为以下语言创建下推自动机

Question

So I was doing practice problems in my book and I spotted this question. Construct an npda accepting the language L on sigma(a,b,c).

L={w: number of a= number of b+1}

so I am interpreting it as it accepts all strings that has one more a then the letter b. I believe that all the states should have a loop that has a transition (c,landa, landa) since we do not really care about the c's. After this I get really confused because there are so many cases to cover since the placement of a's and b's are arbitrary. What is the way to get this problem figured out? Thanks!!

Answer 1

A PDA can use a stack to remember arbitrary amounts of information. This makes PDAs infinitely more capable than finite automata. The key to determining the PDA is figuring out how the stack will be used and then building a PDA around that.

How can we use a stack to ensure the number of a s is equal to the number of b s, plus one? Well, the stack can easily keep track of the running balance of symbols that have been seen. For instance, if we have seen four a s and two b s, our stack might represent this fact by containing aaZ , where Z is the "bottom of stack" symbol. Of course, there are other methods we might use and other representations, but this is a particularly neat one for this class of problem. To fully explain the representation:

1. The stack is initially Z , just the bottom of stack symbol.
2. If we see an a and the top of the stack is a or Z , we add another a .
3. If we see an a and the top of the stack is b , we remove one b .
4. If we see a b and the top of the stack is b or Z , we add another b .
5. If we see a b and the top of the stack is a , we remove one a .
6. If we see a c , leave the stack alone.

If we do this over and over again for all the input, then the content of the stack will be equal to x^m , where x is whichever of a and b occurs more frequently, and m is the absolute value of the difference of the numbers of each symbol.

To accept your language, you must simply recognize the case where the input is exhausted and the stack consists is equal to aZ . This can be done by adding some state(s) and lambda/epsilon transitions to clear the stack and/or enter an accepting state.

Thanks to Peter Leupold for pointing out that the rest of the original answer got the grammar wrong. I made an attempt to fix it and didn't like how long the answer was getting, so I omitted that. I will simply add that another possibility is to produce a CFG for a language and use an algorithm to derive a PDA for it. In this case, for me, giving the PDA directly was a lot less wordy.