Question 1 |

Consider the following logical inferences.

I1: If it rains then the cricket match will not be played.

The cricket match was played.

**Inference**: There was no rain.

I2: If it rains then the cricket match will not be played.

It did not rain.

**Inference**: The cricket match was played.

Which of the following is

**TRUE**?Both I1 and I2 are correct inferences | |

I1 is correct but I2 is not a correct inference | |

I1 is not correct but I2 is a correct inference | |

Both I1 and I2 are not correct inferences |

**GATE CS 2012**

**General Aptitude**

**Discuss it**

Question 1 Explanation:

The cricket match may not be played even if doesn't rain.

Question 2 |

Which of the following is

**TRUE**?Every relation in 3NF is also in BCNF | |

A relation R is in 3NF if every non-prime attribute of R is fully functionally dependent on every
key of R | |

Every relation in BCNF is also in 3NF | |

No relation can be in both BCNF and 3NF |

**GATE CS 2012**

**Database Design(Normal Forms)**

**Discuss it**

Question 3 |

What will be the output of the following C program segment?

char inchar = 'A'; switch (inchar) { case 'A' : printf ("choice A \n") ; case 'B' : printf ("choice B ") ; case 'C' : case 'D' : case 'E' : default: printf ("No Choice") ; }

No choice | |

Choice A | |

Choice A Choice B No choice | |

Program gives no output as it is erroneous |

**GATE CS 2012**

**Discuss it**

Question 3 Explanation:

There is no break statement in case ‘A’. If a case is executed and it doesn’t contain break, then all the subsequent cases are executed until a break statement is found. That is why everything inside the switch is printed.
Try following program as an exercise.

int main() { char inchar = 'A'; switch (inchar) { case 'A' : printf ("choice A \n") ; case 'B' : { printf ("choice B") ; break; } case 'C' : case 'D' : case 'E' : default: printf ("No Choice") ; } }

Question 4 |

Assuming P != NP, which of the following is true ?(A) NP-complete = NP (B) NP-complete P = (C) NP-hard = NP (D) P = NP-complete

A | |

B | |

C | |

D |

**GATE CS 2012**

**Discuss it**

Question 4 Explanation:

The answer is B (no NP-Complete problem can be solved in polynomial time). Because, if one NP-Complete problem can be solved in polynomial time, then all NP problems can solved in polynomial time. If that is the case, then NP and P set become same which contradicts the given condition.

**Related Article:**NP-Completeness | Set 1 (Introduction) P versus NP problem (Wikipedia)

Question 5 |

The worst case running time to search for an element in a balanced in a binary search tree with n2^n elements is

(A) (B) (C) (D)

A | |

B | |

C | |

D |

**GATE CS 2012**

**Discuss it**

Question 5 Explanation:

-> The search time in a binary search tree depends on the form of the tree, that is on the order in which its nodes were inserted. A pathological case: The n nodes were inserted by increasing order of the keys, yielding something like a linear list (but with a worse space consumption), with O(n) search time(in the case of skew tree).
-> A balanced tree is a tree where every leaf is “not more than a certain distance” away from the root than any other leaf.So in balanced tree, the height of the tree is balanced to make distance between root and leafs nodes a low as possible. In a balanced tree, the height of tree is log

_{2}(n). -> So , if a Balanced Binary Search Tree contains n2n elements then Time complexity to search an item: Time Complexity = log(n2^{n}) = log (n) + log(2^{n}) = log (n) +n = O(n) So Answer is C. See http://www.geeksforgeeks.org/data-structures-and-algorithms-set-28/ This solution is contributed by**Nirmal Bharadwaj**Question 6 |

X | |

X+Y | |

X xor Y | |

Y |

**GATE CS 2012**

**Digital Logic & Number representation**

**Discuss it**

Question 6 Explanation:

The value of f(X, Y) is same as X for all input pairs.
We see from truth table –
Column x= f(x,y)
So , f(x,y)=x
Ans is (A) part.

Question 7 |

T he decimal value 0.5 in IEEE single precision floating point representation has

fraction bits of 000…000 and exponent value of 0 | |

fraction bits of 000…000 and exponent value of −1 | |

fraction bits of 100…000 and exponent value of 0 | |

no exact representation |

**GATE CS 2012**

**Number Representation**

**Discuss it**

Question 7 Explanation:

The IEEE 754 standard specifies following distribution of bits:
Sign bit: 1 bit
Exponent width: 8 bits
Significand or Fraction: 24 (23 explicitly stored)
0.5 in base 10 means 1 X 2

^{-1}in base 2. So exponent bits have value -1 and all fraction bits are 0Question 8 |

A process executes the code

fork(); fork(); fork();The total number of child processes created is

3 | |

4 | |

7 | |

8 |

**GATE CS 2012**

**Process Management**

**Discuss it**

Question 8 Explanation:

Let us put some label names for the three lines

fork (); // Line 1 fork (); // Line 2 fork (); // Line 3 L1 // There will be 1 child process created by line 1 / \ L2 L2 // There will be 2 child processes created by line 2 / \ / \ L3 L3 L3 L3 // There will be 4 child processes created by line 3We can also use direct formula to get the number of child processes. With n fork statements, there are always 2^n – 1 child processes. Also see this post for more details.

Question 9 |

Consider the function f(x) = sin(x) in the interval [π/4, 7π/4]. The number and location(s) of the local minima of this function are

One, at π/2 | |

One, at 3π/2 | |

Two, at π/2 and 3π/2 | |

Two, at π/4 and 3π/2 |

**GATE CS 2012**

**Numerical Methods and Calculus**

**Discuss it**

Question 9 Explanation:

Question 10 |

The protocol data unit (PDU) for the application layer in the Internet stack is

Segment | |

Datagram | |

Message | |

Frame |

**GATE CS 2012**

**Application Layer**

**Discuss it**

Question 10 Explanation:

The Protocol Data Unit is the unit of communication at a particular layer.

The Layer 1 (Physical Layer) PDU is the bit or, more generally, symbol The Layer 2 (Data Link Layer) PDU is the frame. The Layer 3 (Network Layer) PDU is the packet. The Layer 4 (Transport Layer) PDU is the segment for TCP or the datagram for UDP. The Layer 5 (Application Layer) PDU is the data or message.

Question 11 |

Let A be the 2 × 2 matrix with elements a11 = a12 = a21 = +1 and a22 = −1. Then the eigenvalues of the matrix A

^{19}areA | |

B | |

C | |

D |

**GATE CS 2012**

**Linear Algebra**

**Discuss it**

Question 11 Explanation:

A = 1 1 1 -1 A^{2}= 2 0 0 2 A^{4}= A^{2}X A^{2}A^{4}= 4 0 0 4 A^{8}= 16 0 0 16 A^{16}= 256 0 0 256 A^{18}= A^{16}X A^{2}A^{18}= 512 0 0 512 A^{19}= 512 512 512 -512 Applying Characteristic polynomial 512-lamda 512 512 -(512+lamda) = 0 -(512-lamda)(512+lamda) - 512 x 512 = 0 lamda^{2 = 2 x 5122 }

Question 12 |

A | |

B | |

C | |

D |

**GATE CS 2012**

**Regular languages and finite automata**

**Discuss it**

Question 12 Explanation:

The given alphabet contains only one symbol {a} and the given NFA accepts all strings with any number of occurrences of ‘a’. In other words, the NFA accepts a+. Therefore complement of the language accepted by automata is empty string.

Question 13 |

What is the correct translation of the following statement into mathematical logic?
“Some real numbers are rational”

A | |

B | |

C | |

D |

**GATE CS 2012**

**Propositional and First Order Logic.**

**Discuss it**

Question 13 Explanation:

(A) "There exist some numbers which are either real OR rational" (B) "All real numbers are rational" (C) "There exist some numbers which are both real AND rational" (D) "There exist some numbers for which rational implies real" (See Propositional Logic for details) Clearly answer C is correct among all

Question 14 |

Given the basic ER and relational models, which of the following is INCORRECT?

An attribute of an entity can have more than one value | |

An attribute of an entity can be composite | |

In a row of a relational table, an attribute can have more than one value | |

In a row of a relational table, an attribute can have exactly one value or a NULL value |

**GATE CS 2012**

**ER and Relational Models**

**Discuss it**

Question 14 Explanation:

The term ‘entity’ belongs to ER model and the term ‘relational table’ belongs to relational model.
A and B both are true. ER model supports both multivalued and composite attributes See this for more details.
(C) is false and (D) is true. In Relation model, an entry in relational table can can have exactly one value or a NULL.

Question 15 |

Which of the following statements are TRUE about an SQL query?

P : An SQL query can contain a HAVING clause even if it does not have a GROUP BY clause Q : An SQL query can contain a HAVING clause only if it has a GROUP BY clause R : All attributes used in the GROUP BY clause must appear in the SELECT clause S : Not all attributes used in the GROUP BY clause need to appear in the SELECT clause

P and R | |

P and S | |

Q and R | |

Q and S |

**GATE CS 2012**

**SQL**

**Discuss it**

Question 15 Explanation:

According to standard SQL answer should be C. Refer
If we talk about different SQL implementations like MySQL, then option B is also right. But in question they seem to be talking about standard SQL not about implementation. For example below is a
P is correct in most of the implementations. HAVING clause can also be used with aggregate function. If we use a HAVING clause without a GROUP BY clause, the HAVING condition applies to all rows that satisfy the search condition. In other words, all rows that satisfy the search condition make up a single group. See this for more details.
S is correct . To verify S, try following queries in SQL.

CREATE TABLE temp ( id INT, name VARCHAR(100) ); INSERT INTO temp VALUES (1, "abc"); INSERT INTO temp VALUES (2, "abc"); INSERT INTO temp VALUES (3, "bcd"); INSERT INTO temp VALUES (4, "cde"); SELECT Count(*) FROM temp GROUP BY name;Output:

count(*) -------- 2 1 1

Question 16 |

The recurrence relation capturing the optimal execution time of the Towers of Hanoi problem with n discs is

T(n) = 2T(n − 2) + 2 | |

T(n) = 2T(n − 1) + n | |

T(n) = 2T(n/2) + 1 | |

T(n) = 2T(n − 1) + 1 |

**GATE CS 2012**

**Discuss it**

Question 16 Explanation:

See question 1 of http://www.geeksforgeeks.org/data-structures-and-algorithms-set-29/

Question 17 |

Let G be a simple undirected planar graph on 10 vertices with 15 edges. If G is a connected graph, then the number of bounded faces in any embedding of G on the plane is equal to

3 | |

4 | |

5 | |

6 |

**GATE CS 2012**

**Graph Theory**

**Discuss it**

Question 17 Explanation:

If the graph is planar, then it must follow below Euler's Formula for planar graphs

```
v - e + f = 2
v is number of vertices
e is number of edges
f is number of faces including bounded and unbounded
10 - 15 + f = 2
f = 7
There is always one unbounded face, so the number of bounded faces = 6
```

Question 18 |

Let w(n) and A(n) denote respectively, the worst case and average case running time of an algorithm executed on an input of size n. which of the following is ALWAYS TRUE?

(A)

(B)

(C)

(D)

(A)

(B)

(C)

(D)

A | |

B | |

C | |

D |

**GATE CS 2012**

**Discuss it**

Question 18 Explanation:

The worst case time complexity is always greater than or same as the average case time complexity.
The term written in Big O notation can always asymptotically same or greater than the term on the other side.

Question 19 |

The amount of ROM needed to implement a 4 bit multiplier is

64 bits | |

128 bits | |

1 Kbits | |

2 Kbits |

**GATE CS 2012**

**Computer Organization and Architecture**

**Discuss it**

Question 19 Explanation:

For a 4 bit multiplier, there are 2

^{4}* 2^{4}combinations, i.e., 2^{8}combinations. Also, Output of a 4 bit multiplier is 8 bits. Thus, the amount of ROM needed = 2^{8}* 8 = 2^{11}= 2048 bits = 2KbitsQuestion 20 |

Which of the following transport layer protocols is used to support electronic mail?

SMTP | |

IP | |

TCP | |

UDP |

**GATE CS 2012**

**Transport Layer**

**Discuss it**

Question 21 |

In the IPv4 addressing format, the number of networks allowed under Class C addresses is

2 ^{14 } | |

2 ^{7} | |

2 ^{21} | |

2 ^{24} |

**GATE CS 2012**

**IP Addressing**

**Discuss it**

Question 21 Explanation:

In class C, 8 bits are reserved for Host Id and 24 bits are reserved for Network Id. Out of these 24 Network Id bits, the leading 3 bits are fixed as 110. So remaining 21 bits can be used for different networks. See this for more details.

Question 22 |

1, 2, 3, 4 | |

1, 2 | |

2, 3, 4 | |

3, 4 |

**GATE CS 2012**

**Undecidability**

**Discuss it**

Question 22 Explanation:

See Question 3 of http://www.geeksforgeeks.org/automata-theory-set-2/

Question 23 |

Given the language L = {ab, aa, baa}, which of the following strings are in L*?

1) abaabaaabaa 2) aaaabaaaa 3) baaaaabaaaab 4) baaaaabaa

1, 2 and 3 | |

2, 3 and 4 | |

1, 2 and 4 | |

1, 3 and 4 |

**GATE CS 2012**

**Regular languages and finite automata**

**Discuss it**

Question 23 Explanation:

See question 2 of http://www.geeksforgeeks.org/automata-theory-set-2/

Question 25 |

**Consider the following transactions with data items P and Q initialized to zero:**

T1: read (P) ; read (Q) ; if P = 0 then Q : = Q + 1 ; write (Q) ; T2: read (Q) ; read (P) ; if Q = 0 then P : = P + 1 ; write (P) ;

**Any non-serial interleaving of T1 and T2 for concurrent execution leads to**

A serializable schedule | |

A schedule that is not conflict serializable | |

A conflict serializable schedule | |

A schedule for which a precedence graph cannot be drawn |

**GATE CS 2012**

**Transactions and concurrency control**

**Discuss it**

Question 25 Explanation:

Question 26 |

The bisection method is applied to compute a zero of the function f(x) = x

^{4}– x^{3}– x^{2}– 4 in the interval [1,9]. The method converges to a solution after ––––– iterations1 | |

3 | |

5 | |

7 |

**GATE CS 2012**

**Numerical Methods and Calculus**

**Discuss it**

Question 26 Explanation:

In bisection method, we calculate the values at extreme points of given interval, if signs of values are opposite, then we find the middle point. Whatever sign we get at middle point, we take the corner point of opposite sign and repeat the process till we get 0.
f(1) < 0 and f(9) > 0
mid = (1 + 9)/2 = 5
f(5) > 0, so zero value lies in [1, 5]
mid = (1+5)/2 = 3
f(3) > 0, so zero value lies in [1, 3]
mid = (1+3)/2 = 2
f(2) = 0

Question 27 |

Let G be a weighted graph with edge weights greater than one and G'be the graph constructed by squaring the weights of edges in G. Let T and T' be the minimum spanning trees of G and G', respectively, with total weights t and t'. Which of the following statements is TRUE?

T' = T with total weight t' = t ^{2} | |

T' = T with total weight t' < t ^{2} | |

T' != T but total weight t' = t ^{2} | |

None of the above |

**GATE CS 2012**

**Graph Minimum Spanning Tree**

**Discuss it**

Question 27 Explanation:

**Squaring the weights of the edges in a weighted graph will not change the minimum spanning tree**. Assume the opposite to obtain a contradiction. If the minimum spanning tree changes then at least one edge from the old graph G in the old minimum spanning tree T must be replaced by a new edge in tree T' from the graph G' with squared edge weights. The new edge from G' must have a lower weight than the edge from G. This implies that there exists some weights C1 and C2 such that C1 < C2 and C12 >= C22. This is a contradiction. Source: http://www.cs.nyu.edu/courses/spring06/V22.0310-001/hw3.htm Sums of squares of two or more numbers is always smaller than square of sum. Example: 2^2 + 2^2 < (4)^2 But

there is one counter example when the graph has only one edge. In that case, the two values are same.

Question 28 |

b'd' | |

b'd' + b'c' | |

b'd' + a'b'c'd' | |

b'd' + b'c' + c'd' |

**GATE CS 2012**

**Digital Logic & Number representation**

**Discuss it**

Question 29 |

Consider the 3 processes, P1, P2 and P3 shown in the table.

Process Arrival time Time Units Required P1 0 5 P2 1 7 P3 3 4The completion order of the 3 processes under the policies FCFS and RR2 (round robin scheduling with CPU quantum of 2 time units) are

FCFS: P1, P2, P3 RR2: P1, P2, P3 | |

FCFS: P1, P3, P2 RR2: P1, P3, P2 | |

FCFS: P1, P2, P3 RR2: P1, P3, P2 | |

FCFS: P1, P3, P2 RR2: P1, P2, P3 |

**GATE CS 2012**

**CPU Scheduling**

**Discuss it**

Question 29 Explanation:

FCFS is clear. In RR, time slot is of 2 units. Processes are assigned in following order p1, p2, p1, p3, p2, p1, p3, p2, p2This question involves the concept of ready queue. At t=2, p2 starts and p1 is sent to the ready queue and at t=3 p3 arrives so then the job p3 is queued in ready queue after p1. So at t=4, again p1 is executed then p3 is executed for first time at t=6.

Question 30 |

**Fetch_And_Add(X,i) is an atomic Read-Modify-Write instruction that reads the value of memory location X, increments it by the value i, and returns the old value of X. It is used in the pseudocode shown below to implement a busy-wait lock. L is an unsigned integer shared variable initialized to 0. The value of 0 corresponds to lock being available, while any non-zero value corresponds to the lock being not available.**

AcquireLock(L){ while (Fetch_And_Add(L,1)) L = 1; } ReleaseLock(L){ L = 0; }

**This implementation**

fails as L can overflow | |

fails as L can take on a non-zero value when the lock is actually available | |

works correctly but may starve some processes | |

works correctly without starvation |

**GATE CS 2012**

**Process Management**

**Discuss it**

Question 30 Explanation:

Take closer look the below while loop.

while (Fetch_And_Add(L,1)) L = 1; // A waiting process can be here just after // the lock is released, and can make L = 1.Consider a situation where a process has just released the lock and made L = 0. Let there be one more process waiting for the lock, means executing the AcquireLock() function. Just after the L was made 0, let the waiting processes executed the line L = 1. Now, the lock is available and L = 1. Since L is 1, the waiting process (and any other future coming processes) can not come out of the while loop. The above problem can be resolved by changing the AcuireLock() to following.

AcquireLock(L){ while (Fetch_And_Add(L,1)) { // Do Nothing } }

Source : http://www.geeksforgeeks.org/operating-systems-set-17/

Question 31 |

Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?

10/21 | |

5/12 | |

2/3 | |

1/6 |

**GATE CS 2012**

**Probability**

**Discuss it**

Question 31 Explanation:

The following are different possibilities
(1,5) (1,6) (2,4) (2,5) (2,6) (3,3) (3,4) (3,5) (3,6)
Plus 1/6 probability that first time 6 is rolled
So total probability is 9/36 + 1/6 = 15/36 = 5/12.

Question 32 |

A n Internet Service Provider (ISP) has the following chunk of CIDR-based IP addresses available with it: 245.248.128.0/20. The ISP wants to give half of this chunk of addresses to Organization A, and a quarter to Organization B, while retaining the remaining with itself. Which of the following is a valid allocation of addresses to A and B?

245.248.136.0/21 and 245.248.128.0/22 | |

245.248.128.0/21 and 245.248.128.0/22 | |

245.248.132.0/22 and 245.248.132.0/21 | |

245.248.136.0/24 and 245.248.132.0/21 |

**GATE CS 2012**

**IP Addressing**

**Discuss it**

Question 32 Explanation:

See Question 4 of http://www.geeksforgeeks.org/computer-networks-set-1/

Question 33 |

Suppose a circular queue of capacity (n – 1) elements is implemented with an array of n elements. Assume that the insertion and deletion operation are carried out using REAR and FRONT as array index variables, respectively. Initially, REAR = FRONT = 0. The conditions to detect queue full and queue empty are

Full: (REAR+1) mod n == FRONT, empty: REAR == FRONT | |

Full: (REAR+1) mod n == FRONT, empty: (FRONT+1) mod n == REAR | |

Full: REAR == FRONT, empty: (REAR+1) mod n == FRONT | |

Full: (FRONT+1) mod n == REAR, empty: REAR == FRONT |

**GATE CS 2012**

**Discuss it**

Question 33 Explanation:

Implementation of Circular Queue :

**Head**- It always points to the location from where next deletion takes place from the queue.**Tail**- It always points to the next empty location in which next insertion will take place in the queue. We will be using wrap around feature since it is a circular queue which is when the tail or head is at the index n-1, next operation will bring them to index 0. In Spite of having capacity of n inside array, we will reserve one empty spot in order to detect the overflow(Queue Full) and underflow(Queue Empty) conditions. The elements in the queue reside in locations Q.head, Q.head + 1, . . . , Q.tail + 1, where we “wrap around” in the sense that location 0 immediately follows location n-1 in a circular order.**Algorithm :**ENQUEUE(Q, x) { if Q.head == Q.tail + 1 error "Queue overflow" Q[Q.tail] = x if Q.tail == Q.length - 1 Q.tail = 0 else Q.tail = Q.tail + 1 } DEQUEUE(Q) { if Q.head == Q.tail error "Queue underflow" x = Q[Q.head] if Q.head == Q.length - 1 Q.head = 0 else Q.head = Q.head + 1 return x }

See http://en.wikipedia.org/wiki/Circular_buffer#Always_Keep_One_Slot_Open

This solution is contributed by**Pranjul Ahuja**Question 34 |

Consider the program given below, in a block-structured pseudo-language with lexical scoping and nesting of procedures permitted.

A | |

B | |

C | |

D |

**GATE CS 2012**

**Principles of Programming Languages**

**Discuss it**

Question 34 Explanation:

Access link is defined as link to activation record of closest lexically enclosing block in program text, so the closest enclosing blocks respectively for A1 ,A2 and A21 are main , main and A2 Since, Activation records are created at procedure entry time and destroyed at procedure exit time.

This solution is contributed by

**Solution:**Link to activation record of closest lexically enclosing block in program text. It depends on the static program text. Here Calling sequence is given as, Main->A1->A2->A21->A1 Now A1,A2 are defined under Main...So A1,A2 Access link are pointed to main A21 is defined under A2 hence its Access link will point to A2 . Reference: http://www.cs.nyu.edu/courses/spring08/G22.2130-001/lectures/lecture-13.htmlThis solution is contributed by

**Nitika Bansal**Question 35 |

How many onto (or surjective) functions are there from an n-element (n >= 2) set to a 2-element set?

2 ^{n} | |

2 ^{n} - 1 | |

2 ^{n} - 2 | |

2(2 ^{n} - 2) |

**GATE CS 2012**

**Combinatorics**

**Discuss it**

Question 35 Explanation:

Total possible number of functions is 2

^{n}. In mathematics, a function f from a set X to a set Y is surjective (or onto), or a surjection, if every element y in Y has a corresponding element x in X such that f(x) = y (Source: http://en.wikipedia.org/wiki/Surjective_function) There are total 2 functions out of 2^{n}that are NOT onto: one that maps to all 1s and other that maps to all 2s. Therefore total number of onto functions is 2^{n}- 2.Question 36 |

Let G be a complete undirected graph on 6 vertices. If vertices of G are labeled, then the number of distinct cycles of length 4 in G is equal to

15 | |

30 | |

45 | |

360 |

**GATE CS 2012**

**Graph Theory**

**Discuss it**

Question 36 Explanation:

There can be total

^{6}C_{4}ways to pick 4 vertices from 6. The value of^{6}C_{4}is 15. Note that the given graph is complete so any 4 vertices can form a cycle. There can be 6 different cycle with 4 vertices. For example, consider 4 vertices as a, b, c and d. The three distinct cycles are cycles should be like this (a, b, c, d,a) (a, b, d, c,a) (a, c, b, d,a) (a, c, d, b,a) (a, d, b, c,a) (a, d, c, b,a) and (a, b, c, d,a) and (a, d, c, b,a) (a, b, d, c,a) and (a, c, d, b,a) (a, c, b, d,a) and (a, d, b, c,a) are same cycles. So total number of distinct cycles is (15*3) = 45. **NOTE**: In original GATE question paper 45 was not an option. In place of 45, there was 90.Question 37 |

A list of n strings, each of length n, is sorted into lexicographic order using the merge-sort algorithm. The worst case running time of this computation is

O(n log n) | |

O(n ^{2} log n) | |

O(n ^{2} + log n) | |

O(n ^{2}) |

**GATE CS 2012**

**Discuss it**

Question 37 Explanation:

When we are sorting an array of n integers, Recurrence relation for Total number of comparisons involved will be,
T(n) = 2T(n/2) + (n) where (n) is the number of comparisons in order to merge 2 sorted subarrays of size n/2.
= (nlog2n)
Instead of integers whose comparison take O(1) time, we are given n strings. We can compare 2 strings in O(n) worst case. Therefore, Total number of comparisons now will be (n2log2n) where each comparison takes O(n) time now.
In general, merge sort makes (nlog2n) comparisons, and runs in (nlog2n) time if each comparison can be done in O(1) time.
See question 5 of http://www.geeksforgeeks.org/data-structures-and-algorithms-set-28/
This solution is contributed by

**Pranjul Ahuja**Question 38 |

Consider the directed graph shown in the figure below. There are multiple shortest paths between vertices S and T. Which one will be reported by Dijstra?s shortest path algorithm? Assume that, in any iteration, the shortest path to a vertex v is updated only when a strictly shorter path to v is discovered.

SDT | |

SBDT | |

SACDT | |

SACET |

**GATE CS 2012**

**Discuss it**

Question 39 |

A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1 indirect block address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and the size of each disk block address is 8 Bytes. The maximum possible file size in this file system is

3 Kbytes | |

35 Kbytes | |

280 Bytes | |

Dependent on the size of the disk |

**GATE CS 2012**

**Input Output Systems**

**Discuss it**

Question 39 Explanation:

Question 40 |

Consider the virtual page reference string
1, 2, 3, 2, 4, 1, 3, 2, 4, 1
On a demand paged virtual memory system running on a computer system that main memory size of 3 pages frames which are initially empty. Let LRU, FIFO and OPTIMAL denote the number of page faults under the corresponding page replacements policy. Then

OPTIMAL < LRU < FIFO | |

OPTIMAL < FIFO < LRU | |

OPTIMAL = LRU | |

OPTIMAL = FIFO |

**GATE CS 2012**

**Memory Management**

**Discuss it**

Question 40 Explanation:

**First In First Out (FIFO)**This is the simplest page replacement algorithm. In this algorithm, operating system keeps track of all pages in the memory in a queue; oldest page is in the front of the queue. When a page needs to be replaced page in the front of the queue is selected for removal.

**Optimal Page replacement:**in this algorithm, pages are replaced which are not used for the longest duration of time in the future. Least Recently Used (LRU) In this algorithm page will be replaced which is least recently used.

**Solution:**the virtual page reference string is 1, 2, 3, 2, 4, 1, 3, 2, 4, 1 size of main memory pages frames is 3. For FIFO: total no of page faults are 6 (depicted in bold and italic) For optimal: total no of page faults are 5 (depicted in bold and italic) For LRU: total no of page faults are 9 (depicted in bold and italic) The Optimal will be 5, FIFO 6 and LRU 9. so, OPTIMAL < FIFO < LRU option (B) is correct answer. See http://www.geeksforgeeks.org/operating-systems-set-5/ This solution is contributed by

**Nitika Bansal**

Question 41 |

Suppose (A, B) and (C,D) are two relation schemas. Let r1 and r2 be the corresponding relation instances. B is a foreign key that refers to C in r2. If data in r1 and r2 satisfy referential integrity constraints, which of the following is ALWAYS TRUE?

A | |

B | |

C | |

D |

**GATE CS 2012**

**ER and Relational Models**

**Discuss it**

Question 41 Explanation:

See Question 3 of http://www.geeksforgeeks.org/database-management-system-set-2/

Question 42 |

Consider a source computer(S) transmitting a file of size 10

^{6}bits to a destination computer(D)over a network of two routers (R_{1}and R_{2}) and three links(L_{1}, L_{2}, and L_{3}). L_{1}connects S to R_{1}; L_{2}connects R_{1}to R_{2}; and L_{3}connects R_{2}to D.Let each link be of length 100 km. Assume signals travel over each link at a speed of 10^{8}meters per second.Assume that the link bandwidth on each link is 1Mbps. Let the file be broken down into 1000 packets each of size 1000 bits. Find the total sum of transmission and propagation delays in transmitting the file from S to D?1005 ms | |

1010 ms | |

3000 ms | |

3003 ms |

**GATE CS 2012**

**Network Layer**

**Discuss it**

Question 42 Explanation:

Question 43 |

Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of the slow start phase is 2 MSS and the threshold at the start of the first transmission is 8 MSS. Assume that a timeout occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.

8 MSS | |

14 MSS | |

7 MSS | |

12 MSS |

**GATE CS 2012**

**Transport Layer**

**Discuss it**

Question 43 Explanation:

See question 2 of http://www.geeksforgeeks.org/computer-networks-set-2/

Question 44 |

Consider the set of strings on {0,1} in which, every substring of 3 symbols has at most two zeros. For example, 001110 and 011001 are in the language, but 100010 is not. All strings of length less than 3 are also in the language. A partially completed DFA that accepts this language is shown below.
The missing arcs in the DFA are

A | |

B | |

C | |

D |

**GATE CS 2012**

**Regular languages and finite automata**

**Discuss it**

Question 44 Explanation:

Question 45 |

The height of a tree is defined as the number of edges on the longest path in the tree. The function shown in the pseudocode below is invoked as height (root) to compute the height of a binary tree rooted at the tree pointer root.
The appropriate expression for the two boxes B1 and B2 are

B1 : (1 + height(n->right)), B2 : (1 + max(h1,h2)) | |

B1 : (height(n->right)), B2 : (1 + max(h1,h2)) | |

B1 : height(n->right), B2 : max(h1,h2) | |

B1 : (1 + height(n->right)), B2 : max(h1,h2) |

**GATE CS 2012**

**Discuss it**

Question 45 Explanation:

The box B1 gets exected when left subtree of n is NULL and right sbtree is not NULL. In this case, height of n will be height of right subtree plus one.
The box B2 gets executed when both left and right sbtrees of n are not NULL. In this case, height of n will be max of heights of left and right sbtrees of n plus 1.

Question 46 |

Register renaming is done in pipelined processors

as an alternative to register allocation at compile time | |

for efficient access to function parameters and local variables | |

to handle certain kinds of hazards | |

as part of address translation |

**GATE CS 2012**

**Computer Organization and Architecture**

**Discuss it**

Question 46 Explanation:

Register renaming is done to avoid data hazards

Question 47 |

Consider a random variable X that takes values +1 and −1 with probability 0.5 each. The values of the cumulative distribution function F(x) at x = −1 and +1 are

0 and 0.5 | |

0 and 1 | |

0.5 and 1 | |

0.25 and 0.75 |

**GATE CS 2012**

**Probability**

**Discuss it**

Question 47 Explanation:

The Cumulative Distribution Function
F(x) = P(X≤x)
F(-1) = P(X≤-1) = P(X=-1) = 0.5
F(+1) = P(X≤+1) = P(X=-1) + (P=+1) = 0.5+0.5 = 1

Question 48 |

Consider the following C program

int a, b, c = 0; void prtFun (void); int main () { static int a = 1; /* line 1 */ prtFun(); a += 1; prtFun(); printf ( "\n %d %d " , a, b) ; } void prtFun (void) { static int a = 2; /* line 2 */ int b = 1; a += ++b; printf (" \n %d %d " , a, b); }What output will be generated by the given code segment?

3 1 4 1 4 2 | |

4 2 6 1 6 1 | |

4 2 6 2 2 0 | |

3 1 5 2 5 2 |

**GATE CS 2012**

**Discuss it**

Question 48 Explanation:

Question 49 |

Consider the above question. What output will be generated by the given code d\segment if:
Line 1 is replaced by “

**auto int a = 1;**” Line 2 is replaced by “**register int a = 2;**”3 1 4 1 4 2 | |

4 2 6 1 6 1 | |

4 2 6 2 2 0 | |

4 2 4 2 2 0 |

**GATE CS 2012**

**Discuss it**

Question 49 Explanation:

See Question 3 of http://www.geeksforgeeks.org/c-language-set-9/

Question 50 |

Consider the following relations A, B, C. How many tuples does the result of the following relational algebra expression contain? Assume that the schema of A U B is the same as that of A.

Table AId Name Age ---------------- 12 Arun 60 15 Shreya 24 99 Rohit 11Table BId Name Age ---------------- 15 Shreya 24 25 Hari 40 98 Rohit 20 99 Rohit 11Table CId Phone Area ----------------- 10 2200 02 99 2100 01

7 | |

4 | |

5 | |

9 |

**GATE CS 2012**

**ER and Relational Models**

**Discuss it**

Question 50 Explanation:

Question 51 |

Table A Id Name Age ---------------- 12 Arun 60 15 Shreya 24 99 Rohit 11 Table B Id Name Age ---------------- 15 Shreya 24 25 Hari 40 98 Rohit 20 99 Rohit 11 Table C Id Phone Area ----------------- 10 2200 02 99 2100 01Consider the above tables A, B and C. How many tuples does the result of the following SQL query contains?

SELECT A.id FROM A WHERE A.age > ALL (SELECT B.age FROM B WHERE B. name = "arun")

4 | |

3 | |

0 | |

1 |

**GATE CS 2012**

**SQL**

**Discuss it**

Question 51 Explanation:

The meaning of “ALL” is the A.Age should be greater than all the values returned by the subquery. There is no entry with name “arun” in table B. So the subquery will return NULL. If a subquery returns NULL, then the condition becomes true for all rows of A (See this for details). So all rows of table A are selected.
Source: http://www.geeksforgeeks.org/database-management-system-set-3/

Question 52 |

For the grammar below, a partial LL(1) parsing table is also presented along with the grammar. Entries that need to be filled are indicated as E1, E2, and E3. is the empty string, $ indicates end of input, and, | separates alternate right hand sides of productions.

A | |

B | |

C | |

D |

**GATE CS 2012**

**Parsing and Syntax directed translation**

**Discuss it**

Question 52 Explanation:

First(X) - It is the set of terminals that begin the strings derivable from X. Follow(X) - It is the set of terminals that can appear immediately to the right of X in some sentential form. Now in the above question, FIRST(S) = { a, b, epsilon} FIRST(A) = FIRST(S) = { a, b, epsilon} FIRST(B) = FIRST(S) = { a, b, epsilon} FOLLOW (A) = { b , a } FOLLOW (S) = { $ } U FOLLOW (A) = { b , a , $ } FOLLOW (B) = FOLLOW (S) = { b ,a , $ } epsilon corresponds to empty string.

Question 53 |

Consider the date same as above question. The appropriate entries for E1, E2, and E3 are

A | |

B | |

C | |

D |

**GATE CS 2012**

**Parsing and Syntax directed translation**

**Discuss it**

Question 53 Explanation:

As we need to find entries E1, E2 and E3 which are against Non terminals S and B, so we will deal with only those productions which have S and B on LHS.
Here representing the parsing table as M[ X , Y ], where X represents rows( Non terminals) and Y represents columns(terminals).

Rule 1: if P --> Q is a production, for each terminal 't' in FIRST(Q) add P-->Q to M [ P , t ] Rule 2 : if epsilon is in FIRST(Q), add P --> Q to M [ P , f ] for each f in FOLLOW(P).For the production rule S--> aAbB, it will be added to parsing table at the position M[ S , FIRST( aAbB ) ], Now FIRST(aAbB) = {a}, hence add S--> aAbB to M[ S , a ] which is E1. For the production rule S--> bAaB, it will be added to parsing table at the position M[ S , FIRST( bAaB ) ], Now FIRST(bAaB) = {b}, hence add S--> bAaB to M[ S , b ] which is E2 For the production rule S--> epsilon , it will be added to parsing table at the position M[ S , FOLLOW(S) ], Now FOLLOW(S) = { a , b , $ }, hence add S --> epsilon to M[ S , a ] and M[ S , b ] which are again E1 and E2 respectively. For the production rule B --> S, it will be added to parsing table at the position M[ B , FIRST( S ) ], Now FIRST(S) also contains epsilon, hence add B --> S to M[ S , FOLLOW(B) ], and FOLLOW(B) contains {$}, i.e. M[ S , $ ] which is E3. epsilon corresponds to empty string.

Question 54 |

A computer has a 256 KByte, 4-way set associative, write back data cache with block size of 32 Bytes. The processor sends 32 bit addresses to the cache controller. Each cache tag directory entry contains, in addition to address tag, 2 valid bits, 1 modified bit and 1 replacement bit.
The number of bits in the tag field of an address is

11 | |

14 | |

16 | |

27 |

**GATE CS 2012**

**Computer Organization and Architecture**

**Discuss it**

Question 54 Explanation:

A set-associative scheme is a hybrid between a fully associative cache, and direct mapped cache. It's considered a reasonable compromise between the complex hardware needed for fully associative caches (which requires parallel searches of all slots), and the simplistic direct-mapped scheme, which may cause collisions of addresses to the same slot (similar to collisions in a hash table). (source: http://www.cs.umd.edu/class/spring2003/cmsc311/Notes/Memory/set.html). Also see http://csillustrated.berkeley.edu/PDFs/handouts/cache-3-associativity-handout.pdf
Number of blocks = Cache-Size/Block-Size = 256 KB / 32 Bytes = 2

^{13}Number of Sets = 2^{13}/ 4 = 2^{11}Tag + Set offset + Byte offset = 32 Tag + 11 + 5 = 32 Tag = 16Question 55 |

Consider the data given in previous question. The size of the cache tag directory is

160 Kbits | |

136 bits | |

40 Kbits | |

32 bits |

**GATE CS 2012**

**Computer Organization and Architecture**

**Discuss it**

Question 55 Explanation:

16 bit address
2 bit valid
1 modified
1 replace
Total bits = 20
20 × no. of blocks
= 160 K bits.

Question 56 |

The cost function for a product in a firm is given by 5q

^{2}, where q is the amount of production. The firm can sell the product at a market price of Rs 50 per unit. The number of units to be produced by the firm such that the profit is maximized is5 | |

10 | |

15 | |

25 |

**GATE CS 2012**

**General Aptitude**

**Discuss it**

Question 56 Explanation:

Profit = Price - Cost = 50q - 5q^{2}The value of above expression is maximum at q = 5.

Question 57 |

Choose the most appropriate alternative from the options given below to complete the following sentence:

**Despite several ––––––––– the mission succeeded in its attempt to resolve the conflict.**attempts | |

setbacks | |

meetings | |

delegations |

**GATE CS 2012**

**English**

**Discuss it**

Question 57 Explanation:

Question 58 |

Which one of the following options is the closest in meaning to the word given below?
Mitigate

Diminish | |

Divulge | |

Dedicate | |

Denote |

**GATE CS 2012**

**English**

**Discuss it**

Question 58 Explanation:

Question 59 |

Choose the grammatically INCORRECT sentence:

They gave us the money back less the service charges of Three Hundred rupees. | |

This country’s expenditure is not less than that of Bangladesh. | |

The committee initially asked for a funding of Fifty Lakh rupees, but later settled for a lesser
sum. | |

This country’s expenditure on educational reforms is very less. |

**GATE CS 2012**

**English**

**Discuss it**

Question 59 Explanation:

Question 60 |

Choose the most appropriate alternative from the options given below to complete the following sentence:

**Suresh’s dog is the one ––––––––– was hurt in the stampede.**that | |

which | |

who | |

whom |

**GATE CS 2012**

**English**

**Discuss it**

Question 60 Explanation:

Question 61 |

Wanted Temporary, Part-time persons for the post of Field Interviewer to conduct personal interviews to collect and collate economic data. Requirements: High School-pass, must be available for Day, Evening and Saturday work. Transportation paid, expenses reimbursed. Which one of the following is the best inference from the above advertisement?

Gender-discriminatory | |

Xenophobic | |

Not designed to make the post attractive | |

Not gender-discriminatory |

**GATE CS 2012**

**English**

**Discuss it**

Question 61 Explanation:

There is no mention of gender. Xenophobia also doesn't fit.

Question 62 |

A political party orders an arch for the entrance to the ground in which the annual convention is being held. The profile of the arch follows the equation

*y = 2x – 0.1x*where y is the height of the arch in meters. The maximum possible height of the arch is^{2}8 meters | |

10 meters | |

12 meters | |

14 meters |

**GATE CS 2012**

**General Aptitude**

**Discuss it**

Question 62 Explanation:

*y = 2x – 0.1x*

^{2}dy/dx = 2 - 0.2x So the value maximizes at 2 - 0.2x = 0 => x = 10 => y = 20 - 10 = 10 metersQuestion 63 |

An automobile plant contracted to buy shock absorbers from two suppliers X and Y. X supplies 60% and Y supplies 40% of the shock absorbers. All shock absorbers are subjected to a quality test. The ones that pass the quality test are considered reliable. Of X’s shock absorbers, 96% are reliable. Of Y’s shock absorbers, 72% are reliable.
The probability that a randomly chosen shock absorber, which is found to be reliable, is made by Y is

0.288 | |

0.334 | |

0.667 | |

0.720 |

**GATE CS 2012**

**General Aptitude**

**Discuss it**

Question 63 Explanation:

Probability that the absorber is reliable = 0.96*0.6 + 0.72*0.4 = 0.576 + 0.288 Probability that the absorber is from y and reliable = (Probability that is made by Y) X (Probability that it is reliable) = 0.4 * 0.72 = 0.288 The probability that randomly picked reliable absorber is from y = (Probability that the absorber is from y and reliable) / (>Probability that the absorber is reliable ) = (0.288)/ (0.576 + 0.288) = 0.334

Question 64 |

Which of the following assertions are CORRECT?
P: Adding 7 to each entry in a list adds 7 to the mean of the list
Q: Adding 7 to each entry in a list adds 7 to the standard deviation of the list
R: Doubling each entry in a list doubles the mean of the list
S: Doubling each entry in a list leaves the standard deviation of the list unchanged

P, Q | |

Q, R | |

P, R | |

R, S |

**GATE CS 2012**

**General Aptitude**

**Discuss it**

Question 64 Explanation:

**Mean is average.**

Let us consider below example [Tex] 2,\ 4,\ 4,\ 4,\ 5,\ 5,\ 7,\ 9 [/Tex] These eight data points have the mean (average) of 5: [Tex] \frac{2 + 4 + 4 + 4 + 5 + 5 + 7 + 9}{8} = 5. [/Tex] When we add 7 to all numbers, mean becomes 12 so P is TRUE. If we double all numbers mean becomes double, so R is also TRUE.References: https://en.wikipedia.org/wiki/Standard_deviation http://staff.argyll.epsb.ca/jreed/math30p/statistics/standardDeviation.htmStandard Deviation is square root of variance. Variance is sum of squares of differences between all numbers and means.Deviation for above example First, calculate the deviations of each data point from the mean, and square the result of each: [Tex] \begin{array}{lll} (2-5)^2 = (-3)^2 = 9 && (5-5)^2 = 0^2 = 0 \\ (4-5)^2 = (-1)^2 = 1 && (5-5)^2 = 0^2 = 0 \\ (4-5)^2 = (-1)^2 = 1 && (7-5)^2 = 2^2 = 4 \\ (4-5)^2 = (-1)^2 = 1 && (9-5)^2 = 4^2 = 16. \\ \end{array} [/Tex] [Tex]variance = \frac{9 + 1 + 1 + 1 + 0 + 0 + 4 + 16}{8} = 4. [/Tex] [Tex]standard deviation = \sqrt{ 4 } = 2 [/Tex] If we add 7 to all numbers, standard deviation won't change as 7 is added to mean also. So Q is FALSE. If we double all entries, standard deviation also becomes double. So S is false.

Question 65 |

Given the sequence of terms, AD CG FK JP, the next term is

OV | |

OW | |

PV | |

PW |

**GATE CS 2012**

**General Aptitude**

**Discuss it**

Question 65 Explanation:

Here 1st letter of each term in order are: A C F J
Now In alphabets, A and C have a difference of 1 alphabet, C and F have a difference of 2, and F and J have a difference of 3. Hence next term here should contain its 1st letter as an alphabet with a difference of 4 after J, i.e O.
Similarly, 2nd letter of each term in order are: D G K P
Now In alphabets, D and G have a difference of 2 alphabets, G and K have a difference of 3, and K and P have a difference of 4. Hence next term here should contain its 2nd letter as an alphabet with a difference of 5 after P, i.e V.

There are 65 questions to complete.