Lecture 3

Another example
If L is regular, so is
Cycle(L) = {a2a1a4a3..a2na2n-1 | a1a2a3a4..a2n-1a2n ∊ L}
Let M be a DFA for L.
M’ will be a DFA for Cycle(L) that uses a state to “remember” what symbol (a) it just
read.
Then, when it reads the next symbol b, M’ will simulate M on input ba. Formally
Q’ = Q ∪ (Q×Σ)
δ’(q,a) = (q,a)
δ’((q,a),b) = δ(q,ba)
Prove by induction on n that
δ’(q,a2a1a4a3..a2na2n-1) = δ(q,a1a2a3a4..a2n-1a2n)
Clearly true when n=0 since δ’(q,ε) = q = δ(q,ε)
Assume true for n-1, so δ’(q,a2a1a4a3..a2n-2a2n-3) = δ(q,a1a2a3a4..a2n-3a2n-2) = p, say
Then δ’(q,a2a1a4a3..a2na2n-1) = δ’(δ’(q,a2a1a4a3..a2n-2a2n-3), a2na2n-1)
= δ(p, a2n-1a2n) = δ(δ(q,a1a2a3a4..a2n-3a2n-2), a2n-1a2n) = δ(q,a1a2a3a4..a2n-1a2n)
We have proved that a2a1a4a3..a2na2n-1 is accepted by M’ if and only if
a1a2a3a4..a2n-1a2n is accepted by M. So Cycle(L) = L(M’) is regular.
Friday, February 8, 13