Maximum hourglass sum possible in a 6*6 array

Question

There is a question on 2D array which says

Given a 6*6 matrix we have to print the largest (maximum) hourglass sum found in the matrix. An hourglass is described as:

Answer 1

Here's some overkill — four different ways to write the 'hourglass sum' function. For reasons outlined in user3629249's comment, I've renamed the functions to hourglass_N (for N in 1..4). Of these variants, hourglass_1() is pretty neat for this particular shape, but hourglass_2() is more readily adaptable to other shapes.

The test code correctly handles matrices with negative maximum sums.

#include 
#include 

enum { ARR_SIZE = 6 };

static int hourglass_1(int a[ARR_SIZE][ARR_SIZE], int i, int j)
{
    assert(i >= 0 && i < ARR_SIZE - 2 && j >= 0 && j < ARR_SIZE - 2);
    int sum = a[i+0][j+0] + a[i+0][j+1] + a[i+0][j+2]
                          + a[i+1][j+1] +
              a[i+2][j+0] + a[i+2][j+1] + a[i+2][j+2];
    return sum;
}

static int hourglass_2(int a[ARR_SIZE][ARR_SIZE], int i, int j)
{
    assert(i >= 0 && i < ARR_SIZE - 2 && j >= 0 && j < ARR_SIZE - 2);
    static const int rows[] = { 0, 0, 0, 1, 2, 2, 2 };
    static const int cols[] = { 0, 1, 2, 1, 0, 1, 2 };
    enum { HG_SIZE = sizeof(rows) / sizeof(rows[0]) };
    int sum = 0;
    for (int k = 0; k < HG_SIZE; k++)
        sum += a[rows[k]+i][cols[k]+j];
    return sum;
}

static int hourglass_3(int a[ARR_SIZE][ARR_SIZE], int i, int j)
{
    assert(i >= 0 && i < ARR_SIZE - 2 && j >= 0 && j < ARR_SIZE - 2);
    int sum = 0;
    for (int i1 = 0; i1 < 3; i1++)
    {
        for (int j1 = 0; j1 < 3; j1++)
        {
            if (i1 == 1)
            {
                sum += a[i + i1][j + j1 + 1];
                break;
            }
            else
                sum += a[i + i1][j + j1];
        }
    }
    return sum;
}

static int hourglass_4(int a[ARR_SIZE][ARR_SIZE], int i, int j)
{
    assert(i >= 0 && i < ARR_SIZE - 2 && j >= 0 && j < ARR_SIZE - 2);
    int n = i + 3;
    int m = j + 3;
    int sum = 0;
    for (int i1 = i; i1 < n; i1++)
    {
        for (int j1 = j; j1 < m; j1++)
        {
            if (i1 == n - 2)
            {
                sum += a[i1][j1 + 1];
                break;
            }
            else
                sum += a[i1][j1];
        }
    }
    return sum;
}

typedef int (*HourGlass)(int arr[ARR_SIZE][ARR_SIZE], int i, int j);

static void test_function(int arr[ARR_SIZE][ARR_SIZE], const char *tag, HourGlass function)
{
    int max_sum = 0;
    int max_row = 0;
    int max_col = 0;
    for (int i = 0; i < (ARR_SIZE-2); i++)
    {
        for (int j = 0; j < (ARR_SIZE-2); j++)
        {
            int n = (*function)(arr, i, j);
            if (n > max_sum || (i == 0 && j == 0))
            {
                max_sum = n;
                max_row = i;
                max_col = j;
            }
        }
    }

    printf("%s: %3d at (r=%d,c=%d)\n", tag, max_sum, max_row, max_col);
}

int main(void)
{
    int arr[ARR_SIZE][ARR_SIZE];

    for (int i = 0; i < ARR_SIZE; i++)
    {
        for (int j = 0; j < ARR_SIZE; j++)
        {
            if (scanf("%d", &arr[i][j]) != 1)
            {
               fprintf(stderr, "Failed to read integer (for row %d, col %d)\n", i, j);
               return 1;
            }
        }
    }

    test_function(arr, "hourglass_1", hourglass_1);
    test_function(arr, "hourglass_2", hourglass_2);
    test_function(arr, "hourglass_3", hourglass_3);
    test_function(arr, "hourglass_4", hourglass_4);

    return 0;
}

For various different data sets, the code produces the correct answer.

Set 1:

1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 2 4 4 0
0 0 0 2 0 0
0 0 1 2 4 0

hourglass_1:  19 at (r=3,c=2)
hourglass_2:  19 at (r=3,c=2)
hourglass_3:  19 at (r=3,c=2)
hourglass_4:  19 at (r=3,c=2)

Set 2:

-1 -1 -1 +0 +0 +0
+0 -1 +0 +0 +0 +0
+1 -1 +1 +0 +0 +0
+0 +0 -2 +4 -4 +0
+0 +0 +0 +2 +0 +0
+0 +0 +1 +2 -4 +0

hourglass_1:   7 at (r=2,c=2)
hourglass_2:   7 at (r=2,c=2)
hourglass_3:   7 at (r=2,c=2)
hourglass_4:   7 at (r=2,c=2)

Set 3:

-7 -2 -9 -7 -4 -4
-6  0 -7 -5 -8 -1
-9 -1 -2 -1 -3 -3
-9 -1 -3 -6 -2 -9
-8 -1 -3 -7 -7 -9
 0 -3 -5 -2 -2 -5

hourglass_1: -18 at (r=2,c=1)
hourglass_2: -18 at (r=2,c=1)
hourglass_3: -18 at (r=2,c=1)
hourglass_4: -18 at (r=2,c=1)

Set 4:

-7 -7  0 -7 -8 -7
-9 -1 -5 -6 -7 -8
-2  0 -7 -7 -6 -3
-5 -3 -1 -6 -3 -1
-3 -5  0 -5  0 -7
-1 -2 -8 -8 -9 -9

hourglass_1: -20 at (r=2,c=0)
hourglass_2: -20 at (r=2,c=0)
hourglass_3: -20 at (r=2,c=0)
hourglass_4: -20 at (r=2,c=0)