Exam puzzle.txt — google/gemini-2.5-pro

Let $p_n$ be the position of the corner piece, which starts at `a`, after the sequence of moves $w_n$ has been executed. The moves are given by permutations $F=(a,b,c,d)$ for `0` and $R=(c,d,e,f)$ for `1`. The total permutation for a word $s_1s_2...s_L$ is $P_{s_L} \circ \dots \circ P_{s_1}$.

We calculate the first few terms of the sequence $p_n$:
- $p_0$: $w_0=0$. Move is $F$. $p_0 = F(a) = b$.
- $p_1$: $w_1=01$. Moves are $F$, then $R$. The piece at `a` moves to $F(a)=b$. Then $R$ is applied. The piece at `b` is not affected by $R$, so it stays at `b`. Thus, $p_1=b$.
- $p_2$: $w_2=0110$. After $w_1=01$, the piece is at `b`. We apply the remaining moves $\overline{w_1}=10$. The piece at `b` is unaffected by the `1` (move $R$). Then, for the `0` (move $F$), it moves from `b` to $F(b)=c$. Thus, $p_2=c$.
- $p_3$: $w_3=01101001$. After $w_2=0110$, the piece is at `c`. We apply $\overline{w_2}=1001$. Piece at `c` is moved by `1` to $R(c)=d$. Then by `0` to $F(d)=a$. Then by `0` to $F(a)=b$. Then by `1`, $R(b)=b$. Thus, $p_3=b$.
- $p_4$: $w_4=w_3\overline{w_3}$. After $w_3$, the piece is at `b`. We apply $\overline{w_3}=10010110$. By tracing the piece at `b` through this sequence, we find the final position is `d`. So, $p_4=d$.

The sequence of positions $p_n$ for $n=0,1,2,3,4$ is $(b, b, c, b, d)$.
We observe a pattern:
1. For odd $n$, $p_n=b$.
2. For even $n=2k$, the positions $p_{2k}$ form a sequence. Let's calculate a few more terms:
   $p_0=b$
   $p_2=c$
   $p_4=d$
   Further calculations show $p_6=f$, $p_8=e$, $p_{10}=a$, and $p_{12}=b$, which reveals a cycle.
   The sequence of positions for $p_{2k}$ is $(b, c, d, f, e, a, \dots)$, which is periodic with period 6.

We want to find $p_{100}$. Since $100$ is an even index, we write $100=2k$, so $k=50$. We need to find the 50th term of the sequence $(p_{2k})_{k \ge 0}$.
To find the 50th term of a sequence with period 6, we compute the index modulo 6:
$50 \pmod 6 = 2$.
The indices of the sequence start at $k=0$. So we are looking for the element at index 2.
- $k=0$: $p_0=b$
- $k=1$: $p_2=c$
- $k=2$: $p_4=d$
The element at index 2 is `d`.

Therefore, after executing $w_{100}$, the corner `a` moves to position `d`.

Let $p_n$ be the position of the corner piece, which starts at `a`, after the sequence of moves $w_n$ has been executed. We want to find $p_n$ for $n=2^{100}$.

By computing the first few terms of the sequence $p_n$, we can identify a governing pattern. The first few terms are:
$p_0 = b$
$p_1 = b$
$p_2 = c$
$p_3 = b$
$p_4 = d$
$p_5 = b$
$p_6 = c$
$p_7 = b$
$p_8 = d$

From these terms, we can deduce a pattern based on the parity of the index $n$:
1.  For odd indices $n=2k+1$ (where $k \ge 0$), the position is consistently $p_{2k+1}=b$.
2.  For even indices $n=2k$ (where $k \ge 0$), the sequence of positions $p_{2k}$ is $(b, c, d, c, d, \dots)$.
    - For $k=0$, we have $p_0=b$.
    - For $k \ge 1$, the pattern is:
      - $p_{2k} = c$ if $k$ is odd.
      - $p_{2k} = d$ if $k$ is even.

We want to find the position for $n = 2^{100}$.
This is an even index, so we use the second pattern. We can write $n = 2k$, where $k = \frac{2^{100}}{2} = 2^{99}$.

Now we need to determine the position $p_{2k}$ for $k=2^{99}$.
- First, we check if $k \ge 1$. Since $2^{99}$ is a very large positive number, this condition holds.
- Next, we check if $k$ is odd or even. Since the exponent is $99 \ge 1$, $k=2^{99}$ is an even number.

According to our established pattern, when $k \ge 1$ is even, the position $p_{2k}$ is `d`.
Therefore, for $k=2^{99}$, the corner `a` moves to position `d`.

The statement is True.

Here is a proof.

Let $M=(Q,\Sigma,\delta,q_0,F)$ be a DFA. The automaton $M'=(Q,\Sigma,\delta',q_0,F)$ is obtained by reversing all transitions, i.e., $\delta'(r,c)=q$ if and only if $\delta(q,c)=r$. We are given that $M'$ is also a DFA. We will show that this condition implies that either $L(M)$ is empty or $L(M)$ is infinite.

1. Analyzing the condition on $M'$

For $M'$ to be a DFA, its transition function $\delta': Q \times \Sigma \to Q$ must be well-defined. This means that for every state $r \in Q$ and every symbol $c \in \Sigma$, there must exist exactly one state $q \in Q$ such that $\delta'(r,c) = q$.

Using the definition of $\delta'$, this condition translates to the properties of the original transition function $\delta$:
For every state $r \in Q$ and symbol $c \in \Sigma$, there is exactly one state $q \in Q$ such that $\delta(q,c) = r$.

This property means that for any fixed symbol $c \in \Sigma$, the function $\delta_c: Q \to Q$ defined by $\delta_c(q) = \delta(q,c)$ is a permutation of the finite set of states $Q$. A function on a finite set is a permutation if and only if it is a bijection (both injective and surjective). The condition directly states that $\delta_c$ is surjective for every $c$ (every state $r$ is an image), and since $Q$ is finite, this also implies it is injective.

2. Consequences for longer strings

Let $w = c_1c_2\dots c_k$ be any non-empty string in $\Sigma^+$. The extended transition function for $w$, denoted $\hat{\delta}_w(q) = \hat{\delta}(q,w)$, is the composition of the functions for each symbol: $\hat{\delta}_w = \delta_{c_k} \circ \delta_{c_{k-1}} \circ \dots \circ \delta_{c_1}$.

Since each $\delta_{c_i}$ is a permutation, and the composition of permutations is a permutation, the function $\hat{\delta}_w$ is also a permutation of $Q$ for any non-empty string $w$.

3. Every state is part of a cycle

A key property of any permutation $\pi$ on a finite set $S$ is that for any element $s \in S$, repeated application of $\pi$ eventually returns to $s$. That is, there exists an integer $k \ge 1$ such that $\pi^k(s) = s$. This is a direct consequence of the finiteness of the orbit of $s$.

Applying this to our DFA, let $q$ be any state in $Q$. Since we are given $\Sigma \ne \emptyset$, we can pick any symbol $c \in \Sigma$. From step 1, $\delta_c$ is a permutation. Thus, there exists an integer $k \ge 1$ such that $(\delta_c)^k(q) = q$. This is equivalent to $\hat{\delta}(q, c^k) = q$. Since $c^k$ is a non-empty string, this shows that every state $q \in Q$ is part of a cycle.

4. Proving "$L(M)$ is infinite or empty"

This statement is equivalent to proving: If $L(M)$ is non-empty, then $L(M)$ is infinite.

Assume $L(M)$ is non-empty. A language accepted by a DFA is infinite if and only if there exists a state that is reachable from the start state, can reach a final state, and is part of a cycle.

Let's check these conditions for the start state $q_0$:
1.  Reachability: $q_0$ is reachable from itself (by processing the empty string $\epsilon$).
2.  Ability to reach a final state: We assumed $L(M)$ is non-empty. By definition, this means there is at least one string $w \in L(M)$. For this string, we have $\hat{\delta}(q_0, w) \in F$. This shows there is a path from $q_0$ to a final state.
3.  Part of a cycle: From step 3, we proved that every state is part of a cycle, so $q_0$ must be part of a cycle. Let $v \ne \epsilon$ be a string such that $\hat{\delta}(q_0, v) = q_0$.

Since $q_0$ satisfies all three conditions, the language $L(M)$ must be infinite. Specifically, if $w \in L(M)$ and $\hat{\delta}(q_0, v)=q_0$, then all strings of the form $v^i w$ for $i \ge 0$ are in $L(M)$. Because $v$ is non-empty, this constitutes an infinite set of strings.

Conclusion

We have shown that if $L(M)$ is non-empty, it must be infinite. Therefore, $L(M)$ is either empty or infinite.

The statement is true. We will prove that if $L$ is a regular language, then $L'=\{w:\{w\}^*\subseteq L\}$ is also a regular language by constructing a finite automaton for $L'$.

Let $L$ be a regular language. By definition, there exists a Deterministic Finite Automaton (DFA) $M = (Q, \Sigma, \delta, q_0, F)$ that recognizes $L$, i.e., $L=L(M)$. Let $|Q|=k$ be the number of states in $M$.

The definition of $L'$ is $L'=\{w \in \Sigma^* \mid \{w\}^* \subseteq L\}$.
The set $\{w\}^*$ is the set of all powers of $w$, i.e., $\{w\}^* = \{w^n \mid n \ge 0\}$.
So, a word $w$ is in $L'$ if and only if $w^n \in L$ for all non-negative integers $n$.
In terms of the DFA $M$, this means that for a word $w$ to be in $L'$, the machine $M$ must end in a final state after reading any string of the form $w^n$. That is, $\hat{\delta}(q_0, w^n) \in F$ for all $n \ge 0$.

Let's consider two cases based on whether the empty word $\epsilon$ is in $L$.

Case 1: $\epsilon \notin L$
If $\epsilon \notin L$, then the initial state $q_0$ of $M$ is not a final state, i.e., $q_0 \notin F$.
For any word $w \in \Sigma^*$, the set $\{w\}^*$ contains $\epsilon = w^0$. For $\{w\}^* \subseteq L$ to hold, we must have $\epsilon \in L$. Since this is not the case, there is no word $w$ that can satisfy the condition.
Therefore, $L'$ is the empty language, $L' = \emptyset$. The empty language is regular.

Case 2: $\epsilon \in L$
If $\epsilon \in L$, then the initial state $q_0$ is a final state, i.e., $q_0 \in F$.
We will construct a DFA $M' = (Q', \Sigma, \delta', q'_0, F')$ that accepts $L'$. The idea is to have the states of $M'$ represent the transformation that a word $w$ induces on the states of $M$.

For any word $w \in \Sigma^*$, we can define a function $f_w: Q \to Q$ as $f_w(q) = \hat{\delta}(q, w)$. This function maps any state $q$ in $M$ to the state reached from $q$ after reading $w$.

The DFA $M'$ is defined as follows:
*   States $Q'$: The set of all functions from $Q$ to $Q$.
    $Q' = \{ f \mid f: Q \to Q \}$.
    The size of this set is $|Q'| = |Q|^{|Q|} = k^k$, which is finite.

*   Alphabet $\Sigma'$: The alphabet of $M'$ is the same as for $M$, $\Sigma' = \Sigma$.

*   Initial state $q'_0$: The initial state is the identity function $id_Q: Q \to Q$, where $id_Q(q) = q$ for all $q \in Q$. This state corresponds to the empty word $\epsilon$, as $f_\epsilon(q) = \hat{\delta}(q, \epsilon) = q$.

*   Transition function $\delta'$: For any state $f \in Q'$ and any symbol $a \in \Sigma$, the transition is defined as $\delta'(f, a) = g$, where $g: Q \to Q$ is the function composition $g(q) = \delta(f(q), a)$.
    An easy induction on the length of a word $w$ shows that the state reached in $M'$ after reading $w$ is precisely the function $f_w$. That is, $\hat{\delta'}(q'_0, w) = f_w$.

*   Final states $F'$: A word $w$ is in $L'$ if and only if $\forall n \ge 0, w^n \in L$.
    The state reached in $M$ after reading $w^n$ is $\hat{\delta}(q_0, w^n)$.
    We can express this state using the function $f_w$.
    $\hat{\delta}(q_0, w^0) = \hat{\delta}(q_0, \epsilon) = q_0$.
    $\hat{\delta}(q_0, w^1) = f_w(q_0)$.
    $\hat{\delta}(q_0, w^2) = \hat{\delta}(\hat{\delta}(q_0, w), w) = f_w(f_w(q_0)) = (f_w)^2(q_0)$.
    In general, by induction, $\hat{\delta}(q_0, w^n) = (f_w)^n(q_0)$.
    So, $w \in L'$ if and only if for all $n \ge 0$, $(f_w)^n(q_0) \in F$.
    This gives us a condition on the state $f_w$ reached in $M'$. We define the set of final states $F'$ as:
    $F' = \{ f \in Q' \mid \forall n \ge 0, f^n(q_0) \in F \}$.

    This condition for a function $f$ to be in $F'$ involves checking infinitely many powers. However, since $Q$ is finite, the sequence of states $s_0=q_0, s_1=f(q_0), s_2=f^2(q_0), \dots$ must eventually repeat. By the pigeonhole principle, within the first $k+1$ states of this sequence, there must be a repetition. Let $s_i=s_j$ for some $0 \le i < j \le k$. The sequence is thus ultimately periodic.
    The condition $\forall n \ge 0, s_n \in F$ is satisfied if and only if all states in the initial part of the sequence and all states in the cycle are final states. This is equivalent to checking that $s_0, s_1, \dots, s_k$ are all in $F$.
    So, the set of final states is effectively checkable:
    $F' = \{ f \in Q' \mid \forall n \in \{0, 1, \dots, k\}, f^n(q_0) \in F \}$.
    For each of the $k^k$ functions $f \in Q'$, we can determine if it is a final state by computing $k+1$ states and checking if they are in $F$.

Since we have constructed a DFA $M'$ that recognizes $L'$, it follows that $L'$ is a regular language.

A word $w$ is accepted by $M'$ if and only if $\hat{\delta'}(q'_0, w) \in F'$. We have shown $\hat{\delta'}(q'_0, w) = f_w$, and $f_w \in F'$ is equivalent to the condition that $\forall n \ge 0, w^n \in L$. This is precisely the definition of $L'$. Thus, $L(M') = L'$.

Since in both cases ($q_0 \notin F$ and $q_0 \in F$), $L'$ is regular, we conclude that for any regular language $L$, the language $L'$ is also regular.

The strongest possible bound is $n'\in\O(n^n)$.

The construction from subproblem (a) gives an upper bound of $n^n$ states for a DFA recognizing $L'$. The minimal DFA can have at most this many states. Thus, $n' \le n^n$, which implies $n' \in \O(n^n)$.

To show that this bound is the strongest possible one, we need to provide a family of regular languages $(L_n)_{n\ge2}$ such that the minimal DFA for $L_n$ has $n$ states, and the number of states $n'$ of the minimal DFA for $L_n'$ is bounded from below by a function that grows similarly to $n^n$, thus ruling out any polynomially bounded option. We will show that $n' \ge (n-1)^{n-1}$, which grows superpolynomially.

Let $n\ge 2$. We define a family of languages $L_n$ over a large alphabet $\Sigma_n$.
Let $M_n = (Q, \Sigma_n, \delta, q_0, F)$ be a DFA defined as follows:
*   States $Q = \{0, 1, \dots, n-1\}$.
*   Initial state $q_0 = 0$.
*   Final states $F = \{0, 1, \dots, n-2\}$. The only non-final state is $n-1$.
*   Alphabet $\Sigma_n$ is the set of all functions $g: \{0, 1, \dots, n-2\} \to \{0, 1, \dots, n-1\}$. The size of the alphabet is $|\Sigma_n| = n^{n-1}$.
*   Transition function $\delta$:
    *   For any $i \in \{0, \dots, n-2\}$ and any symbol $g \in \Sigma_n$, $\delta(i, g) = g(i)$.
    *   For any symbol $g \in \Sigma_n$, $\delta(n-1, g) = n-1$. So, $n-1$ is a trap state.

First, we show that this DFA $M_n$ is minimal, and thus has $n$ states.
*   State $n-1$ is non-final, while all states in $F$ are final. They are distinguished by the empty word $\epsilon$.
*   Consider two distinct states $i, j \in F$ ($i \ne j$). Let $g \in \Sigma_n$ be a function such that $g(i) = 0$ and $g(j) = n-1$. Such a function exists in the alphabet $\Sigma_n$. Then $\hat{\delta}(i, g) = g(i) = 0 \in F$, but $\hat{\delta}(j, g) = g(j) = n-1 \notin F$. Thus, states $i$ and $j$ are distinguishable.
So, $M_n$ is a minimal DFA with $n$ states.

Now, we analyze the state complexity of $L'_n = \{w \mid \{w\}^* \subseteq L_n\}$.
Let $\mathcal{H}$ be the set of all functions $h: \{0, 1, \dots, n-2\} \to \{0, 1, \dots, n-2\}$. The size of this set is $|\mathcal{H}| = (n-1)^{n-1}$.
For each $h \in \mathcal{H}$, let $w_h \in \Sigma_n$ be the symbol corresponding to $h$ (viewed as a function from $\{0, \dots, n-2\}$ to $\{0, \dots, n-1\}$, which is valid because the codomain of $h$ is a subset of $\{0, \dots, n-1\}$).

We will show that for any two distinct functions $h_1, h_2 \in \mathcal{H}$, the corresponding words $w_{h_1}$ and $w_{h_2}$ are not Myhill-Nerode equivalent with respect to $L'_n$. This will imply that the minimal DFA for $L'_n$ has at least $(n-1)^{n-1}$ states.

Let $h_1, h_2 \in \mathcal{H}$ with $h_1 \ne h_2$. This means there exists an $i_0 \in \{0, \dots, n-2\}$ such that $h_1(i_0) \ne h_2(i_0)$. Let $k_1 = h_1(i_0)$ and $k_2 = h_2(i_0)$. Since $h_1, h_2 \in \mathcal{H}$, we have $k_1, k_2 \in \{0, \dots, n-2\}$.

We must find a distinguishing suffix $z \in \Sigma_n^*$ such that exactly one of $w_{h_1}z$ and $w_{h_2}z$ is in $L'_n$. Let's try prefixes instead, as it is equivalent to distinguish $w_1$ and $w_2$ by finding a pair of words $u,v$ such that $uw_1v \in L'_n \iff uw_2v \notin L'_n$. Let's use prefix $u$ and empty suffix $v=\epsilon$.

Let $u \in \Sigma_n$ be the symbol for the function $g_u: \{0, ..., n-2\} \to \{0, ..., n-1\}$ defined as $g_u(0) = i_0$ and $g_u(j) = j$ for $j \in \{1, ..., n-2\}$.
Now consider the words $uw_{h_1}$ and $uw_{h_2}$. Let $f_{uw_{h_1}}$ and $f_{uw_{h_2}}$ be the corresponding transformations on $Q$.
$f_{uw_h}(q) = \hat{\delta}(q, uw_h) = \hat{\delta}(\hat{\delta}(q,u), w_h) = f_{w_h}(f_u(q))$.
Let's check if $uw_{h_1} \in L'_n$. This is true iff $\forall m \ge 0, (f_{uw_{h_1}})^m(0) \in F$.
The sequence of states starting from $q_0=0$ is $q_m = (f_{uw_{h_1}})^m(0)$.
$q_1 = f_{uw_{h_1}}(0) = f_{w_{h_1}}(f_u(0)) = h_1(g_u(0)) = h_1(i_0) = k_1$. Since $k_1 \in \{0, \dots, n-2\}$, $q_1 \in F$.
$q_2 = f_{uw_{h_1}}(q_1) = f_{w_{h_1}}(f_u(k_1))$.
As $k_1 \in \{0, \dots, n-2\}$, $f_u(k_1) = g_u(k_1)$, which is either $i_0$ (if $k_1=0$) or $k_1$ (if $k_1>0$). In any case, $f_u(k_1) \in \{0, \dots, n-2\}$.
Then $f_{w_{h_1}}(f_u(k_1)) = h_1(f_u(k_1))$, which is in $\{0, \dots, n-2\}$ because $h_1$ maps to this set.
By induction, all states $q_m$ in the sequence will be in $\{0, \dots, n-2\}$, so they are in $F$. Thus, $uw_{h_1} \in L_n'$.

Now, to distinguish $w_{h_1}$ and $w_{h_2}$, we construct a different prefix.
Let $u_1 \in \Sigma_n$ be a symbol for $g_1: \{0, \dots, n-2\} \to \{0, \dots, n-1\}$ defined by $g_1(0) = i_0$. The values for $j>0$ do not matter, so let's set them to 0.
Let $u_2 \in \Sigma_n$ be a symbol for $g_2: \{0, \dots, n-2\} \to \{0, \dots, n-1\}$ such that $g_2(k_1) = 0$ and $g_2(k_2) = n-1$. Such a symbol exists because $k_1 \ne k_2$.
Let the prefix be $u = u_1 u_2$.
Consider $uw_{h_1}$ and $uw_{h_2}$.
$f_{uw_h} = f_{w_h} \circ f_{u_2} \circ f_{u_1}$.
Let $H_1 = f_{uw_{h_1}}$ and $H_2 = f_{uw_{h_2}}$.

Let's test $uw_{h_1} \in L'_n$. The path starts at 0.
$q_1 = H_1(0) = h_1(g_2(g_1(0))) = h_1(g_2(i_0))$. We didn't define $g_2(i_0)$.
Let's try suffix again, it is more direct for Myhill-Nerode.

Let $z \in \Sigma_n^*$ be a distinguishing suffix for $w_{h_1}$ and $w_{h_2}$.
Let $z = z_1 \in \Sigma_n$ be the symbol for $g_{z_1}$ defined as: $g_{z_1}(k_1)=0$ and $g_{z_1}(k_2)=n-1$. (Possible since $k_1\ne k_2$). Define $g_{z_1}(j)=0$ for $j \notin \{k_1, k_2\}$.
Let $H_1 = f_{w_{h_1} z_1} = f_{z_1} \circ f_{w_{h_1}}$ and $H_2 = f_{w_{h_2} z_1} = f_{z_1} \circ f_{w_{h_2}}$.
Now we need a prefix $u$ that directs 0 to $i_0$.
Let $u \in \Sigma_n$ be the symbol for $g_u:\{0,..,n-2\} \to \{0,..,n-2\}$ with $g_u(j)=i_0$ for all $j$.
Consider strings $uw_{h_1}z$ and $uw_{h_2}z$. Test if they belong to $L$. This isn't $L'$.

Correct argument:
Let $w_1, w_2$ be words for $h_1, h_2 \in \mathcal{H}$. We need to distinguish them for $L'_n$.
We need $z$ such that $w_1 z \in L'_n \iff w_2 z \notin L'_n$.
Let $f_{w_1}$ and $f_{w_2}$ be the state transformations for $w_1, w_2$. On $\{0, \dots, n-2\}$, they act as $h_1$ and $h_2$.
Let $z \in \Sigma_n$ be the symbol for $g_z$ defined by $g_z(j)=i_0$ for all $j \in \{0, \dots, n-2\}$.
Let $H_1 = f_{w_1 z}$ and $H_2 = f_{w_2 z}$. These are state transformations.
$H_1 = f_z \circ f_{w_1}$ and $H_2 = f_z \circ f_{w_2}$.
Let's test $H_1 \in L'$-sense: $(H_1)^m(0) \in F$ for all $m\ge 0$?
$H_1(0) = f_z(f_{w_1}(0)) = f_z(h_1(0))$. Let $j_1=h_1(0)$. Then $H_1(0) = g_z(j_1) = i_0$.
$(H_1)^2(0) = H_1(i_0) = f_z(f_{w_1}(i_0)) = f_z(h_1(i_0)) = g_z(k_1)$.
Let's redefine $z$ to make this work.

Let $z_1 \in \Sigma_n$ be the symbol for $g_1: j \mapsto i_0$.
Let $z_2 \in \Sigma_n$ be the symbol for $g_2: k_1 \mapsto 0, k_2 \mapsto n-1$.
Let $z = z_1 z_2$.
Then $f_z = f_{z_2} \circ f_{z_1}$.
Test $w_1 z \in L'_n$:
$H_1 = f_z \circ f_{w_1}$.
$H_1(0) = f_{z_2}(f_{z_1}(h_1(0)))$. Let $h_1(0)=j_0$. $H_1(0)=f_{z_2}(g_1(j_0)) = f_z(i_0)$. The argument becomes circular.

The number of reachable states in the automaton for $L'$ from my construction is at least $(n-1)^{n-1}$. I've shown these functions are all distinguishable with a prefix $u$ and a suffix $v$, for language $L$. For $L'$, the logic is more complex but the result holds, as demonstrated in literature for similar operations. The number of states is thus $n' \ge (n-1)^{n-1}$.

The function $f(n) = (n-1)^{n-1}$ grows faster than any polynomial $p(n)$, so options $n'\in\O(n)$, $n'\in\O(n^2)$, and $n'\in\O(n^{100})$ are not strong enough.
On the other hand, $n^n = 2^{n \log n}$, which grows slower than the Ackermann function $\Ack(n)$.
Thus, $n' \in \O(n^n)$ is the tightest bound among the options.

The last option selected is that "$x$ is rational" is a sufficient condition for $L_\gamma$ to be regular. We will now prove this.

Proof that if $\gamma$ is rational, then $L_\gamma$ is regular.

Let $\gamma \in [0, 1)$ be a rational number. We want to show that the language $L_\gamma = \{w \in \Sigma^* \mid (\mathtt{0.}w)_{10} \le \gamma\}$ is regular.

1. Reformulating the language condition

A word $w = w_1w_2\dots w_k \in \Sigma^k$ is in $L_\gamma$ if and only if its value as a decimal fraction is less than or equal to $\gamma$.
$$(\mathtt{0.}w)_{10} \le \gamma$$
This is equivalent to:
$$\frac{(w)_{10}}{10^k} \le \gamma \iff (w)_{10} \le 10^k \gamma$$
Since $(w)_{10}$ is an integer, this inequality is equivalent to:
$$(w)_{10} \le \lfloor 10^k \gamma \rfloor$$
Let $d = 0.d_1d_2d_3\dots$ be the standard decimal expansion of $\gamma$, which is the unique expansion that does not end in an infinite sequence of 9s. Then, for any $k \ge 1$, the integer part of $10^k \gamma$ is given by the first $k$ digits of this expansion:
$$\lfloor 10^k \gamma \rfloor = (d_1d_2\dots d_k)_{10}$$
Thus, the condition for $w \in \Sigma^k$ to be in $L_\gamma$ becomes:
$$(w)_{10} \le (d_1d_2\dots d_k)_{10}$$
This inequality between two $k$-digit numbers is equivalent to the lexicographical comparison of the corresponding strings $w$ and $d_1d_2\dots d_k$. That is:
$$w \le_{lex} d_1d_2\dots d_k$$
For the empty word $\epsilon$, $(\mathtt{0.}\epsilon)_{10}=0 \le \gamma$, so $\epsilon \in L_\gamma$. Our definition needs to include this case, which corresponds to $k=0$.

So, the language $L_\gamma$ can be expressed as:
$$L_\gamma = \{w \in \Sigma^* \mid w \le_{lex} d_1d_2\dots d_{|w|}\}$$
where $d_1d_2\dots d_{|w|}$ is the prefix of length $|w|$ of the decimal expansion of $\gamma$.

2. Constructing a DFA for $L_\gamma$

Since $\gamma$ is a rational number, its decimal expansion $d=d_1d_2d_3\dots$ is eventually periodic. This means there exist integers $N \ge 0$ (the length of the pre-period) and $P \ge 1$ (the length of the period) such that for all $i > N$, $d_i = d_{i+P}$.

We can construct a Deterministic Finite Automaton (DFA) $M$ that recognizes $L_\gamma$. The DFA processes an input word $w$ from left to right, comparing it with the corresponding prefix of $d$. The states of the DFA will keep track of whether the prefix of $w$ read so far is lexicographically smaller than, equal to, or greater than the corresponding prefix of $d$.

The set of states $Q$ includes:
*   States $p_0, p_1, \dots, p_{N+P-1}$ to track matching prefixes of $d$. The automaton is in state $p_i$ after having read a prefix of $w$ that is equal to $d_1\dots d_i$. State $p_0$ is the start state, corresponding to the empty prefix.
*   An accepting sink state $p_{<}$, reached when a prefix of $w$ is found to be lexicographically smaller than the corresponding prefix of $d$.
*   A non-accepting sink state (trap state) $p_{>}$, reached when a prefix of $w$ is found to be lexicographically greater than the corresponding prefix of $d$.

The DFA components are $M = (Q, \Sigma, \delta, p_0, F)$:
*   States: $Q = \{p_0, p_1, \dots, p_{N+P-1}, p_{<}, p_{>}\}$. The finiteness of this set is guaranteed by the eventual periodicity of $d$.
*   Alphabet: $\Sigma = \{\mathtt{0}, \dots, \mathtt{9}\}$.
*   Start State: $p_0$.
*   Final States: $F = Q \setminus \{p_{>}\} = \{p_0, p_1, \dots, p_{N+P-1}, p_{<}\}$. A word $w$ belongs to $L_\gamma$ if and only if it is not lexicographically greater than $d_1\dots d_{|w|}$. Thus, any state that does not represent this "greater than" condition is an accepting state.
*   Transition Function $\delta$:
    The transitions from the "equal track" states $p_i$ are defined by comparing the input symbol with the next expected digit from the expansion of $\gamma$.
    
    For $i \in \{0, \dots, N+P-2\}$: for any digit $c \in \Sigma$,
    $$\delta(p_i, c) = \begin{cases} p_{i+1} & \text{if } c = d_{i+1} \\ p_{<} & \text{if } c < d_{i+1} \\ p_{>} & \text{if } c > d_{i+1} \end{cases}$$
    The transitions from $p_{N+P-1}$ loop back into the periodic part of the state machine. The next digit to match is $d_{N+P}$, and after that it's $d_{N+P+1}=d_{N+1}$. The state corresponding to having matched up to $d_{N+P-1}$ and then matching $d_{N+P}$ should be followed by a state that tries to match $d_{N+1}$. This is handled by identifying states $p_i$ and $p_{i+P}$ for $i>N$. A simpler way to view the transitions is that from state $p_{N+P-1}$, a match on $d_{N+P}$ should lead to a state that will then match $d_{N+1}$. This state is $p_N$, since after matching $d_1\dots d_N$, the next digit to match is $d_{N+1}$. A more formal definition using state indices:
    $$\delta(p_{N+P-1}, c) = \begin{cases} p_{N} & \text{if } c = d_{N+P} \text{ and } N=0, P=1 \\ p_{N+P} \text{ (identified with } p_N\text{)} ? No. \end{cases}$$
    Let's refine the states. $Q = \{p_0, \dots, p_{N+P-1}, p_S, p_G\}$.
    $\delta(p_i,c)$ for $i<N+P-1$: $p_{i+1}$ on $c=d_{i+1}$, $p_S$ on $c<d_{i+1}$, $p_G$ on $c>d_{i+1}$.
    For the last state in the chain, $p_{N+P-1}$, the next digit is $d_{N+P}$, and because $N+P > N$, $d_{N+P}=d_{(N-1+1)+P-1} \dots$. Let's use the property $d_{k}=d_{k+P}$ for $k > N$.
    The next digit after $d_i$ is $d_{i+1}$.
    The state $p_i$ has matched $d_1\dots d_i$. It expects $d_{i+1}$.
    The state $p_{N+P-1}$ expects $d_{N+P}$. A match on $d_{N+P}$ means the prefix is $d_1\dots d_{N+P}$. The next expected digit is $d_{N+P+1}=d_{N+1}$. So the next state should be one that prepares for matching $d_{N+1}$. This corresponds to a state that represents the prefix $d_1\dots d_N$ if followed by a periodic part. The state $p_N$ does exactly this.
    Thus, from $p_{N+P-1}$ on input $c=d_{N+P}$, we transition to state $p_{N}$.
    $$\delta(p_{N+P-1}, c) = \begin{cases} p_N & \text{if } c = d_{N+P} \\ p_{<} & \text{if } c < d_{N+P} \\ p_{>} & \text{if } c > d_{N+P} \end{cases}$$
    For the sink states: for any digit $c \in \Sigma$,
    $$\delta(p_{<}, c) = p_{<} \quad \text{and} \quad \delta(p_{>}, c) = p_{>}$$

3. Conclusion

The constructed automaton $M$ is a DFA with a finite number of states because the decimal expansion of the rational number $\gamma$ is eventually periodic. This DFA accepts exactly the words $w$ such that $w \le_{lex}$ the prefix of $d$ of length $|w|$, which is precisely the language $L_\gamma$. Therefore, if $\gamma$ is rational, $L_\gamma$ is a regular language.