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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.

1. 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.
2. 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

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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

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(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 $0,0.s/s$

sir, here in 0,s/0.s..

'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..
@praveen sir ,pls verify once s is content of stack or only stack symbol
nitish is right.
In the diagram of a PDA which accepts by empty stack, there's no accept state, right?
@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.

Part(b)

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

• $(q_0,0, Z) \vdash (q_0,0Z)$
• $(q_0,0, Z) \vdash (q_0,00Z)$
• $(q_0,0, 0) \vdash(q_0,000)$
• $(q_0,0,0) \vdash (q_0,00)$
• $(q_0,1,0) \vdash (q_1,\epsilon)$
• $(q_1,1,0) \vdash (q_1,\epsilon)$
• $(q_0,\epsilon, Z) \vdash (q_0,\epsilon)$
• $(q_1,\epsilon, Z) \vdash (q_1,\epsilon)$
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i dont think it is maintainning nm2n this...

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.
right sir 00001 accepted which should'nt but if i remove last 2 transition then how can i maintain empty stack property?
String will be accepted if stack is empty- not otherwise. You don't have to do anything. Non-determinism will take care of it..
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 $is reached and top of stack is s then also accept it. in the solution why the same state q0 is maintained in both the transitions?? @Arjun sir. I have a doubt here. δ(q1,∊, z0) |---- (q1,∊) . This transition is required, right? otherwise, how stack top symbol will be popped? And if it is not popped then how stack will be emptied? How it i is maintaining n ≤ m ≤ 2n this ... edited by @Puja, It is maintaining n ≤ m ≤ 2n by using the 2nd choices for the first 2 IDs (in Bold)- (q0,0,z0) |---- (q0,0z0) or (q0,00z0) (q0,0,0) |---- (q0,00) or (q0,0000) They have been used to bound the n values to a maximum of 2n, using the non-determinism as asked in the question. It should also have δ(q1,∊, z0) |---- (q1,∊) for the 2nd state for empty stack acceptance. I think the second transition must be of the form$\delta(q_0,0,0)=\{(q_0,00),(q_0,000)\}$instead of$\delta(q_0,0,0)=\{(q_0,00),(q_0,0000)\}$@Arjun Sir , I think this answer needs review this accepts string 00000011 which it should not be @saurav04 What ayush said is true... Transition in the answer is pushing 3 0's onto stack while we should be pushing only 2 0's. Your answer would've been correct if they have asked for n<=m<=3n.. Please check it Just drawing what's written in answer by @@saurav04 sir. The question asks for a 3 state PDA so this shouldn't be an acceptable answer isn't it? @In that case one can simply add another state q2 from q1 where ∊,z/∊. a) b)$\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.

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What software did you use to construct this?
This is mentioned in Peter Linz's book: http://www.jflap.org/

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