Geraud Senizergues (2001) proved that the equivalence problem for deterministic PDA (i.e. given two deterministic PDA A and B, is L(A)=L(B)?) is decidable. For nondeterministic PDA, equivalence is undecidable.

If we can prove this model is equivalent to deterministic PDA the problem decidable. But it can't be so difficult for 4 points :).

]]>i think i saw in the forum claims that it is decidable ( maybe even from yonathan, don't remember ),

but if it's true, i can take G that derives sigma* , and then i get a decider for the question L(G) = sigma*

and we know it isn't decidable ..

if we cannot compare 2 CFL, how is this problem decidable ?

]]>Given the following decision problem:

Input: Non-Deterministic PDA r that accepts via an empty stack

Output: Does r have the smallest number of states of all PDAs that accepts the language L(r) ?

Which class does this problem belong to ?

a. R

b. RE

c. co-RE

d. None of the above

I think we can enumerate all PDAs with less states than r, but comparing 2 CFLs isn't a decidable problem (at least due to my knowledge).

The correct answer is R, how can I compare those CFLs ?

Thanks

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