如何构造 L={a^nb^m where n<=m<=2n} 的下推自动机？

Question

It should be constructed without using 2 stacks. I tried it but I couldn't do it without 2 stacks.

Answer 1

Here's the strategy: we can easily make a PDA that accepts a^nb^n, and we can easily make one that accepts a^nb^2n. Our language is a superset of these languages that also accepts anything with a number of b in between n and 2n. We can make use of nondeterminism to allow this as follows: for every a we put onto the stack, we can nondeterministically decide whether to consume one or two b before popping the a. If our NPDA chooses to consume one each time, we get a^nb^n. If it choose to consume two each time, we get a^nb^2n. If it chooses some of both, we get a number of b in between these extremes. We only accept when we exhaust the input with an empty stack.

Q    s    S    Q'    S'    Comment

q0   e    e    qA    e     Allow empty string to be accepted

q0   a    x    q0    ax    Count a and push onto stack
q0   e    x    q1    x     Transition to counting b

q1   b    ax   q1    x     Mark off a single b for each a
q1   b    ax   q2    x     Mark off the first of two b for this a
q1   e    e    qA    e     Allow string in language to be accepted

q2   b    x    q1    x     Mark off the second of two b for this a

In this PDA we have q0 as the initial state and qA as the accepting state. Processing on aabbb :

   q0, aabbb, e 
-> q0, abbb, a 
-> q0, bbb, aa
-> q1, bbb, aa
-> q1, bb, a
-> q2, b, e
-> q1, e, e
-> qA, e, e

Of course there are many parsings that don't lead to qA but in an NPDA we accept if there is at least one.

Answer 2

This image include graphical push down automata of this Language