Question

Let T(n) be the runtime of the following RaiseToPower() function that computes baseVal raised to the n-th power. int RaiseToPower (int baseval, int n){ int resultval; if (n == 0) { resultval = 1; } else { resultval = baseval - RaiseToPower (baseval, n-1 ); return resultval; } Which of the following is correct about T(n)? There are more than one correct answers. Select one or more: a. T(n) = CO, if n=0. (CO is a small constant variable.) b. T(n) also varies with the first parameter, namely, int baseVal. c. T(n) = C1 +T(n-1), if n>1. (C1 is a small constant) How many times will the highlighted line need to be executed? for (int i =0; i<n-1; i++) for (int j=0; j<n; j++) cout <<i + j <<endl; Select one: a. n*(n-1) b. n*n c. n*(n+1) 0 d. (n-1)*(n+1)

Answer 1

Q->1:

Given code:

1. int RaiseToPower(int baseval, int n){
2.     int resultval;
3.     if (n == 0){
4.         resultval = 1;
5.     }
6.     else{
7.         resultval = baseval * RaiseToPower(baseval, n - 1);
8.     }
9. return resultval;
10. }

Options:

a. T(n) = C0, if n=0, (C0 is a small constant variable)

True, this is because of condition at line 3 which sets resultval to 1 and at line 9 it will be returned.

b) T(n) also varies with the first parameter, namely int baseVal

False, in programming we assume that multiplication is a constant time operation.

c) T(n) = C1 + T(n-1), if n>1 (C1 is a small constant)

True, this is because this function calls itself recursively n-1 times, and nth time it returns in C2 time.

So, a and c are correct

.

Q->2: How many times will highlighted line need to be executed?

Code:

1. for (int i = 0; i < n - 1; i++)
2.     for (int j = 0; j < n; j++)
3.         cout << i + j << endl;

Answer:

a. n*(n-1)

This is because,loop at line 1 executes n-1 times

Then loop at line 2 executes n times