TEST SERIES

Question

As we know that, PDA=Finite Automata + one Stack. Assume a modified PDA M1 in which, in place of stack if we use counter (where the counter is able to count the number of occurrences of terminals in input string), i.e., PDA M1 = Finite Automata + one counter.
Consider the given below languages:
L1={a^nb^n |n>0}
L2={wwr| w ϵ {a,b}*}
L3= {w | na(w) = nb(w)}
Select the correct option.

 

I think the answer is L1 and L3, But the answer given is only L3.

in L1 we can maintain counter it will increment for occurrences of a, and when b starts it will decrement and after seeing first "b" if any "a" comes then machine will goto dead state.

Comments

Not very sure bt probably because of this

say the string is abbbaa

Now, na(w) = nb(w) here

But, for a if you increment counter, and for b you decrement, the counter will go 0 after ab and won't decrease later for following occurrences of b

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if in counter you can increment and decrement then it is nothing but stack with increment as push and decrement as pop so FA+ stack = PDA.