I think the language is not regular as it's infinite but if the case would be m+n<100 then it would be regular.

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I think the language is not regular as it's infinite but if the case would be m+n<100 then it would be regular.

+1

L={${1^n}$ ${0^n}$ ${0^m}$} ==> {${1^n}$ ${0^{n+m}}$ ; n+m>100}

I think it is regular language...

PS:: now i think it is not regular as here no of 1's should be less than or equal to no of 0's which require comparison.

I think it is regular language...

PS:: now i think it is not regular as here no of 1's should be less than or equal to no of 0's which require comparison.

+1 vote

language which require comparison or storing of value is not regular.

its not regular as it require comparison for equality check between 1^n0^n

here no of 1's should be less than or equal to no of 0's which require comparison.

if the condition is m+n<100 in that case this will become finite hence regular.

its not regular as it require comparison for equality check between 1^n0^n

here no of 1's should be less than or equal to no of 0's which require comparison.

if the condition is m+n<100 in that case this will become finite hence regular.

0

Yes you are right.

here no of 1's should be less than or equal to no of 0's which require comparison.

:)

0

@cyberscam

if the condition is m+n<100 in that case this will become finite hence regular.

How it will be regular ?..here also we r doing compariosio noe

Please suggest

if the condition is m+n<100 in that case this will become finite hence regular.

How it will be regular ?..here also we r doing compariosio noe

Please suggest

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