The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
+16 votes
02. Choose the correct alternatives (more than one may be correct) and write the corresponding letters only:

Which of the following regular expression identities is/are TRUE?

(a) $r^{(^*)} =r^*$

(b) $(r^*s^*)=(r+s)^*$

(c) $(r+s)^* = r^* + s^*$

(d) $r^*s^* = r^*+s^*$
asked in Theory of Computation by Veteran (69k points)
edited by | 729 views

r(*) = r*



pls tell the difference between above two

2 Answers

+25 votes
Best answer
(a) is the answer

(b) RHS generates $\Sigma^*$ while LHS can't generate strings where $r$ comes after $s$ like $sr, srr$ etc. LHS $\subset$ RHS.

(c) LHS generates $\Sigma^*$ while RHS can't generate strings where $r$ comes after an $s$. RHS $\subset$ LHS.

(d) LHS contains all strings where after an $s$, no $r$ comes. RHS contains all strings of either $r$ or $s$ but no combination of them. So, RHS $\subset$ LHS.
answered by Veteran (346k points)
selected by

Can you please give any source which contain  r(*)=r*.Previous year question book gave answer b

and their b is (r*s*)*=(r+s)*

search question no (XVII)

(r*s*)*=(r+s)* but in case of  (r*s*) !(r+s)* 


@Anu No. I don't have a proof. In fact I choose i only because all other options are false. In case (b) is with a *, I would choose (b) as there is ambuiguity over (a). I doubt if they meant (empty string*) with (*) in which case (a) is false.

**** Disclaimer : I am not sure that its right or wrong . So if its wrong please don`t blame me :P

   I think  r() !=  r∗  ,

let take a example A(B*) . In this case (B*) will be evaluated first because its with in the bracket . And now consider this R(*) then  (*) will be evaluated first but it contain only ∅ and  ∅*= ∈ .             So R.(∈) = R . peace 


^But why you took {}* and not $\epsilon^*$ ? That seems more appropriate. $\epsilon^* = \epsilon$, so answer would be still same.
yes answer might be same but i think ( for safety purpose i say " i think " what if i got wrong :D ) it`s not right thing to do .

suppose L= {} this language means two things either there is no final state or final state is unreachable from initial state .

And for L={∈} this language means without any input we can reach the final state ,So ∅ and ∈  is very much different . But in this case ∅*=∈*=∈ . peace

@Amitabh Tiwari 1 @Arjun Sir, here in this question mentioned in link

r(*)=r* is false, but in this question it's true. So, what is it True or False?

In this question , it is  r(*)=r*  and in your given link r(*)=r* 

moreover , here option B) is  (r*S*) = (r + s)* .. there option B) is  (r*S*)* = (r + s)* .. 

as you know  (r*S*) != (r + s)* and other options also false so option A is chosen here.

These 2 questions are Not same ..option B is different

see this is 1992 paper,  #xvii see here 

and option A) here r(*) = r 

option B, C and D all are false here .

@Bikram, makes sense, Sir! Thanks.
a is partially correct i think.. As per this question option 1 is false. Please clarify.

–1 vote
B, C, D are wrong so,answer must be A.
answered by Active (1.6k points)

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

33,687 questions
40,230 answers
38,783 users