_p SAT, then if SAT is in P, A is in P. But without that, we cannot guarantee NP membership. However any language reducible to SAT can be decided using SAT oracle; but not known. In general, a language that reduces to an NP-complete problem need not be in NP; could be undecidable? Wait many-one reducible (poly-time) to SAT. If A is undecidable, that would imply undecidable ≤_p SAT, but SAT is decidable, so can't happen (since reduction must be computable, if A were undecidable then we could decide it using SAT decider and reduction, making it decidable). So A must be decidable. But does that mean A ∈ NP? Not necessarily; could be in P, but also could be in coNP? But NP contains P, but does it contain all decidable languages? No. So we need to check whether we can prove A ∈ NP. Since SAT is NP-complete, and many-one reduction is composition: If A ≤_p SAT, then there exists poly-time computable f such that x ∈ A iff f(x) ∈ SAT. To decide A, we could compute f(x) and then run a nondeterministic polynomial time algorithm for SAT: guess assignment and verify; total NP time. So A ∈ NP. Indeed for any language many-one reducible to an NP language, A ∈ NP? Let's think: If L ≤_p L' and L' ∈ NP, does that imply L ∈ NP? Yes, because you can transform with f and then use verifier for L'. So L ∈ NP. So option1 true.

Option2: B ∈ NP. We have SAT ≤_p B. Since SAT is NP-complete, any language that SAT reduces to need not be in NP (it could be harder). That's a hardness reduction: B is at least as hard as SAT. Could be NP-complete (if also in NP) or NP-hard but not in NP (e.g., language that is coNP-hard). Since we don't know, cannot guarantee B ∈ NP. So false.

Option3: B is NP-hard. Since SAT ≤_p B, B is NP-hard (by definition). Indeed yes, B is NP-hard. So true.

Option4: A ≤_p B. Since A ≤_p SAT and SAT ≤_p B, transitivity of polynomial many-one reductions implies A ≤_p B. So true.

Thus answers: 1 true, 2 false, 3 true, 4 true => (1,0,1,1). The assistant gave (1,0,1,1). That's correct.

Subproblem e: L ⊆ Σ* # Σ* a language such that L ∈ P. L contains words of form u#v, with u,v ∈ Σ*. Which of following languages are in NP?

Option1: L (the original). Since L ∈ P, and P ⊆ NP, so yes, L ∈ NP. So true.

Option2: { u | ∃ v ∈ Σ*: u#v ∈ L }. This is the projection of L onto first component (existential quantifier). If L is in P, the language of u such that exists a v with u#v in L is NP because you can guess v and verify membership of u#v in L in poly time. So it is in NP. Also could be in P? Not necessarily; may be NP-complete (like SAT). But at least NP. So true.

Option3: { u | ∃ v ∈ Σ*: u#v ∈ L ∧ |u| = |v| }. Similar but with additional condition that lengths equal; you can guess v of same length as u (polynomially many possibilities). You can guess v and verify both conditions in poly time. So in NP. So true.

Thus all three are in NP. So answer (1,1,1). The assistant hasn't answered yet; we need to output that.

Check that no subtlety: Existence quantifier over strings of arbitrary length; guess v of length equal to u (bounded by length of u). That's polynomially bounded (since length of v equals length of u, so size of certificate is O(|u|)). So NP.

Thus final answer:

<option id="1">1</option>
<option id="2">1</option>
<option id="3">1</option>

We must provide only that.

<option id="1">1</option>
<option id="2">1</option>
<option id="3">1</option>

Sei $L\subseteq \Sigma^*\{\#\}\Sigma^*$ eine Sprache, sodass $L\in\P$; $L$ enthält also Wörter der Form $u\#v$, mit $u,v\in\Sigma^*$. Welche der folgenden Sprachen sind in $\NP$?

wrong answer for choice (1)

Prompt

system:You are an AI assistant specializing in theoretical computer science. Your task is to answer a series of exam questions, providing accurate and well-reasoned responses that demonstrate a deep understanding of theoretical computer science concepts.

# Instructions

You must follow the given answer format.

If the exercises contains yes or no questions for you to answer, there will be one <option> element for each such question. You must answer these with either 0 or 1.

For longer answers, there will be <answer> elements. If the answer must follow a specific template, a corresponding <answer-template> element defines this template, and you MUST follow it precisely when providing your <answer>, as any deviations from the template will cause your answer to be invalid. A '?' symbol indicates a place for you to insert a short text, while a '???' symbol marks places for you to insert multi-line answers. To define the precise syntax to use, an <example> element may be provided. You MUST follow this syntax. However, note that the answers in the example may not be correct answers to the question.

Some questions relate to specific lecture material. In this case, the material will be provided in <context> elements for you to reference.

## Formatting
Always follow the format of the given exercise. If you are writing longer text, you can put text in *italics* or make it **bold**. Formulas should be given as LaTeX and enclosed properly, i.e. $...$ or \[...\].

user:# Exercise: Quiz: Berechenbarkeit und Komplexität

**Teilaufgaben (a-e):** Für diese Fragen ist eine Begründung nicht gefordert. Sie erhalten die Punkte auf eine Teilaufgabe genau dann, wenn Sie alle Antwortmöglichkeiten korrekt angekreuzt haben.
**Es ist immer mindestens eine Antwortmöglichkeit richtig.**
In dieser Aufgabe verwenden wir durchgehend das Alphabet $\Sigma := \{0, 1\}$.

## subproblem a (2.0 points)
Sei $L$ eine kontextfreie Sprache. Welche Aussagen sind wahr?

<options>
<option id="1">$L$ ist entscheidbar</option>
<option id="2">$\overline{L}$ ist entscheidbar</option>
<option id="3">$L$ ist semi-entscheidbar</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem b (2.0 points)
Seien $L_1,L_2\subseteq\Sigma^*$ entscheidbare Sprachen. Welche der folgenden Sprachen sind entscheidbar?

<options>
<option id="1">$L_1\cap L_2$</option>
<option id="2">$L_1\setminus L_2$</option>
<option id="3">$L_1L_2$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem c (2.0 points)
Sei $L\subseteq\{0\}^*$ unentscheidbar. Welche der folgenden Funktionen $f:\N\rightarrow\N$ sind berechenbar?

<options>
<option id="1">$f(n):=|L\cap \Sigma^n|$</option>
<option id="2">$f(n):=\min\{k\in\N: 0^k\in L\}$</option>
<option id="3">$f(n):=1$, falls $0^n\in L$, sonst $f(n):=0$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">0</option>
<option id="2">0</option>
<option id="3">0</option>

user:
## subproblem d (2.0 points)
Seien $A,B\subseteq\Sigma^*$ Sprachen mit $A\le_p\SAT\le_p B$. Welche Aussagen sind wahr?

<options>
<option id="1">$A\in\NP$</option>
<option id="2">$B\in\NP$</option>
<option id="3">$B$ ist NP-hart</option>
<option id="4">$A\le_p B$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>
<option id="4">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">0</option>
<option id="3">1</option>
<option id="4">1</option>

user:
## subproblem e (2.0 points)
Sei $L\subseteq \Sigma^*\{\#\}\Sigma^*$ eine Sprache, sodass $L\in\P$; $L$ enthält also Wörter der Form $u\#v$, mit $u,v\in\Sigma^*$. Welche der folgenden Sprachen sind in $\NP$?

<options>
<option id="1">$L$</option>
<option id="2">$\{u\mid\exists v\in\Sigma^*:u\#v\in L\}$</option>
<option id="3">$\{u\mid\exists v\in\Sigma^*:u\#v\in L\wedge |u|=|v|\}$</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<option id="3">0 or 1</option>

assistant:<option id="1">1</option>
<option id="2">1</option>
<option id="3">1</option>

user:
## subproblem f (4.0 points)
Sei $L\in\NP$. Zeigen oder widerlegen Sie: $L^*\in\NP$.

<options>
<option id="1">Ja, Beweis:</option>
<option id="2">Nein, Gegenbeispiel:</option>
</options>

### answer format
Give your answer in the following format:

<option id="1">0 or 1</option>
<option id="2">0 or 1</option>
<answer id="3">
...
</answer>

We need to answer subproblem f. Options: 1 "Ja, Beweis:", 2 "Nein, Gegenbeispiel:" Each expects 0 or 1 to indicate if statement true? The question: "Sei L in NP. Zeigen oder widerlegen Sie: L* in NP." So we need to decide if statement is true. Is the closure of NP under Kleene star? I recall NP is not known to be closed under Kleene star; in fact, if NP is closed under star then it would imply something? Let's think: For any language L in NP, is L* also in NP? L* is set of concatenations of zero or more strings from L. The verification: given input w, we need to check if it can be partitioned into k substrings each in L (for some k). This seems like a "NP" problem: you could nondeterministically guess the partition points (which are at most |w| many) and verify each piece is in L using NP verifier. The total verification may be polytime? Need to verify each piece in L using NP verifier, which may be nondeterministic. But we can simulate all nondeterminism in one chain? In NP, you have one certificate of polynomial length; you could encode partition points plus certificates for each piece. Number of pieces could be up to |w|, so total certificate length could be O(|w|^2) maybe? Each piece has its own certificate length polynomial in its length. Summing yields O(|w|^2). That's still polynomial. So seems L* is in NP. Indeed NP is closed under concatenation (since you can nondeterministically guess split point and verify both). Inductively, repeated concatenations (k times) gives closure under finite concatenations. For unbounded Kleene star, we need to guess variable number of pieces; but we can guess each split and each certificate, total length O(n^2). That's okay. So closure holds. I think indeed NP is closed under Kleene star. Let's verify: Reason: Given w, nondeterministically guess a decomposition w = w1 ... wk where each wi in L, with k ≤ |w|. Provide certificates c_i for each wi; total length sum_i |c_i| <= k * p(max|wi|) ≤ |w| * p(|w|) = O(|w|^2). That's polynomial. The verifier checks the splits and each certificate using the L verifier in polynomial time (overall O(|w| * p(|w|))). So L* ∈ NP.

Thus option 1 "Ja, Beweis:" should be 1, option2 "Nein, Gegenbeispiel:" should be 0.

Also we need to provide an answer in <answer id="3"> with explanation? The format: options then <answer id="3"> ... </answer>. Provide proof.

Let's craft answer.

<option id="1">1</option>
<option id="2">0</option>
<answer id="3">
**Beweis, dass $L^* \in \NP$ für jedes $L\in\NP$**

Sei $V$ ein polynomieller Verifizierer für $L$.  
Für ein Wort $w\in\Sigma^*$ der Länge $n$ konstruieren wir einen Verifizierer $V^{*}$ für $L^{*}$ wie folgt.

*Zertifikat.*  
Das Zertifikat besteht aus  

1. einer Folge von Trennpositionen $0 = i_0 < i_1 < i_2 < \dots < i_k = n$ (wobei $k\le n$), die die Zerlegung   
   $$w = w_1 w_2 \dots w_k,\qqu w_j = w[i_{j-1}+1\ ..\ i_j]$$  
   definiert, und  

2. für jedes Teilwort $w_j$ ein Zertifikat $c_j$ für $V$, das beweist, dass $w_j\in L$Da $k\le n$ und jedes $c_j$ nur polynomielle Länge in $|w_j|$ (also $\le n$) besitzt, hat das gesamte Zertifikat Gesamtlänge  
\[
\underbrace{O(n)}_{\text{Trennpositionen}} + \underbrace{k\cdot \operatorname{poly}(n)}_{\sum |c_j|} \;=\; O\!\bigl(n\cdot\operatorname{poly}(n)\bigr)=\operatorname{poly}(n).
\]

*Verifikation.*  
Der Verifizierer $V^{*}$ prüft in polynomialer Zeit:

1. dass die angegebenen Trennpositionen tatsächlich eine Zerlegung von $w$ ergeben,
2. für jedes $j=1,\dots,k$, dass $V$ das Teilwort $w_j$ zusammen mit dem zugehörigen Zertifikat $c_j$ akzeptiert.

Jede einzelne Ausführung von $V$ benötigt Zeit $\operatorname{poly}(|w_j|)\le\operatorname{poly}(n)$, also insgesamt  
\[
k\cdot\operatorname{poly}(n) \le n\cdot\operatorname{poly}(n)=\operatorname{poly}(n)
\]
Zeit.

Damit existiert ein polynomieller Verifizierer für $L^{*}$, also liegt $L^{*}$ in $\NP$.

\[
\boxed{L\in\NP \;\Longrightarrow\; L^{*}\in\NP}
\]

Damit ist die Aussage „$L^{*}\in\NP$“ richtig.  
</answer>

Sei $L\in\NP$. Zeigen oder widerlegen Sie: $L^*\in\NP$.

Beweis, dass $L^* \in \NP$ für jedes $L\in\NP$

Sei $V$ ein polynomieller Verifizierer für $L$.  
Für ein Wort $w\in\Sigma^*$ der Länge $n$ konstruieren wir einen Verifizierer $V^{*}$ für $L^{*}$ wie folgt.

Zertifikat.  
Das Zertifikat besteht aus  

1. einer Folge von Trennpositionen $0 = i_0 < i_1 < i_2 < \dots < i_k = n$ (wobei $k\le n$), die die Zerlegung   
   $$w = w_1 w_2 \dots w_k,\qqu w_j = w[i_{j-1}+1\ ..\ i_j]$$  
   definiert, und  

2. für jedes Teilwort $w_j$ ein Zertifikat $c_j$ für $V$, das beweist, dass $w_j\in L$Da $k\le n$ und jedes $c_j$ nur polynomielle Länge in $|w_j|$ (also $\le n$) besitzt, hat das gesamte Zertifikat Gesamtlänge  
$$\underbrace{O(n)}_{\text{Trennpositionen}} + \underbrace{k\cdot \operatorname{poly}(n)}_{\sum |c_j|} \;=\; O\!\bigl(n\cdot\operatorname{poly}(n)\bigr)=\operatorname{poly}(n).$$

Verifikation.  
Der Verifizierer $V^{*}$ prüft in polynomialer Zeit:

1. dass die angegebenen Trennpositionen tatsächlich eine Zerlegung von $w$ ergeben,
2. für jedes $j=1,\dots,k$, dass $V$ das Teilwort $w_j$ zusammen mit dem zugehörigen Zertifikat $c_j$ akzeptiert.

Jede einzelne Ausführung von $V$ benötigt Zeit $\operatorname{poly}(|w_j|)\le\operatorname{poly}(n)$, also insgesamt  
$$k\cdot\operatorname{poly}(n) \le n\cdot\operatorname{poly}(n)=\operatorname{poly}(n)$$
Zeit.

Damit existiert ein polynomieller Verifizierer für $L^{*}$, also liegt $L^{*}$ in $\NP$.

$$\boxed{L\in\NP \;\Longrightarrow\; L^{*}\in\NP}$$

Damit ist die Aussage „$L^{*}\in\NP$“ richtig.