Re: Q on associative binary operation

12 Aug 2003

      Actually, strike that...
The last row can only be (d,c,c,d).
I had an off-by-one in the check_assoc subroutine.

It should be:

sub check_assoc {
  my $op = shift;
  for(my $i=0;$i<4;$i++){
  for(my $j=0;$j<4;$j++){
  for(my $k=0;$k<4;$k++){
    if( $op->[ $op->[$i][$j]] [ $k            ]
     != $op->[ $i           ] [ $op->[$j][$k] ] )
    {
      return 0;
    }
  } } }
  return 1;
}

On Tue, Aug 12, 2003 at 03:04:41PM -0400, BillyGOTO wrote:
...
"For my ally is Perl, and a powerful ally it is."
On Tue, Aug 12, 2003 at 02:06:43AM -0700, Sarad AV wrote:
...
hi,
how do we complete this table
Table shown may be completed to define 'associative'
binary operation * on S={a,b,c,d}. Assume this is
possible and compute the missing entries
...
*|a|b|c|d
---------
a|a|b|c|d
---------
b|b|a|c|d
---------
c|c|d|c|d
---------
d| | | |
Lucky you!  There are only 256 possibilities.
There are four solutions:
The last row can be any of:
d c c a
d c c b
d c c c
d c c d
...
#!/usr/bin/perl -w
use strict;
my $optbl = [
    [0,1,2,3],
    [1,0,2,3],
    [2,3,2,3],
];
for(my $i=0; $i<0x100; $i++){
    $optbl->[3] = [
        ($i>>0)&0x3,
        ($i>>2)&0x3,
        ($i>>4)&0x3,
        ($i>>6)&0x3,
    ];
    if(&check_assoc($optbl)){
        for(join(',',@{$optbl->[3]})){
            tr/0123/abcd/;
            print "$_\n";
        }
    }
}
sub check_assoc {
    my $op = shift;
    for(my $i=0;$i<3;$i++){
    for(my $j=0;$j<3;$j++){
    for(my $k=0;$k<3;$k++){
        if( $op->[ $op->[$i][$j]] [ $k            ]
         != $op->[ $i           ] [ $op->[$j][$k] ] )
        {
            return 0;
        }
    } } }
    return 1;
}