GATE-CS-2006

Question 1

Consider the polynomial p(x) = a0 + a1x + a2x² + a3x³ , where ai ≠ 0 ∀i. The minimum number of multiplications needed to evaluate p on an input x is:

• A. 3
• B. 4
• C. 6
• D. 9

Answer: (A)

Background Explanation : Horner's rule for polynomial division is an algorithm used to simplify the process of evaluating a polynomial f(x) at a certain value x = x0 by dividing the polynomial into monomials (polynomials of the 1st degree). Each monomial involves a maximum of one multiplication and one addition processes. The result obtained from one monomial is added to the result obtained from the next monomial and so forth in an accumulative addition fashion. To explain the above, let is re-write the polynomial in its expanded form; f(x0) = a0 + a1x0+ a2x0^2+ ... + anx0^n This can, also, be written as: f(x0) = a0 + x0(a1+ x0(a2+ x0(a3+ ... + (an-1 + anx0)....) The algorithm proposed by this rule is based on evaluating the monomials formed above starting from the one in the inner-most parenthesis and move out to evaluate the monomials in the outer parenthesis. Solution : Using Horner's Rule, we can write the polynomial as following a0 + (a1 + (a2 + a3x)x)x In the above form, we need to do only 3 multiplications

p = a3 X x    ------------ (1)

q = (a2 + p) X x  ---------(2)

r = (a1 + q) X x  ---------(3)

result = a0 + r 

Reference : http://www.geeksforgeeks.org/horners-method-polynomial-evaluation/ This solution is contributed by Nitika Bansal.

Question 2

Let X, Y, Z be sets of sizes x, y and z respectively. Let W = X x Y. Let E be the set of all subsets of W. The number of functions from Z to E is:

• A. z^2xy
• B. z x 2 ^xy
• C. z^{2x + y}
• D. 2^xyz

Answer: (D)

Number of functions from a set A of size m to set B of size n is $n^m$, because each of the m elements of A has n choices for mapping. Now here $m = |Z| = z$, and $n = |E| = 2^{xy}$ because number of subsets of a set of size n is $2^n$, and here set W has size of $xy$.
So number of functions from Z to E = $(2^{xy})^z = 2^{xyz}$. So option (D) is correct. Source: http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2006.html

Question 3

The set {1, 2, 3, 5, 7, 8, 9} under multiplication modulo 10 is not a group. Given below are four plausible reasons. Which one of them is false?

• A. It is not closed
• B. 2 does not have an inverse
• C. 3 does not have an inverse
• D. 8 does not have an inverse

Answer: (C)

A is not closed under multiplication as we may get 0 after multiplication and 0 is not present in set. 2 doesn't have an inverse as there is no x such that (2*x) mod 10 is 1. 3 has an inverse as (3*7) mod 10 is 1. 8 doesn't have an inverse as there is no x such that (2*x) mod 10 is 1.

Question 4

A relation R is defined on ordered pairs of integers as follows: (x,y) R(u,v) if x < u and y > v. Then R is: Then R is:

• A. Neither a Partial Order nor an Equivalence Relation
• B. A Partial Order but not a Total Order
• C. A Total Order
• D. An Equivalence Relation

Answer: (A)

An equivalence relation on a set x is a subset of x*x, i.e., a collection R of ordered pairs of elements of x, satisfying certain properties. Write “x R y" to mean (x,y) is an element of R, and we say "x is related to y," then the properties are: 1. Reflexive: a R a for all a Є R, 2. Symmetric: a R b implies that b R a for all a,b Є R 3. Transitive: a R b and b R c imply a R c for all a,b,c Є R. An partial order relation on a set x is a subset of x*x, i.e., a collection R of ordered pairs of elements of x, satisfying certain properties. Write “x R y" to mean (x,y) is an element of R, and we say "x is related to y," then the properties are: 1. Reflexive: a R a for all a Є R, 2. Anti-Symmetric: a R b and b R a implies that for all a,b Є R 3. Transitive: a R b and b R c imply a R c for all a,b,c Є R. An total order relation a set x is a subset of x*x, i.e., a collection R of ordered pairs of elements of x, satisfying certain properties. Write “x R y" to mean (x,y) is an element of R, and we say "x is related to y," then the properties are: 1. Reflexive: a R a for all a Є R, 2. Anti-Symmetric: a R b implies that b R a for all a,b Є R 3. Transitive: a R b and b R c imply a R c for all a,b,c Є R. 4. Comparability : either a R b or b R a for all a,b Є R. As given in question, a relation R is defined on ordered pairs of integers as follows: (x,y) R(u,v) if x < u and y > v , reflexive property is not satisfied here, because there is > or < relationship between (x ,y) pair set and (u,v) pair set . Other way , if there would have been x <= u and y>= v (or x=u and y=v) kind of relation among elements of sets then reflexive property could have been satisfied. Since reflexive property in not satisfied here , so given relation can not be equivalence, partial orderor total order relation. So, option (A) is correct. This solution is contributed by Nirmal Bharadwaj.

Question 5

For which one of the following reasons does Internet Protocol (IP) use the timeto- live (TTL) field in the IP datagram header

• A. Ensure packets reach destination within that time
• B. Discard packets that reach later than that time
• C. Prevent packets from looping indefinitely
• D. Limit the time for which a packet gets queued in intermediate routers.

Answer: (C)

following are lines from wikipedia Time to live (TTL) or hop limit is a mechanism that limits the lifespan or lifetime of data in a computer or network. TTL may be implemented as a counter or timestamp attached to or embedded in the data. Once the prescribed event count or timespan has elapsed, data is discarded. In computer networking, TTL prevents a data packet from circulating indefinitely.

Question 6

Consider three CPU-intensive processes, which require 10, 20 and 30 time units and arrive at times 0, 2 and 6, respectively. How many context switches are needed if the operating system implements a shortest remaining time first scheduling algorithm? Do not count the context switches at time zero and at the end.

• A. 1
• B. 2
• C. 3
• D. 4

Answer: (B)

Shortest remaining time, also known as shortest remaining time first (SRTF), is a scheduling method that is a pre-emptive version of shortest job next scheduling. In this scheduling algorithm, the process with the smallest amount of time remaining until completion is selected to execute. Since the currently executing process is the one with the shortest amount of time remaining by definition, and since that time should only reduce as execution progresses, processes will always run until they complete or a new process is added that requires a smaller amount of time. Solution: Let three process be P0, P1 and P2 with arrival times 0, 2 and 6 respectively and CPU burst times 10, 20 and 30 respectively. At time 0, P0 is the only available process so it runs. At time 2, P1 arrives, but P0 has the shortest remaining time, so it continues. At time 6, P2 also arrives, but P0 still has the shortest remaining time, so it continues. At time 10, P1 is scheduled as it is the shortest remaining time process. At time 30, P2 is scheduled. Only two context switches are needed. P0 to P1 and P1 to P2. See question 1 of http://www.geeksforgeeks.org/operating-systems-set-14/ This solution is contributed by Nitika Bansal Watch GeeksforGeeks Video Explanation :

Question 7

Consider the following grammar.

S -> S * E
S -> E
E -> F + E
E -> F
F -> id

Consider the following LR(0) items corresponding to the grammar above.

(i) S -> S * .E
(ii) E -> F. + E
(iii) E -> F + .E 

Given the items above, which two of them will appear in the same set in the canonical sets-of-items for the grammar?

• A. (i) and (ii)
• B. (ii) and (iii)
• C. (i) and (iii)
• D. None of the above

Answer: (D)

Let's make the LR(0) set of items. First we need to augment the grammar with the production rule S' -> .S , then we need to find closure of items in a set to complete a set. Below are the LR(0) sets of items.

Question 8

You are given a free running clock with a duty cycle of 50% and a digital waveform f which changes only at the negative edge of the clock. Which one of the following circuits (using clocked D flip-flops) will delay the phase of f by 180°? 

• A. A
• B. B
• C. C
• D. D

Answer: (C)

We assume the D flip-flop to be negative edge triggered.  In option (A), during the negative edge of the clock, first flip-flop inverts complement of ‘f’(we get f as the output). But, the complement of the output of first flip-flop(i.e. f') is given as the input to the second flip-flop. The second flip flop is enabled by 'clk'. The output at the second flip flop is f'+90 degrees (as +ve edged clk at output delays it by 90 degrees). Thus f is delayed by 270 degrees. So, A is not the correct option.  Following the above procedures as in (A) we will get:In option (B) and (D), the output is ‘f’. But, we want inverted ‘f’ as the output. So, (B) and (D) can’t be the answer. In option (C), the first flip-flop is activated by ‘clk’. So, the output of first flip-flop has the same phase as ‘f’. But, the second flip-flop is enabled by complement of ‘clk’. Since the clock ‘clk’ has a duty cycle of 50% , we get the output having phase delay of 180 degrees.  Therefore, (C) is the correct answer. 

Question 9

A CPU has 24-bit instructions. A program starts at address 300 (in decimal). Which one of the following is a legal program counter (all values in decimal)?

• A. 400
• B. 500
• C. 600
• D. 700

Answer: (C)

Each address is multiple of 3 as the starting address is 300 and is each instruction consists of 24 bit, i.e., 3 byte.
Thus, in the given options the valid counter will be the one which is the multiple of 3. Out of the options we can see that only 600 satisfies the condition.
Therefore, it is 600.

Question 10

In a binary max heap containing n numbers, the smallest element can be found in time

• A. O(n)
• B. O(Logn)
• C. O(LogLogn)
• D. O(1)

Answer: (A)

See Question 1 of http://www.geeksforgeeks.org/data-structures-and-algorithms-set-7/