Self Doubt in Regular Language TOC

Question

L = {w | no of 0's in w != no of 1's in w and w Ɛ(0+1)*}
What is the Language L?

a. Regular

b. CFL

C. CSL

Comments

B. CFL

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B. CFL (because we need to count number of 0's and then need to compare it with no. of 1's using stack)

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Its B...

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B option is correct

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infact it is DCFL.

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Yes, because DCFLs are closed under complementation and $\overline{L}$ is nothing but language of strings having equal number of 0's and 1's

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@just_bhavana, you are correct Language will be DCFL, but $\bar{L}$ is not {$0^n$$1^n$ | n >= 0}. Here there is no specific order like all the 1's are followed by all the 0's. OR all the 0's is followed by all the 1's. 1 and 0 can occur in any order.

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Yes, { 0ⁿ1ⁿ|n≥0 } it is not able to accept   010101....   or any order of 0 and 1 with equal number of  0's and 1's

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what about {1ⁿ0ⁿ ​​​|  n>=1}

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It means there must be atleast one string  forming 10 but it still not able to from any order f 0 and 1 it will form 10,1100,111000, ..

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Diksha m not sying this is language but i was saying she miss these strings atleast since 0 and 1 are equal.

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yes, right

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yes @hemant, thanks for the correction. It will simply accept all strings over {0,1} having equal number of 0's and 1's

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Complement will n0 = n1.