Recursion with Double Call (C++) - c++

Question

Can someone please explain to me how exactly this adds up to 26?
I get confused by the 'double call'. Maybe I just don't understand recursion as well as I think I do.
#include <iostream>
using namespace std;
int rec(int * N, int max) {
if (max < 0)
return 0;
return N[max] + rec(N, max - 1) + rec(N, max - 2);
}
int main() {
const int max = 5;
int N[] = { 1, 2, 3, 4, 5 };
int f = rec(N, max - 1);
cout << f << endl;
return 0;
}

Answer 1

int f = rec(N, 4)
= N[4] + rec(N, 3) + rec(N, 2)
= 5 + (N[3] + rec(N, 2) + rec(N, 1)) + (N[2] + rec(N, 1) + rec(N, 0))
= 5 + (4 + (N[2] + rec(N, 1) + rec(N, 0)) + (N[1] + rec(N, 0) + rec(N, -1)) + (3 + (N[1] + rec(N, 0) + rec(N, -1)) + (N[0] + rec(N, -1) + rec(N, -2)))
= 5 + (4 + (3 + rec(N, 1) + rec(N, 0)) + (2 + rec(N, 0) + 0) + (3 + (2 + rec(N, 0) + 0) + (1 + 0 + 0))
= 5 + (4 + (3 + (N[1] + rec(N, 0) + rec(N, -1)) + (N[0] + rec(N, -1) + rec(N, -2))) + (2 + (N[0] + rec(N, -1) + rec(N, -2)) + 0) + (3 + (2 + (N[0] + rec(N, -1) + rec(N, -2)) + 0) + 1)
= 5 + (4 + (3 + (2 + (N[0] + rec(N, -1) + rec(N, -2)) + 0) + (1 + 0 + 0)) + (2 + (1 + 0 + 0) + 0) + (3 + (2 + (1 + 0 + 0) + 0) + 1))
= 5 + (4 + (3 + (2 + (1 + 0 + 0) + 0) + 1) + (2 + 1 + 0) + (3 + (2 + 1 + 0) + 1))
= 5 + (4 + (3 + (2 + 1 + 0) + 1) + 3 + (3 + 3 + 1))
= 5 + (4 + (3 + 3 + 1) + 3 + 7)
= 5 + (4 + 7 + 10)
= 5 + 21
= 26
And as suggested above, the following code generates the entire expression that, when evaluated, equals 26. It can also be played with to generate step-by-step work.
#include <iostream>
#include <string>
using namespace std;
string rec(string * N, int max) {
if (max < 0)
return "0";
auto a = rec(N, max - 1), b = rec(N, max - 2);
return "(" + N[max] + " + " + a + " + " + b + ")";
}
int main() {
const int max = 5;
string N[] = { "1", "2", "3", "4", "5" };
auto f = rec(N, max - 1);
cout << f << endl;
return 0;
}