Question

a^n b^n c^m d^n e^n Is it CFG?

Comments

Ok, That's the point. I was missing. Now I got it.
Thanks

---

Yeah this is absolutely correct! All we care is the no. of a,b,d and e must be equal.

First we will push a and b. Then we will skip all the c. For every d we will pop b. If no of B = no. of D then only we can see a’s on the stack and we can use them to count e. So if no of a = no. of e then we can see the stack symbol which we will use to reach the final state.

All this can be done using PDA(DPDA) hence it is CFG.

Any suggestion is welcomed😉

---

Draw a PDA for the language to validate your point.

Answer 1

I think it is cfl : push a

pop b for all a

skip c

push d

and pop e for all d

please correct me if i am wrong

Comments

According to  your logic aaabbbcddee is accepted . But language can not genrate this.

---

yes,

got it ,thanks