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Ksju [112]
3 years ago
15

What is the big-O performance estimate of the following function?

Computers and Technology
2 answers:
larisa [96]3 years ago
8 0

Answer:

The time complexity of the code is O(log₇n).

Explanation:

The i is updated by 7*i.On each iteration i is multiplied by 7.So on finding the time complexity of the code given above it will come out to  be log base 7.

When we divide the input by 2 the time complexity is log base 2.

So on dividing it by 7 we get the time complexity of log base 7.

Digiron [165]3 years ago
4 0

Answer:

Since the loop index variable i is initialized to 0, this function will result in infinite execution for all values of n>=0.

If instead the loop index variable is initialized to 1, Big O performance estimate of the code is O(log n) with logarithm base as 7.

Explanation:

Given function:

int f (n) {

int sum = 0;

for (i = 0; i < n; i = 7 * i)

sum += i;

return sum;

} // end f

The complexity is determined by the for loop. Starting value of the index variable i is 0.  At each iteration the value of the index variable i is multiplied by 7 (i=7*i). But multiplication by 7 still leaves the value of the index variable as 0. So loop condition i<n will fail for all n>=0 and the loop will continue indefinitely since the termination condition will never be achieved.

However if the loop index variable i is initialized to 1 instead, the loop will run O(log n) times with 7 as the base of the logarithm due to successive multiplication of index variable by 7 at each iteration.

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Given an int variable n that has been initialized to a positive value and, in addition, int variables k and total that have alre
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Answer:

The c++ program to implement the while loop is given below.

#include <iostream>

using namespace std;

int main() {

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Explanation:

The program begins with the declaration of integer variables.  

int k, n, total;

This is followed by initialization of these variables.

k=1, n=4, total=0;

The while loop runs over the variable k which is initialized to 1. The loop runs till value of k reaches the value of integer n.

First, cube of k is computed and added to the variable total.

After first execution of the loop, total is initialized to the cube of 1.

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After first execution of the loop, k is incremented from 1 to 2.

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{

total = total + ( k * k * k );

k = k + 1 ;

   }

When the value of k reaches the value of integer n, the cube of n is calculated and added to the variable, total.

When k is incremented, it becomes more than n and hence, loop gets terminated.

As the return type of main is int, the program terminates with the statement shown below.

return 0;

No output is displayed as it is not mentioned in the question.

No user input is taken as it is mentioned that integer variables are already initialized.

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