This function uses a curious mix of iteration and recursion: function F(n) if n < 1...

This function uses a curious mix of iteration and recursion:

function F(n)

    if n < 1

        return 1

    t <- 0

    for i <- 0 to n

        for j <- i to n

            t <- t + j

    return t + F(n-1)

What is the number of basic operations (additions and subtractions) performed? Answer: Θ(n³)

Could you tell me how I can solve this problem??

Expert Solution

This is a recursive function. so, follow these steps for calculating it's time complexity.

total amount of work done per one recursive call:
--------------------------------------------------
There are two for loops
outer for loop iterates n times
inner for loop iterates n times
so, total number of operations = n^2 = O(n^2)

size of the recursive call: n-1.  (because of F(n-1))
number of recursive calls inside F(n) = 1. (single recursive calls to F(n-1)

so, overall time complexity T(n) = 1*T(n-1) + O(n^2)
T(n) = T(n-1) + O(n^2)

let's solve this recurrence relation:
--------------------------------------
T(n)
= T(n-1) + n^2
= T(n-2) + (n-1)^2 + n^2
= T(n-3) + (n-2)^2 + (n-1)^2 + n^2
= T(1) + 2^2 + ... + (n-2)^2 + (n-1)^2 + n^2
= 1 + 2^2 + ... + (n-2)^2 + (n-1)^2 + n^2
formula for sum of squares of first n numbers is n(n+1)(2n+1)/6
ignore constance terms.
so, time complexity is O(n^3)

Answer: O(n^3)

venereology answered 2 years ago

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