Push down automata

Question

Which solution is correct one?

I wish I solved it correctly (sol1)

Comments

You are correct I think

Answer 1

One Approach 

1 .If you are in state Qo and a is the input and Zo is on top of the stack then push aa instead of single a

2 .If you are in state Qo and a is the input and a is on top of the stack then push aa instead of single a

3 .If you are in state Qo and b is the input and a is on top  of the stack then get the state Q1 and  pop a 

Pop one a for one b because we have already pushed 2 aa's instead of one a .

As long as you are in state Q2 keep on poping off  a for every b.

Once the string is empty you can accept it

Second Approach

instead of pushing 2 a's push only one a in state Qo and once you see a "b" make a transition in state Q2 and for every one "a" pop off 2 b's.

(for first b don't do anything and in case of second b pop off a) Do it alternatively 

Both approach's  work well....

Answer 2

Step1:Push 'aa'  instead 'a', whenever 'a' comes on q₀(Start) .

Step2: Pop one 'a' whenever 'b' comes and change the state from q₀ to q₁.

Step3: For all remaining 'b' repeat the same procedure staying on q₁.

Step4: When ^ comes and if TOS is Z₀, than go to the accept state(End).