Following questions have been asked in GATE CS 2014 exam.

**1) The number of distinct minimum spanning trees for the weighted graph below is ____
**

Answer: 6

Highlighted (in green) are the edges picked to make a MST. In the right side of MST, we could either pick edge ‘a’ or ‘b’. In the left side, we could either pick ‘c’ or ‘d’ or ‘e’ in MST.

There are 2 options for one edge to be picked and 3 options for another edge to be picked. Therefore, total 2*3 possible MSTs.

**2) Consider the following rooted tree with the vertex P labeled as root**

**The order in which the nodes are visited during in-order traversal is**

(A) SQPTRWUV

(B) SQPTURWV

(C) SQPTWUVR

(D) SQPTRUWV

Answer: (A)

The only confusion in this question is, there are 3 children of R. So when should R appear – after U or after R? There are two possibilities: SQPTRWUV and SQPTWURV. Only 1st possibility is present as an option A, the IInd possibility is not there. Therefore option A is the right answer.

**3) Let A be a square matrix of size n x n. Consider the following program. What is the expected output?**

C = 100 for i = 1 to n do for j = 1 to n do { Temp = A[i][j] + C A[i][j] = A[j][i] A[j][i] = Temp - C } for i = 1 to n do for j = 1 to n do Output(A[i][j]);

**(A)** The matrix A itself

**(B)** Transpose of matrix A

**(C)** Adding 100 to the upper diagonal elements and subtracting 100 from diagonal elements of A

**(D)** None of the above

Answer: A

If we take look at the inner statements of first loops, we can notice that the statements swap A[i][j] and A[j][i] for all i and j.

Since the loop runs for all elements, every element A[l][m] would be swapped twice, once for i = l and j = m and then for i = m and j = l. Swapping twice means the matrix doesn’t change.

**4) The minimum number of arithmetic operations required to evaluate the polynomial P(X) = X ^{5} + 4X^{3} + 6X + 5 for a given value of X using only one temporary variable.**

**(A)**6

**(B)**7

**(C)**8

**(D)**9

Answer: B

We can parenthesize the polynomial to minimize the number of operations (See Horner’s Method). We get X(X^{2}(X^{2} + 4) + 6) + 5 after parenthesization.

Following is sequence of operations to be used. Note that we are allowed to use only one variable. res = X*X res = res + 4 res = X*res res = X*res res = res + 6 res = X*res res = res + 5

**5) You have an array of n elements. Suppose you implement quicksort by always choosing the central element of the array as the pivot. Then the tightest upper bound for the worst case performance is**

(A) O(n^{2})

(B) O(nLogn)

(C) Θ(nLogn)

(D) O(n^{3})

Answer: (A)

The middle element may always be an extreme element (minimum or maximum) in sorted order, therefore time complexity in worst case becomes O(n^{2})

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