Push-down automaton for a language with number of 1s double than 0s

Question

This questions is taken from a complexity theory course.

Design a push-down automaton that recognize $L = x \in \{0,1\}^* $ such that $x$ contains a number of ones double than a number of zeros.

Example: x = 011011, 100011111, 011011110.

My approach

I tried to push on the stack a symbol "A" every time I read "0", than pop "A" from the stack when I read two consecutive "1" but this approach works not for every string $x$ in the language. Maybe the final PDA is non-deterministic. Any hints?

score 1 · Answer 1 · answered Nov 05 '17 at 02:00

Consider this algorithm:

Initialize a variable $x := 0$.
Scan the string, left to right. If you see $1$, set $x := x + 1$, if you see $0$, set $x := x - 2$.
When you are done, accept iff $x = 0$.

It should be obvious that it recognizes your language. Can you figure out a way to implement it replacing the counter with the stack?

score 1 · Answer 2 · answered Jan 14 '18 at 19:55

Here's an algorithm for what I think could be a solution:

program_recL(entry) is
    stack: type_stack
    while read(entry) != $ do
        if read(entry) == 0 then
            if empty(stack) then
                push(stack, 0)
                push(stack, 0)
            else
                if top(stack) == 0 then
                    push(stack, 0)
                    push(stack, 0)
                else
                    pop(stack)
                    push(stack, 0)
                endif
            endif
        else
            if empty(stack) then
                push(stack, 1)
            else
                if top(stack) == 1 then
                    push(stack, 1)
                else
                    pop(stack)
                endif
            endif
        endif

        next(entry)
    endwhile

    if is_empty(stack) then
        accept()
    endif
end

The basic idea here is that every 0 has x2 value for each 1, meaning we need to push two 0s every time we read a 0 on our entry. This will translate into needing two 1s to get a 0 popped from the stack, in other words, a number of 1s double than a number of 0s.

user79739 · Answer 3 · 2017-11-05T02:02:52.773

-2

You need a 2-stack PDA to solve this problem.

Since the language you've given is non-context free, it's more powerful than a single stack PDA. More specifically, your language the union of many CFLs and non-CFLs. The inputs that seem to be working with your way can be recognized by a single stack PDA. e.g.: S -> 0A11 A -> S which is the language {0^n1^2n, n>=0} S -> 11A0 A -> S which is the language {1^2n1^n, n>=0} . . . However, there is no way to describe others with a context-free grammar.

edited Nov 05 '17 at 02:02

answered Nov 05 '17 at 01:56

user79739

17
2

1

No, you don't need a "2-stack PDA" (friends call him a Turing Machine, by the way). Consider this grammar: $S \rightarrow \varepsilon$, $S \rightarrow S0S1S1S$ and cyclic variants. One arrow is obvious, the other is structural induction. – quicksort Nov 05 '17 at 02:17

Push-down automaton for a language with number of 1s double than 0s

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