GATE 2003
GATE 2003
Question 1 
Consider the following C function.
float f(float x, int y) { float p, s; int i; for (s=1, p=1, i=1; i < y; i ++) { p*= x/i; s+=p; } return s; }
For large values of y, the return value of the function f best approximates
X^{y}  
e^{x}  
ln(1+x)  
X^{x} 
S = S+P
Iteration 1:
P=1; i=1; S=1
P=x
S = 1+x
Iteration 2:
P=x; S = 1+x; i=2
P = x * x/2 = x^{2}/2
Iteration 3:
P = x^{2}/2; S = 1+x+x^{2}/2; i=3
P = (x^{2}/2)(x/3) = x^{3}/6
S = 1 + x + x^{2}/2 + x^{3}/6
Continue upto n
Then f( ) returns:
S = 1 + x/1 + x^{2}/2⋅1 + x^{3}/3⋅2 + ...
= 1 + x/1! + x^{2}/2! + x^{3}/3! + ... + x^{n}/n!
= e^{x}
Question 2 
Assume the following C variable declaration
int *A [10], B[10][10];
Of the following expressions
I. A[2] II. A[2][3] III. B[1] IV. B[2][3]
which will not give compiletime errors if used as left hand sides of assignment statements in a C program?
I, II, and IV only  
II, III, and IV only  
II and IV only  
IV only 
ii) A[2][3] This results an integer, no error will come.
iii) B[1] is a base address of an array. This will not be changed it will result a compile time error.
iv) B[2][3] This also results an integer. No error will come.
Question 3 
Let P(E) denote the probability of the event E. Given P(A) = 1, P(B) = 1/2, the values of P(AB) and P(BA) respectively are
1/4, 1/2  
1/2, 1/4  
1/2, 1  
1, 1/2 
P(A/B) = P(A∩B)/P(B)
= P(A)⋅P(B)/P(B) (consider P(A), P(B) are two independent events)
= 1
P(B/A) = P(B∩A)/P(A)
= P(B)⋅P(A)/P(A)
= 1/2
Question 4 
Let A be a sequence of 8 distinct integers sorted in ascending order. How many distinct pairs of sequences, B and C are there such that (i) each is sorted in ascending order, (ii) B has 5 and C has 3 elements, and (iii) the result of merging B and C gives A?
2  
30  
56  
256 
→ If we are pick 3 elements from 8 sequence integers then remaining 5 elements are already in ascending order. After merging these elements then it gives A.
→ No. of possibilities of choosing 8 elements from total of 8 = ^{8}C_{3}
= 8!/3!5!
= 8 * 7
= 56
Question 5 
n couples are invited to a party with the condition that every husband should be accompanied by his wife. However, a wife need not be accompanied by her husband. The number of different gatherings possible at the party is
i) Both husband and wife comes
ii) Only wife comes
iii) Both are not come
The no. of different gatherings possible at party is
= 3 * 3 * 3 * 3 * ... n times
= 3^{n}
Question 6 
Let T(n) be the number of different binary search trees on n distinct elements. Then , where x is
n – k + 1  
n – k  
n – k – 1  
n – k – 2 
Question 7 
Consider the set ∑* of all strings over the alphabet ∑ = {0, 1}. ∑* with the concatenation operator for strings
does not form a group  
forms a noncommutative group  
does not have a right identity element  
forms a group if the empty string is removed from Σ*

→ To perform concatenation with the given set can result a Monoid and it follows the property of closure, associativity and consists of identity element.
Question 8 
Let G be an arbitrary graph with n nodes and k components. If a vertex is removed from G, the number of components in the resultant graph must necessarily lie between
k and n  
k – 1 and k + 1  
k – 1 and n – 1  
k + 1 and n – k 
If a vertex is removed then it results that all the components are also be disconnected. So removal can create (n1) components.
Question 9 
Assuming all numbers are in 2's complement representation, which of the following numbers is divisible by 11111011?
11100111  
11100100  
11010111  
11011011 
MSB bit is '1' then all numbers are negative
1's complement = 00000100
2's complement = 00000100 + 00000001 = 00000101 = 5
(A) 11100111  (25)_{10}
(B) 11100100  (28)_{10 (C) 11010111  (41)10 (D) 11011011  (37)10 Answer: Option A (25 is divisible by 5)}
Question 10 
For a pipelined CPU with a single ALU, consider the following situations
I. The j + 1st instruction uses the result of the jth instruction as an operand II. The execution of a conditional jump instruction III. The jth and j + 1st instructions require the ALU at the same time
Which of the above can cause a hazard?
I and II only  
II and III only  
III only  
All the three 
II is belongs to the Control hazard.
III is belongs to the Structural hazard.
→ Hazards are the problems with the instruction pipeline in CPU micro architectures.
Question 11 
Consider an array multiplier for multiplying two n bit numbers. If each gate in the circuit has a unit delay, the total delay of the multiplier is
Θ (1)  
Θ (log n)  
Θ (n)  
Θ (n^{2}) 
The total time is n+knk = Θ(n)
Question 12 
Ram and Shyam have been asked to show that a certain problem Π is NPcomplete. Ram shows a polynomial time reduction from the 3SAT problem to Π, and Shyam shows a polynomial time reduction from Π to 3SAT. Which of the following can be inferred from these reductions?
Π is NPhard but not NPcomplete
 
Π is in NP, but is not NPcomplete
 
Π is NPcomplete  
Π is neither NPhard, nor in NP

Question 13 
Nobody knows yet if P = NP. Consider the language L defined as follows:
Which of the following statements is true?
L is recursive  
L is recursively enumerable but not recursive  
L is not recursively enumerable  
Whether L is recursive or not will be known after we find out if P = NP 
P = NP (or) P != NP
→ If P=NP then L=(0+1)* which is regular, then it is recursive.
→ If P!=NP then L becomes ɸ which is also regular, then it is recursive.
So, finally L is recursive.
Question 14 
The regular expression 0*(10*)* denotes the same set as
(1*0)*1*  
0+(0+10)*  
(0+1)*10(0+1)*  
None of the above 
Option (B) and (C) doesn't generate 11.
Question 15 
If the strings of a language L can be effectively enumerated in lexicographic (i.e., alphabetic) order, which of the following statements is true?
L is necessarily finite
 
L is regular but not necessarily finite  
L is context free but not necessarily regular  
L is recursive but not necessarily context free 
The give 'L' is recursive but not necessarily context free.
Question 16 
Which of the following suffices to convert an arbitrary CFG to an LL(1) grammar?
Removing left recursion alone  
Factoring the grammar alone  
Removing left recursion and factoring the grammar  
None of the above 
To convert an arbitrary CFG to an LL(1) grammar we need to remove the left recursion and as well as left factoring without that we cannot convert.
Question 17 
Assume that the SLR parser for a grammar G has n_{1} states and the LALR parser for G has n_{2} states. The relationship between n_{1} and n_{2} is:
n_{1} is necessarily less than n_{2}  
n_{1} is necessarily equal to n_{2}
 
n_{1} is necessarily greater than n_{2}
 
None of the above 
Question 18 
In a bottomup evaluation of a syntax directed definition, inherited attributes can
always be evaluated  
be evaluated only if the definition is Lattributed  
be evaluated only if the definition has synthesized attributes  
never be evaluated 
LAttributed definitions are a class of syntax directed definitions whose attributes can be evaluated by a single traversal of the parsetree.
Question 19 
Suppose the numbers 7, 5, 1, 8, 3, 6, 0, 9, 4, 2 are inserted in that order into an initially empty binary search tree. The binary search tree uses the usual ordering on natural numbers. What is the inorder traversal sequence of the resultant tree?
7 5 1 0 3 2 4 6 8 9  
0 2 4 3 1 6 5 9 8 7  
0 1 2 3 4 5 6 7 8 9  
9 8 6 4 2 3 0 1 5 7 
Inorder: 0 1 2 3 4 5 6 7 8 9
Question 20 
Consider the following three claims
I. (n + k)^{m} = Θ(n^{m}), where k and m are constants II. 2^{n+1} = O(2^{n}) III. 2^{2n+1} = O(2^{n})
Which of these claims are correct?
I and II
 
I and III  
II and III  
I, II, and III 
Which is true by considering leading ordered term present in polynomial expression.
II) 2^{n+1} = Θ(n^{m}) → True
2^{n}×2^{n} can't be written as Θ(2^{n})
So, this is False.
Question 21 
Consider the following graph
Among the following sequences:
(I) a b e g h f (II) a b f e h g (III) a b f h g e (IV) a f g h b e
Which are depth first traversals of the above graph?
I, II and IV only  
I and IV only  
II, III and IV only  
I, III and IV only 
II) a → b → f → e (✖️)
III) a → b → f → h → g → e (✔️)
IV) a → f → g → h → b → e (✔️)
Question 22 
The usual Θ(n^{2}) implementation of Insertion Sort to sort an array uses linear search to identify the position where an element is to be inserted into the already sorted part of the array. If, instead, we use binary search to identify the position, the worst case running time will
remain Θ(n^{2})  
become Θ(n (log n)^{2})  
become Θ(n log n)  
become Θ(n) 
Instead, linear search use binary search then (log n) will be the worst case time complexity of binary search and performing n swaps to place an element in right position for the corresponding n elements
i.e., n×(logn+n)
Θ((n×logn)+n^{2})
Θ(n^{2})
Remains same.