GATE CSE 2000 | Question: 8

Question

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.

• A. 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.
• B. 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

Comments

A Thing to note here  is  s is anything  or like any symbol on the top of the stack so whatever the top of the stack .S represents in that way not any special symbol of the language

 

AND

the transition on qo is    1,s/1.s 

Answer 1

(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 $0,0.s/s$

Comments

nitish is right.

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In the diagram of a PDA which accepts by empty stack, there's no accept state, right?

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@shraddha there should be one more transition on q2, eps| stackSymbol| eps, but seems it's some hypothetical PDA and they don't have any stack symbol at the bottom of the stack. so once input is a member of the language and when input is finished stack will be empty.

Answer 2

Part(b)

($ \epsilon$ is used to denote pop operation, $Z$ is the starting symbol on stack)

• $(q_0,0, Z) \vdash (q_0,0Z)$
• $(q_0,0, Z) \vdash (q_0,00Z)$
• $(q_0,0, 0) \vdash(q_0,000)$
• $(q_0,0,0) \vdash (q_0,00)$
• $(q_0,1,0) \vdash (q_1,\epsilon)$
• $(q_1,1,0) \vdash (q_1,\epsilon)$
• $(q_0,\epsilon, Z) \vdash (q_0,\epsilon)$
• $(q_1,\epsilon, Z) \vdash (q_1,\epsilon)$

Comments

i dont think it is maintainning n≤m≤2n this...

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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.

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

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

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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.

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in the solution why the same state q0 is maintained in both the transitions??

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@Arjun sir. I have a doubt here.

 δ(q₁,∊, z₀)   |---- (q₁,∊) . 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?

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How it i is maintaining n ≤ m ≤ 2n this ...

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@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(q₁,∊, z₀)   |---- (q₁,∊) for the 2nd state for empty stack acceptance.

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I think the second transition must be of the form

$\delta(q_0,0,0)=\{(q_0,00),(q_0,000)\}$ instead of $\delta(q_0,0,0)=\{(q_0,00),(q_0,0000)\}$

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@Arjun

Sir , I think this answer needs review this accepts string 00000011 which it should not be

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@saurav04

What ayush said is true...

Transition in the answer is pushing 3 0's onto stack while we should be pushing only 2 0's.

Your answer would've been correct if they have asked for n<=m<=3n..

Please check it

 

 

 

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Just drawing what's written in answer by  @@saurav04 sir.

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The question asks for a 3 state PDA so this shouldn't be an acceptable answer isn't it?

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@avraw In that case one can simply add another state q2 from q1 where ∊,z/∊.

Answer 3

a)

b)

$\lambda$ in the stack part is used to indicate "whatever be the input". And the additional $(\lambda,Z,\lambda)$ was added to pop the initial symbol from the stack.

Comments

What software did you use to construct this?

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This is mentioned in Peter Linz's book: http://www.jflap.org/

Answer 4

a   

on the state q0 edge from q0 to q0 add 0,s/0.s0,s/0.s
and on state q1 edge from q1 to q1 add 0,0.s/s

 

 

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