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I've tried quite a large amount of examples, starting from 1 bit all the way up to 5 bits yet I couldn't put my finger on any recurring pattern of words that are accepted.

the DFA is as such:

the dfa

The only language I could come up with that sort of covers this, is:

$L=\{w: \#_1(w)\geq 1 \; \& \; ends\;with\;even\;no.\;of\; '0'\;or\;ends\;with\;'1'\}$

yet this formulation for $L$ seems really off to me.

is there a better way to formulate this? or perhaps my answer is incorrect?

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1 Answer 1

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Your answer is correct. $L = \{ w \; | \; w \text{ has at least one 1 and ends with an even number of 0s or ends with a 1 } \;\}$

Sketch of Proof in one direction: Because there's at least one 1, somewhere we find a 1 and get to $q_2$. To end in $q_2$, either the last word was a 1; or it was zero and last state was $q_3$. Everytime we're in $q_3$, we have an odd number of zeros at the end.

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  • $\begingroup$ This is an answer I went through, but the “length” of the definition felt odd to me, though I see the logic behind this definition. I suppose this is the correct answer indeed, thank you! $\endgroup$
    – Aishgadol
    Commented Jan 3 at 7:50
  • $\begingroup$ Its the same answer the asker suggested. $\endgroup$ Commented Jan 3 at 8:18

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