Question

Q5 (25pts) Consider the code:
int foo(int N){
if (N <= 3) return 2;
int res1 = 3*foo(N-4);
int res2 = foo(N-2);
return res1-res2;
}
a) (6 points) Write the recurrence formula for the time complexity of this function (including the base cases) for N>=0. You do NOT need to solve it.
b) (5 points) Draw the tree that shows the function calls performed in order to compute foo(8) (the root will be foo(8) and it will have a child for each recursive call.)
c) (4pts) In the tree, show the return values for function calls (2pts) and fill in the answer below
foo(8) = _____________________________________________ (2 out of 4 pts)
d) (7pts) Write code for foo_mem, the memoized version of foo (use a solution array to look-up and store results of recursive calls). Comment [O6]: NOT posted
5
e) (3pts) Write the wrapper function that calls the foo_mem function.

Answer 1

a)

b)

c)

foo(8) = 3*foo(4) - foo(6)

foo(6) = 3*foo(2) - foo(4)

foo(4) = 3*foo(0) - foo(2) = 3*2 - 2 = 4

foo(6) = 3*2 - 4 = 2

  foo(8) = 3*4 - 2 = 10

foo(8) = 10

d)

int foo_mem(int N) {

    int arr[N+1];

    arr[0] = arr[1] = arr[2] = arr[3] = 2;

    for (int i = 4; i <= N; i++)

        arr[i] = 3*arr[i-4] - arr[i-2];

    return arr[N];

}

e)

#include<iostream>

using namespace std;

int foo_mem(int N) {

    int arr[N+1];

    arr[0] = arr[1] = arr[2] = arr[3] = 2;

    for (int i = 4; i <= N; i++)

        arr[i] = 3*arr[i-4] - arr[i-2];

    return arr[N];

}

int main() {

    cout << foo_mem(8);

    return 0;

}