Unexpected output in 2D array [duplicate] - c

Question

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Maximum hourglass sum possible in a 6*6 array
(3 answers)
Closed 3 years ago.
Given a 6x6 2D Array, :
1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
We define an hourglass in to be a subset of values with indices falling in this pattern in this graphical representation:
a b c
d
e f g
There are 16 hourglasses in arr, and an hourglass sum is the sum of an hourglass' values. Calculate the hourglass sum for every hourglass in arr, then print the maximum hourglass sum.
For example, given the 2D array:
-9 -9 -9 1 1 1
0 -9 0 4 3 2
-9 -9 -9 1 2 3
0 0 8 6 6 0
0 0 0 -2 0 0
0 0 1 2 4 0
We calculate the following 16 hourglass values:
-63, -34, -9, 12,
-10, 0, 28, 23,
-27, -11, -2, 10,
9, 17, 25, 18
This is the code I have written
#include<stdio.h>
const int M=6;
const int N=6;
int hourglassSum(int arr_rows, int arr_columns, int arr[M][N]) {
int rows,columns;
rows=arr_rows-(arr_rows/3);
columns=arr_columns-(arr_columns/3);
int a[columns-1][rows-1];
int min_r,min_c,max_r,max_c,sum;
sum=0;
for(int k=0;k<columns;k++)
{
min_c=k;
max_c=k+2;
for(int l=0;l<rows;l++)
{
min_r=l;
max_r=l+2;
sum=0;
for(int i=min_c;i<=max_c;i++)
{
if(max_c>=arr_columns)
break;
for(int j=min_r;j<=max_r;j++)
{
if(max_r>=arr_rows)
break;
if(i!=min_c && i!=max_c)
{
int no=j+1;
sum +=arr[i][no];
break;
}
sum += arr[i][j];
}
}
a[min_c][min_r]=sum;
}
}
int max=-111;
for(int b=0;b<columns;b++)
{
for(int c=0;c<rows;c++)
{
if(max<a[b][c])
max=a[b][c];
}
}
for(int i=0;i<columns;i++)
{
for(int j=0;j<rows;j++)
printf("%d ",a[i][j]);
printf("\n");
}
return max;
}
int main()
{
int arr[6][6];
int arr_rows=6;
int arr_columns=6;
for(int i=0;i<arr_columns;i++)
for(int j=0;j<arr_rows;j++)
scanf("%d",&arr[i][j]);
int result=hourglassSum(arr_rows,arr_columns,arr);
printf("\n%d",result);
}
and I gave input as
-9 -9 -9 1 1 1
0 -9 0 4 3 2
-9 -9 -9 1 2 3
0 0 8 6 6 0
0 0 0 -2 0 0
0 0 1 2 4 0
and expected output is
-63 -34 -9 12
-10 0 28 23
-27 -11 -2 10
9 17 25 18
but my output is
-63 -34 -9 -10
-10 0 28 -27
-27 -11 -2 9
9 17 25 18
What's wrong?

Answer 1

Your calculations:
rows=arr_rows-(arr_rows/3);
columns=arr_columns-(arr_columns/3);
happen to produce the correct answer because 6 / 3 is 2, which is the correct value to subtract. If the matrix was 12x12, you'd be omitting rows and columns — you should be subtracting 2 because of the shape of the hourglass (or subtract hourglass_width - 1 and hourglass_height - 1 for the general case of a non-square hourglass).
Using int max=-111; is dangerous — again, it happens to work with the sample data, but it isn't a general solution. Use the value in a[0][0] as the initial maximum. It might be a lucky guess and actually the maximum, but it will be replaced by another larger value if there is one.
You should eliminate M and N and use:
int hourglassSum(int arr_rows, int arr_columns, int arr[arr_rows][arr_columns]) {
This would allow you to pass other sizes of array to the code.
However, the primary problem is that you make the a matrix too small.
You should be using int a[columns][rows]. (Arrays in C are normally a[rows][columns], but if you're consistent, you can reverse them and the difference doesn't matter when rows == columns anyway.) With the undersized matrix, I got the result you got; with the correctly sized matrix, I got the result expected.
#include <stdio.h>
static int hourglassSum(int arr_rows, int arr_columns, int arr[arr_rows][arr_columns])
{
int rows = arr_rows - 2;
int columns = arr_columns - 2;
int a[columns][rows];
for (int k = 0; k < columns; k++)
{
int min_c = k;
int max_c = k + 2;
for (int l = 0; l < rows; l++)
{
int min_r = l;
int max_r = l + 2;
int sum = 0;
for (int i = min_c; i <= max_c; i++)
{
if (max_c >= arr_columns)
break;
for (int j = min_r; j <= max_r; j++)
{
if (max_r >= arr_rows)
break;
if (i != min_c && i != max_c)
{
int no = j + 1;
sum += arr[i][no];
break;
}
sum += arr[i][j];
}
}
a[k][l] = sum;
}
}
int max = a[0][0];
for (int b = 0; b < columns; b++)
{
for (int c = 0; c < rows; c++)
{
if (max < a[b][c])
max = a[b][c];
}
}
for (int i = 0; i < columns; i++)
{
for (int j = 0; j < rows; j++)
printf(" %3d", a[i][j]);
printf("\n");
}
return max;
}
int main(void)
{
int arr[6][6];
int arr_rows = 6;
int arr_columns = 6;
for (int i = 0; i < arr_columns; i++)
{
for (int j = 0; j < arr_rows; j++)
{
if (scanf("%d", &arr[i][j]) != 1)
{
fprintf(stderr, "failed to read an integer\n");
return 1;
}
}
}
int result = hourglassSum(arr_rows, arr_columns, arr);
printf("\n%d\n", result);
return 0;
}
Output:
-63 -34 -9 12
-10 0 28 23
-27 -11 -2 10
9 17 25 18
28
Note that this is a more-or-less minimal fixup to the code in the question. My own code would look different in a number of respects (different variable names, more functions, etc.).
One other change (not made above): you don't actually need the array a; you can simply record the current maximum after you calculate the hourglass sum for each position.
For the alternative input:
-1 1 -1 0 0 0
0 -1 0 0 0 0
-1 -1 -1 0 0 0
0 -9 2 -4 -4 0
-7 0 0 -2 0 0
0 0 -1 -2 -4 0
I get the output:
-5 -2 -2 0
-9 -13 -6 -8
-19 -2 -7 -6
-8 -14 -15 -14
0
This produces the answer 0 which you say is expected (and I agree that 0 is what should be expected).

Answer 2

To output the maximum sum there is no need to create an auxiliary array. All you need is to calculate correctly the number of iterations for two nested loops.
In your function these calculations and the declaration of the array
rows=arr_rows-(arr_rows/3);
columns=arr_columns-(arr_columns/3);
int a[columns-1][rows-1];
are incorrect. For example it is unclear why the array has the number of rows equal to the value of columns - 1.
The function can be written simpler.
Here you are.
#include <stdio.h>
enum { M = 6, N = 6 };
long long int hourglassSum( int ( *a )[N], size_t m )
{
long long int max_sum = 0;
const size_t SIZE = 3;
if ( !( m < SIZE ) )
{
max_sum = ( long long int )
a[0][0] + a[0][1] + a[0][2] +
a[1][1] +
a[2][0] + a[2][1] + a[2][2];
for ( size_t i = 0; i + SIZE - 1 < m; i++ )
{
for ( size_t j = i == 0 ? 1 : 0; j + SIZE - 1 < N; j++ )
{
long long int sum = ( long long int )
a[i][j] + a[i][j+1] + a[i][j+2] +
a[i+1][j+1] +
a[i+2][j] + a[i+2][j+1] + a[i+2][j+2];
// printf( "%lld\n", sum );
if ( max_sum < sum ) max_sum = sum;
}
}
}
return max_sum;
}
int main(void)
{
int a[M][N] =
{
{ -9, -9, -9, 1, 1, 1 },
{ 0, -9, 0, 4, 3, 2 },
{ -9, -9, -9, 1, 2, 3 },
{ 0, 0, 8, 6, 6, 0 },
{ 0, 0, 0, -2, 0, 0 },
{ 0, 0, 1, 2, 4, 0 }
};
long long int sum = hourglassSum( a, M );
printf( "Hourglass Sum = %lld\n", sum );
return 0;
}
The program output is
Hourglass Sum = 28