91影视

a)

Firstly, if a string $x=x_1{x}_2路路路x_t$ of length t is accepted by M , and there exists a sequence of states$,q_{0}i,q_{1}i,...,q_{i}tsuchthat{q}_{0}i=q_0,q_{i}j=未\left(q_{i}j-1,xj\right),and{q}_{i}t鈭�F.$This implies that x can be accepted by machine one based on the sequence of states . $,q_{0}i,q_{1}i,...,q_{i}tsuchthat{q}_{0}i=q_0,q_{i}j=未\left(q_{i}j-1,xj\right),and{q}_{i}t鈭�F.$Thus, Secondly, if a $stringy=y1y2路路路xt$of length t is accepted by machine of its state that follows the provided algorithm.

let $\left[qi0\right],\left[qi1\right],...,\left[qit\right]$ be the set of $,q_{0}i,q_{1}i,...,q_{i}tsuchthat{q}_{0}i=q_0,q_{i}j=未\left(q_{i}j-1,xj\right),and{q}_{i}t鈭�F.$states such that $,q_{0}i,q_{1}i,...,q_{i}tsuchthat{q}_{0}i=q_0,q_{i}j=未\left(q_{i}j-1,xj\right),and{q}_{i}t鈭�F.$

By induction, we can show that when y is input to M, the corresponding sequence of states visited by $M,say{r}_0,r_1,r_2,...,r_t,willsatisfy{r}_0=q_0,andrj鈭�\left[qij\right]forallj.$ Now,$as\left[qit\right]鈭�F0and\left[qit\right]鈯�F.$

Thus$,r_t鈭�F,$ so that $L\left(M_0\right)鈯�L\left(M\right).$

In conclusion, $L\left(M\right)=L\left(M_0\right),$so that$Mand{M}_0$ , $M,say{r}_0,r_1,r_2,...,r_t,willsatisfy{r}_0=q_0,andrj鈭�\left[qij\right]forallj.$are equivalent.

Hence, M and M^' are equivalent is proved..

b)

Let $未\left(q_0,x\right)$ denote the state of M after reading X when M starts from q₀ .

By induction, we can show that for two distinct states q and r in the undirected graph, q and r are connected by an edge if and only if there exists strings $未\left(q_0,x\right)=q,未\left(q_0,y\right)=r$such that $未\left(q_0,x\right)=q,未\left(q_0,y\right)=r$, x and y are distinguishable by localid="1660807511606" $L\left(M\right).$

Based on the above result [q] will store the all states q₀ such that for all x and y with $未\left(q_0,x\right)=qand未\left(q_0,y\right)=q_0,xandy$are indistinguishable. Also, for any $q_0鈭�\left[q\right],$we observe that $q_0=\left[q\right],$.

In other words, $M,say{r}_0,r_1,r_2,...,r_t,willsatisfy{r}_0=q_0,andrj鈭�\left[qij\right]forallj.$

The distinct set of states [q] in Q₀ forms a partition of Q. By picking one string x with $未\left(q_0,x\right)鈭�q$for each distinct [q], the resulting $\left|Q_0\right|$ strings are pair wise distinguishable by L(M) .

By Myhill-Nerode theorem, any DFA recognizing must have at least [q] states. And it has $\left|Q_0\right|$ states and $L\left(M\right)=L\left(M_0\right),M_0$ is a minimal.

Hence, No DFA with fewer states recognizes the same language is proved.

c)

Let $\left|Q\right|=n.M,say{r}_0,r_1,r_2,...,r_t,willsatisfy{r}_0=q_0,andrj鈭�\left[qij\right]forallj.$.

Step one is to takes at most O$\left(n3+n2\left|危\right|\right)time$,

By using the brute force connectivity algorithm. For Step three, it takes $O\left(n^2\right)$ time. For Step four, it repeats Steps five and six for at most $O\left(n^2\right)$times, each repetition takes at most O(n 2 |危|) time. For Step seven, it is done in localid="1660123076236" $O\left(n^2\right)$time. For Step eight, we first check if $\left[q\right]=\left[r\right]$for each pair of , which requires $O\left(n^3\right)$time; this naturally gives a partition of Q' s states, and the partition can be stored in a table; then, we construct the final deterministic finite automaton, which takes an additional $O\left(n^2\left|危\right|\right)$ time. In total, the time required for constructing deterministic finite automaton, is $O\left(n^4\left|危\right|\right)$, which is polynomial in the length of the input.