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Match the following NFAs with the regular expressions they correspond to:

1. $\epsilon + 0\left(01^*1+00\right)^*01^*$

2. $\epsilon + 0\left(10^*1+00\right)^*0$

3. $\epsilon + 0\left(10^*1+10\right)^*1$

4. $\epsilon + 0\left(10^*1+10\right)^*10^*$

1. $P-2, Q-1, R-3, S-4$
2. $P-1, Q-3, R-2, S-4$
3. $P-1, Q-2, R-3, S-4$
4. $P-3, Q-2, R-1, S-4$
edited | 2.7k views
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S is not accepting 01011010. How is c possible?
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Please someone provide reference for conversion of FA to R.E.
+2
you can get the correct answer by looking at end of Final state incoming edges:

Incoming edges to FS for FA is

P.  01*

Q.  0

R. 1

S. 10*

P-1

Q-2

R-3

S-4

But if two FA having same incoming edges in that case you have check for remaining part of Regular Expression :)

$S-4$ is confirmed

$R-3$ is true coz everything it accepts ends with $1$; this is made mandatory only by $3$
this rules out option B and option D

use string $01010$ and compare $P$ Vs $Q$; this makes $Q-2$ as confirmed.

Hence, option C is correct.

edited
+2
1.   0010 should be accepted , only P is possible for it...

2. 0110 should be accepted , only Q is possible for it...

4 . 010 should be accepted.. Only possible by S.

So option C is Right..
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the minimum value accepted by P is  0

but that isnt accepted by any of the options

where did i go wrong ?
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The minimum value accepted by p is not 0. It's empty string.

Moreover p doesn't even accept "0".

Second minimal length string is "00".
0
@nymeria

in the diagram

only on input 0 it goes to accepting state right ?

to accept empty string the start state must be an accepting state , which is not so
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@A_i_\$_h In case of P, start state is the accepting state too.

Lets go step by step by step, eliminating the given options:

Min string accepted by P is 00 , therefore 3, and 4 cant be the RE for it as they say 01 must be included.

Min string accepted by Q is 00 , therefore again 3, 4 cant be RE for Q for the same reason mentioned above.

So by now, we left only 2 options, answer has to be A or C.

2nd min accepted by P is 001, but RE in 2 says string start and end with 0. Therefore RE for P is 1 making 2 as RE for Q.

0
nice!
Correct Ans is (C)

Trace the given regular expressions with the diagrams
for these kind of qst its better to go with trail and error (i.e by taking a valid string and by checking each FA but not to go with method) bcz this method takes lots of time which will kill around 5 mins

and i did like that (trail and error)
it takes
me only 2 mins.

OPTION IS C
edited

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