GATE CSE 1991 | Question: 03,xiii

Question

Let $r=1(1+0)^*, s=11^*0 \text{ and } t=1^*0 $ be three regular expressions. Which one of the following is true?

• A. $L(s) \subseteq L(r)$ and $L(s) \subseteq L(t)$

• B. $L(r) \subseteq L(s)$ and $L(s) \subseteq L(t)$

• C. $L(s) \subseteq L(t)$ and $L(s) \subseteq L(r)$

• D. $L(t) \subseteq L(s)$ and $L(s) \subseteq L(r)$

• E. None of the above

Comments

L(r) = $1(1+0)^*$
L(s) = $1^+0$
L(t) = $1^*0$
(A) and (C) are answer!

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R=1(1+0)*.  L=1{e,0,1,01,10,00,11,...}=> L={1,10,11,101,110,100,111,..}.           S=11*0. L=1{e,1,11,111,1111,..}0 => L={1,11,111,111,1111,..}0. L={10,110,1110,11110, ...}.                                                                                                                                       T=1*0. L={e,1,11,111,1111,..}0. => L={0,10,110,1110,11110,….}.   Now see, S is subset of T, and S and T Both are subset of R.

Answer 1

Answer is A and C.

To know the answer let us check all the options:

• A. $L(s)\subseteq L(r)$:  strings generated by $s$ are any numbers of $1$'s followed by one $0$, i.e., $10,110,1110,1110,\dots$. Strings generated by $r$ are is $1$ followed by any combination of $0$ or $1$, i.e., $1,10,11,1110,101,110 \ldots$. This shows that all the strings that can be generated by $s$, can also be generated by $r$ it means $L(s)\subseteq L(r)$ is true.
  $L(s)\subseteq L(t)$: here strings generated by $t$ are any numbers of $1$ (here $1^*$ means we have strings as $\epsilon,1,11,111,\dots$) followed by only one $0$, i.e., $0,10,110,1110, \dots$. So we can see that all the strings that are present in $s$ can also be generated by $t$, hence $L(s) \subseteq L(t)$ which shows that option A is true.
   
• B. $L(r)\subseteq L(s)$: this is false because string $1$ which can be generated by $r$, cannot be generated by $s$.
   
• C. Same as option A.
   
• D.  $L(t) \subseteq L(s)$: this is false because string $0$ which can be generated by $t$,cannot be generated by $s$.

Comments

Both 's' and 't' looks same..

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t can generate only zero while S cannot thus T is superset of S and not subet .Thus only c is correct.

 it should be strictly subset in the all cases above.

please correct if i am wrog? @ArjunSir

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@ankyAS Options A and C are same so how only C can be the answer? 

 

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$ L(r) \supseteq  L(t) \supseteq L(s) $.

Answer 2

If we take simpler strings,

t accepts 0, s doesn't. But both accept 10. So s is a subset of t.

r accepts 1, s doesn't. But both accept 10. So, s is a subset of r.

So, options A and C, as both are the same.

Answer 3

option A and C are the same and are the answers as well.

Answer 4

L(s) ⊆ L(r), because 'r' generates all strings which 's' does but 'r' also generates '101' which 's' does not generate.
L(s) ⊆ L(t), because 't' generates all the strings which 's' generates but 't' also generates '0' which 's' do not generates.