Tell me why this program is equal to 9?

[the_d]

can someone tell me why this program is equal to 9? I have been sitting for 2

hours trying to figure out how the program gets and when doing it by hand I

get something different. Here is the program:

#include <stdio.h>

int main()
{
int i, j;
int a=0;
for (i=0;i<3;i++)
for (j=0;j<2;j++)
a+=2*i+j-1;
printf("%d\n", a);
return 0;
}

a=9

 

[whatta]

what you mean why?
i=0, j=0: 2*i+j-1=-1, a=-1
i=0, j=1: 2*i+j-1=0, a=-1
i=1, j=0: 2*i+j-1=1, a=0
i=1, j=1: 2*i+j-1=2, a=2
i=2, j=0: 2*i+j-1=3, a=5
i=2, j=1: 2*i+j-1=4, a=9

 

[the_d]

can u explain why you alternated the values in i and j and in what sequence you did it in please?

 

[D H]

can u explain why you alternated the values in i and j and in what sequence you did it in please?

They represent

  for (i=0;i<3;i++)
   for (j=0;j<2;j++)

In other words, whatta executed the program by hand.

Mathematically, your code is calculating

$$a=\sum_{i=0}^2\sum_{j=0}^1 2i+j-1$$

First evaluate the inner sum (the sum over $j$). Note that the term $2i-1$ is a constant in this inner sum. Using

$$\sum_{k=0}^n c = (n+1)*c$$

$$\sum_{k=0}^n k = \frac{k*(k+1)}2$$

The inner sum evaluates to

$$\sum_{j=0}^1 2i+j-1 = 2*(2i-1)+\frac{1*2}2 = 4i-1$$

The outer sum thus becomes

$$a=\sum_{i=0}^2 4i-1 = 4\frac{2*3}{2}-3*1 = 12-3 = 9$$

 

[whatta]

or, in the rant straight from mighty wikipedia: click

 

[verty]

Perhaps it makes more sense with while loops:

int main()
{
	int i, j;
	int a=0;
	
	i = 0;
	while (i < 3) {
		j = 0;
		while (j < 2) {
			a += 2*i+j-1;
			j++;
		}
		i++;
	}
	printf("%d\n", a);
	return 0;
}