Test: Dynamic Programming & Divide-And-Conquer - Computer Science Engineering (CSE) MCQ

Kadane algorithm is used to find:

Four matrices M1, M2, M3 and M4 of dimensions pxq, qxr, rxs and sxt respectively can be multiplied is several ways with different number of total scalar multiplications. For example, when multiplied as ((M1 X M2) X (M3 X M4)), the total number of multiplications is pqr + rst + prt. When multiplied as (((M1 X M2) X M3) X M4), the total number of scalar multiplications is pqr + prs + pst. If p = 10, q = 100, r = 20, s = 5 and t = 80, then the number of scalar multiplications needed is

The subset-sum problem is defined as follows. Given a set of n positive integers, S = {a1 ,a2 ,a3 ,…,an} and positive integer W, is there a subset of S whose elements sum to W? A dynamic program for solving this problem uses a 2-dimensional Boolean array X, with n rows and W+1 columns. X[i, j],1 <= i <= n, 0 <= j <= W, is TRUE if and only if there is a subset of {a1 ,a2 ,...,ai} whose elements sum to j. Which of the following is valid for 2 <= i <= n and ai <= j <= W?

In the above question, which entry of the array X, if TRUE, implies that there is a subset whose elements sum to W?

A sub-sequence of a given sequence is just the given sequence with some elements (possibly none or all) left out. We are given two sequences X[m] and Y[n] of lengths m and n respectively, with indexes of X and Y starting from 0. We wish to find the length of the longest common sub-sequence(LCS) of X[m] and Y[n] as l(m,n), where an incomplete recursive definition for the function l(i,j) to compute the length of The LCS of X[m] and Y[n] is given below:

l(i,j) = 0, if either i=0 or j=0
       = expr1, if i,j > 0 and X[i-1] = Y[j-1]
       = expr2, if i,j > 0 and X[i-1] != Y[j-1]

Consider two strings A = "qpqrr" and B = "pqprqrp". Let x be the length of the longest common subsequence (not necessarily contiguous) between A and B and let y be the number of such longest common subsequences between A and B. Then x + 10y = ___.

Let A1, A2, A3, and A4 be four matrices of dimensions 10 x 5, 5 x 20, 20 x 10, and 10 x 5, respectively. The minimum number of scalar multiplications required to find the product A1A2A3A4 using the basic matrix multiplication method is

Which of the following algorithms is NOT a divide & conquer algorithm by nature?

Consider the following C program

Q. What does the program compute?

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

Maximum Subarray Sum problem is to find the subarray with maximum sum. For example, given an array {12, -13, -5, 25, -20, 30, 10}, the maximum subarray sum is 45. The naive solution for this problem is to calculate sum of all subarrays starting with every element and return the maximum of all. We can solve this using Divide and Conquer, what will be the worst case time complexity using Divide and Conquer.

Consider a situation where you don't have function to calculate power (pow() function in C) and you need to calculate x^n where x can be any number and n is a positive integer. What can be the best possible time complexity of your power function?

Consider the problem of searching an element x in an array 'arr[]' of size n. The problem can be solved in O(Logn) time if. 1) Array is sorted 2) Array is sorted and rotated by k. k is given to you and k <= n 3) Array is sorted and rotated by k. k is NOT given to you and k <= n 4) Array is not sorted

The secant method is used to find the root of an equation f(x) = 0. It is started from two distinct estimates x_a and x_b for the root. It is an iterative procedure involving linear interpolation to a root. The iteration stops if f(x_b) is very small and then x_b is the solution. The procedure is given below. Observe that there is an expression which is missing and is marked by? Which is the suitable expression that is to be put in place of? So that it follows all steps of the secant method? 

Initialize: x_a, x_b, ε, N // ε = convergence indicator
f_b = f(x_b) i = 0
while (i < N and |f_b| > ε) do
i = i + 1             // update counter
x_t = ?                // missing expression for
                        // intermediate value
x_a = x_b             // reset x_a
x_b = x_t              // reset x_b
f_b = f(x_b)           // function value at new x_b

end while
if |f_b| > ε
then            // loop is terminated with i = N
write “Non-convergence”
else
write “return x_b”
end if

Suppose you are provided with the following function declaration in the C programming language.

int partition (int a[ ], int n);

The function treats the first element of a[] as a pivot, and rearranges the array so that all elements less than or equal to the pivot is in the left part of the array, and all elements greater than the pivot is in the right part. In addition, it moves the pivot so that the pivot is the last element of the left part. The return value is the number of elements in the left part. The following partially given function in the C programming language is used to find the kth smallest element in an array a[ ] of size n using the partition function. We assume k ≤ n

int kth_smallest (int a[], int n, int k) 
{ 
int left_end = partition (a, n); 
if (left_end+1==k) 
{ 
    return a [left_end]; 
} 
if (left_end+1 > k) 
{ 
    return kth_smallest (____________________); 
} 
else
{ 
    return kth_smallest (____________________); 
    } 
} 
 

Q.
The missing argument lists are respectively

Consider the problem of computing min-max in an unsorted array where min and max are minimum and maximum elements of array. Algorithm A1 can compute min-max in a1 comparisons without divide and conquer. Algorithm A2 can compute min-max in a2 comparisons by scanning the array linearly. What could be the relation between a1 and a2 considering the worst case scenarios?

Consider the C program below.
#include <stdio.h> 
int *A, stkTop; 
int stkFunc (int opcode, int val) 
{ 
    static int size=0, stkTop=0; 
    switch (opcode) 
    { 
    case -1: 
        size = val; 
        break; 
    case 0: 
        if (stkTop < size ) A[stkTop++]=val; 
        break; 
    default: 
        if (stkTop) return A[--stkTop]; 
    } 
    return -1; 
} 
int main() 
{ 
    int B[20]; 
    A=B; 
    stkTop = -1; 
    stkFunc (-1, 10); 
    stkFunc (0, 5); 
    stkFunc (0, 10); 
    printf ("%d\n", stkFunc(1, 0)+ stkFunc(1, 0)); 
} 

The value printed by the above program is ___________