Question 1
If ROAD is written as URDG, then SWAN should be written as:
 A VXDQ B VZDQ C VZDP D UXDQ
General Aptitude    GATE-CS-2015 (Set 3)
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Question 1 Explanation:
In ROAD to URDG transformation, every character is replaced by next third character in alphabetic sequence. (R by U, O by R, A by D and D by G) Applying same, SWAN becomes VZDQ
 Question 2
The Tamil version of ________ John Abraham-starrer Madras Cafe _____ cleared by the Censor Board with no cuts last week, but the film’s distributors ______ no takers among the exhibitors for a release in Tamil Nadu _________ this Friday.
 A Mr., was, found, on B a, was, found, at C the, was, found, on D a, being, find at
English    GATE-CS-2015 (Set 3)
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Question 2 Explanation:
Refer use of the here.
 Question 3
Extreme focus on syllabus and studying for tests has become such a dominant concern of Indian students that they close their minds to anything ________ to the requirements of the exam.
 A related B extraneous C outside D useful
English    GATE-CS-2015 (Set 3)
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Question 3 Explanation:
 Question 4
Select the pair that best expresses a relationship similar to that expressed in the pair: Children : Pediatrician
 A Adult : Orthopaedist B Females : Gynaecologist C Kidney : Nephrologist D Skin : Dermatologist
English    GATE-CS-2015 (Set 3)
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Question 4 Explanation:
Pediatrician is a doctor for children. Gynaecologist is a doctor for Females.
 Question 5
A function f(x) is linear and has a value of 29 at x = –2 and 39 at x = 3. Find its value at x = 5.
 A 59 B 45 C 43 D 35
General Aptitude    GATE-CS-2015 (Set 3)
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Question 5 Explanation:
f(x) = 2x+33
 Question 6
Alexander turned his attention towards India, since he had conquered Persia. Which one of the statements below is logically valid and can be inferred from the above sentence?
 A Alexander would not have turned his attention towards India had he not conquered Persia B Alexander was not ready to rest on his laurels, and wanted to march to India C Alexander was completely in control of his army and could command it to move towards India. D Since Alexander’s kingdom extended to India borders after the conquest of Persia, he was keen to move further
General Aptitude    GATE-CS-2015 (Set 3)
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Question 6 Explanation:
Conquering Persia was the reason for Alexander turning his attention towards India
 Question 7
The head of a newly formed government desires to appoint five of the six selected members P, Q, R, S, T and U to portfolios of Home, Power, Defence, Telecom, and Finance. U does not want any portfolio if S gets one of the five. R wants either Home or Finance or no portfolio. Q says that if S gets either Power of telecom, then she must get the other one. T insists on a portfolio if P gets one Which is the valid distribution of portfolios?
 A P-Home, Q-Power, R-Defence, S-Telecom, T-Finance B R-Home, S-Power, P-Defence, Q-Telecom, T-Finance C P-Home, Q-Power, T-Defence, S-Telecom, U-Finance D Q-Home, U-Power, T-Defence, R-Telecom, P-Finance
General Aptitude    GATE-CS-2015 (Set 3)
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Question 7 Explanation:
Six selected members are P, Q, R, S, T and U

Portfolios are Home, Power, Defence, Telecom, and Finance.

U does not want any portfolio if S gets one of the five.

R wants either Home or Finance or no portfolio.

Q says that if S gets either Power of telecom, then she must
get the other one.

T insists on a portfolio if P gets one Which is the valid
distribution of portfolios?

A) P-Home, Q-Power, R-Defence, S-Telecom, T-Finance
Not valid as R doesn't get home or finance here

R-Home, S-Power, P-Defence, Q-Telecom, T-Finance
Is valid and satisfies all conditions.

C) P-Home, Q-Power, T-Defence, S-Telecom, U-Finance
Not valid as U is not satisfied. Condition for U is, S should not
get any portfolio.

D) Q-Home, U-Power, T-Defence, R-Telecom, P-Finance
Like A, not valid as R doesn't get home or finance here


 Question 8
Choose the most appropriate equation for the function drawn as a thick line, in the plot below.
 A A B B C C D D
General Aptitude    GATE-CS-2015 (Set 3)
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Question 8 Explanation:
The problem can be easily solved by trying point (2, -1)
on thick line, i.e., x = 2, y = -1.

None of the options except (B) satisfy above values. 
 Question 9
Most experts feel that in spite of possessing all the technical skills required to be a batsman of the highest order, he is unlikely to be so due to lack of requisite temperament. He was guilty of throwing away his wicket several times after working hard to lay a strong foundation. His critics pointed out that until he addressed this problem, success at the highest level will continue to elude him. Which of the statements(s) below is/are logically valid and can be inferred from the above passage?
(i) He was already a successful batsman at the highest level.
(ii) He has to improve his temperament in order to become a great batsman.
(iii) He failed to make many of his good starts count.
(iv) Improving his technical skills will guarantee success.
 A (iii) and (iv) B (ii) and (iii) C (i), (ii) and (iii) D (ii) only
General Aptitude    GATE-CS-2015 (Set 3)
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Question 9 Explanation:
(i) is incorrect "He was guilty of throwing away his wicket several times after working hard to lay a strong foundation. " (ii) is correct, "His critics pointed out that until he addressed this problem, success at the highest level will continue to elude him." (iii) is correct, "His critics pointed out that until he addressed this problem, success at the highest level will continue to elude him." (iv) is incorrect "Temperament is also required"
 Question 10
The exports and imports (in crores of Rs.) of a country from the year 2000 to 2007 are given in the following bar chart. In which year is the combined percentage increase in imports and exports the highest?
 A 2004 B 2005 C 2006 D 2007
General Aptitude    GATE-CS-2015 (Set 3)
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Question 10 Explanation:
In 2006 export increased from 70 yo 100 and import increased from 90 to 120.

% increase in import = 30/70 = 42.8 %
% increase in export = 30/90 = 33.33 %

Combined % increase in 2006 is more than any other year.  
 Question 11
The maximum number of processes that can be in Ready state for a computer system with n CPUs is
 A n B n2 C 2n D Independent of n
GATE-CS-2015 (Set 3)    CPU Scheduling
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Question 11 Explanation:
The size of ready queue doesn't depend on number of processes. A single processor system may have a large number of processes waiting in ready queue.
 Question 12
Let # be a binary operator defined as X # Y = X′ + Y′ where X and Y are Boolean variables. Consider the following two statements.
S1: (P # Q) # R = P # (Q # R)
S2: Q # R = R # Q 
Which of the following is/are true for the Boolean variables P, Q and R?
 A Only S1 is True B Only S2 is True C Both S1 and S2 are True D Neither S1 nor S2 are True
Propositional and First Order Logic.    GATE-CS-2015 (Set 3)
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Question 12 Explanation:
S2 is true, as X' + Y' = Y' + X'

S1 is false.
Let P = 1, Q = 1, R = 0, we get different results
(P # Q) # R = (P' + Q')' + R' = (0 + 0)' + 1 = 1 + 1 = 1
P # (Q # R) = P' + (Q' + R')' = 0 + (0 + 1)' = 0 + 0 = 0


 Question 13
Consider the following relation
  Cinema (theater, address, capacity)
Which of the following options will be needed at the end of the SQL query
SELECT P1. address
FROM Cinema P1 
Such that it always finds the addresses of theaters with maximum capacity?
 A WHERE P1. Capacity> = All (select P2. Capacity from Cinema P2) B WHERE P1. Capacity> = Any (select P2. Capacity from Cinema P2) C WHERE P1. Capacity > All (select max(P2. Capacity) from Cinema P2) D WHERE P1. Capacity > Any (select max (P2. Capacity) from Cinema P2)
SQL    GATE-CS-2015 (Set 3)
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Question 13 Explanation:
When the ALL condition is followed by a list, the optimizer expands the initial condition to all elements of the list and strings them together with AND operators. When the ANY condition is followed by a list, the optimizer expands the initial condition to all elements of the list and strings them together with OR operators, as shown below. Source: http://oracle-base.com/articles/misc/all-any-some-comparison-conditions-in-sql.php
 Question 14
The equality above remains correct if X is replace by
 A Only I B Only II C I or III or IV but not II D II or III or IV but not I
Analysis of Algorithms    GATE-CS-2015 (Set 3)
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Question 14 Explanation:
X = Sum of the cubes of {1, 2, 3, .. n| X = n2 (n+1)2 / 4
 Question 15
The number of 4 digit numbers having their digits in non-decreasing order (from left to right) constructed by using the digits belonging to the set {1, 2, 3} is ____.
 A 12 B 13 C 14 D 15
Combinatorics    GATE-CS-2015 (Set 3)
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Question 15 Explanation:
{1, 1, 1, 1} {1, 1, 1, 2} {1, 1, 1, 3} {1, 1, 2, 2} {1, 1, 2, 3} {1, 1, 3, 3} {1, 2, 2, 2} {1, 2, 2, 3} {1, 2, 3, 3} {1, 3, 3, 3} {2, 2, 2, 2} {2, 2, 2, 3} {2, 2, 3, 3} {2, 3, 3, 3} {3, 3, 3, 3}
 Question 16
Consider a CSMA/CD network that transmits data at a rate of 100 Mbps (108 bits per second) over a 1 km (kilometer) cable with no repeaters. If the minimum frame size required for this network is 1250 bytes, what is the signal speed (km/sec) in the cable?
 A 8000 B 10000 C 16000 D 20000
Data Link Layer    GATE-CS-2015 (Set 3)
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Question 16 Explanation:
Data should be transmitted at the rate of 100 Mbps.
Transmission Time >= 2*Propagation Time
=> 1250*8 / (100 * 10^6) <= 2*length/signal_speed
=> signal_speed  <= (2 * 10^3 * 100 * 10^6) / (1250 * 8)
<= 2 * 10 * (10 ^ 3) km/sec
<= 20000
Refer http://pet.ece.iisc.ernet.in/course/E2223/Problems.pdf for more details.
 Question 17
Consider the following C program segment.
# include <stdio.h>
int main( )
{
char s1[7] = "1234", *p;
p = s1 + 2;
*p = '0' ;
printf ("%s", s1);
}
What will be printed by the program?
 A 12 B 120400 C 1204 D 1034
String    GATE-CS-2015 (Set 3)
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Question 17 Explanation:
    char s1[7] = "1234", *p;
p = s1 + 2;    // p holds address of character 3
*p = '0' ;   // memory at s1 + 3 now becomes 0
printf ("%s", s1);  // All characters are printed

 Question 18
In a web server, ten WebPages are stored with the URLs of the form http://www.yourname.com/var.html; where, var is a different number from 1 to 10 for each Webpage. Suppose, the client stores the Webpage with var = 1 (say W1) in local machine, edits and then tests. Rest of the WebPages remains on the web server. W1 contains several relative URLs of the form “var.html” referring to the other WebPages. Which one of the following statements needs to be added in W1, so that all the relative URLs in W1 refer to the appropriate WebPages on the web server?
 A B C D
HTML and XML    GATE-CS-2015 (Set 3)
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Question 18 Explanation:
The tag specifies the base URL/target for all relative URLs in a document. There can be at maximum one element in a document, and it must be inside the element. Source: http://www.w3schools.com/tags/tag_base.asp
 Question 19
 A 0 B 1/2 C 1 D ∞
Numerical Methods and Calculus    GATE-CS-2015 (Set 3)
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Question 19 Explanation:
This can be solved using L'Hôpital's rule that uses derivatives to help evaluate limits involving indeterminate forms. Since [Tex] \lim_{x \to c}f(x)=\lim_{x \to c}g(x)=\infty, and \lim_{x\to c}\frac{f'(x)}{g'(x)} exists[/Tex] We get [Tex] \lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}. [/Tex] [Tex] \lim_{x\to \infty}\frac{1 + x^2}{e^x} = \lim_{x\to \infty}\frac{2x}{e^x} = \lim_{x\to \infty}\frac{2}{e^x} = 0 [/Tex]
 Question 20
Two processes X and Y need to access a critical section. Consider the following synchronization construct used by both the processes. Here, varP and varQ are shared variables and both are initialized to false. Which one of the following statements is true?
 A The proposed solution prevents deadlock but fails to guarantee mutual exclusion B The proposed solution guarantees mutual exclusion but fails to prevent deadlock C The proposed solution guarantees mutual exclusion and prevents deadlock D The proposed solution fails to prevent deadlock and fails to guarantee mutual exclusion
Process Management    GATE-CS-2015 (Set 3)
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Question 20 Explanation:
When both processes try to enter critical section simultaneously,both are allowed to do so since both shared variables varP and varQ are true.So, clearly there is NO mutual exclusion. Also, deadlock is prevented because mutual exclusion is one of the four conditions to be satisfied for deadlock to happen.Hence, answer is A.
 Question 21
Consider a software program that is artificially seeded with 100 faults. While testing this program, 159 faults are detected, out of which 75 faults are from those artificially seeded faults. Assuming that both real and seeded faults are of same nature and have same distribution, the estimated number of undetected real faults is ____________.
 A 28 B 175 C 56 D 84
Software Engineering    GATE-CS-2015 (Set 3)
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Question 21 Explanation:
Total faults detected = 159
Real faults detected among all detected faults = 159 - 75
= 84
Since probability distribution is same, total number of real
faults is (100/75)*84 = 112

Undetected real faults = 112- 84 = 28 
Another Solution : 75% of faults are detected because 75 artificially seeded faults are detected out of 100. Given that the total faults detected = 159 => Real faults detected among all detected faults = 159 – 75= 84 Since probability distribution is same, total number of real faults is (100/75)*84 = 112 Therefore undetected real faults = 112-84 = 28. So, option (A) is correct one. This solution is contributed by Nitika Bansal.
 Question 22
The result evaluating the postfix expression 10 5 + 60 6 / * 8 – is
 A 284 B 213 C 142 D 71
Stack    GATE-CS-2015 (Set 3)
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Question 22 Explanation:
 Question 23
While inserting the elements 71, 65, 84, 69, 67, 83 in an empty binary search tree (BST) in the sequence shown, the element in the lowest level is
 A 65 B 67 C 69 D 83
Binary Search Trees    GATE-CS-2015 (Set 3)
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Question 23 Explanation:
Here is The Insertion Algorithm For a Binary Search Tree :
Insert(Root,key)
{
if(Root is NULL)
Create a Node with value as key and return
Else if(Root.key <= key)
Insert(Root.left,key)
Else
Insert(Root.right,key)
}

Creating the BST one by one using the above algorithm in the below given image : This solution is contributed by Pranjul Ahuja.
 Question 24
Consider a machine with a byte addressable main memory of 220 bytes, block size of 16 bytes and a direct mapped cache having 212 cache lines. Let the addresses of two consecutive bytes in main memory be (E201F)16 and (E2020)16. What are the tag and cache line address (in hex) for main memory address (E201F)16?
 A E, 201 B F, 201 C E, E20 D 2, 01F
Computer Organization and Architecture    GATE-CS-2015 (Set 3)
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Question 24 Explanation:
Block Size = 16 bytes
Block Offset = 4

No. of sets or cache lines = 212
Number of index bits = 12

Size of main memory = 220
Number of tag bits = 20 - 12 - 4 = 4

Let us consider the hex address E201F
Tag lines = First 4 bits = E (in hex)
Cache lines = Next 12 bits  = 201 (In Hex) 
Refer http://virtual-labs.ac.in/labs/cse10/dmc.html
 Question 25
In the given matrix, one of the eigenvalues is 1. the eigenvectors corresponding to the eigenvalue 1 are
⎡ 1 -1  2 ⎤
⎢ 0  1  0 ⎥
⎣ 1  2  1 ⎦
 A A B B C C D D
Linear Algebra    GATE-CS-2015 (Set 3)
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Question 25 Explanation:
Let z represents the eigenvalues.
And let the given matrix be A (square matrix of order 3 x3)

The characteristic equation for this is :

AX = zX ( X is the required eigenvector )
AX - zX = 0
[ A - z I ] [X] = 0 ( I is an identity matrix of order 3 )

put z = 1 ( because one of the eigenvalue is 1 )

[ A - 1 I ] [X] = 0

The resultant matrix is :

[ 0 -1 2 ] [x1] [0]
| 0 0 0 ] |x2] =|0|
[ 1 2 0 ] |x3] [0]

Multiplying thr above matrices and getting the equations as:

-x2 + 2x3 = 0 ----------------(1)
x1 + 2x2 = 0-----------------(2)

now let x1 = k, then x2 and x3 will be -k/2 and -k/4
respectively.

hence eigenvector X = { (k , -k/2, -k/4) } where k != 0

put k = -4c ( c is also a constant, not equal to zero ),
we get X = { ( -4c, 2c, 1c ) }, i.e. { c ( -4, 2, 1 ) }

Hence option B.
 Question 26
Among simple LR (SLR), canonical LR, and look-ahead LR (LALR), which of the following pairs identify the method that is very easy to implement and the method that is the most powerful, in that order?
 A SLR, LALR B Canonical LR, LALR C SLR, canonical LR D LALR, canonical LR
Parsing and Syntax directed translation    GATE-CS-2015 (Set 3)
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Question 26 Explanation:
SLR parser is a type of LR parser with small parse tables and a relatively simple parser generator algorithm. Canonical LR parser or LR(1) parser is an LR(k) parser for k=1, i.e. with a single lookahead terminal. It can handle all deterministic context-free languages. LALR parser or Look-Ahead LR parser is a simplified version of a canonical LR parser,
 Question 27
Given a hash table T with 25 slots that stores 2000 elements, the load factor α for T is __________
 A 80 B 0.0125 C 8000 D 1.25
Hash    GATE-CS-2015 (Set 3)
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Question 27 Explanation:
load factor = (no. of elements) / (no. of table slots) = 2000/25 = 80
 Question 28
Let T be the language represented by the regular expression Σ∗0011Σ∗ where Σ = {0, 1}. What is the minimum number of states in a DFA that recognizes L' (complement of L)?
 A 4 B 5 C 6 D 8
Regular languages and finite automata    GATE-CS-2015 (Set 3)
Discuss it

Question 28 Explanation:
The given regular expression matches with all strings that contain 0011. The complement should match with all strings except the strings with 0011 as substring. Following are some interesting facts/observations from this question. 1) Complement of a regular language is also regular. 2) Since complement is regular, it is always possible to make a DFA for complement. 3) A DFA that accepts its complement is obtained from the above DFA by changing all non-accepting states to accepting states and vice versa as done in this question. Below is DFA for the complement. We can get below DFA by first drawing DFA of regular language for strings with substring as 0011. After drawing DFA, we can invert all states (single circle to double circle and vice versa) Reference : http://www.cs.odu.edu/~toida/nerzic/390teched/regular/fa/complement.html
 Question 29
Consider the following array of elements. 〈89, 19, 50, 17, 12, 15, 2, 5, 7, 11, 6, 9, 100〉. The minimum number of interchanges needed to convert it into a max-heap is
 A 4 B 5 C 2 D 3
Heap    GATE-CS-2015 (Set 3)
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Question 29 Explanation:
〈89, 19, 50, 17, 12, 15, 2, 5, 7, 11, 6, 9, 100〉

89
/         \
19          50
/  \        /  \
17    12     15    2
/  \   / \   /  \
5   7  11  6  9  100

Minimum number of swaps required to convert above tree
to Max heap is 3. Below are 3 swap operations.
Swap 100 with 15
Swap 100 with 50
Swap 100 with 89

100
/         \
19          89
/  \        /  \
17    12     50    5
/  \   / \   /  \
7   11  6  9  2   15
 Question 30
Consider the relation X(P, Q, R, S, T, U) with the following set of functional dependencies
F = {
{P, R} → {S,T},
{P, S, U} → {Q, R}
}
Which of the following is the trivial functional dependency in F+ is closure of F?
 A {P,R}→{S,T} B {P,R}→{R,T} C {P,S}→{S} D {P,S,U}→{Q}
Database Design(Normal Forms)    GATE-CS-2015 (Set 3)
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Question 30 Explanation:
A functional dependency X -> Y is trivial if Y is a subset of X.
 Question 31
Consider a software project with the following information domain characteristic for calculation of function point metric.
  Number of external inputs (I) = 30
Number of external output (O) = 60
Number of external inquiries (E) = 23
Number of files (F) = 08
Number of external interfaces (N) = 02 
It is given that the complexity weighting factors for I, O, E, F and N are 4, 5, 4, 10 and 7, respectively. It is also given that, out of fourteen value adjustment factors that influence the development effort, four factors are not applicable, each of he other four factors have value 3, and each of the remaining factors have value 4. The computed value of function point metric is ____________
 A 612.06 B 212.05 C 305.09 D 806.9
Software Engineering    GATE-CS-2015 (Set 3)
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Question 31 Explanation:
Function point metrics provide a standardized method for measuring the various functions of a software application
The value of function point metric = UPF * VAF

Here,
UPF: Unadjusted Function Point (UFP) count

UPF  = 4*30 + 60*5 + 23*4 + 8*10 + 7*2 = 606

VAF = (TDI * 0.01) + 0.65

Here TDI is Total Degree of Influence
TDI = 3*4 + 0*4 + 4*6 = 36

VAF = (TDI * 0.01) + 0.65
= 36*0.01 + 0.65
= 0.36 + 0.65
= 1.01

FP = UPF * VAF
= 1.01 * 606
= 612.06 
Refer https://cs.uwaterloo.ca/~apidduck/CS846/Seminars/abbas.pdf
 Question 32
Consider the following statements.
I. TCP connections are full duplex.
II. TCP has no option for selective acknowledgment
III. TCP connections are message streams.
 A Only I is correct B Only I and II are correct C Only II and III are correct D All of I, II and III are correct
Transport Layer    GATE-CS-2015 (Set 3)
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Question 32 Explanation:

1. In TCP, as sender and receiver can send segments at the same time, It is FULL-DUPLEX.
2. TCP has options for selective acknowledgement. With selective acknowledgments (SACKs), the data receiver can inform the sender about all segments that have arrived successfully, so the sender need retransmit only the segments that have actually been lost.
3. As each BYTE is counted in TCP segment, and Sequence number of First BYTE is kept into header, TCP is BYTE stream protocol.
So, only 1st is correct and other incorrect. Reference : https://tools.ietf.org/html/rfc2018 This solution is contributed by sandeep pandey.
 Question 33
Suppose U is the power set of the set S = {1,2,3,4,5,6}. For any T ∈ U, let |T| denote the number of elements in T and T′ denote the complement of T. For any T, R ∈ U, let T\R be the set of all elements in T which are not in R. Which one of the following is true?
 A A B B C C D D
Propositional and First Order Logic.    GATE-CS-2015 (Set 3)
Discuss it

Question 33 Explanation:
D is true, it can be seen by drawing Venn Diagram.

A is false,  Take an example like X = {1, 2, 3, 4},
X' = {5, 6}, |X| is not same as |X'|.

B is false, as any two subsets of size 5 of U
would definitely have some common elements.

C is false,  Take an example like X = {1, 2}
Y = {3, 4, 5}, X\Y = {1, 2}.
 Question 34
In a room there are only two types of people, namely Type 1 and Type 2. Type 1 people always tell the truth and Type 2 people always lie. You give a fair coin to a person in that room, without knowing which type he is from and tell him to toss it and hide the result from you till you ask for it. Upon asking, the person replies the following:
“The result of the toss is head if and only if I am telling the truth.”
Which of the following options is correct?
 A The result is head B The result is tail C If the person is of Type 2, then the result is tail D If the person is of Type 1, then the result is tai
Propositional and First Order Logic.    GATE-CS-2015 (Set 3)
Discuss it

Question 34 Explanation:
“The result of the toss is head if and only
if I am telling the truth.”

If the person is of Type 1 who always tell truth, then result must be head.

If the person is of Type 2 who always tell lie, then result must be head.

Negation of a sentence of the form "X is true if and only if Y is true" is
"Either X is true and Y is false, or X is false and Y is true."

Refer: http://math.stackexchange.com/questions/10435/negation-of-if-and-only-if

Which means "Either toss is head and I am not telling truth, or toss is tail
and I am telling truth".

Since the person always lie, it is "Either toss is head and I am not telling truth"

 Question 35
Consider a binary tree T that has 200 leaf nodes. Then, the number of nodes in T that have exactly two children are _________.
 A 199 B 200 C Any number between 0 and 199 D Any number between 100 and 200
GATE-CS-2015 (Set 3)
Discuss it

Question 35 Explanation:
This can be proved using Handshaking Lemma. Refer below post to see complete proof. Handshaking Lemma and Interesting Tree Properties
 Question 36
Consider the following C program.
# include
int main( )
{
static int a[] = {10, 20, 30, 40, 50};
static int *p[] = {a, a+3, a+4, a+1, a+2};
int **ptr = p;
ptr++;
printf("%d%d", ptr - p, **ptr};
}

The output of the program is _________
 A 140 B 120 C 100 D 40
Discuss it

Question 36 Explanation:
In order to simplify programs involving complex operations on pointers, we suggest you to draw proper diagrams in order to avoid silly mistakes. Let’s assume that integer is of 4 Bytes and Pointer size is also 4 Bytes. Let’s assume array a Base address is 1000. Array name actually holds the array base address. Let’s assume array p Base address is 2000.   Double Pointer ptr Base Address is 3000.   Now ptr is actually pointing to the first element of array p. ptr++ will make it point to the next element of array p. Its value will then change to 2004. One of the Rule of Pointer Arithmetic is that When you subtract two pointers, as long as they point into the same array, the result is the number of elements separating them. ptr is pointing to the second element and  p is pointing to the first element so ptr-p will be equal to 1(Excluding the element to which ptr is pointing). Now ptr = 2004 -----> *(2004) = 1012 ----> *(1012) ----> 40 which is the final answer.   This solution is contributed by Pranjul Ahuja. .
 Question 37
Assume that a mergesort algorithm in the worst case takes 30 seconds for an input of size 64. Which of the following most closely approximates the maximum input size of a problem that can be solved in 6 minutes?
 A 256 B 512 C 1024 D 2048
Sorting    GATE-CS-2015 (Set 3)    MergeSort
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Question 37 Explanation:
Time complexity of merge sort is Θ(nLogn)

c*64Log64 is 30
c*64*6 is 30
c is 5/64

For time 6 minutes

5/64*nLogn = 6*60

nLogn = 72*64 = 512 * 9

n = 512. 
 Question 38
Consider a network connecting two systems located 8000 kilometers apart. The bandwidth of the network is 500 × 106 bits per second. The propagation speed of the media is 4 × 106 meters per second. It is needed to design a Go-Back-N sliding window protocol for this network. The average packet size is 107 bits. The network is to be used to its full capacity. Assume that processing delays at nodes are negligible. Then, the minimum size in bits of he sequence number field has to be ________.
 A 2 B 4 C 8 D 16
Transport Layer    GATE-CS-2015 (Set 3)
Discuss it

Question 38 Explanation:
Propagation time = (8000 * 1000)/ (4 * 10^6)
= 2 seconds

Total round trip propagation time = 4 seconds

Transmission time for one packet = (packet size) / (bandwidth)
= (10^7) / (500 * 10^6)
= 0.02 seconds

Total number of packets that can be transferred before an
acknowledgement comes back = 4 / 0.02 = 200

Maximum possible window size is 200.

In Go-Back-N, maximum sequence number should be one more than
window size.

So total 201 sequence numbers are needed. 201 different sequence
numbers can be represented using 8 bits.
 Question 39
Consider the following partial Schedule S involving two transactions T1 and T2. Only the read and the write operations have been shown. The read operation on data item P is denoted by read(P) and the write operation on data item P is denoted by write(P). Suppose that the transaction T1 fails immediately after time instance 9. Which one of the following statements is correct?
 A T2 must be aborted and then both T1 and T2 must be re-started to ensure transaction atomicity B Schedule S is non-recoverable and cannot ensure transaction atomicity C Only T2 must be aborted and then re-started to ensure transaction atomicity D Schedule S is recoverable and can ensure atomicity and nothing else needs to be done
Transactions and concurrency control    GATE-CS-2015 (Set 3)
Discuss it

Question 39 Explanation:
if transaction fails, atomicity requires effect of transaction to be undone. Durability states that once transaction commits, its change cannot be undone (without running another, compensating, transaction). Recoverable schedule: A schedules exactly where, for every set of transaction Ti and Tj. If Tj reads a data items previously written by Ti, then the commit operation of Ti precedes the commit operation of Tj. Aborting involves undoing the operations and redoing them since by the time stamp it is aborted. Option (A): T2 must be aborted and then both T1 and T2 must be re-started to ensure transaction atomicity. It is incorrect because it says abort transaction T2 and then redo all the operations. But there is no gaurantee that it will succeed this time as again T1 may be fail. Option(B): Schedule S is non-recoverable and cannot ensure transaction atomicity. Correct, it is by definition an irrecoverable schedule so now even if we start to undo the actions one by one(after t1 fails) in order to ensure transaction atomicity. Still we cannot undo a commited transaction. hence this schedule is irrecoverable by definition and also not atomic since it leaves the database in an inconsistent state. Simply dirty read so nonrecoverable. Option (C): Only T2 must be aborted and then re-started to ensure transaction atomicity. It is incorrect because it says abort only transaction T2 and then redo all the T2 operations. But this is dirty read problem as it is reading the data item A which is written by T1 and T1 is not committed. Again it will be the dirty read problem. So incorrect. Option (D): Schedule S is recoverable and can ensure transaction atomicity and nothing else needs to be done. Incorrect, it is clearly saying that schedule s is recoverable but it is irrecoverable because T2 read the data item A which is written by T1 and T1 failed and rollback, at the rollback T1 start undo all operations and modified the value of A with previous value but T2 is already committed so T2 can't change the read value of A which was earlier taken from T1.   This solution is contributed by Nitika Bansa.
 Question 40
Consider the following two C code segments. Y and X are one and two dimensional arrays of size n and n × n respectively, where 2 ≤ n ≤ 10. Assume that in both code segments, elements of Y are initialized to 0 and each element X[i][j] of array X is initialized to i + j. Further assume that when stored in main memory all elements of X are in same main memory page frame.
Code segment 1:
// initialize elements of Y to 0
// initialize elements X[i][j] of X to i+j
for (i = 0; i < n; i++)
y[i] + = X[0][i];

Code segment 2:
// initialize elements of Y to 0
// initialize elements X[i][j] of X to i+j
for (i = 0; i < n; i++)
y[i] + = X[i][0];

Which of the following statements is/are correct?
S1: Final contents of array Y will be same in both code segments.
S2: Elements of array X accessed inside the for loop shown in
code segment 1 are contiguous in main memory.
S3: Elements of array X accessed inside the for loop shown in
code segment 2 are contiguous in main memory. 
 A Only S2 is correct B Only S3 is correct C Only S1 and S2 are correct D Only S1 and S3 are correct
GATE-CS-2015 (Set 3)
Discuss it

Question 40 Explanation:
In C, 2D arrays are stored in row major order. Therefore, S2 is correct, but S3 is not correct.
 Question 41
Consider the following grammar G.
  S → F ⎪ H
F → p ⎪ c
H → d ⎪ c 
Where S, F and H are non-terminal symbols, p, d and c are terminal symbols. Which of the following statement(s) is/are correct?
S1: LL(1) can parse all strings that are generated using grammar G.
S2: LR(1) can parse all strings that are generated using grammar G. 
 A Only S1 B Only S2 C Both S1 and S2 D Neither S1 and S2
Parsing and Syntax directed translation    GATE-CS-2015 (Set 3)
Discuss it

Question 41 Explanation:
The given grammar is ambiguous as there are two possible leftmost derivations for string "c".

First Leftmost Derivation
S → F
F → c

Second Leftmost Derivation
S → H
H → c  
An Ambiguous grammar can neither be LL(1) nor LR(1)
 Question 42
Which of the following languages are context-free?
L1 = {ambnanbm ⎪ m, n ≥ 1}
L2 = {ambnambn ⎪ m, n ≥ 1}
L3 = {ambn ⎪ m = 2n + 1} 
 A L1 and L2 only B L1 and L3 only C L2 and L3 only D L3 only
Context free languages and Push-down automata    GATE-CS-2015 (Set 3)
Discuss it

Question 42 Explanation:
We can build a push down automata for L1 and L3, but can not build push down automata for L@. Note that a PDA can uses a stack. L1 and L3 can be identified using a single stack, but L2 can't be.
 Question 43
If the following system has non-trivial solution,
  px + qy + rz = 0
qx + ry + pz = 0
rx + py + qz = 0 
then which one of the following options is True?
 A p – q + r = 0 or p = q = –r B p + q – r = 0 or p = –q = r C p + q + r = 0 or p = q = r D p – q + r = 0 or p = –q = –r
Linear Algebra    GATE-CS-2015 (Set 3)
Discuss it

Question 43 Explanation:
For non-trivial solution, |A| should be equal to 0 Hence, Now solve it using matrix rules: (p+q+r) [(q-r)(p-q) - (r-p) (r-p) ] = 0 Either (p+q+r) = 0 or [(q-r)(p-q) - (r-p) (r-p) = 0 From (p+q+r) =0, you can clearly say that option C is the correct one. and for more precise answer, let’s solve second equation: [(q-r)(p-q) - (r-p) (r-p) = 0 (q-r)(p-q) = (r-p) (r-p) and only p = q = r satisfies this equation. So option C is correct one. This explanation has been contributed by Nitika Bansal.
 Question 44
For the processes listed in the following table, which of the following scheduling schemes will give the lowest average turnaround time?
Process    Arrival Time    Processing Time
A              0              3
B              1              6
C              4              4
D              6              2
 A First Come First Serve B Non-preemptive Shortest Job First C Shortest Remaining Time D Round Robin with Quantum value two
GATE-CS-2015 (Set 3)    CPU Scheduling
Discuss it

Question 44 Explanation:
Turnaround time is the total time taken between the submission of a program/process/thread/task (Linux) for execution and the return of the complete output to the customer/user. Turnaround Time = Completion Time - Arrival Time. FCFS = First Come First Serve (A, B, C, D) SJF = Non-preemptive Shortest Job First (A, B, D, C) SRT = Shortest Remaining Time (A(3), B(1), C(4), D(2), B(5)) RR = Round Robin with Quantum value 2 (A(2), B(2), A(1),C(2),B(2),D(2),C(2),B(2)
Pr  Arr.Time  P.Time   FCFS     SJF      SRT    RR
A      0       3      3-0=3    3-0=3    3-0=3   5-0=5
B      1       6      9-1=8    9-1=8    15-1=14 15-1=14
C      4       4      13-4=9   15-4=11  8-4=4   13-4=9
D      6       2      15-6=9   11-6=5   10-6=4  11-6=5

Average                7.25     6.75     6.25    8.25

Shortest Remaining Time produces minimum average turn-around time.
 Question 45
Consider the equation (43)x = (y3)8 where x and y are unknown. The number of possible solutions is ________.
 A 3 B 4 C 5 D 6
Number Representation    GATE-CS-2015 (Set 3)
Discuss it

Question 45 Explanation:
3 + 4x = 3 + 8y where 0 <= y <= 7
and x >= 5 (because the number represented in base x is 34)

x = 2y  and 0 <= y <= 7

The following are possible solutions
y = 3, 4, 5, 6, 7
x = 6, 8, 10, 12, 14
 Question 46
Since it is a network that uses switch, every packet goes through two links, one from source to switch and other from switch to destination. Since there are 10000 bits and packet size is 5000, two packets are sent. Transmission time for each packet is 5000 / 1077 bits per second links. Each link has a propagation delay of 20 microseconds. The switch begins forwarding a packet 35 microseconds after it receives the same. If 10000 bits of data are to be transmitted between the two hosts using a packet size of 5000 bits, the time elapsed between the transmission of the first bit of data and the reception of the last bit of the data in microseconds is _________.
 A 1075 B 1575 C 2220 D 2200
Misc Topics in Computer Networks    GATE-CS-2015 (Set 3)
Discuss it

Question 46 Explanation:
Sender host transmits first packet to switch, the transmission time is 5000/107 which is 500 microseconds. After 500 microseconds, the second packet is transmitted. The first packet reaches destination in 500 + 35 + 20 + 20 + 500 = 1075 microseconds. While the first packet is traveling to destination, the second packet starts its journey after 500 microseconds and rest of the time taken by second packet overlaps with first packet. So overall time is 1075 + 500 = 1575.
 Question 47
Suppose Xi for i = 1, 2, 3 are independent and identically distributed random variables whose probability mass functions are Pr[Xi = 0] = Pr[Xi = 1] = 1/2 for i = 1, 2, 3. Define another random variable Y = X1 X2 ⊕ X3, where ⊕ denotes XOR. Then Pr[Y = 0 ⎪ X3 = 0] = ____________.
 A 0.75 B 0.5 C 0.85 D 0.25
Probability    GATE-CS-2015 (Set 3)
Discuss it

Question 47 Explanation:

P (A|B) = P (A∩B) / P (B)

P (Y=0 | X3=0) = P(Y=0 ∩X3=0) / P(X3=0)

P(X3=0) = 1⁄2

Y = X1X2 ⊕ X3

The number of possibilities where Y = 0 can be obtained by constructing a table

From the above table, P(Y=0 ∩X3=0) = 3/8 And P (X3=0) = 1⁄2

P (Y=0 | X3=0) = P(Y=0 ∩X3=0) / P(X3=0) = (3/8) / (1/2) = 3⁄4 = 0.75

This solution is contributed by Anil Saikrishna Devarasetty .

Another Solution :

It is given X3 = 0. Y can only be 0 when X1 X2 is 0. X1 X2 become 0 for X1 = 1, X2 = 0, X1 = X2 = 0 and X1 = 0, X = 1 So the probability is = 0.5*0.5*3 = 0.75
 Question 48
In the network 200.10.11.144/27, the fourth octet (in decimal) of the last IP address of the network which can be assigned to a host is ________
 A 158 B 255 C 222 D 223
Network Layer    GATE-CS-2015 (Set 3)
Discuss it

Question 48 Explanation:
The last or fourth octet of network address is 144
144 in binary is 10010000.

The first three bits of this octal are fixed as 100,
the remaining bits can get maximum value as 11111.
So the maximum possible last octal IP address is
10011111 which is 159.

and can't be assigned to a host. So the maximum possible
last octal in a host IP is 10011110 which is 158.

The maximum possible network address that can be assigned
is 200.10.11.158/31 which has last octet as 158.

 Question 49
Consider the following recursive C function. If get(6) function is being called in main() then how many times will the get() function be invoked before returning to the main()?
void get (int n)
{
if (n < 1) return;
get(n-1);
get(n-3);
printf("%d", n);
}

 A 15 B 25 C 35 D 45
Recursion    GATE-CS-2015 (Set 3)
Discuss it

Question 49 Explanation:
                              get(6) [25 Calls]
/      \
[17 Calls] get(5)       get(3) [7 Calls]
/     \
get(4)    get(2)[5 Calls]
/    \
[7 Calls] get(3)  get(1)[3 Calls]
/     \
get(2)   get(0)
/    \
[3 Calls]get(1)  get(-1)
/  \
get(0) get(-2)
We can verify the same by running below program. 1 # include int count = 0; void get (int n) { count++; if (n < 1) return; get(n-1); get(n-3); } int main() { get(6); printf("%d ", count); } [/sourcecode] Output: 25
 Question 50
Let G be connected undirected graph of 100 vertices and 300 edges. The weight of a minimum spanning tree of G is 500. When the weight of each edge of G is increased by five, the weight of a minimum spanning tree becomes ________.
 A 1000 B 995 C 2000 D 1995
Graph Minimum Spanning Tree    GATE-CS-2015 (Set 3)
Discuss it

Question 50 Explanation:
Since there are 100 vertices, there must be 99 edges in Minimum Spanning Tree (MST). When weight of every edge is increased by 5, the increment in weight of MST is = 99 * 5 = 495 So new weight of MST is 500 + 495 which is 995
 Question 51
Let R be a relation on the set of ordered pairs of positive integers such that ((p, q), (r, s)) ∈ R if and only if p–s = q–r. Which one of the following is true about R?
 A Both reflexive and symmetric B Reflexive but not symmetric C Not reflexive but symmetric D Neither reflexive nor symmetric
Set Theory & Algebra    GATE-CS-2015 (Set 3)
Discuss it

Question 51 Explanation:
((p, q), (r, s)) ∈ R if and only if p–s = q–r

(p, q) is not related to (p, q)

as p-q is not same as q-p.

The relation is symmetric because if p–s = q–r, then s-q = s-p. 
 Question 52
Let f(n) = n and g(n) = n(1+sin n), where n is a positive integer. Which of the following statements is/are correct?
I.  f(n) = O(g(n))
II. f(n) = Ω(g(n))  
 A Only I B Only II C Both I and II D Neither I nor II
Analysis of Algorithms (Recurrences)    GATE-CS-2015 (Set 3)
Discuss it

Question 52 Explanation:
The value of sine function varies from -1 to 1.

For sin = -1 or any other negative value, I becomes false.

For sin = 1 or any other negative value, II becomes false
 Question 53
The total number of prime implicants of the function f(w, x, y, z) = Σ(0, 2, 4, 5, 6, 10) is ________.
 A 2 B 3 C 4 D 5
Digital Logic & Number representation    GATE-CS-2015 (Set 3)
Discuss it

Question 53 Explanation:
Red,Blue and Green together make total three prime implicant.
As we know "A prime implicant of a function is an implicant that cannot be covered by a more general (more reduced - meaning with fewer literals) implicant. (Wikipedia)" we've to see only prime implicants . We can make group of four 1's as in the figure(green group) , a group of two 1's in blue and red .As we can't reduce it further so this is minimal representation of problem and hence total number of prime implicant =3. So answer (B) part.
 Question 54
Given the function F = P′ + QR, where F is a function in three Boolean variables P, Q and R and P′ = !P, consider the following statements.
  S1: F = Σ (4, 5, 6)
S2: F = Σ (0, 1, 2, 3, 7)
S3: F = Π (4, 5, 6)
S4: F = Π (0, 1, 2, 3, 7) 
Which of the following is true?
 A S1-False, S2-True, S3-True, S4-False B S1-True, S2-False, S3-False, S4-True C S1-False, S2-False, S3-True, S4-True D S1-True, S2-True, S3-False, S4-False
Digital Logic & Number representation    GATE-CS-2015 (Set 3)
Discuss it

Question 54 Explanation:
After drawing K map of F = P + QR ,we can find out S2 and S3 are TRUE. Alternate Explanation :
P Q R = (!P + QR)

0 => 0 0 0 => 1+0.0 =1 =>Σ

1 => 0 0 1 => 1+0.1 =1 =>Σ

2 => 0 1 0 => 1+1.0 =1 =>Σ

3 => 0 1 1 => 1+1.1 =1 =>Σ

4 => 1 0 0 => 0+0.0=0 =>Π

5 => 1 0 1 => 0+0.1=0 =>Π

6 => 1 1 0 => 0+1.0=0 =>Π

7 => 1 1 1 => 0+1.1=1 =>Σ

as sigma means 1 and pi means 0
therefore Σ = (0,1,2,3,7)
Π = (4,5,6)

 Question 55
 A A B B C C D D
Numerical Methods and Calculus    GATE-CS-2015 (Set 3)
Discuss it

 Question 56
Consider B+ tree in which the search key is 12 bytes long, block size is 1024 bytes, record pointer is 10 bytes long and block pointer is 8 bytes long. The maximum number of keys that can be accommodated in each non-leaf node of the tree is
 A 49 B 50 C 51 D 52
B and B+ Trees    GATE-CS-2015 (Set 3)
Discuss it

Question 56 Explanation:
Let m be the order of B+ tree

m(8)+(m-1)12 <= 1024
[Note that record pointer is not needed in non-leaf nodes]

m <= 51

Since maximum order is 51, maximum number of keys is 50. 
 Question 57
Consider the following code sequence having five instructions I1 to I5. Each of these instructions has the following format.
    OP Ri, Rj, Rk 
where operation OP is performed on contents of registers Rj and Rk and the result is stored in register Ri.
   I1 : ADD R1, R2, R3
I2 : MUL R7, R1, R3
I3 : SUB R4, R1, R5
I4 : ADD R3, R2, R4
I5 : MUL R7, R8, R9 
Consider the following three statements:
S1: There is an anti-dependence between instructions I2 and I5.
S2: There is an anti-dependence between instructions I2 and I4.
S3: Within an instruction pipeline an anti-dependence always
creates one or more stalls. 
Which one of above statements is/are correct?
 A Only S1 is true B Only S2 is true C Only S1 and S2 are true D Only S2 and S3 are true
Computer Organization and Architecture    GATE-CS-2015 (Set 3)
Discuss it

Question 57 Explanation:
The given instructions can be written as below:
I1: R1 = R2 + R3
I2: R7 = R1 * R3
I3: R4 = R1 - R5
I4: R3 = R2 + R4
I5: R7 = R8 * R9 
An anti-dependency, also known as write-after-read (WAR), occurs when an instruction requires a value that is later updated.
S1: There is an anti-dependence between instructions I2 and I5.
False, I2 and I5 don't form any write after read situation.
They both write R7.

S2: There is an anti-dependence between instructions I2 and I4.
True, I2 reads R3 and I4 writes it.

S3: Within an instruction pipeline an anti-dependence always
creates one or more stalls.
Anti-dependency can be removed by renaming variables.
See following example.
1. B = 3
2. A = B + 1
3. B = 7
Renaming of variables could remove the dependency.
1. B = 3
N. B2 = B
2. A = B2 + 1
3. B = 7
 Question 58
Consider the following C program:
# include <stdio.h>
int main( )
{
int i, j, k = 0;
j = 2 * 3 / 4 + 2.0 / 5 + 8 / 5;
k  -= --j;
for (i = 0; i < 5; i++)
{
switch(i + k)
{
case 1:
case 2: printf("\n%d", i + k);
case 3: printf("\n%d", i + k);
default: printf("\n%d", i + k);
}
}
return 0;
}

The number of times printf statement is executed is __________.
 A 8 B 9 C 10 D 11
Misc    GATE-CS-2015 (Set 3)
Discuss it

Question 58 Explanation:
The following statement makes j = 2
 j = 2 * 3 / 4 + 2.0 / 5 + 8 / 5; 
The following statement makes k = -1.

k  -= --j; 
There is one important thing to note in switch is, there is no break. Let count of printf statements be 'count'
For i = 0, the value of i+k becomes -1, default block
is executed, count = 1.
For i = 1, the value of i+k becomes 0, default block
is executed, count = 2.
For i = 2, the value of i+k becomes 1, all blocks are
executed as there is no break, count = 5
For i = 3, the value of i+k becomes 2, three blocks
after case 1: are executed, count = 8
For i = 4, the value of i+k becomes 3, two blocks
are executed, count = 10 
 Question 59
Suppose c = 〈c[0], ... , c[k – 1]〉 is an array of length k, where all the entries are from the set {0, 1}. For any positive integers a and n, consider the following pseudocode.
DOSOMETHING (c, a, n)
z ← 1
for i ← 0 to k – 1
do z ← z2 mod n
if c[i] = 1
then z ← (z × a) mod n
return z 
If k = 4, c = 〈1, 0, 1, 1〉, a = 2 and n = 8, then the output of DOSOMETHING(c, a, n) is ____________.
 A 0 B 1 C 2 D 3
Misc    GATE-CS-2015 (Set 3)
Discuss it

Question 59 Explanation:
DOSOMETHING (c, a, n)
z ← 1
for i ← 0 to k – 1
do z ← z2 mod n
if c[i] = 1
then z ← (z × a) mod n
return z 
If k = 4, c = 〈1, 0, 1, 1〉, a = 2 and n = 8, then the output of DOSOMETHING(c, a, n) is ____________.
For i = 0, z = 1 mod 8 = 1, since c[0] = 1, z = 1*2 mod 8 = 2.

For i = 1, z = 2*2 mod 8 = 1, since c[1] = 0, z remains 4.

For i = 2, z = 16 mod 8 = 0 
Once z becomes 0, none of the statements inside DOSOMETHING() can make it non-zero.
 Question 60
The velocity v (in kilometer/minute) of a motorbike which starts from rest, is given at fixed intervals of time t(in minutes) as follows:
t   2  4  6  8  10  12  14  16 18 20
v  10  18 25 29 32  20  11  5  2  0 
The approximate distance (in kilometers) rounded to two places of decimals covered in 20 minutes using Simpson’s 1/3rd rule is _________.
 A 309.33 B 105.33 C 110 D 405.6
Numerical Methods and Calculus    GATE-CS-2015 (Set 3)
Discuss it

Question 60 Explanation:
 Question 61
Consider the following reservation table for a pipeline having three stages S1, S2 and S3.
     Time -->
-----------------------------
1    2   3    4     5
-----------------------------
S1  | X  |   |   |    |  X |
S2  |    | X |   | X  |    |
S3  |    |   | X |    |    |
The minimum average latency (MAL) is __________
 A 3 B 2 C 1 D 4
Computer Organization and Architecture    GATE-CS-2015 (Set 3)
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Question 61 Explanation:
S1 | X | Y |   |   | X | Y | X | Y |   |   | X | Y |
S2 |   | X | Y | X | Y |   |   | X | Y | X | Y |   |
S3 |   |   | X | Y |   |   |   |   | X | Y |   |   |

We can interleave instructions like the above
pattern.

Latency between X and Y is 1.

Latency between fist and second X is 5.

The pattern repeats after that.
So, MAL is (1 + 5)/2;


 Question 62
Consider the following policies for preventing deadlock in a system with mutually exclusive resources.
I. Processes should acquire all their resources at the
beginning of execution.  If any resource is not
available, all resources acquired so far are released.
II. The resources are numbered uniquely, and processes are
allowed to request for resources only in increasing
resource numbers.
III. The resources are numbered uniquely, and processes are
allowed to request  for resources only in decreasing
resource numbers.
IV. The resources are numbered uniquely. A process is allowed
to request only for a resource with resource number larger
than its currently held resources.
Which of the above policies can be used for preventing deadlock?
 A Any one of I and III but not II or IV B Any one of I, III and IV but not II C Any one of II and III but not I or IV D Any one of I, II, III and IV
Discuss it

Question 62 Explanation:
If Ist is followed, then hold and wait will never happen. II, III and IV are similar. If any of these is followed, cyclic wait will not be possible.
 Question 63
Language L1 is polynomial time reducible to language L2. Language L3 is polynomial time reducible to L2, which in turn is polynomial time reducible to language L4. Which of the following is/are True?
I. If L4 ∈ P, L2 ∈ P
II. If L1 ∈ P or L3 ∈ P, then L2 ∈ P
III. L1 ∈ P, if and only if L3 ∈ P
IV. If L4 ∈ P, then L1 ∈ P and L3 ∈ P 
 A II only B III only C I and IV only D I only
NP Complete    GATE-CS-2015 (Set 3)
Discuss it

Question 63 Explanation:
 Question 64
Consider the following C program. The output of the program is __________.
# include <stdio.h>
int f1(void);
int f2(void);
int f3(void);
int x = 10;
int main()
{
int x = 1;
x += f1() + f2() + f3() + f2();
pirntf("%d", x);
return 0;
}

int f1()
{
int x = 25;
x++;
return x;
}

int f2( )
{
static int x = 50;
x++;
return x;
}

int f3( )
{
x *= 10;
return x;
}

 A 230 B 131 C 231 D 330
GATE-CS-2015 (Set 3)
Discuss it

Question 64 Explanation:
x += f1() + f2() + f3() + f2();

x = x +  f1() + f2() + f3() + f2();

f1() returns 26

f2() returns 51

f3() returns 100

second call to f2() returns 52
[Note x is static in f2()]

x = 1 + 26 + 51 + 100 + 52  = 230 
 Question 65
Consider three software items: Program-X, Control Flow Diagram of Program-Y and Control Flow Diagram of Program-Z as shown below The values of McCabe’s Cyclomatic complexity of Program-X, Program-Y and Program-Z respectively are
 A 4, 4, 7 B 3, 4, 7 C 4, 4, 8 D 4, 3, 8
Software Engineering    GATE-CS-2015 (Set 3)
Discuss it

Question 65 Explanation:
The cyclomatic complexity of a structured program[a] is defined
with reference to the control flow graph of the program, a directed
graph containing the basic blocks of the program, with an edge
between two basic blocks if control may pass from the first to the
second. The complexity M is then defined as.

M = E − N + 2P,

where

E = the number of edges of the graph.
N = the number of nodes of the graph.
P = the number of connected components.

Source: http://en.wikipedia.org/wiki/Cyclomatic_complexity

For first program X, E = 11, N = 9, P = 1,  So M = 11-9+2*1 = 4
For second program Y, E = 10, N = 8, p = 1, So M = 10-8+2*1 = 4
For Third program X, E = 22, N = 17, p = 1, So M = 22-17+2*1 = 7`
There are 65 questions to complete.

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