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Is L={1n0n0m | m+n>100} a regular language?

edited | 108 views
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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.
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I think it is a regular language because 0n0m will makes it regular

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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.
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If we talk about complement of given Language that would be L' ={1n0n0m | m+n<=100}

Now it is regular and we know that regular is closed under regular . That why ,I think language should be regular .

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Mishra how its complement becomes regular??
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Since it is finite  . but please varify what i said about complement of language .
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Its complement is {(0+1)*-L}

Exactly extracting what is L complement is little difficult..but it is definitely not like m+n<100

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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.
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Yes you are right.

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

:)

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@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

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Mayankprakash,

If m+n<100 then there is boundation, So we can find all combinations of m and n and we can draw corresponding FA
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m+n<100  regular

m-n<100  not regular

@verma ashish  am i right?

correct me if i am wrong
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Yes m-n<100 ==> m<100+n

This is not regular