如何构造n(a)小于或等于2n(b)的PDA

Question

所以这就是我卡住的地方，我必须构建一个 PDA 来接受来自 {a,b}* 的单词，条件 n(a) 小于或等于 2n(b)

Answer 1

Put yourself into the frame of mind of a pushdown automaton. All you know how to do is read input and the stack, and then push/pop and change state based on what you see. If you start reading a string and need to tell whether it's in the language, what can you do?

It seems to me the best we can do starting out is to remember how many a s we are reading. Until we see b s, there's no reason to do anything else. That is, simply read a s and push them on the stack. The following rules should suffice:

Q    i    s    Q'    s'
--   --   --   --    --
q0   a    Z    q0    aZ
q0   a    a    q0    aa

Now, what happens when we see a b ? If we want at least twice as many b s as a s, we can't just cross of a s for each b , since that would give at least as many b s as a s but not at least twice as many.

What if we cross off two a s for each b ? Well, that gives us at least half as many b s as a s, which is the wrong direction. This suggests crossing off one a for every two b s, which turns out to be correct. The rules:

Q    i    s    Q'    s'
--   --   --   --    --
q0   b    a    q1    a
q1   b    a    q2    -
q2   b    a    q1    a

Note we created a new state q2 which has one rule in common with q0 but not the ones for reading a s. Indeed, once we start reading b s, we don't want to allow any more a s to be read. Leaving out the rules will crash the automaton and reject the string.

If we have exactly twice as many b s as a s, we will end up in state q2 with no more input and an empty stack. If we have additional b s, we need a rule to allow them. It suffices to add a self loop on state q2:

Q    i    s    Q'    s'
--   --   --   --    --
q2   b    Z    q2    Z

Taken together, these states and transitions should be pretty close to a working PDA for your language. Other options would have been to write a CFG first and then apply an algorithm to convert the CFG to a PDA, for example, a top-down or bottom-up parser.