ECA_TOC_Q10

Question

Suppose L={1^m | m=2x+4y, x,y>= 0} is any regular language over Σ = {1}.

Which of the following is the regular expression r such that L(r)=L.

(A) 1*    (B) (11)*    (C) (111)*     (D) none

Answer 1

The ans should be B. i.e. (11)* The no of 1s should always be even so in this way they can always be represented as:2x+4y where x and y are >=0

(11)*={epsilon,11,1111,11111 so on} epsilon if x=y=0

for eg: for 11 x=1 y=0 for 1111 x=0 y=1 111111x=1 y=1 and so on....

Comments

@anjali007 can't we take both x and y as 0?

then m=0

1^0=1 ......... then

 please clear my doubt..!

 

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yes it is (11)*={epsilon,11,1111,11111 so on} so both can be zero

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@BOB 1^0=1 but here the referrence is no of 1s

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@BOB it's not math in toc 1^0 = number of one is zero
                                               1^2 = number of ones are two