# JUNE 2013 - Paper 2

1:

Match the following:

a. Good quality              i. Program does  not fail for a
specified time in a given
environment'
b. Correctness              ii. Meets the functional
requirements
c. Predictable                iii. Meets both functional and
non-functional requirements ,
d. Reliable                    iv.  Process is under statistical
control

Codes
a   b   c  d
 A. iii     ii      iv     i B. ii      iii     iv      i C. i     ii     iv     iii D. i      ii     iii      iv Answer Report Discuss Option: B Explanation : "Quality" should be used as a measure of functional/aesthetic utility to our consumer, and not as a measure of defects. Really, it should just be assumed that defects are generally absent. This should just be implied in what it means to be a Professional.  So answer is 'B Click on Discuss to view users comments. Write your comments here:
2:
When the following code is executed , what will be the value of x and y ?

int x =1, y= 0;
y=x++;
 A. 2,1 B. 2,2 C. 1,1 D. 1,2 Answer Report Discuss Option: A Explanation : Here y will be assigned with value of x first and after that x will be incremented. So  x=2 and  y=1    Click on Discuss to view users comments. Sanjeev said: (7:47pm on Tuesday 3rd December 2013) answer should be A - 2,1FROM THE BOOK C LANGUAGE - Kernihan and Ritchie------C provides two unusual operators for incrementing and decrementing variables. The increment operator adds 1 to its operand, while the decrement operator -- subtracts 1. We have frequently used to increment variables, as in if (c == '\\n') nl; The unusual aspect is that and -- may be used either as prefix operators (before the variable, as in n), or postfix operators (after the variable: n ). In both cases, the effect is to increment n. But the expression n increments n before its value is used, while n increments n after its value has been used. This means that in a context where the value is being used, not just the effect, n and n are different. If n is 5, then x = n ; sets x to 5, but x = n; 44 sets x to 6. In both cases, n becomes 6. Write your comments here:
3:
In substitution, a character in the plaintext is always changed to the same character in the ciphertext, regardless of its position in the text
 A. polyalphabetic B. mono alphabetic C. transpositional D. multialphabetic Answer Report Discuss Option: B Explanation : In Symmetric-Key Cryptography , we use two main approaches: • Substitution ciphers • Transposition Ciphers Substitution ciphers  is of 2 types   Mono-alphabetic: In mono-alphabetic substitution, a character in the plaintext is always changed to the same character in the cipertext regardless of its position in the text. Example: If character ‘A’ in the plaintext must be changed to character ‘D’, every character ‘A’ is changed to character ‘D’, regardless of its position in the text  Poly-alphabetic: In poly-alphabetic substitution, each occurrence of a character can have a different substitute. The relationship between a character in the plaintext to a character in the ciphertext is one-to-many. Example: Character ‘A’ can be changed to ‘D’ in the beginning of the text, but it could be changed to ‘N’ at the middle.   Click on Discuss to view users comments. Write your comments here:
4:
The equivalent productidn rules corresponding to the production rules
Sα1Sα2β1 | β2
 A. S → β1 | β2   A   → α1A | α2 A | λ B. S → β1 | β2  | β1 A | β2 A A   → α1A | α2 A C. S → β1 | β2   A   → α1A | α2 A D. S → β1 | β2  | β1A |  β2A A   → α1A | α2 A  | λ Answer Report Discuss Option: D Explanation : S -->  S+S | SS | S*|a | (S)   ---------------------     S α form   β form   Left-Recursive-Rules  Non-Left-Recursive-Rules  We can write like S ---> Sα1 | Sα2 | Sα3 | β1 | β2 Rules to convert in equivalent Non-recursive grammar: S ---> β1 | β2 Z ---> α1 | α2 | α3 Z ---> α1Z | α2Z | α3Z S ---> β1Z | β2Z Click on Discuss to view users comments. komal said: (11:26am on Tuesday 24th September 2013) as in option d,a tends to lemda then after putting lemda we get equivalent ans as given by s, in option b,a does not contain lemdaso it give beta1 alpha1 A which is not similar to s Write your comments here: → Sα1 | Sα2 | β1 | β2     '>
5:

Using data p=3, q=11, n=pq, d=7 in RSA algorithm find the cipher text of  the given plain text SUZANNE

 A. BUTAEEZ B. SUZANNE C. XYZABCD D. ABCDXYZ Answer Report Discuss Option: A Explanation : Encryption RSA algorithm Cipher text = ((plaintext)^e)mod n From the given problem we know p =3,q = 11, d =7, n=33 find z :: z = ((p-1)*(q-1)) => 2*10 = 20 Find 'e' or 'd' using the given method : (de) mod z = 1 Therefore (7e)mod 20 = 1 => with trail and error giving e =3, => (7*3)mod 20 = Finding e=3 we go to the encryption by converting the given plain text equivalent number starting a=1,b=2...z=26 Therefore SUZANNE = 19,21,26,1,14,14,5 taking single character and apply encryption (19^3) mod 33 = 28 ie (28-26) = 2 => B (21^3) mod 33 = 21=> U (26^3)mod 33 = 20=> T (1^3)mod 33 = 1=> A (14^3)mod 33 = 5=>E (5^3)mod 33 = 26 => Z Click on Discuss to view users comments. Write your comments here:

## Suggest an improvement

This section provides solution for UGC NET Computer Science Solved Papers June 2013. This can also be used as UGC NET June 2013 Answer Keys for Computer Science Exams. You can download UGC NET free answer keys from this section. Any candidate who is preparing for UGC NET Computer Science Exam should revise UGC NET Computer Science Solved Question Papers before appearing for final exams.