Pushdown Autometa


Design NPDA for language L = {a^n b^2n |n>=1}


this is a classic problem just like the PDA for language L={a^nb^n|n>=1}
here we can form a PDA where in the stack we will PUSH a ‘x’ for every single a and for b, we will take alternate 2 b’s where for the first b we can just skip PUSHor POP operation from the stack and POP a ‘x’ from the stack for the alternate 2nd b.So, basically we willbe able to count 2 b’s for a single ‘a’ and just the language L={a^nb^2n|n>=1} can be satisfied.