# GATE CSE 2021 Set 1 | Question: 51

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In a  pushdown automaton $P=(Q, \Sigma, \Gamma, \delta, q_0, F)$, a transition of the form,

where $p,q \in Q$, $a \in \Sigma \cup \{ \epsilon \}$, and $X,Y \in \Gamma \cup \{ \epsilon \}$, represents $$(q,Y) \in \delta(p,a,X).$$  Consider the following pushdown automaton over the input alphabet $\Sigma = \{a,b\}$ and stack alphabet $\Gamma = \{ \#, A\}$.

The number of strings of length $100$ accepted by the above pushdown automaton is ___________

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First transition puts $\#$ on stack top. Then in $q_1$, they have a transition labelled $a, \epsilon → A$. Shouldn’t it be $a, \# → A$ as stack top has $\#$? And then there would need to be an extra transition that does $a, A → AA$.

What am I missing?
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yeah its confusing for sure but what they mean is without seeing anything on stack you can push A on reading a.

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$L = \{ a^n b^m | n > m \ \& \ n+m=100 \}$

$L = { a^{51} b^{49} ,….. , a^{99}b, a^{100}}$

$|L| = 50$

for reaching q3 from q2 there must be “A” on the top of the stack, hence $n > m$.
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In the given automata the transition from q0 to q1 pushes ‘#’ on the stack. As there is no transition from q1 that sees ‘#’ on stack top. I believe that the given automata doesn’t accept any language.

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@Mk Utkarsh can you please point to a standard problem set that has your kind of PDA interpretation (i.e. “without seeing anything on stack, push A” given in a transition diagram)? I am confused because I haven’t seen anything like this before...atleast I don’t remember. But I’m seeing a lot of people use your interpretation of PDA spec.

@mayankso I gave the same answer but not sure.
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As there is no transition from q1 that sees ‘#’ on stack top. I believe that the given automata doesn’t accept any language.

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Thanks bro. TIL
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I felt the same thing. The first transition pushes # into the stack, there are no transitions with # on top of the stack again. I also gave zero as answer. Donno how others are getting 50.

Pushdown automata accept string of

(a^n b^n | a>b and a+b=100)

Conditions to accept string is a is greater then b and a+b = 100

So only 50 string can generate by this automata

a will be (51 to 100) because a can not less then b and b (0 to 49) always less then a

Example a^99b^1 to a^51b^49

Ans is 50

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How

a>b in PDA

Please explain where I am missing
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In given PDA, for transition from q$_{2}$ to q$_{3}$(final state), it requires symbol A to be at top. It is possible only when a > b.

A detailed analysis of the given Automaton(skip if you’re comfortable with PDAs):

1. PDAs accept strings either by emptying the stack or by reaching the final state after reading the entire input tape. In this example, the transitions from initial state $q_0\to q_1$  is marked by pushing $\#$ into the stack and is the only transition that involves $\#$ and can’t be popped out again. So, there is no chance of acceptance by emptying the stack rather it’s only by reaching the final state.
2. $a’s$ can only be accepted in state $q_1$ and $b’s$ only by state $q_2$ this implies the string must be of the form $\{a^mb^n\}$ for now.
3. To read $b’s$ from the input tape, we’ve to pop-out $A$ each time which’s obtainable from accepting a’s. So, the number $b’s<a’s$ for all the $b’s$ to be accepted in this state.
4. To reach the final state we need at least one $A$ to make the transition to the final state although it’s funny about $A$ not being spent.

Note: Reading $\epsilon$ from stack doesn’t use up any of the stack contents just like $\epsilon$ transitions in a conventional FAs.

The given automaton accepts strings of the form $\{a^mb^n, \text{where }n\geq m+1\}$. And, it’s given the question that $m+n=100$.

So the possible set values in the form of $(m,n)$ are $\{(100,0),(99,1)…(51,49)\}$ which in total there are $50$. Try for $(m=50, n=50)$ to better understand why it isn’t accepted.

ago
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