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| 395 views

$1.L_{1}=0^{*}10^{*}10^{*}$

$2.L_{2}=aaaa^{*}\left ( bb \right )^{*}$

$3.L_{3}=\left ( 0+1 \right )^{*}00\left ( 0+1 \right )^{*}+\left ( 0+1 \right )^{*}11\left ( 0+1 \right )^{*}$

$4.L_{4}=b^{*}(ab^{*}ab^{*}ab^{*})^{*}$

by Boss (15.9k points)
edited by
+1
For $L_4$, b is not accepted.
0
sir $L_{4}=(b^{*}ab^{*}ab^{*}ab^{*})^{+}?$
0

can we write L1 as (110* + 0*110*+ 0*10*1+10*10*) ?

I know L1=0*10*10*  covers it , but still is above is also correct?

0
No, that also won't accept b rt? We should allow no a's because 0 is divisible by 3.
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Sir , i am not getting you .

If you are saying that my $L_{4}$ should accept $b$,then my $L_{4}$ is actually accepting $b$

$L_{4}=b^{1}(b^{*}ab^{*}ab^{*}ab^{*})^{0}b^{0}$

Actually i too though that $n(a)=0$ should be accpeted as 0 is divisible by 3
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yes, sorry I missed that. But you can avoid b* at end.
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okk sir , thank you :)
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thank you Sir
0
for L4 why not   b*(ab*ab*ab*)*   ?
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@diksha , yes you are correct.Thanks
0

L4 can also be expressed as (b + ab*ab*a)*

+1 vote
solution

Ans1.   0*10*10*
by (129 points)

.........

by Boss (34.4k points)
edited

1