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My textbook asks:

Consider the sets of strings on $\{0,1\}$ where the 4th symbol from the right end is different from the leftmost symbol. Construct an accepting FSA.

The answer it provides is below. However, I don't understand how you can have one transition for 0,1 and another for 1 from the same state. What does this mean? answer to the problem

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The FA in the image is an NFA as clearly a DFA cannot have more than one move for a symbol from a particular state. Also, it doesn't have any moves for state $q_5$ nor for $q_{10}$.

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  • $\begingroup$ But in state q1, why are there two options for where to go if the symbol is '1'? How does this work? $\endgroup$ – Casper C. Mar 12 at 5:02
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    $\begingroup$ Question asks to construct an accepting FSA(Finite State Automata) so it can be a DFA or a NFA, hence answer given as NFA. As it's easier to draw for this question. $\endgroup$ – simple-sober Mar 12 at 5:11
  • $\begingroup$ @CasperC. You seem to be unfamiliar with NFAs, which are FSAs precisely allowing two or more transitions from the same state with the same label. A NFA accepts a string if there is some way to move from the starting state to a final state using the input string. Note that there might be multiple paths you can take, but as long as at least one path reaches a final state, the string is accepted. $\endgroup$ – chi Mar 12 at 9:10
  • $\begingroup$ Thank you very much $\endgroup$ – Casper C. Mar 12 at 19:58

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