is it context free?

Question

is it cfl L = { a^n b^n c^m  |  n>=1 , m=2n  } ?

i did like for stack push and pop  " a & b " are coverd  and  later we are having stack empty? can we write this language as

a^n b^n c^2n  ?

Answer 1

I think a pda cannot be drawn for this language,a stack can be used to check dependency between a and b ( push for a &pop for b ) but for implementing dependency with c a tm will be needed thus it's nt a cfl

And the language is same as a^nb^nc^2n

Comments

Why can't we not go like this??
Push for n no of 'a'
Push for n no of 'b'( after changing state)
Now pop a and b for every c

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I thought of the same logic, but its wrong. Here's why:

It will also accept abbbcccc. That is, even if  #a != #b and #a + #b =  #c , your logic will hold. But we only want to accept if #a == #b.

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Thanks @weevil

Answer 2

This is not CFL as , aⁿbⁿc²ⁿ  can be written as aⁿbⁿcⁿcⁿ . So , if we use a stack , on seeing 'a' we will push into stack , on seeing 'b' , we will pop . but , then when 'c' will come , we will not have any 'a' on stack to be compared. 

So , we can not design a PDA for the language. Hence , it is not CFL.

Comments

Well that’s not the right logic, we can do something like: push ‘a’ and ‘b’ whenever we see them and start popping whenever we see ‘c’.

But that shall also accept the string “abbcccccc” where:

2 * total no. of a and b = 2 * 3  = 6 which is equal to the number of c’s

BUT no. of a’s and b’s are not equal. Hence this logic shall also be wrong.

Answer 3

It is nt cfl but this is cfl L=a^p b^q c^r where r=p+q here p may be equals to q or may not

Answer 4

Yes, this is a CFL ( deterministic one ). There are two approaches, one right other wrong. The wrong approach is to push 2 a's for every input a. This will accept languages like aabccc.  The right approach is to push a for every a, then on seeing b's pop the    'a'   and push a   'b' and thereafter for every   'c'    pop a    'b'.