Let n be an integer. How can I write finite automata for the language L?
L = {W∈{$0,1,2$}*| The $n_{th}$ from last letter in w is $0$}.
(Please suggest answers; not hints.)

Using Regex, I could write it as $[012]*0[012](n - 1)$ - replacing n - 1 with the actual integer value. But I'm not sure of a good program to diagram it with.


1 Answer 1


It is possible to prove that such a DFA would always have at least $2^n$ states (the idea of the proof is given here, but for a two-letters alphabet).

This bound can be reached: consider $A = (Q, \delta, q_0, F)$ where:

  • $Q = \{0,1\}^n$: to each state corresponds a word of length $n$ over the alphabet $\{0,1\}$;
  • $q_0 = 1^n$;
  • $F = \{0u\mid u\in \{0,1\}^{n-1}\}$: states that begin with $0$;
  • for $u\in \{0,1\}^n$, $u = u_1…u_n$, $\delta(u, 0) = u_2…u_n0$ and $\delta(u,1) = \delta(u, 2) = u_2…u_n1$.

The idea is that each state tells you what are the last $n$ letters you read (at least for words of length $\geqslant n$), with $2$'s being associated with $1$'s (we just need to know the positions of $0$'s).

Proof left to you that $L(A) = \{w\in\{0,1,2\}^*\mid w = u0v, \text{ with }|v| = n-1\}$.

  • $\begingroup$ Say, how can I write this as an NFA if I wanted? @Nathaniel? $\endgroup$
    – NiStack
    Commented Apr 6, 2023 at 20:01
  • 1
    $\begingroup$ @NitayStack Add a cycling transition with any letter on the initial state, then a transition labeled with $0$ to the second state, then a line of $n-1$ states, each one with a transition to the following, labeled with any letter, the last one being the only ending state. $\endgroup$
    – Nathaniel
    Commented Apr 6, 2023 at 20:36

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