The maximum length of shortest accepted strings for direction-determinate two-way finite automata

There are as many pairs $(P,R)$ as pairs $(Q^{-}\setminus P,R)$, where $|R|=|P|+1$. The number of pairs of the latter form is equal to the number of subsets of $Q$ of size $\ell+1$, that is, $\binom{k+\ell}{\ell+1}$. ∎

It is claimed that the automaton $A$, executed on the string $w$, eventually arrives to each symbol $a_{i}$ in the state $r^{(i)}_{m_{i}+1}$. This is proved by induction on $i$.

Base case $i=1$: the first transition (1a) moves the automaton to the state $r^{(1)}_{1}$. The first pair $(P^{(1)},R^{(1)})$ is $(\varnothing,\{1\})$, and so $r^{(1)}_{1}=r^{(1)}_{m_{1}+1}$.

Induction step. Assume that the automaton comes to the symbol $a_{i}$ in the state $r^{(i)}_{m_{i}+1}$. Then it makes a transition (1c) to the right in the state $r^{(i+1)}_{1}$. Then it executes the sequence of transitions (1b), (1d), defined by the pair $(P_{i+1},R_{i+1})$, moving back and forth between $a_{i+1}$ and $a_{i}$, and passing through the states $p^{(i+1)}_{1}$, $r^{(i+1)}_{2}$, $p^{(i+1)}_{2}$, …$r^{(i+1)}_{m_{i+1}}$, $p^{(i+1)}_{m_{i+1}}$, $r^{(i+1)}_{m_{i+1}+1}$. And so it comes to the symbol $a_{i+1}$ in the state $r^{(i+1)}_{m_{i+1}+1}$, as shown in Figure 2.

In the end, the automaton comes to the last symbol $a_{N-1}$ in the state $r^{(N-1)}_{m_{N-1}+1}$. Then it makes a transition (1c) and moves to the right end-marker in the state $r^{(N)}_{1}$. And this is the accepting state $r^{(N)}_{m_{N}+1}$, because the last pair $(P^{(N)},R^{(N)})$ is $(\varnothing,\{k\})$. Therefore, the string $w$ is accepted. ∎

Let the automaton $A$ accept some string that starts from some symbol $a_{i}$. The transition from the initial configuration leads the automaton to the state $r^{(1)}_{1}$ at the first symbol $a_{i}$. As $(P^{(1)},R^{(1)})=(\varnothing,\{1\})$, the state $r^{(1)}_{1}$ is $1$.

Transitions by the symbol $a_{i}$ are defined only in states from $R^{(i)}\cup P^{(i+1)}$, and hence $1\in R^{(i)}$, for otherwise the automaton immediately rejects. If there is at least one more state in $R^{(i)}$, then the transition in the state $1$ by $a_{i}$ moves the automaton to the left. Then the automaton returns to the left end-marker, and then either loops or rejects, because there is only one transition defined there. Therefore, there are no other states in $R^{(i)}$ besides $1$, and so, $(P^{(i)},R^{(i)})=(\varnothing,\{1\})=(P^{(1)},R^{(1)})$, which implies $i=1$. ∎

Let a string accepted by $A$ end with a symbol $a_{i}$. To accept, the automaton should move from $a_{i}$ to the right using the transition (1c), and it arrives to the right end-marker in the state $r^{(i+1)}_{1}$. As the only accepting state is $k$, and the automaton rejects at the right end-marker in all other states, this state must be $r^{(i+1)}_{1}=k$. Because the state $r^{(i+1)}_{1}$ is the least in $R^{(i+1)}$, it follows that $R^{(i+1)}=\{k\}$ and $P^{(i+1)}=\varnothing$. Therefore, this is the last pair, and $i=N-1$. ∎

The proof is by contradiction. Suppose that $A$ accepts a string that contains a substring $a_{i}a_{j}$, with $j>i+1$. In order to accept, the automaton should eventually reach this symbol $a_{j}$ for the first time, moving to it from the symbol $a_{i}$. To make this transition, the automaton should be at $a_{i}$ in some state from $Q^{+}$ (indeed, if it were in the state from $Q^{-}$, then it would have been at $a_{j}$ already at the previous step). Then the automaton must use the transition (1c) to move from $a_{i}$ to $a_{j}$, and this transition leads to the state $r^{(i+1)}_{1}$. For the computation to go onward, this state should lie in $R^{(j)}$. Moreover, the state $r^{(i+1)}_{1}$ should be the least in $R^{(j)}$, for otherwise the pair $(P^{(j)},R^{(j)})$ would be less than the pair $(P^{(i+1)},R^{(i+1)})$. Also $r^{(i+1)}_{1}$ cannot be the only state in $R^{(j)}$: if not, then $(P^{(j)},R^{(j)})$ would either coincide with or be less than $(P^{(i+1)},R^{(i+1)})$.

It can be concluded that $r^{(i+1)}_{1}=r^{(j)}_{1}$, and the next transition from this state leads to the state $p^{(j)}_{1}$, moving to the symbol $a_{i}$. For the automaton to have a transition in the state $p^{(j)}_{1}$ at $a_{i}$, this state should belong to $P^{(i+1)}$. In addition, it should be the least among the state in $P^{(i+1)}$, because if there were a lesser state $p$, then the second term in the sequence for $(P^{(i+1)},R^{(i+1)})$ would be $-p$, and this pair would be greater than $(P^{(j)},R^{(j)})$. This leads to the equality $p^{(j)}_{1}=p^{(i+1)}_{1}$.

By analogous arguments, one can prove that the sequences for $(P^{(j)},R^{(j)})$ and for $(P^{(i+1)},R^{(i+1)})$ must coincide and continue infinitely. This is impossible, because the numbers of states increase, and there finitely many of them. ∎

The automata and the shortest strings they accept are constructed inductively; for small values of $n$ they are given in Figure 3.

For the inductive proof to work, the following set of properties is ensured for every $n$.

The first observation is that Theorem 3 follows from this claim. Let $n\geqslant 2$, and let $A_{n}$ and $w_{n}$ be an automaton and a string that satisfy the conditions in the claim. Then $A_{n}$ is supplemented with an initial state $n$, a set of accepting states $\{1\}$ and a single transition by the left end-marker: from the state $n$ to the state $n$; no transitions by the right end-marker are defined. The resulting automaton $A^{\prime}_{n}$ becomes a valid 2DFA, and it accepts the string $w_{n}$ as follows: from the initial state at $\vdash$ it moves to the first symbol of $w_{n}$ in the state $n$, then, by the first point of the claim, the automaton eventually leaves $w_{n}$ to the right in the state $1$, and thus arrives to the right end-marker $\dashv$ in an accepting state.

To see that every string accepted by $A^{\prime}_{n}$ is of length at least $|w_{n}|$, let $u$ be any accepted string. It is not empty, because on the empty string the automaton steps on the right end-marker in the state $n$ and rejects. Then, after the first step the automaton $A^{\prime}_{n}$ is at the first symbol of $u$ in the state $n$. It cannot return to $\vdash$, because it has already used the only transition at this label, and if it ever comes back, it will reject or loop. Also the automaton cannot come to $\dashv$ in states other than $1$. In order to accept, it must arrive to $\dashv$ in the state $1$, and this is the first and the only time when it leaves the string $u$. Then, by the second point of the claim, the length of $u$ cannot be less than the length of $w_{n}$.

It remains to prove the claim, which is done by induction on $n$.

Base case: $n=2$.

The automaton $A_{2}=(\Sigma_{2},Q_{2},\delta_{2})$ for $n=2$ is constructed as follows. The alphabet is $\Sigma_{2}=\{a,b\}$, and the set of states is $Q_{2}=\{1,2\}$. The transition function is defined by

The string $w_{2}$ is $ab$, and the computation of $A_{2}$ on $w_{2}$ is presented in Figure 3 (top left). To be precise, computations starting in the state $2$ either at $a$ or at $b$ both end by leaving the string to the right in the state $1$, as claimed. There are only two shorter non-empty strings: $a$ and $b$. If the automaton starts on the string $a$ in the state $2$, then it moves to the right in the state $2$; on $b$, it moves to the left in the state $1$. In either case, it does not go to the right in the state $1$. Thus, the second point of the claim is satisfied. The length of the string is $|w_{2}|=2=3\cdot 2^{0}-1$.

Induction step: $n\to n+1$.

Let an $n$-state 2DFA $A_{n}=(\Sigma_{n},Q_{n},\delta_{n})$ and a string $w_{n}\in\Sigma_{n}^{*}$ satisfy the claim. The $(n+1)$-state automaton $A_{n+1}$ satisfying the claim is constructed as follows. Let $A_{n+1}=(\Sigma_{n+1},Q_{n+1},\delta_{n+1})$.

• •

  Its alphabet is $\Sigma_{n+1}=\overrightarrow{\Sigma_{n}}\cup\overleftarrow{\Sigma_{n}}\cup\{\#\}$, where $\overrightarrow{\Sigma_{n}}=\{\,\overrightarrow{a}\mid a\in\Sigma_{n}\,\}$ and $\overleftarrow{\Sigma_{n}}=\{\,\overleftarrow{a}\mid a\in\Sigma_{n}\,\}$

• •

  The set of states is $Q_{n+1}=Q_{n}\cup\{n+1\}=\{1,\ldots,n+1\}$.

• •

  The transition function is defined as follows. In the new state $n+1$, the automaton moves by all symbols with arrows in the directions pointed by the arrows.

  	$\displaystyle\delta_{n+1}(n+1,\overrightarrow{a})$	$\displaystyle=(n+1,+1),$	$\displaystyle\text{for }a\in\Sigma$
  	$\displaystyle\delta_{n+1}(n+1,\overleftarrow{a})$	$\displaystyle=(n+1,-1),$	$\displaystyle\text{for }a\in\Sigma$
  In all old states $1,\ldots,n$, on symbols with arrows, the new automaton works in the same way as the automaton $A_{n}$ on the corresponding symbols without arrows.
  	$\displaystyle\delta_{n+1}(i,\overrightarrow{a})=\delta_{n+1}(i,\overleftarrow{a})$	$\displaystyle=\delta_{n}(i,a),$	$\displaystyle\text{for }a\in\Sigma\text{ and }i\in\{1,\ldots,n\}$
  By the new separator symbol $\#$, only two transitions are defined. In the state $n+1$, the automaton moves to the left in the state $n$, thus starting the automaton $A_{n}$ on the substring to the left.
  	$\displaystyle\delta_{n+1}(n+1,\#)$	$\displaystyle=(n,-1)$
  And if the automaton gets to $\#$ in the state $1$ (which happens after concluding the simulation of $A_{n}$ on the substring to the left), then the automaton moves to the right in the state $n$ to start the simulation of $A_{n}$ also on the substring to the right of the separator $\#$.
  	$\displaystyle\delta_{n+1}(1,\#)$	$\displaystyle=(n,+1)$

  The rest of transitions are undefined.

Note that once the automaton $A_{n+1}$ leaves the state $n+1$, it never returns to it, because there are no transitions to $n+1$ from any other state. Let $h\colon(\overrightarrow{\Sigma_{n}}\cup\overleftarrow{\Sigma_{n}})^{*}\to\Sigma_{n}^{*}$ be a string homomorphism which removes the arrow from the top of every symbol, that is, $h(\overrightarrow{a})=h(\overleftarrow{a})=a$ for all $a\in\Sigma_{n}$. The automaton $A_{n+1}$ works in the states $1,\ldots,n$ on symbols from $\overrightarrow{\Sigma_{n}}\cup\overleftarrow{\Sigma_{n}}$ as $A_{n}$ works on the corresponding symbols from $\Sigma_{n}$. Then, if $h(w)=w_{n}$ for some $w\in(\overrightarrow{\Sigma_{n}}\cup\overleftarrow{\Sigma_{n}})^{*}$, it follows that the automaton $A_{n+1}$, having started in the state $n$ at any symbol of $w$, eventually leaves the string $w$ by moving to the right in the state $1$. Furthermore, if $|w|<|w_{n}|$ for some string $w\in(\overrightarrow{\Sigma_{n}}\cup\overleftarrow{\Sigma_{n}})^{*}$, then the automaton $A_{n+1}$, having started in the state $n$ at any symbol of $w$, cannot leave the string by moving to the right in the state $1$.

The string $w_{n+1}$ is defined as $\overrightarrow{w_{n}}\#\overleftarrow{w_{n}}$, where $\overrightarrow{a_{1}\ldots a_{\ell}}=\overrightarrow{a_{1}}\ldots\overrightarrow{a_{\ell}}$ and $\overleftarrow{a_{1}\ldots a_{\ell}}=\overleftarrow{a_{1}}\ldots\overleftarrow{a_{\ell}}$ for every string $a_{1}\ldots a_{\ell}\in\Sigma_{n}^{*}$. The length of $w_{n+1}$ is $|w_{n+1}|=2|w_{n}|+1=2(3\cdot 2^{n-2}-1)+1=3\cdot 2^{n-1}-1$, as desired.

First, it is proved that the automaton $A_{n+1}$ works on the string $w_{n+1}$ as stated in the first point of the claim. Let $A_{n+1}$ start its computation on the string $w_{n+1}$ at any symbol in the state $n+1$, as shown in Figure 4. By the symbols in $\overrightarrow{w_{n}}$, the automaton moves to the right, maintaining the state $n+1$; by the symbols in $\overleftarrow{w_{n}}$, it moves to the left in $n+1$. Thus, wherever the automaton begins, it eventually arrives to the separator $\#$ in the state $n+1$. Next, the automaton moves to the last symbol of $\overrightarrow{w_{n}}$ in the state $n$. Since $h(\overrightarrow{w_{n}})=w_{n}$, the automaton $A_{n+1}$ operates on $\overrightarrow{w_{n}}$ as $A_{n}$ on $w_{n}$, and leaves $\overrightarrow{w_{n}}$ by a transition to the right in the state $1$. Then $A_{n+1}$ arrives to the separator $\#$ again, now in the state $1$, and moves to the first symbol of $\overleftarrow{w_{n}}$ in the state $n$. As $h(\overleftarrow{w_{n}})=w_{n}$, the automaton $A_{n+1}$ works as $A_{n}$ on $w_{n}$, and leaves $\overleftarrow{w_{n}}$ (and the whole string $w_{n+1}$) by moving to the right in the state $1$.

Turning to the second point of the claim, it should be proved that computations of a certain form are impossible on any strings shorter than $w_{n+1}$. Let $w\in\Sigma_{n+1}^{*}$ be a string, and let there be a position in $w$, such that the automaton $A_{n+1}$, having started at this position in the state $n+1$, eventually leaves the string $w$ by a transition to the right in the state $1$. It is claimed that $|w|\geqslant|w_{n+1}|$.

Consider the computation of $A_{n+1}$ leading out of $w$ to the right in the state $1$. It begins in the state $n+1$, and the automaton maintains the state $n+1$ at all symbols except $\#$. In order to reach the state $1$, there should be a moment in the computation on $w$ when the automaton arrives at some symbol $\#$ in the state $n+1$. Let $u$ be the prefix of $w$ to the left of this $\#$, and let $v$ be the suffix to the right of this $\#$; note that the substrings $u$ and $v$ may contain more symbols $\#$. It is sufficient to prove that $|u|\geqslant|w_{n}|$ and $|v|\geqslant|w_{n}|$.

Consider first the case of the suffix $v$. Let $v_{0}$ be the longest suffix of $v$ that does not contain the symbol $\#$; then the symbol preceding $v_{0}$ in $w$ is the separator $\#$, as shown in Figure 5. Once the automaton $A_{n+1}$ steps from the last $\#$ in $w$ to the right, it arrives to the first symbol of $v_{0}$ in the state $n$ (by the unique transition to the right at $\#$). The string $v_{0}$ cannot be empty, because $n\neq 1$. Once the automaton is inside $v_{0}$, it cannot return to $\#$ anymore, since it has already used the only transition to the right from $\#$, and cannot use it again without looping. Therefore, the automaton $A_{n+1}$ starts on the string $v_{0}\in(\Sigma_{n+1}\setminus\{\#\})^{*}$ in the state $n$, and, operating as $A_{n}$, eventually leaves this string to the right in the state $1$. Then $|v_{0}|\geqslant|w_{n}|$ by the induction hypothesis, and hence $|v|\geqslant|w_{n}|$.

Now consider the prefix $u$. Once the automaton $A_{n+1}$ comes in the state $n+1$ to the separator $\#$ between $u$ and $v$, it moves to the last symbol of $u$ in the state $n$. In order to leave the string $u$ to the right and proceed further, it must return to the separator $\#$ in the state $1$, because there are no transitions by any states $\{2,\ldots,n\}$ at this separator. If there are no symbols $\#$ in $u$, or if there are some, but the automaton does not reach them, then the entire computation of $A_{n+1}$ on $u$ takes place on a certain suffix of $u$ that does not contain $\#$, as illustrated in Figure 6. This computation follows a computation of $A_{n}$ on a string from $\Sigma_{n}^{*}$. Then, by the induction hypothesis, this suffix is not shorter than $w_{n}$, and therefore $|u|\geqslant|w_{n}|$.

The remaining case is when the automaton comes to some symbol $\#$ inside the string $u$. Let $u_{0}$ be the maximal suffix of $u$ not containing any symbols $\#$, as in Figure 7. The automaton $A_{n+1}$ visits the separator $\#$ to the left of $u_{0}$, and then immediately moves from this separator back to the first symbol of $u_{0}$ in the state $n$ (the string $u_{0}$ is non-empty, because it is followed by $\#$, which has no transitions in the state $n$). Returning back to $\#$ to the left of $u_{0}$ is not an option, since the unique transition by $\#$ to the right has been used already. Therefore, the automaton leaves $u_{0}$ by a transition to the right, and comes to the separator $\#$ between $u$ and $v$. In order to continue the computation, it should come there in the state $1$. By the induction hypothesis for this computation on $u_{0}$, the length of $u_{0}$ is at least $|w_{n}|$. Then the length of the entire $u$ is also at least $|w_{n}|$.

This confirms that $|w|=|u|+1+|v|\geqslant|w_{n}|+1+|w_{n}|=|w_{n+1}|$ and completes the proof. ∎