| CS-GATE-2013 PAPER
Q. No. 1 – 25 Carry One Mark Each
1. Consider an undirected random graph of eight vertices. The probability that there
is an edge between a pair of vertices is ½. What is the expected number of
unordered cycles of length three?
(A) 1/8 (B) 1 (C) 7 (D) 8
Ans. (C)
Exp:
1
P(edge)
2
=
Number of ways we can choose the vertices out of 8 is
3 c 8
(Three edges in each cycle)
Expected number of unordered cycles of length 3 =
3
3
C
1
8 7
2
× =
2. Which of the following statements is/are TRUE for undirected graphs?
P: Number of odd degree vertices is even.
Q: Sum of degrees of all vertices is even.
(A) P only (B) Q only
(C) Both P and Q (D) Neither P nor Q
Ans: (C)
Exp: Q: Sum of degrees of all vertices = 2 × (number of edges)
3. Function f is known at the following points:
x 0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3.0
f(x) 0 0.09 0.36 0.81 1.44 2.25 3.24 4.41 5.76 7.29 9.00
The val
ue of ( ) 3
0
f x dx computed using the trapezoidal rule is
(A) 8.983 (B) 9.003 (C) 9.017 (D) 9.045
Ans: (D)
Exp: ( ) ( ) ( ) ( ( ) ( ) ( )) 3
0 10 1 2 9
0
h
f x dx f x f x 2 f x f x ... f x
2
= + + + + +
( ) 0.3
9.00 2 25.65 9.045
2
=
+ =
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4. Which one of the following functions is continuous at x = 3 ?
( ) ( )
2, if x 3
A f x x 1, if x 3
x 3
, if x 3
3
=
= − >
+ <
( ) ( ) 4, if x 3
B f x
8 x if x 3
=
=
− ¹
( ) ( ) x 3, if x 3
C f x
x 4 if x 3
+ £
=
− >
( ) ( ) 3
1
D f x , if x 3
x 27
= ¹
−
Ans: (A)
Exp: ( ) ( ) ( )
x 3 x 3
lim f x lim x 1 2 f 3
® + ® +
= − = =
( ) ( )
( )
x 3 x 3
x 3
lim f x lim 2 f 3
3
f x is continuous at x 3
® − ® −
+
= = =
\ =
5. Which one of the following expressions does NOT represent exclusive NOR of x
and y?
(A) xy + x ' y ' (B) x Å y ' (C) x 'Å y (D) x 'Å y '
Ans: (D)
Exp: (A) x y = xy + x y
(B) x Å y = xy + x y = xy + x y = x y
(C) x Å y = (x) y + x y = x y + xy = x y
(D) x Å y = (x) y + x y = x Å y
6. In a k-way set associative cache, the cache is divided into v sets, each of which
consists of k lines. The lines of a set are placed in sequence one after another.
The lines in set s are sequenced before the lines in set (s+1). The main memory
blocks are numbered 0 onwards. The main memory block numbered j must be
mapped to any one of the cache lines from
(A) (j mod v) *k to (j mod v) *k + (k − 1)
(B) (j mod v) to (j mod v) + (k − 1)
(C) (j mod k) to ( j mod k) + (v − 1)
(D) (j mod k) * v to ( j mod k) * v + (v − 1)
Ans: (B)
Exp: Set number in the cache = (main memory block number) MOD number of sets in
the cache.
As the lines in the set are placed in sequence, we can have the lines from 0 to K
– 1 in the set.
Number of sets = v
Main memory block number = j
First line = (j mod v); last line = (j Mod v) + (k – 1)
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7. What is the time complexity of Bellman-Ford single-source shortest path
algorithm on a complete graph of n vertices?
(A) ( 2 ) Q n (B) ( 2 ) Q n logn (C) ( 3 ) Q n (D) ( 3 ) Q n logn
Ans: (C)
Exp: Bellman-ford time complexity: Q( V × E )
For complete graph:
n(n 1)
E
2
−
=
3
V n
n(n 1)
n (n )
2
=
−
\ Q × = Q
8. Which of the following statements are TRUE?
(1) The problem of determining whether there exists a cycle in an undirected
graph is in P.
(2) The problem of determining whether there exists a cycle in an undirected
graph is in NP.
(3) If a problem A is NP-Complete, there exists a non-deterministic polynomial
time algorithm to solve A.
(A) 1,2 and 3 (B) 1 and 2 only (C) 2 and 3 only (D) 1 and 3 only
Ans: (A)
Exp: 1. Cycle detection using DFS: 2 O(V + E) = O(V ) and it is polynomial problem
2. Every P-problem is NP (sin ce P Ì NP)
3. NP − complete Î NP
Hence, NP-complete can be solved in non-deterministic polynomial time
9. Which of the following statements is/are FALSE?
(1) For every non-deterministic Turing machine, there exists an equivalent
deterministic Turing machine.
(2) Turing recognizable languages are closed under union and complementation.
(3) Turing decidable languages are closed under intersection and
complementation
(4) Turing recognizable languages are closed under union and intersection.
(A) 1 and 4 only (B) 1 and 3 only (C) 2 only (D) 3 only
Ans: (C)
Exp: (1) NTM @ DTM
(2) RELs are closed under union & but not complementation
(3) Turing decidable languages are recursive and recursive languages are closed
under intersection and complementation
(4) RELs are closed under union & intersection but not under complementation
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10. Three concurrent processes X, Y, and Z execute three different code segments
that access and update certain shared variables. Process X executes the P
operation (i.e., wait) on semaphores a, b and c; process Y executes the P
operation on semaphores b, c and d; process Z executes the P operation on
semaphores c, d, and a before entering the respective code segments. After
completing the execution of its code segment, each process invokes the V
operation (i.e., signal) on its three semaphores. All semaphores are binary
semaphores initialized to one. Which one of the following represents a deadlockfree
order of invoking the P operations by the processes?
(A) X :P (a )P (b)P (c) Y :P (b )P (c)P (d) Z :P (c )P (d)P (a)
(B) X :P (b )P (a)P (c) Y :P (b )P (c)P (d) Z :P (a )P (c)P (d)
(C) X :P (b )P (a)P (c) Y :P (c )P (b)P (d) Z :P (a )P (c)P (d)
(D) X :P (a )P (b)P (c) Y :P (c )P (b)P (d) Z :P (c )P (d)P (a)
Ans: (B)
Exp: Suppose X performs P(b) and preempts, Y gets chance, but cannot do its first
wait i.e., P(b), so waits for X, now Z gets the chance and performs P(a) and
preempts, next X gets chance. X cannot continue as wait on ‘a’ is done by Z
already, so X waits for Z. At this time Z can continue its operations as down on c
and d. Once Z finishes, X can do its operations and so Y. In any of execution
order of X, Y, Z one process can continue and finish, such that waiting is not
circular. In options (A),(C) and (D) we can easily find circular wait, thus deadlock
11. An index is clustered, if
(A) it is on a set of fields that form a candidate key.
(B) it is on a set of fields that include the primary key.
(C) the data records of the file are organized in the same order as the data
entries of the index.
(D) the data records of the file are organized not in the same order as the data
entries of the index.
Ans: (C)
Exp: Clustered index is built on ordering non key field and hence if the index is
clustered then the data records of the file are organized in the same order as the
data entries of the index.
12. Assume that source S and destination D are connected through two intermediate
routers labeled R. Determine how many times each packet has to visit the
network layer and the data link layer during a transmission from S to D.
(A) Network layer – 4 times and Data link layer-4 times
(B) Network layer – 4 times and Data link layer-3 times
(C) Network layer – 4 times and Data link layer-6 times
(D) Network layer – 2 times and Data link layer-6 times
S R R D
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Ans: (C)
Exp:
From above given diagram, its early visible that packet will visit network layer 4
times, once at each node [S, R, R, D] and packet will visit Data Link layer 6
times. One time at S and one time at D, then two times for each intermediate
router R as data link layer is used for link to link communication.
Once at packet reaches R and goes up from physical –DL-Network and second
time when packet coming out of router in order Network – DL- Physical
13. The transport layer protocols used for real time multimedia, file transfer, DNS
and email, respectively are
(A) TCP, UDP, UDP and TCP (B) UDP, TCP, TCP and UDP
(C) UDP, TCP, UDP and TCP (D) TCP, UDP, TCP and UDP
Ans: (C)
Exp: Real time multimedia needs connectionless service, so under lying transport layer
protocol used is UDP
File transfer rums over TCP protocol with port no-21
DNS runs over UDP protocol within port no-53
Email needs SMTP protocol which runs over TCP protocol within port no – 25
14. Using public key cryptography, X adds a digital signature s to message M,
encrypts <M, s >, and sends it to Y, where it is decrypted. Which one of the
following sequences of keys is used for the operations?
(A) Encryption: X’s private key followed by Y’s private key; Decryption: X’s public
key followed by Y’s public key
(B) Encryption: X’s private key followed by Y’s public key; Decryption: X’s public
key followed by Y’s private key
(C) Encryption: X’s public key followed by Y’s private key; Decryption: Y’s public
key followed by X’s private key
(D) Encryption: X’s private key followed by Y’s public key; Decryption: Y’s private
key followed by X’s public key
Ans: (D)
Exp:
X Y
X public
X private
−
−
Y public
Y private
−
−
Application
Transport
Network
Datalink
Physical
Network
DataLink
Physical
Network
DataLink
Physical
Application
Transport
Network
Datalink
Physical
( )
Source
S R R
( )
Destination
D
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15. Match the problem domains in Group I with the solution technologies in Group
II.
Group I Group II
(p) Services oriented computing (1) Interoperability
(q) Heterogeneous communicating systems (2) BPMN
(R) Information representation (3) Publish-find bind
(S) Process description (4) XML
(A) P – 1, Q – 2, R – 3, S – 4 (B) P – 3, Q – 4, R – 2, S – 1
(C) P – 3, Q – 1, R – 4, S – 2 (D) P – 4, Q – 3, R – 2, S – 1
Ans: (C)
16. A scheduling algorithm assigns priority proportional to the waiting time of a
process. Every process starts with priority zero(the lowest priority). The
scheduler re-evaluates the process priorities every T time units and decides the
next process to schedule. Which one of the following is TRUE if the processes
have no I/O operations and all arrive at time zero?
(A) This algorithm is equivalent to the first-come-first-serve algorithm.
(B) This algorithm is equivalent to the round-robin algorithm.
(C) This algorithm is equivalent to the shortest-job-first algorithm.
(D) This algorithm is equivalent to the shortest-remaining-time-first algorithm.
Ans: (B)
Exp: The given scheduling definition takes two parameters, one is dynamically
assigned process priority and the other is ‘T’ time unit to re-evaluate the process
priorities.
This dynamically assigned priority will be deciding processes order in ready queue
of round robin algorithm whose time quantum is same as ‘T’ time units. As all the
processes are arriving at the same time, they will be given same priority but soon
after first ‘T’ time burst remaining processes will get higher priorities
17. What is the maximum number of reduce moves that can be taken by a bottomup
parser for a grammar with no epsilon- and unit-production
(i.e., of type A ® Î and A ® a) to parse a string with n tokens?
(A) n/2 (B) n-1 (C) 2n-1 (D) 2n
Source has to encrypt with its pr ivate key for
forming Digital signature for Authentication.
Encryption
source has to encrypt the M, with Y ' s
public key to send it confidentially
Destination Y has to decrypt first
Decryption wit
s
h its private key, then decrypt
using source public key
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Ans: (C)
Exp: string = abcd
2× (4) − 1 = 7 reductions
2n − 1reductions required
Note : Unit productions is given as A ® a, it was typo
Above reductions are not in reverse of RMD but when they are reduced in bottom
up parsing we will get same number of reductions.
18. Consider the languages { } 1 2 L = F and L = a . Which one of the following
represents * *
1 2 1 L L UL ?
(A) {Î} (B) F (C) a * (D) {e, a}
Ans: (A)
EXP: Concatenation of empty language with any language will give the empty
language and 1 L * = F* =Î. Hence * *
1 2 1 L L UL ={Î}
19. Which one of the following is the tightest upper bound that represents the time
complexity of inserting an object into a binary search tree of n nodes?
(A) O(1) (B) O(log n) (C) O(n) (D) O(n log n)
Ans: (C)
Exp: For skewed binary search tree on n nodes, the tightest upper bound to insert a
node is O(n)
20. Which one of the following is the tightest upper bound that represents the
number of swaps required to sort n numbers using selection sort?
(A) O(log n) (B) O(n) (C) O(n log n) (D) O(n2)
S
YD
XCD
ABCD
ABCd
ABcd
Abcd
abcd
(7)
(6 : Y ® XC)
(5 : X ® AB)
(4 :D ® d)
(3 :C ® c)
(2 :B ® b)
(1: A ® a)
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Ans: (B)
Exp: The maximum number of swaps that takes place in selection sort on n numbers is
n
21. In the following truth table, V = 1 if and only if the input is valid.
Inputs Outputs
D0 D1 D2 D3 X0 X1 V
0 0 0 0 X X 0
1 0 0 0 0 0 1
0 1 0 0 1 1
1 X 1 0 0 1
X X X 1 1 1 1
What function does the truth table represent?
(A) Priority encoder (B) Decoder
(C) Multiplexer (D) Demultiplexer
Ans: (A)
Exp: 4 to 2 priority encoder.
22. The smallest integer than can be represented by an 8-bit number in 2’s
complement form is
(A) -256 (B) -128 (C) -127 (D) 0
Ans: (B)
Exp: −28−1 = −128. Range is -2(n-1) to +2(n-1)-1
23. Which one of the following does NOT equal
2
2
2
1 x x
1 y y ?
1 z z
( )
( )
( )
( )
1 x x 1 x 1
A 1 y y 1 y 1
1 z z 1 z 1
+ +
+ +
+ +
( )
2
2
2
1 x 1 x 1
B 1 y 1 y 1
1 z 1 z 1
+ +
+ +
+ +
( )
2 2
2 2
2
0 x y x x
C 0 y z y z
1 z z
− −
− − ( )
2 2
2 2
2
2 x y x y
D 2 y z y z
1 z z
+ +
+ +
Ans: (A)
If matrix B is obtained from matrix A by replacing the lth row by itself plus k
times the mth row, for l ¹ m then det(B)=det(A). With this property given
matrix is equal to the matrices given in options (B),(C) and (D).
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24. Suppose p is number of cars per minute passing through a certain road junction
between 5 PM and 6PM, and p has a Poisson distribution with mean 3. What is
the probability of observing fewer than 3 cars during any given minute in this
interval?
(A) ( 3 ) 8 / 2e (B) ( 3 ) 9 / 2e (C) ( 3 ) 17 / 2e (D) ( 3 ) 26 / 2e
Ans: (C)
Exp:
( ) ( ) ( ) ( )
( )
0 1 2
3
3 3
3
3
P p 3 P p 0 P p 1 P p 2
e e e
where 3
0! 1! 2!
e 9
e e 3
2
9 17
e 1 3
2 2e
−μ −μ −μ
−
− −
−
< = = + = + =
μ μ μ
= + + μ =
×
= + × +
= + + =
25. A binary operation Å on a set of integers is defined as 2 2 x Å y = x + y . Which one
of the following statements is TRUE aboutÅ ?
(A) Commutative but not associative (B) Both commutative and associative
(C) Associative but not commutative (D) Neither commutative nor associative
Ans: (A)
Exp:
2 2 2 2 x y x y y x y x
commutative
Å = + = + = Å
\
Not associative, since, for example
(1 Å 2) Å 3 ¹ 1 Å (2 Å 3)
Q. No. 26 – 55 Carry Two Marks Each
26. Which one of the following is NOT logically equivalent to ¬$x ("y (a) Ù "z (b))?
(A) "x ($z (¬b) ® "y (a)) (B) "x ("z (b) ® $y (¬a))
(C) "x ("y (a) ® $z (¬b)) (D) "x ($y (¬a) ® $z (¬b))
Ans: (A)
Exp: ¬$x ("y (a) Ù "z (b)) º " ×
"y (a) ®$z (¬b) option "C"
( )
( ) ( )
( ) ( )
( ) ( )
P q p q
x z y option "B"
p q q p
x y z option "D"
p q p q
¬ Ù º ¬
º "
" b ® $ ¬a
º ¬ ¬
º "
" a ® $ ¬b
¬ Ù º ¬
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27. A RAM chip has a capacity of 1024 words of 8 bits each (1K × 8) . The number of
2 × 4 decoders with enable line needed to construct a
16K × 16 RAM from 1K × 8 RAM is
(A) 4 (B) 5 (C) 6 (D) 7
Ans: (B)
Exp. RAM chip size = 1k × 8
1024 words of 8 bits each
RAM to construct =16k × 16
Number of chips required =
16k 16
16 2
1k 8
×
= ×
×
[16 chips vertically with each
having 2 chips horizontally]
So to select one chip out of 16 vertical chips, we need 4 x 16 decoder.
Available decoder is – 2 x 4 decoder
To be constructed is 4 x 16 decoder
So we need 5, 2 x 4 decoder in total to construct 4 x 16 decoder.
28. Consider an instruction pipeline with five stages without any branch prediction:
Fetch Instruction (FI), Decode Instruction (DI), Fetch Operand (FO), Execute
Instruction (EI) and Write Operand (WO). The stage delays for FI, DI, FO, EI and
WO are 5 ns, 7 ns, 10 ns, 8 ns and 6 ns, respectively. There are intermediate
storage buffers after each stage and the delay of each buffer is 1 ns. A program
consisting of 12 instructions 1 2 3 12 I , I , I ,......I is executed in this pipelined
processor. Instruction 4 I is the only branch instruction and its branch target is 9 I .
If the branch is taken during the execution of this program, the time (in ns)
needed to complete the program is
(A) 132 (B) 165 (C) 176 (D) 328
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Ans: (C)
Exp: Total clock slots taken are 16. Each slot will take maximum of {5, 7, 10, 8 ,7}
=10.
Hence total slots for all the instructions =
16 × 10 + 16 (pipeline delay) = 176
29. Consider the following operation along with Enqueue and Dequeue operations on
queues, where k is a global parameter
( )
( ) ( )
( )
MultiDequeue Q {
m k
while Q is not empty and m 0 {
Dequeue Q
m m 1
}
}
=
>
= −
What is the worst case time complexity of a sequence of n queue operations on
an initially empty queue?
(A) Q (n) (B) Q (n + k) (C) Q (nk) (D) ( 2 ) Q n
Ans: (C)
30. The preorder traversal sequence of a binary search tree is 30, 20, 10, 15, 25, 23,
39, 35, 42. Which one of the following is the postorder traversal sequence of the
same tree?
(A) 10,20,15,23,25,35,42,39,30 (B) 15,10,25,23,20,42,35,39,30
(C) 15,20,10,23,25,42,35,39,30 (D) 15,10,23,25,20,35,42,39,30
Ans: (D)
Exp:
Pr eorder : 30,20,10,15,25,23,39,35,42
Inorder :10,15,20,23,25,30,35,39,42
30
20 39
10
15
25
23
35 42
BST :
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31. What is the return value of f (p,p) if the value of p is initialized to 5 before the
call? Note that the first parameter is passed by reference, whereas the second
parameter is passed by value.
( )
( )
( )
int f int & x, int c {
c c 1;
if c 0 return 1;
x x 1;
return f x,c * x;
}
= −
==
= +
(A) 3024 (B) 6561 (C) 55440 (D) 161051
Ans: (B)
Exp:
32. Which of the following is/are undecidable?
1. G is a CFG. Is L (G) = F?
2. G is a CFG. IS L (G) = S * ?
3. M is a Turning machine. Is L(M) regular?
4. A is a DFA and N is a NFA. Is L (A) = L (N) ?
(A) 3 only (B) 3 and 4 only (C) 1, 2 and 3 only (D) 2 and 3 only
Ans: (D)
Exp: There is an algorithm to check whether the given CFG is empty, finite or infinite
and also to convert NFA to DFA hence 1 and 4 are decidable.
33. Consider the following two sets of LR(1) items of an LR(1) grammar
X c.X, c / d X c.X, $
X .cX, c / d X .cX, $
X .d, c / d X .d, $
® ®
® ®
® ®
Which of the following statements related to merging of the two sets in the
corresponding LALR parser is/are FALSE?
( )
( )
( )
( )
( )
( )
( )
( )
( )
f x, 5
1
f x, 4 x
2
f x, 3 x
3
f x, 2 x
4
f x, 1 x
×
×
×
×
9 6 7 8 9
9
9 × 9
9 × 9 × 9
9 9 9 9
6561
× × ×
=
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1. Cannot be merged since look aheads are different
2. Can be merged but will result in S–R conflict
3. Can be merged but will result in R–R conflict
4. Cannot be merged since goto on c will lead to two different sets
(A) 1 only (B) 2 only (C) 1 and 4 only (D) 1, 2, 3 and 4
Ans: (D)
Exp:
1. Merging of two states depends on core part (production rule with dot
operator), not on look aheads.
2. The two states are not containing Reduce item ,So after merging, the merged
state can not contain any S-R conflict
3. As there is no Reduce item in any of the state, so can’t have R-R conflict.
4. Merging of stats does not depend on further goto on any terminal.
So all statements are false.
34. A certain computation generates two arrays a and b such that
a
i = f (i) for 0 £ i < n and b
i = g(a
i ) for 0 £ i < n. Suppose this computation is
decomposed into two concurrent processes X and Y such that X computes the
array a and Y computes the array b. The processes employ two binary
semaphores R and S, both initialized to zero. The array a is shared by the two
processes. The structures of the processes are shown below.
( ) ( )
( ) ( )
( ) ( )
Pr ocess X; Pr ocess Y;
private i; private i;
for i 0; i n; i { for i 0; i n; i {
a i f i ; EntryY R, S ;
ExitX R, S ; b i g a i ;
} }
= < + + = < + +
=
=
Which one of the following represents the CORRECT implementations of ExitX
and EntryY?
x c.X, c / d
x .cX, c / d
x .d,c / d
®
®
®
x c.X,$
x .cX,$
x .d,$
®
®
®
x c.X, c / d /$
x .cX,c / d /$
x .d, c / d /$
®
®
®
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(A) ( )
( )
( )
( )
( )
( )
ExitX R, S {
P R ;
V S ;
}
EntryY R, S {
P S ;
V R ;
}
(B) ( )
( )
( )
( )
( )
( )
ExitX R, S {
V R ;
V S ;
}
EntryY R, S {
P R ;
P S ;
}
(C) ( )
( )
( )
( )
( )
( )
ExitX R, S {
P S ;
V R ;
}
EntryY R, S {
V S ;
P R ;
}
(D) ( )
( )
( )
( )
( )
( )
ExitX R, S {
V R ;
P S ;
}
EntryY R, S {
V S ;
P R ;
}
Ans: (B)
35. The following figure represents access graphs of two modules M1 and M2. The
filled circles represent methods and the unfilled circles represent attributes. IF
method m is moved to module M2 keeping the attributes where they are, what
can we say about the average cohesion and coupling between modules in the
system of two modules?
(A) There is no change.
(B) Average cohesion goes up but coupling is reduced
(C) Average cohesion goes down and coupling also reduces
(D) Average cohesion and coupling increase
Ans: (B)
36. In an IPv4 datagram, the M bit is 0, the value of HLEN is 10, the value of total
length is 400 and the fragment offset value is 300. The position of the datagram,
the sequence numbers of the first and the last bytes of the payload, respectively
are
(A) Last fragment, 2400 and 2789 (B) First fragment, 2400 and 2759
(C) Last fragment, 2400 and 2759 (D) Middle fragment, 300 and 689
Module M1 Module M2
m
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Ans: (C)
Exp: M= 0 – Means there is no fragment after this, i.e. Last fragment
HLEN=10 - So header length is 4×10=40, as 4 is constant scale factor
Total Length = 400(40 Byte Header + 360 Byte Payload)
Fragment Offset = 300, that means 300×8 Byte = 2400 bytes are before this last
fragment
So the position of datagram is last fragment
Sequence number of First Byte of Payload = 2400 (as 0 to 2399 Sequence no are
used)
Sequence number of Last Byte of Payload = 2400+360-1=2759
37. Determine the maximum length of cable (in km) for transmitting data at a rate of
500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal
speed in the cable to be 2,00,000 km/s
(A) 1 (B) 2 (C) 2.5 (D) 5
Ans: (B)
Exp:
500 × 106 bits − − − − − −1 sec
8 4
4
4 8 4
5
5
4 4
5 10 10 1
10 bits sec sec
10 5 10 5 10
1 sec 2 10 km
1 2 10
sec 4km
5 10 5 10
4
Maximum length of cable 2 km
2
×
\ − − − − − − = =
× ×
− − − − − − ×
×
\ − − − − − =
× ×
\ = =
38. Consider the following relational schema.
Students(rollno: integer, sname: string)
Courses(courseno: integer, cname: string)
Registration(rollno: integer, courseno; integer, percent: real)
Which of the following queries are equivalent to this query in English?
“Find the distinct names of all students who score more than 90% in the course
numbered 107”
(I) SELECT DISTINCT S.sname
FROM Students as S, Registration as R
WHERE R.rollno=S.rollno AND R.Courseno=107 AND R.percent>90
(II) ( ( )) sname courseno 107 percent 90 Re gistration Students = P s Ù >
(III) {T | $S Î Students, $R Î Re gistration (S.rol lno = R.rollno Ù
R.courseno = 107 Ù R.percent > 90 Ù T.sname = S.name)}
(IV) { ( )} N R P R N R P P < S >| $S $R < S ,S >Î StudentsÙ < S ,107,R >Î Re gistration Ù R > 90
(A) I, II, III and IV (B) I, II and III only
(C) I, II and IV only (D) II, III and IV only
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Ans: (A)
Exp: Four queries given in SQL, RA, TRC and DRC in four statements respectively
retrieve the required information.
39. A shared variable x, initialized to zero, is operated on by four concurrent processes
W, X, Y, Z as follows. Each of the processes W and X reads x from memory,
increments by one, stores it to memory, and then terminates. Each of the processes
Y and Z reads x from memory, decrements by two, stores it to memory, and then
terminates. Each process before reading x invokes the P operation (i.e., wait) on a
counting semaphore S and invokes the V operation (i.e., signal) on the semaphore S
after storing x to memory. Semaphore S is initialized to two. What is the maximum
possible value of x after all processes complete execution?
(A) –2 (B) –1 (C) 1 (D) 2
Ans: (D)
Exp:
W X Y Z
1 R (x) R (x) R (x) R (x)
2 x++ x++ x=x-2; x=x-2;
3 w(x) w(x) w(x) w(x)
R(x) is to read x from memory, w(x) is to store x in memory
(I) ( ) 1 w x 0
W is Preempted
(II) ( ) 1 2 3 Y , Y , Y x −2
Y is completed
(III) ( ) 1 2 3 Z ,Z ,Z x −4
Z is completed
(IV)
( ) 2 3 W ,W x 1
It increments local copy of x and stores & W is completed
(V) ( ) 1 2 3 X ,X ,X x 2
X is completed
Maximum value of x = 2
40. Consider the DFA given below.
Which of the following are FALSE?
1
0 0
0,1
1
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1. Complement of L(A) is context–free
2. L (A) = L ((11 * 0 + 0) (0 + 1) * 0 *1 *)
3. For the language accepted by A, A is the minimal DFA
4. A accepts all strings over {0, 1} of length at least 2
(A) 1 and 3 only (B) 2 and 4 only (C) 2 and 3 only (D) 3 and 4 only
Ans: (D)
Exp:
(1) L(A) is regular, its complement is also regular and if it is regular it is also
context free.
(2) L (A) = (11* 0 + 0) (0 + 1) * 0 * 1 * = 1 * 0 (0 + 1) *
Language has all strings where each string contains ‘0’.
(3) A is not minimal, it can be constructed with 2 states
(4) Language has all strings, where each string contains ‘0’. (atleast length one)
41. Consider the following languages
{ }
{ }
p q r
1
p q r
2
L 0 1 0 |p,q,r 0
L 0 1 0 |p,q,r 0, p r
= ³
= ³ ¹
Which one of the following statements is FALSE?
(A) 2 L is context–free
(B) 1 2 L L is context–free
(C) Complement of 2 L is recursive
(D) Complement of 1 L is context–free but not regular
Ans: (D)
Exp: { P q r }
1 L = 0 1 0 p,q,r ³ 0 is regular
{ P q r }
2 L = 0 1 0 p,q,r ³ 0, p ¹ r is CFL
(A) ( ) 2 L is CFL True
(B) ( ) 1 2 L Ç L = CFL True
(C) ( ) 2 L complement is recursive True
(D) 1 ( ) 1 1 L complement is CFL but not regular False as L isregular L is regular
1
0
0, 1
0
1
A :
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42. Consider the following function
( )
( )
( )
( )
int unknown int n {
int i, j, k 0;
for i n / 2; i n; i
for j 2; j n; j j * 2
k k n / 2;
return k ;
}
=
= <= + +
= <= =
= +
(A) ( 2 ) Q n (B) ( 2 ) Q n logn (C) ( 3 ) Q n (D) ( 3 ) Q n logn
Ans: (B)
Exp:
n n n
i , 1, 2, n
2 2 2
= + + − − − − −
Repeats
n n
to n 1 times
2 2
= +
( )
( )
2 3 4 J 2, 2 , 2 , 2 , n
k nlogn
n
k k
2
= − − − − −
= Q
= +
n n n
k logn times logn
2 2 2
= + + − − − − =
( 2 )
n n n n
logn logn logn 1 times
2 2 2 2
n n
1 . logn
2 2
n logn
= + + − − − − +
= +
= Q
43. The number of elements that can be sorted in Q(logn) time using heap sort is
(A) Q(1) (B) Q( log n) (C)
log n
log log n
Q
(D) Q(log n)
Ans: (A)
Exp: After constructing a max-heap in the heap sort , the time to extract
maximum element and then heapifying the heap takes Q (log n) time by which
we could say that Q (log n) time is required to correctly place an element in
sorted array. If Q(logn) time is taken to sort using heap sort, then number of
elements that can be sorted is constant which is Q(1)
44. Consider a hard disk with 16 recording surfaces (0 − 15) having 16384 cylinders
(0 − 16383) and each cylinder contains 64 sectors (0 − 63) . Data storage capacity
in each sector is 512 bytes. Data are organized cylinder–wise and the addressing
format is <cylinder no., sector no.>. A file of size 42797 KB is stored in the disk
and the starting disk location of the file is <1200, 9, 40>. What is the cylinder
number of the last sector of the file, if it is stored in a contiguous manner?
(A) 1281 (B) 1282 (C) 1283 (D) 1284
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Ans: (D)
42797 1024
42797 KB 85594 sec tors
512
×
º =
Starting is 1200,9,40 contains total 24 + (6 × 64) = 408 sec tors
Next, 1201, --------, 1283 cylinders contains total 1024 × 83 = 84992 sec tors
( each cylinder contains 16 64 1024 sec tors)
Total 408 84992 85400 sec tors
× =
\ = + =
\ The required cylinder number is 1284 which will contain the last sector of
the file
45. Consider the following sequence of micro–operations
MBR PC
MAR X
PC Y
Memory MBR
¬
¬
¬
¬
Which one of the following is a possible operation performed by this sequence?
(A) Instruction fetch (B) Operand fetch
(C) Conditional branch (D) Initiation of interrupt service
Ans: (D)
Exp: PC content is stored in memory via MBR and PC gets new address from Y. It
represents a function call (routine), which is matching with interrupt service
initiation
46. The line graph L(G) of a simple graph G is defined as follows:
• There is exactly one vertex v(e) in L(G) for each edge e in G.
• For any two edges e and e’ in G, L(G) has an edge between v(e) and v(e’), if
and only if e and e’ are incident with the same vertex in G.
Which of the following statements is/are TRUE?
(P) The line graph of a cycle is a cycle.
(Q) The line graph of a clique is a clique.
(R) The line graph of a planar graph is planar.
(S) The line graph of a tree is a tree.
(A) P only (B) P and R only (C) R only (D) P, Q and S only
Ans: (B)
Exp: P) The line graph of a cycle is a cycle
( ) 4 V e
( ) 1 V e
( ) 3 V e
( ) 2 V e
a b
d c
( ) 1 V e
( ) 2 V e
( ) 3 V e
( ) 4 L (G) : V e
is also cycle graph
G
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R) Line graph of planar graph is planar
S)
The line graph of a tree need not be tree.
47. What is the logical translation of the following statement?
“None of my friends are perfect.”
(A) $x (F (x) Ù ¬P (x)) (B) $x (¬F (x) Ù P (x))
(C) $x (¬F (x) Ù ¬P (x)) (D) ¬$x (F (x) Ù P (x))
Ans: (D)
Exp: “None of my friends are perfect”
( ( ) ( ))
( ( ) ( ))
( ( ) ( ))
x F x P x
x F x P x
x F x P x
= " ®¬
= " ¬ Ú ¬
= ¬$ Ù
Common Data Questions: 48 & 49
The procedure given below is required to find and replace certain characters inside an
input character string supplied in array A. The characters to be replaced are
supplied in array oldc, while their respective replacement characters are supplied
in array newc. Array A has a fixed length of five characters, while arrays oldc and
newc contain three characters each. However, the procedure is flawed
( )
( )
( )
( )
void find _ and _ replace char * A, char * oldc, char *newc {
for int i 0; i 5; i
for int j 0; j 3; j
if A i oldc j A i newc j ;
}
= < + +
= < + +
= =
=
The procedure is tested with the following four test cases
(1) oldc = "abc ", newc = "dab" (2) oldc = " cde", newc = "bcd"
(3) oldc = "bca", newc = "cda" (4) oldc = "abc ", newc = "bac "
48. The tester now tests the program on all input strings of length five consisting of
characters ‘a’, ‘b’, ‘c’, ‘d’ and ‘e’ with duplicates allowed. If the tester carries out
this testing with the four test cases given above, how many test cases will be able to
capture the flaw?
(A) Only one (B) Only two (C) Only three (D) All four
( ) 2 V e
c d
( ) 3 V e
( ) 1 V e
( ) 2 ( ) V e 3 V e
a
( ) 1 V e
b L (G) :
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Ans: (B)
Exp: Flaw in this given procedure is that one character of Array ‘A’ can be replaced by
more than one character of newc array, which should not be so.Test case (3) and
(4) identifies this flaw as they are containing ‘oldc’ and ‘newc’ array characters
arranged in specific manner. Following string can reflect flaw, if tested by test
case (3).
initially i= j= 0
Next i= 0 & j=1
Likewise single character ‘b’ in A is replaced by ‘c’ and then by ‘d’.
Same way test case (4) can also catch the flaw.
49. If array A is made to hold the string “abcde”, which of the above four test cases
will be successful in exposing the flaw in this procedure?
(A) None (B) 2 only (C) 3 and 4 only (D) 4 only
Ans: (C)
Exp: Now for string “abcde” in array A, both test case (3) and (4) will be successful in
finding the flaw, as explained in above question.
Common Data Questions: 50 & 51
The following code segment is executed on a processor which allows only register
operands in its instructions. Each instruction can have almost two source operands
and one destination operand. Assume that all variables are dead after this code
segment
( )
c a b;
d c * a;
e c a;
x c * c;
if x a {
y a* a;
}
else {
d d * d;
e e * e;
}
= +
=
= +
=
>
=
=
=
A ="bc da" oldc ="bc a"
j 0
=
i 0
=
newc =" c da"
j 0
=
b = b so replaced by c
A ="c c da" oldc ="bc a"
j 1
=
i 0
=
newc =" c da"
j 1
=
c = c so replaced by d
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50. Suppose the instruction set architecture of the processor has only two registers. The
only allowed compiler optimization is code motion, which moves statements from one
place to another while preserving correctness. What is the minimum number of spills
to memory in the compiled code?
(A) 0 (B) 1 (C) 2 (D) 3
Ans: (C)
Exp:
c = a + b; 2 1 2 R ¬ R + R
d = c * a; 2 1
spill ¬ R *R
e = c + a;
2 2 1
spill ¬ R + R
x = c * c; 2 2 2 R ¬ R *R
if (x > a) 2 1 CMP R R
JNG xxx (Jump if not
greater)
{y = a* a;} 1 1 1 R R *R
goto yyy
¬
else
{
d d * d;
e e * e;
}
=
=
1 1
2 2
1 1 1
2 2 2
xxx : R spill
R spill
R R *R
R R *R
yyy : Exit
¬
¬
¬
¬
In the above code total number of spills to memory is 2
51. What is the minimum number of registers needed in the instruction set architecture
of the processor to compile this code segment without any spill to memory? Do not
apply any optimization other than optimizing register allocation
(A) 3 (B) 4 (C) 5 (D) 6
Ans: (B)
Exp:
c = a + b; 2 1 2 R ¬ R + R
d = c * a; 3 2 1 R ¬ R *R
e = c + a;
4 2 1 R ¬ R + R
x = c * c; 2 2 2 R ¬ R *R
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if (x > a) 2 1 CMP R R
JNG xxx (Jump if not
greater)
{y = a* a;} 1 1 1 R R *R
goto yyy
¬
else
{
d d * d;
e e * e;
}
=
=
3 3 3
4 4 4
xxx : R R *R
R R *R
yyy : Exit
¬
¬
In the above code minimum number of registers are used = 4
Linked Answer Questions: Q.52 to Q.55 Carry Two Marks Each
Statement for Linked Answer Questions: 52 & 53
Relation R has eight attributes ABCDEFGH. Fields of R contain only atomic values.
F = {CH ® G, A ® BC, B ® CFH, E ® A, F ® EG} is a set of functional
dependencies (FDs) so that F+ is exactly the set of FDs that hold for R
52. How many candidate keys does the relation R have?
(A) 3 (B) 4 (C) 5 (D) 6
Ans: (B)
Exp: Candidate keys are AD, BD, ED and FD
53. The relation R is
(A) in INF, but not in 2NF (B) in 2NF, but not in 3NF
(C) in 3NF, but not in BCNF (D) in BCNF
Ans: (A)
Exp: A ® BC,B ® CFH and F ® EG are partial dependencies. Hence it is in 1NF but
not in 2NF
Statement for Linked Answer Questions: 54 & 55
A computer uses 46–bit virtual address, 32–bit physical address, and a three–level
paged page table organization. The page table base register stores the base
address of the first–level table ( ) 1 T ,which occupies exactly one page. Each entry of
1 T stores the base address of a page of the second–level table ( ) 2 T . Each entry of 2 T
stores the base address of a page of the third–level table ( ) 3 T Each entry of 3 T stores
a page table entry (PTE). The PTE is 32 bits in size. The processor used in the
computer has a 1 MB 16 way set associative virtually indexed physically tagged cache.
The cache block size is 64 bytes.
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54. What is the size of a page in KB in this computer?
(A) 2 (B) 4 (C) 8 (D) 16
Ans: (B)
Exp: As it is virtually indexed
Set size = 16 × 64 Bytes =210
20
10
10
2
number of sets 2
2
= =
Index of cache bits will be used as frame bits.
Frame size = 212=4K bytes
55. What is the minimum number of page colours needed to guarantee that no two
synonyms map to different sets in the processor cache of this computer?
(A) 2 (B) 4 (C) 8 (D) 16
Ans:
Q. No. 56 – 60 Carry One Mark Each
56. Complete the sentence:
Universalism is to particularism as diffuseness is to ________
(A) specificity (B) neutrality (C) generality (D) adaptation
Ans: (A)
The relation is that of antonyms
57. Were you a bird, you ___________ in the sky.
(A) would fly (B) shall fly
(C) should fly (D) shall have flown
Ans: (A)
58. Which one of the following options is the closest in meaning to the word given
below?
Nadir
(A) Highest (B) Lowest (C) Medium (D) Integration
32 − bits
TAG Index
12
20
32 − bits
20 1MB =2 byte
20 12
F.No offset
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Ans: (B)
Nadir in the lowest point on a curve
59. Choose the grammatically INCORRECT sentence:
(A) He is of Asian origin
(B) They belonged to Africa
(C) She is an European
(D) They migrated from India to Australia
Ans: (C)
60. What will be the maximum sum of 44, 42, 40, ... ?
(A) 502 (B) 504 (C) 506 (D) 500
Ans: (C)
The maximum sum is the sum of 44, 42,- - - - -2.
The sum of ‘n’ terms of an AP
( ) n
2a n 1 d
2
=
+ −
In this case, n = 22, a = 2 and d = 2
\ Sum =11
4 + 21× 2 = 11× 46 = 506
Q. No. 61 – 65 Carry Two Marks Each
61. Out of all the 2-digit integers between 1 and 100, a 2-digit number has to be
selected at random. What is the probability that the selected number is not
divisible by 7?
(A) 13/90 (B) 12/90 (C) 78/90 (D) 77/90
Ans: (D)
The number of 2 digit multiples of 7 = 13
\ Probability of choosing a number
Not divisible by
90 13 77
7
90 90
−
= =
62. A tourist covers half of his journey by train at 60 km/h, half of the remainder by
bus at 30 km/h and the rest by cycle at 10 km/h. The average of the tourist in
km/h during his entire journey is
(A) 36 (B) 30 (C) 24 (D) 18
Ans: (C)
Let the total distance covered be ‘D’
Now, average speed =
D
Total time taken
D 1 120
24km / hr
D D D 1 1 1 5
2 4 4 120 120 40
60 30 10
= = = =
+ +
+ +
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63. Find the sum of the expression
1 1 1 1
.....
1 2 2 3 3 4 80 81
+ + + +
+ + + +
(A) 7 (B) 8 (C) 9 (D) 10
Ans: (B)
The expression can be written as
( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2
2 1 3 2 81 80
1 2 2 3 80 81
− − −
+ + − − − − −
+ + +
( ) ( )
( )
2 1 1 2 ( 81 80)( 81 80)
1 2 80 81
− + − +
= + − − − − − − +
+ +
64. The current erection cost of a structure is Rs. 13,200. If the labour wages per
day increase by 1/5 of the current wages and the working hours decrease by
1/24 of the current period, then the new cost of erection in Rs. is
(A) 16,500 (B) 15,180 (C) 11,000 (D) 10,120
Ans: (B)
Let ‘W’ be the labour wages, and ‘T’ be the working hours.
Now, total cost is a function of W× T
Increase in wages = 20%
\ Revised wages = 1.2 W
Decrease in labour time =
100
%
24
1 23
Re vised time 1 T T
24 24
23
Re vised Total cos t 1.2 WT 1.15WT
24
1.15 13200 15180
\ = − =
\ = × =
= × =
65. After several defeats in wars, Robert Bruce went in exile and wanted to commit
suicide. Just before committing suicide, he came across a spider attempting
tirelessly to have its net. Time and again, the spider failed but that did not deter
it to refrain from making attempts. Such attempts by the spider made Bruce
curious. Thus, Bruce started observing the near-impossible goal of the spider to
have the net. Ultimately, the spider succeeded in having its net despite several
failures. Such act of the spider encouraged Bruce not to commit suicide. And
then, Bruce went back again and won many a battle, and the rest is history.
Which one of the following assertions is best supported by the above information?
(A) Failure is the pillar of success
(B) Honesty is the best policy
(C) Life begins and ends with adventures
(D) No adversity justifies giving up hope
Ans: (D)
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