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I'm trying to prove the if $𝑀$ is a $π‘˜$-state synchronizable DFA, then it has a synchronizing sequence of length at most $π‘˜^3$. first I have to prove that if a string syncs two states, then its length is at most $k^2$.

The question was answered here.

The proof of this a standard shrinking argument: if such a word is longer than $π‘˜^2$, then during the runs from $π‘ž_1,π‘ž_2$, a pair of states repeats, and we can shrink $𝑀$.

But how to know if the repeating string in both paths from $π‘ž_1,π‘ž_2$ to some states like $q_i$ is the same? and shrinking the string won't damage the synchronization?

To make it more clear I drew the below picture.

enter image description here

The string $abcd$ takes $π‘ž_1,π‘ž_2$ to $q_i$ but after eliminating the loops, string $acd$ takes $π‘ž_1$ to $q_i$ and string $abd$ takes $π‘ž_2$ to $q_i$, thus removing the loops will break the synchronization.

Any suggestions would be greatly appreciated, thanks.

Edit: This is the visualisation I drew from Hendrik Jan's helpful answer:

Consider two arbitrary states $q_1$ and $q_2$ where $𝛿(q_1, w)=𝛿(q_2, w)=q_i$.

Suppose $|w|=k,\ w=v_1…v_k$ then: enter image description here The loop happens when $𝛿((q_1^j, q_2^j),\ v_{j+1})=(q_1^i, q_2^i)$.

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I am afraid that "a pair of states repeats" is a little ambiguous.

Both computations (from $q_1$ and $q_2$) have to be considered in parallel. So we get a pair of "synchronized" computations $(q_1,q_2), (q^1_1,q^1_2), (q^2_1,q^2_2), \dots, (q_i,q_i)$. If in such a sequence-pair a state-pair repeats we can remove the sequence in between (from both components).

That is also consistent with the $k^2$ bound, as that is the total number of state pairs.

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  • $\begingroup$ This was very helpful! thanks. I would upvote but my reputation won't let me :) $\endgroup$
    – ladypary
    Jun 16, 2021 at 19:50

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