's' is not any symbol, it means anything that is on top of stack, no matter what is it???...beacuse if it is not so we require more transitions..

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+20 votes

A push down automation (pda) is given in the following extended notation of finite state diagram:

The nodes denote the states while the edges denote the moves of the pda. The edge labels are of the form $d$, $s/s'$ where $d$ is the input symbol read and $s, s'$ are the stack contents before and after the move. For example the edge labeled $1, s/1.s$ denotes the move from state $q_0$ to $q_0$ in which the input symbol $1$ is read and pushed to the stack.

- Introduce two edges with appropriate labels in the above diagram so that the resulting pda accepts the language $\left\{x2x^{R} \mid x \in \left\{0,1\right\}^*,x^{R} \text{ denotes reverse of x}\right\}$, by empty stack.
- Describe a non-deterministic pda with three states in the above notation that accept the language $\left\{0^{n}1^{m} \mid n \leq m \leq 2n\right\}$ by empty stack

+16 votes

**a)** $x2x^R$

Say for some word $0112110$ we have to push every thing into the stack till $2$ . then we get $1$ then $1$ will be at top of stack so pop it or if get $0$ then $0$ will at top of stack so pop it. For any word of language it is applicable. $2$ is a mark that tell now we have to pop $0$ for $0$ and $1$ for $1$.

So, on the edge $q_0$ to $q_0$ add $0,s/0.s$

and on edge $q_1$ to $q_1$ add $0,0.s/s$

+11 votes

part(b) e is epsilon(to denote pop operation)

(q0,0,z0) |---- (q0,0z0) or (q0,00z0)

(q0,0,0) |---- (q0,00) or (q0,0000)

(q0,1,0) |---- (q1,e)

(q1,1,0) |---- (q1,e)

+3

The no. of 0's in stack is between no. of 0's in string and its double (all possible due to non-determinism). Now, the string is accepted if the no. of 1's in string matches the no. of 0's on stack.

@saurav you have to remove the last 2 transitions. Otherwise strings like 00001 will be accepted.

@saurav you have to remove the last 2 transitions. Otherwise strings like 00001 will be accepted.

0

right sir 00001 accepted which should'nt but if i remove last 2 transition then how can i maintain empty stack property?

+6

String will be accepted if stack is empty- not otherwise. You don't have to do anything. Non-determinism will take care of it..

0

can we do it like this: for each 0 push two zeroes.When a 1 comes pop 2 0s.If stack becomes empty and there are still 1 left keep on reading it or in other words do nothing just read them till input gets over.Or if $ is reached and top of stack is s then also accept it.

0

@Arjun sir. I have a doubt here.

δ(q_{1},∊, z_{0}) |---- (q_{1},∊) . This transition is required, right? otherwise, how stack top symbol will be popped? And if it is not popped then how stack will be emptied?

0

@Puja,

It is maintaining n ≤ m ≤ 2n by using the 2nd choices for the first 2 IDs (in Bold)-

(q0,0,z0) |---- (q0,0z0) or *(q0,00z0)*

(q0,0,0) |---- (q0,00) or **(q0,0000)**

They have been used to bound the n values to a maximum of 2n, using the non-determinism as asked in the question.

It should also have δ(q_{1},∊, z_{0}) |---- (q_{1},∊) for the 2nd state for empty stack acceptance.

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