How many equivalence classes in the RL relation for {w in {a, b}* | (#a(w) mod m) = ((#b(w)+1) mod m)}

Question

How many equivalence classes in the RL relation for

{w in {a, b}* | (#a(w) mod m) = ((#b(w)+1) mod m)}

I am looking at a past test question which gives me the options

m(m+1)
2m
m^2
m^2+1
infinite

However, i claim that its m, and I came up with an automaton that I believe accepts this language which contains 3 states (for m=3).

Am I right?

Your DFA fails to accept the string "bba" (it goes from state a2, to a3, to a1, back to a3, runs out of input and is not in an accepting state). So, your DFA is no good. — Patrick87, Jul 02 '20 at 19:38
@Patrick87 the string bba is not in the language. The language is all words with one more a than b (mod m) — Abraham Murciano Benzadon, Jul 02 '20 at 19:41

score 3 · Accepted Answer · answered Jul 02 '20 at 19:49

Actually you're right. To see this, observe that the difference of #a(w) and #b(w), #a(w) - #b(w) modulo m, is all that matters; and there are only m possible values of this difference modulo m. So, m states are always sufficient to accept a language of this form: simply make the state corresponding to the appropriate difference the accepting state.

In your DFA, a2 corresponds to a difference of zero, a1 to a difference of one and a3 to a difference of two.

How many equivalence classes in the RL relation for {w in {a, b}* | (#a(w) mod m) = ((#b(w)+1) mod m)}

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