Co-occurrence matrix of a binary data (DIPS Homework #3.)

com.c

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define BUFFER 128

int main(int argc, char* argv[])
{
	char buf[BUFFER];
	char* ptr;
	int i, j, k, l, row, col, dx, dy, z, y;
	int **input, **output;

	// com [no. of rows] [no. of columns] [deltaX] [deltaY] [Z] [Y] <return>
	// for instance, com 3 4 1 0 0 1
	if(argc != 7)
	{
		printf("usage: %s [no. of rows] [no. of columns] [deltaX] [deltaY] [Z] [Y]\n", argv[0]);
		printf("\tfor instance, %s 3 4 1 0 0 1\n", argv[0]);
		return -1;
	}

	// initialize
	row = atoi(argv[1]);
	col = atoi(argv[2]);
	dx = atoi(argv[3]);
	dy = atoi(argv[4]);
	z = atoi(argv[5]);
	y = atoi(argv[6]);

	// malloc
	input = (int**)malloc(sizeof(int) * row);
	for(i = 0; i < row; i++)
	{
		input[i] = (int*)malloc(sizeof(int) * col);
	}

	output = (int**)malloc(sizeof(int) * row);
	for(i = 0; i < row; i++)
	{
		output[i] = (int*)malloc(sizeof(int) * col);
		memset(output[i], 0, sizeof(int) * col);
	}

	// input array
	for(i = 0; i < row; i++)
	{
		j = 0;
		gets(buf);

		ptr = strtok(buf, " ");
		input[i][j++] = atoi(ptr);

		while(ptr = strtok(NULL, " "))
		{
			input[i][j++] = atoi(ptr);
		}
	}

	// co-occurrence matrix of a binary data
	// 1, if I(i,j) = Z and I(i+dx,j+dy) = Y
	// 0, otherwise
	
	// C△x,△y (Z,Y) matrix
	printf("\nC[%d,%d](%d,%d) matrix is like this.\n", dx, dy, z, y);

	for(i = 0; i < row; i++)
	{
		for(j = 0; j < col; j++)
		{
			if((i+dx < row && y+dy < col) && 
				(input[i][j] == z && input[i+dx][j+dy] == y))
			{
				output[z][y]++;
			}
		}
	}

	for(k = 0; k < row; k++)
	{
		for(l = 0; l < col; l++)
		{
			printf("%d ", output[k][l]);
		}

		memset(output[k], 0, sizeof(int) * col);
		printf("\n");
	}
	
	// C△x,△y matrix
	printf("\nCOM[%d,%d] matrix is like this.\n", dx, dy);

	for(i = 0; i < row; i++)
	{
		for(j = 0; j < col; j++)
		{
			// i => Z, j => Y
			for(k = 0; k < row; k++)
			{
				for(l = 0; l < col; l++)
				{
					// k => i, l => j
					if((k+dx < row && l+dy < col) &&
						(input[k][l] == i && input[k+dx][l+dy] == j))
					{
						output[i][j]++;
					}
				}
			}	
		}
	}

	for(k = 0; k < row; k++)
	{
		for(l = 0; l < col; l++)
		{
			printf("%d ", output[k][l]);
		}

		printf("\n");
	}

	// free
	for(i = 0; i < row; i++)
	{
		free(input[i]);
	}
	free(input);
	
	for(i = 0; i < row; i++)
	{
		free(output[i]);
	}
	free(output);

	return 0;
}