Can anybody suggest me a better/shorter method for this pattern? - c

Question

4 4 4 4 4 4 4
4 3 3 3 3 3 4
4 3 2 2 2 3 4
4 3 2 1 2 3 4
4 3 2 2 2 3 4
4 3 3 3 3 3 4
4 4 4 4 4 4 4
I want a shorter code or a more smarter code for this pattern.My code works fine but it is bit long.
MY CODE
#include <stdio.h>
int main()
{
int n,i,j;
scanf("%d",&n);
for(i=n;i>=1;i--)
{
for(j=n;j>i;j--)
{
printf("%d ",j);
}
for(j=1;j<(2*i)-1;j++)
{
printf("%d ",i);
}
for(j=i;j<=n;j++)
{
printf("%d ",j);
}
printf("\n");
}
for(i=2;i<=n;i++)
{
for(j=n;j>i;j--)
{
printf("%d ",j);
}
for(j=1;j<(2*i)-1;j++)
{
printf("%d ",i);
}
for(j=i;j<=n;j++)
{
printf("%d ",j);
}
printf("\n");
}
return 0;
}

Answer 1

How about like this:
#include <stdio.h>
#include <stdlib.h>
int max (int a, int b) {
return a > b ? a : b;
}
void pattern (int width) {
int count = width;
int digits = 1;
while (count /= 10)
++digits;
for (int line = 0; line < (2*width-1); ++line) {
for (int col = 0; col < (2*width-1); ++col) {
int val = max (abs (1+line-width)+1, abs (1+col-width)+1);
if (col > 0)
putc (' ', stdout);
printf ("%*d", digits, val);
}
puts ("");
}
}
int main (int argc, char* argv []) {
int width = argc > 1 ? atoi (argv [1]) : 4;
pattern (width);
}
The idea is to loop through all positions, calculate the distance to the middle position in both X and Y direction, and take the maximum of that.

Answer 2

I can suggest the following solution. Enjoy!:)
#include <stdio.h>
#include <limits.h>
#include <stdlib.h>
int main()
{
while ( 1 )
{
printf( "Enter a non-negative number (0 - exit): " );
int n;
if ( ( scanf( "%d", &n ) != 1 ) || ( n <= 0 ) ) break;
if ( INT_MAX / 2 < n )
{
n = INT_MAX / 2;
}
int width = 1;
for ( int tmp = n; tmp /= 10; ) ++width;
putchar( '\n' );
int m = 2 * n - 1;
for ( int i = 0; i < m; i++ )
{
for ( int j = 0; j < m; j++ )
{
int value1 = abs( n - i - 1 ) + 1;
int value2 = abs( n - j - 1 ) + 1;
printf( "%*d ", width, value1 < value2 ? value2 : value1 );
}
putchar( '\n' );
}
putchar( '\n' );
}
return 0;
}
The program output might look like
Enter a non-negative number (0 - exit): 10
10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10
10 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 10
10 9 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 10
10 9 8 7 7 7 7 7 7 7 7 7 7 7 7 7 8 9 10
10 9 8 7 6 6 6 6 6 6 6 6 6 6 6 7 8 9 10
10 9 8 7 6 5 5 5 5 5 5 5 5 5 6 7 8 9 10
10 9 8 7 6 5 4 4 4 4 4 4 4 5 6 7 8 9 10
10 9 8 7 6 5 4 3 3 3 3 3 4 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 2 2 3 4 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 2 2 3 4 5 6 7 8 9 10
10 9 8 7 6 5 4 3 3 3 3 3 4 5 6 7 8 9 10
10 9 8 7 6 5 4 4 4 4 4 4 4 5 6 7 8 9 10
10 9 8 7 6 5 5 5 5 5 5 5 5 5 6 7 8 9 10
10 9 8 7 6 6 6 6 6 6 6 6 6 6 6 7 8 9 10
10 9 8 7 7 7 7 7 7 7 7 7 7 7 7 7 8 9 10
10 9 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9 10
10 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 10
10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10
Enter a non-negative number (0 - exit): 0
In fact there are used only two loops that output the pattern itself.
for ( int i = 0; i < m; i++ )
{
for ( int j = 0; j < m; j++ )
{
int value1 = abs( n - i - 1 ) + 1;
int value2 = abs( n - j - 1 ) + 1;
printf( "%*d ", width, value1 < value2 ? value2 : value1 );
}
putchar( '\n' );
}