Question

Show L = { w | w is an element of {a, b, c, d, e, f}* such that the number of a's + number of b's = number of c's + number of d's = number of e's + number of f's } is not context-free

Answer 1

Now some important points for a language to be context free

1. It must have A PDA accepting the language

2. A PDA means push and pop operations should be clear.

3. Language can do atmost one comparison only

Now i will show how it voilates all condition

let L = aabbcdddeffe (a+b=c+d=e+f=4)

Steps Of PDA

1. Push all a's and b's into stack

2. To make sure that number of a's + number of b's = number of c's + number of d's we will have to pop the stack on each occurence of c and d as input.

3. If after all c and d in input sequence ended then stack need to be empty beacuse in this example stack contain aabb (i.e stack have 4 elements ) when  c and d come so if total c+d =4 then only it is equal to a+b then all a and b are pop out

4.If stack is empty then a+b=c+d else not

5. Now we donot know about count of a+b or c+d so now how will you check whether e+f = a+b = c+d as stack is empty

So we can only compare a+b = c+d or not as after that we lost the value (i.e 4)  and then we don't know whether to push or pop e and f

Point 1 :

So here when push and pop are not clear a PDA is not possible so Given L is not A context -free

Point 2:

I said we can do only 1 comparison but here we are 2 comparison so it not possible for a context free language

Thank You

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