in Theory of Computation edited by
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56 votes
56 votes

The regular expression $0^*(10^*)^*$ denotes the same set as

  1. $(1^*0)^*1^*$
  2. $0+(0+10)^*$
  3. $(0+1)^*10(0+1)^*$
  4. None of the above
in Theory of Computation edited by
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4 Comments

P(QP)* = (PQ)*P
here P= 0*, Q= 1
0*(10*)* = (0*1)*0*
2
2
010 is generated by the expression in the question but how it will be generated by option a
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1
areey take (1*0)*1*   as   (1*0)(1*0) = (0)(10) = 010...tadaaa

Inshort (1*0) taken 2 times and 1* taken as Epsilon.
1
1

12 Answers

1 vote
1 vote
B) can not generate 01 and

C) can not generate 00

so option A is the correct answer
1 vote
1 vote
Can we do it like this,

In the given RE,

0*(10*)*

If we put *= epsilon, it generates nothing.

But in option A and C if we put * =epsilon, they generate 0 and 10 respectively ,

So, we can say that option A is correct
0 votes
0 votes
Given regular expression is  0*(10*)*

A: (1*0)*1*
All strings that can be generated from given regular expression
can also be generated from this.

B: 0 + (0 + 10)*  and C: (0 + 1)* 10(0 + 1)*
We can generate  11 from given regular expression which is not 
possible with B and C

C: (0 + 1)* 10(0 + 1)*
Not possible as we can produce {epsilon} from the given Regular 
Expression but not from C
0 votes
0 votes

option (A) and the given expression generates string 1

but option b anc c is not genrate 

2 Comments

option (A) and the given expression generates string 1

But only by comparing single string we can't declare that two regular expressions are same.

0
0
we can declare by only one string also
–1
–1
Answer:

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