1. L = {w | w ∈ {a,b}, na(w) >= nb(w)+1}
case 1 : first a's come and then b's
push a's
pop a's for every b's , if there is one or more than one a's still then condition satisfied
case 2: first b's come and then a's
push b's
pop b's for every a's , if there is one or more than one a's still then condition satisfied
therefore DCFL
2. L = {aibj | i ≠ 2j+1}
Here only one case - first a's come and then b's
push a's
for every b pop one 2 a's
if a single a is left then dont accpet, other everything can be accepted
Therefore DCFL
3. L = {ambn | m=2n+1}
Here only one case - first a's come and then b's
push a's
for every b pop 2 a's
if a single a is left then accept
Therefore DCFL
eg : if n=2 then 2n+1 = 4+1 =5
m=2n+1, therefore m=5
now for 2 m's one n is popped ..therefore now n=1 and m=3
again for 2 m's one n is popped....therefore now n=0 and m=1
Still one m is left therefore condition satisfied