I think L1 is not regular

L2 is regular

L3 is regular

L2 is regular

L3 is regular

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0 votes

check whether given language is regular or not

1) ** (a ^{n}) ^{n} **where n ≥1

2) ** ( a ^{m }) **

3) w= { **(( a ^{2 }) ^{n}) * (( a^{n })^{2} )*** } where n ≥ 1

+1

2 and 3 regular ...1st one is not regular..

Language i.e accepted by first is

{a,a^4,a^9,.......} ,as its not in AP so you cant make any FA..

second accept language as. if i consider m>=1

{a,aa,aaa,aaaa....}=a*

third accept language as

{€,a^2,a^4,a^6.......} for this you can make FA..

Language i.e accepted by first is

{a,a^4,a^9,.......} ,as its not in AP so you cant make any FA..

second accept language as. if i consider m>=1

{a,aa,aaa,aaaa....}=a*

third accept language as

{€,a^2,a^4,a^6.......} for this you can make FA..

0

can you please explain L3 briefly...

how {€,a^2,a^4,a^6.......} this series came from { **(( a ^{2 }) ^{n}) * (( a^{n })^{2} )*** }

0

@rahul rathod

when n=1 what will be the strings generated by the language

{€,a^2,a^4...}.{€,a^2,a^4...}={€,a^2,a^4...}

same way if you put n=2 you will get

{€,a^4,a^8, ... }.{€,a^4,a^8...}={€,a^4,a^8...}

so further all languages generated will be a subset of language generated by n=1, so languages generated is

{€,a^2,a^4,a^6,a^8....}

when n=1 what will be the strings generated by the language

{€,a^2,a^4...}.{€,a^2,a^4...}={€,a^2,a^4...}

same way if you put n=2 you will get

{€,a^4,a^8, ... }.{€,a^4,a^8...}={€,a^4,a^8...}

so further all languages generated will be a subset of language generated by n=1, so languages generated is

{€,a^2,a^4,a^6,a^8....}

+1 vote

L1 is not regular we will not have the string in A.P only a,a^4,a^9,......

L2 is regular i.e if m=2 a^2,a^4,a^8,.....

L3 is also regular only i.e p1= **(( a ^{2 }) ^{n}) * is regular **

p2=**(( a ^{n })^{2} )* is also regular only**

concatenation of two regular is regular only.

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