UGC NET CSE | August 2016 | Part 2 | Question: 31

Question

The number of strings of length 4 that are generated by the regular expression $(0+ 1+ | 2+ 3+ )^{*}$, where | is an alternation character and ${+, ^{*}}$ are quantification characters, is :

• A. $08$
• B. $09$
• C. $10$
• D. $12$

Comments

i did not understand this question ??

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Please edit the question + is superscript on 0, 1, 2 & 3.

It is closure of numbers.

Answer 1

Given RegExp = (0⁺1⁺|2⁺3⁺)^∗

We can assume this exp as (A⁺|B⁺)^*  . We can form the string with A(01) or B(23) or both(01,23). So below are the 10 possible 4 char strings.

With 01 --> 0111, 0011, 0001, 0101

With 23 --> 2333, 2233, 2223, 2323

Using both 01 and 23 --> 0123, 2301

Hope it helps.. Thanks :)

Answer 2

I think answer is (D).

Number of sequences are as:

0001, 0011, 0111, 0123, 0122, 0012, 0112, 2223, 2233, 2333, 2301, 2330

 

 

Comments

0122, 0012, 0112, 2330 are not possible.. either 0&1 come together or they both don't come.. same in the case of 2&3.

0001, 0011, 0111, 0101

2223, 2233, 2333, 2323

0123, 2301

So, option C