PDA Examples
1. an+mbncm n,m>=1
it is not regular language but DCFL as well as CFL
2. ambn+mcn n,m>=1
it is not regular language but DCFL as well as CFL
3. ambncn+m n,m>=1
it is not regular language but DCFL as well as CFL
4. ambmcndn n,m>=1
it is not regular language but DCFL as well as CFL
5. ambncmdn n,m>=1
it is not regular language as well as not CFL because by the time you reaches c's there will b'son the top of the stacks ,you can't match a's
there fore if a language is not CFL then that willl not beDCFL or regular.
6. ambncndm n,m>=1
it is not regular language but DCFL as well as CFL
7. ambicmdk i,j,m>=1
it is not regular language but DCFL as well as CFL
8. 8. ambn / m>n
it is not regular language but DCFL as well as CFL
here push a's and when we see b's pop a's and then at least there should be one a should remain and go to final state.
8. anb2n / n>=1
here for every "a" push two a's and when see any b's pop a's.
it is not regular language but DCFL as well as CFL
9. anbn^2 / n>=1
it is not regular language
i can't put bn^2 in loop and i can't popup for one loop ,one a's
therefore it is not possible to give any PDA either deterministic or Non deterministic.
∴ not CFL
10. wwr /w∈(a,b)*
not regular not DCFL but CFL(NDCFL)
11. anbncm /n>m
not regular
for such language PDA is not possible because when you pop all a's using b's then at that time only c's will be remain and there will nothing to compair with c's i.e(m).
12. anbncndn /n<=1010
since language is bounded (i.e finit ) and every finite language is regular therefore DCFL that means CFL.
13. XcY / X,Y(0,1)*
it is nothing but set of all string containing c as a sub string
so it is regular
therefore DCFL that means CFL.
14. aibjckdl /i=k or j=l
CFL as well as DCFL
15. w /w∈{a.b} ,na(w)=nb(w)+1
DCFL as well as CFL