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Construct the NPDA for the language: L={w:w∈{a,b}^*,the number of a' s is at least the number of b' s }

Fulla
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2 Answers2

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Our strategy can be this:

  • input a, stack a/Z: push a
  • input a, stack b: pop
  • input b, stack a: pop
  • input b, stack b/Z: push b
  • accept if no additional input and stack a/Z

Why does this work? If we have more a than b, we end up with a on the stack. If the numbers of a and b are the same, we end up with Z on the stack. If there are more b than a, we end up with b on the stack. So we accept either a or Z on top when the input is exhausted.

q    e    s    q'   s'
q0   a    Z    q0   aZ
q0   a    a    q0   aa
q0   a    b    q0   -
q0   b    Z    q0   bZ
q0   b    a    q0   -
q0   b    b    q0   bb
q0   -    a    q1   a
q0   -    Z    q1   Z
q1   -    a    q1   -

This PDA ends in q1 with an empty stack if the input string has at least as many a as b.

Patrick87
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L = {w:w∈{a,b}^*}

So here in this case the Stack would start with z and then afterwards pushing the symbol 'a' and then popping the symbol 'b'.

And, also the minimal string would be string 'ab';

So the transitions function in this case would be as follows: 

δ(q0, a, z) = (q0, az) => a,z/az
δ(q0, b, a) = (q1, ε)  => b,a/ε
δ(q1, ε, z) = (q2, z)  => ε/z

Here the state q0 would be the start state and state q2 would be the final state.
Deep Patel
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