Toc -Regular or cfl

Question

{0^p0^q+ 1^p1^q  |  0<=p<=q } Regular or Cfl?

My confusion is there is comparison involved between p and q ?

Answer 1

Yes it should be regular AND NO COMPARISON .

And its regular expression is (0*+1*)
Blindly if we see than it seems to be cfl but it is not .becoz we can generate null, any no of zeros ,any no of ones BUT ZEROS and ONES (BOTH)CAN'T BE GENERATED .

SO ITS REGULAR .

Comments

L =0^p0^q + 1^p1^q  |  0<=p<=q 

   = e + 0 + 00 + 000 + 0000....0 + 1 + 11 +111 +1111 +......

so language should be regular..

no need of comparison between p and q. 

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Yes vijaycs u are correct now edited .

Answer 2

Since comparison is there on same alphabet so there is no  need to remember anything so no stack is need .

so language is Regular. i.e. (0* + 1*)