Matrix elements between two diagonals-CodePudding

I need to make a matrix that would print number X between two diagonals, and the rest is number Y.

Right diagonal is: 'i==j' Left diagonal is: 'i j==n-1'

The problem here is to select elements that are between them.

Picture of matrix:

Code:

#include <stdio.h>

int main() {
  int i, j, n = 5, x = 4, y = 6;
  int mat[100][100];
  for (i = 0; i < n; i  )
    for (j = 0; j < n; j  ) {
      if (i == j || i   j >= n - 1 || j < i) mat[i][j] = y;
      else mat[i][j] = x;
    }
  for (i = 0; i < n; i  ) {
    for (j = 0; j < n; j  )
      printf("%d ", mat[i][j]);
    printf("\n");
  }
  return 0;
}

OUTPUT:

6 4 4 4 6

6 6 4 6 6

6 6 6 6 6

6 6 6 6 6

6 6 6 6 6

Could you please help me to fix the remaining part of matrix? I just need to make number X between the main diagonals in the lower part of matrix.

CodePudding user response：

The upper and lower triangles can be isolated with separate tests:

#include <stdio.h>

int main() {
    int i, j, n = 5, x = 4, y = 6;
    int mat[n][n];
    for (i = 0; i < n; i  ) {
        for (j = 0; j < n; j  ) {
            if (j > i && j < n - i - 1) {
                /* upper triangle */
                mat[i][j] = x;
            } else
            if (j < i && j > n - i - 1) {
                /* lower triangle */
                mat[i][j] = x;
            } else {
                /* everywhere else */
                mat[i][j] = y;
            }
        }
    }
    for (i = 0; i < n; i  ) {
        for (j = 0; j < n; j  )
            printf("%d ", mat[i][j]);
        printf("\n");
    }
    return 0;
}

Since you set the same value in both triangles, you can combine the tests as a single expression:

#include <stdio.h>

int main() {
    int i, j, n = 5, x = 4, y = 6;
    int mat[n][n];
    for (i = 0; i < n; i  ) {
        for (j = 0; j < n; j  ) {
            if ((j > i && j < n - i - 1) || (j < i && j > n - i - 1))
                mat[i][j] = x;
            else
                mat[i][j] = y;
        }
    }
    for (i = 0; i < n; i  ) {
        for (j = 0; j < n; j  )
            printf("%d ", mat[i][j]);
        printf("\n");
    }
    return 0;
}

CodePudding user response：

You can express the diagonals in terms of i and n, like you did. Just instead, bring j to the left of the expression. I have written them out as d1 and d2:

#include <stdio.h>
#include <sys/param.h>

int main() {
  int i, j, n = 5, x = 4, y = 6;
  int mat[n][n];
  for (i = 0; i < n; i  )
    for (j = 0; j < n; j  ) {
      int d1 = i;
      int d2 = n - i - 1;
      if (j > MIN(d1, d2) && j < MAX(d1, d2)) mat[i][j] = x;
      else mat[i][j] = y;
    }
  for (i = 0; i < n; i  ) {
    for (j = 0; j < n; j  )
      printf("%d ", mat[i][j]);
    printf("\n");
  }
  return 0;
}

I have taken MIN and MAX from sys/param.h, but you might want to define those yourself, whatever suits your needs.

Note that I have also inverted the assignment of x and y to make it easier to explain from the picture: If j on line i is between d1 and d2, then make it x, otherwise make it y.

CodePudding user response：

Simply count inward from the left and right margin, appropriate to the row. When you reach the centre, reverse the direction of counting. Simple.

int main() {
    int i, j, n = 11, x = 4, y = 6;

    int cnt = 0;

    int mat[100][100];
    for (i = 0; i < n; i  ) {
        for (j = 0; j < n; j  )
            mat[i][j] = ( j>cnt && j<(n-cnt-1) ) ? x : y;
        cnt  = i < n/2 ? 1 : -1;
    }

    for (i = 0; i < n; i  ) {
        for (j = 0; j < n; j  )
            printf("%d ", mat[i][j]);
        printf("\n");
    }
    return 0;
}

Since the pattern is symmetric both horizontally and vertically, why not populate more cells at the same time:

int main() {
    int i, j, n = 11, x = 4, y = 6;

    int mat[100][100];
    for (i = 0; i < (n 1)/2; i  )
        for (j = 0; j < (n 1)/2; j  )
            mat[i    ][j    ] =
            mat[n-1-i][j    ] =
            mat[i    ][n-1-j] =
            mat[n-1-i][n-1-j] =
            ( j>i && j<(n-1-i) ) ? x : y;

    for (i = 0; i < n; i  ) {
        for (j = 0; j < n; j  )
            printf("%d ", mat[i][j]);
        printf("\n");
    }
    return 0;
}

Since the stated objective is to print the pattern, there's no need for the matrix and separate operations:

int main() {
    int cnt = 0, n = 11;

    for( int i = 0; i < n; i   ) {
        for ( int j = 0; j < n; j   )
            putchar( "64"[ j>cnt && j<(n-cnt-1) ] );
        cnt  = i < n/2 ? 1 : -1;
        puts( "" );
    }

    return 0;
}

And, a final version that "works outward from the centre" (with the proviso that the values are all small):

#include <string.h>

int main() {
    const uint8_t n = 11, x = 4, y = 6;
    uint8_t mat[n][n];

    memset( mat, y, sizeof mat );
    for( int r = 1, w = 1, c = n>>1; r < (n 1)/2; r  , w  = 2 ) {
        memset( &mat[c-r][c-r 1], x, w );
        memset( &mat[c r][c-r 1], x, w );
    }

// Alternatively as a single statement without needing braces:
//  for( int r = 1, w = 1, c = n>>1; r < (n 1)/2; r  , w  = 2 )
//      memcpy( &mat[c r][c-r 1], memset( &mat[c-r][c-r 1], x, w ), w );

    for( int i = 0; i < n; i   ) {
        for( int j = 0; j < n; j   )
            printf("%d ", mat[i][j]);
        puts( "" );
    }

    return 0;
}

So many ways to skin a cat...