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Re: [vox-tech] [OT] Binary Representation Challenge

# Re: [vox-tech] [OT] Binary Representation Challenge

Tim Riley wrote:

On Tue, 2005-09-20 at 12:02, Alex Mandel wrote:

I realize this might be not be a challenge for some of you.
--
I need to make a list of all possible permutations given 9 options and that you can choose any number of options at once.
I've figured out using nCr statistics that this is 511 choices, but now I need to represent them in 2^9 binary code: 000000001, 000000010 etc

The following code should work:

/* binary_count.c					*/
/* ---------------------------------------------------- */
/* By Tim Riley						*/
/* ---------------------------------------------------- */
/* Compile: cc -Wall binary_count.c -o binary_count	*/
/* ---------------------------------------------------- */

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

#define NUMBER_BINARY_DIGITS	16

/* From "The C Programming Language" by Kernighan and Ritchie */
/* ---------------------------------------------------------- */
unsigned getbits( unsigned integer, unsigned position, unsigned n )
{
return ( integer >> ( position + 1 - n ) ) & ~( ~0 << n );
}

char *integer2binary( unsigned integer )
{
static char binary[ NUMBER_BINARY_DIGITS + 1 ] = {0};
char *pointer;
int i;

pointer = binary;
for ( i = 0; i < NUMBER_BINARY_DIGITS; i++, pointer++ )
*pointer = '0';

pointer = binary + NUMBER_BINARY_DIGITS - 1;

for( i = 0; i < NUMBER_BINARY_DIGITS; i++, pointer-- )
{
if ( getbits( integer, i, 1 ) ) *pointer = '1';

}

return binary;
}

int main( int argc, char **argv )
{
unsigned count_up_to;
unsigned count;

if ( argc != 2 )
{
fprintf( stderr, "Usage: %s count_up_to\n", argv[ 0 ] );
return 1;
}

count_up_to = atoi( argv[ 1 ] );

for ( count = 0; count < count_up_to; count++ )
printf( "%s\n", integer2binary( count ) );

return 0;
}

Python is a little more compact.

def int2bin(n,digits=9):
return "".join(map(lambda y:str((n>>y)&1), range(digits-1, -1, 				-1)))
for i in range(512):
print int2bin(i)

Bruce

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