L = {a^nb^m : m ≥ n, mn 为偶数的 PDA

Question

Design a PDA for the following language

L = {a^nb^m : m ≥ n, mn is even}.

Answer 1

Let's start with a PDA for a^nb^m where m >= n. A PDA can push an a to the stack for every a it sees, pop ab for every b it sees, and if it runs out of b while there are still a on the stack, it rejects.

Now, what else do we need to do to exclude the case where m - n is odd? Well, m - n is odd means we have some b left over in the input. We can simply modify our accepting state so that on further b, it moves to a new state (encoding odd b) and then back to the accepting state on the next b, encoding the requirement that the residual b must be even.

A full PDA might look like this:

q    s    S    q'    S'
q0   a    Z    q0    aZ
q0   a    ax   q0    aax
q0   b    Zx   q2    Z
q0   b    ax   q1    x
q1   b    ax   q1    x
q1   b    Z    q2    Z
q2   b    Z    q1    Z

Check that to see if it works, there might be some bugs. The way to read this is:

From state q, on input s, with stack configuration S, the PDA can transition to state q' and update the stack configuration to S'.