[Solved]: Synchronizing sequence and Synchronizable DFA

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Problem Detail: I am trying to prove problem 1.59 in Sipser’s book: Introduction to the theory of computation , 2nd Edition.

Let $M=(Q,Sigma,delta,q_0,A)$ be a DFA and let $q’$ be a state of $M$ called its “home”. A Synchronizing sequence for $M$ and $q’$ is a string $sin Sigma^*$ where $delta (q,s)=q’$ for every $qin Q$. (We actually have extended $delta$ to strings so that $delta(q,s)$ equals the state where $M$ ends up when $M$ starts at state $q$ and reads input $s$). Say that $M$ is Synchronizable if it has a synchronizing sequence for some state $q’$. Prove that, if $M$ is a $k$-state synchronizable DFA, then it has a synchronizing sequence of length at most $k^3$. Moreover, can you improve upon this bound?

I’m more interested in proving that the synchronized sequence is of length of at most $k^3$ then trying to improve upon this bound. I tried to prove (with no success) that there exists $win Sigma^*$ which $|w| leq k^2 $ so that: $delta(q_1,w)=delta(q_2,w)$ for two distinct states in $M$: $q_1,q_2in Q$ (thus, $w$ can be read from two states in the automaton and get to the same final state).
If I prove it, I could construct a word $w$ which will be a synchronizing sequence in $M$ and $|w|leq k^3$ as required. Any suggestions?

Asked By : DaniaG.

Answered By : Shaull

I think you are on the right track: Consider two states $q_1,q_2$. We claim the following: If there exists a word $w$ such that $delta(q_1,w)=delta(q_2,w)$, then there is such a word $w$ of length at most $k^2$. The proof of this a standard shrinking argument: if such a word is longer than $k^2$, then during the runs from $q_1,q_2$, a pair of states repeats, and we can shrink $w$. Now, since you assume the existence of a synchronizing word for all states, you can proceed to construct a word that synchronizes all the states, pair by pair.
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Question Source : http://cs.stackexchange.com/questions/18692