Context free or not ?

Question

$L = \{ a^i b^j c^k d^l \mid i+k=j+l \}$
Is L context free? 
If yes, then draw PDA. If no, why?

Comments

https://gateoverflow.in/183838/dcfl-or-not

Answer 1

Consider for the case  $i >= j$

Push A for each a
Pop A for each b
Push C for each c
Pop C first and then A for each d. 

Accept if stack is empty when input finishes. 

For the case $i < j$

Push A for each a
Pop A for each b until stack becomes empty
Push B for each b after that
Pop B for each c until stack becomes empty
Push C for each c after that
Pop C first and then B for each d.

Accept if stack is empty when input finishes. 

So, clearly $L$ is CFL. But we can combine the two cases and there is no guess needed making $L$ a DCFL. 

Comments

AWESOME 

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Yes well explained sir. Thank u

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difference in answer if (i=k and j=l)  given or if (i=j or j=l ) given

Answer 2

The language can be like a^n b^m c^n d^m. and cannot put in one stack. So it is not CFL

Comments

#SresthaGoel: No, it is CFL...If it had been {a^i b^j c^k d^l |i=k & j=l } then it wud be a^mb^nc^md^n but here sum must be equal...So it CFL

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Yes. DCFL.

Steps would be:

1)Push a's

2)push b's

3)For C-for every C pop B untill b's is present.If B is over push C.

4)Pop C or A for every D.

5) Check for corner cases.

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Verify this DPDA?

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right