Using the recurrence relation derived in the previous post, here’s a little C program to print Pascal’s triangle. Why? Because I felt like it.
/*
* Print Pascal's triangle up to a given size.
*
* Usage: pascalsTriangle [n]
*
* n = 10 by default.
*
* Example:
* ./pascalsTriangle 6
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
*/
#include <stdio.h>
#include <stdlib.h>
/** For the last row of Pascal's triangle that we will print, figure out how many digits is required for one of the central values.
*
* This function and printrow both use a recurrence relationship to compute the binomial coefficient:
*
* \binom{n,k} = \binom{n,k-1} (n - k + 1)/k
*
*/
static int biggest( int n ) {
int binom = 1;
for ( int k = 1 ; k < n/2 ; k++ ) {
binom *= (n - k + 1);
binom /= k;
}
int len = snprintf( NULL, 0, "%d", binom );
return len;
}
/// Print a row of Pascal's triangle
static void printrow( int n, int m, int spaces ) {
int binom = 1;
int leading = (m - n - 1) * spaces/2;
printf( "%*s%*d", leading, "", spaces, 1 );
for ( int k = 1 ; k < n ; k++ ) {
binom *= (n - k + 1);
binom /= k;
printf( "%*d", spaces, binom );
}
if ( n > 0 ) {
printf( "%*d", spaces, 1 );
}
printf( "\n" );
}
int main( int argc, char ** argv ) {
int m = 10;
if ( argc == 2 ) {
m = atoi( argv[1] );
}
if ( m <= 0 ) {
fprintf( stderr, "invalid size: %d\n", m );
return 1;
}
int spaces = biggest( m ) + 1;
for ( int n = 0 ; n < m ; n++ ) {
printrow( n, m, spaces );
}
return 0;
}
// vim: et ts=3 sw=3
This was just for fun, but I was pleased with the way that I easily computed the required spacing to make it all line up nicely (at least when using a monospaced font.)
I didn’t thoroughly test the output for correctness, but it looks right, and I didn’t find any errors spot checking using “sum of elements above”.
