Give a big-O estimate the number of operations, where an operation is a comparison or a multiplication, used in this segment of an algorithm (ignoring comparisons used to test the conditions in the for loops, where a1, a2, . . . , an are positive real numbers)

Answer and Explanation:

For the first iteration of i for loop 1 to n, the j for loop will run from 2 to n times. i.e. n-1 times.

For the second iteration of i for loop, the j for loop will run from 3 to n times. i.e. n-2 times.

From the third to the last iteration of i for loop, the j for loop will run n-1 to n times. i.e. 2 times.

From the second to the last iteration of i for loop, the j for loop will run from n to n times. i.e. 1 time.

For the last iteration of i for loop, the j for loop will run 0 times because i+1 >n.

Hence the equation looks like below:

1 + 2 + 3 + + (n-2) + (n-1) = n(n-1)/2

So the number of total iterations is n(n-1)/2.

There are two operations per loop, i.e. Comparison and Multiplication, so the iteration is 2 * n(n-1)/2 = n ^2 - n

So f(n) = n ^ 2 - n

f(n) <= n ^ 2 for n > 1

Hence, The algorithm is O(n^2) with C = 1 and k = 1.

4) Suppose n = 1, then i = 2

n = 2, then i = 4

n = 3, then i = 6

n = 4, then i = 8

So the value of i is doubled for each iteration of while loop.

So i = 2 ^ n

i is growing at the rate of a power of 2, so the number of operations is log(2n)

Hence the algorithm is O(log(n)).

try this:

go tos official website.

click on your profile pic at the top right of the screen and click 'edit profile'.

click on "preferences" and press "year of birth".

and then just maybe try to reset whatever error made before by typing in the correct age details again. hope this !