note
Rhose
Ok, I ran out of time working on the "brute force" checking, but here is the shell of a program which should reduce the number of numbers which need to be checked to a small enough set. Sorry I could not complete the program (as it was actually starting to be fun. *Smiles*)
<p>
<code>
#!/usr/bin/perl
use strict;
use warnings;
use constant MATRIX_DIM => 3; # Set your matrix size here
use constant MAX_NUM => (10 ** MATRIX_DIM) - 1;
# Since you need to find 2 times the matrix width distinct numbers, you cannot
# have a divisor greater than the maximum number divided by that product.
#
# For example, a 3 x 3 matrix has a maximum number of 999. You need 6 distinct
# numbers to "solve" the matrix, so the maximum possible divisor would be 166.
# 166 would result in 166, 332, 498, 664, 830, and 996.
#
my $Divisor = int(MAX_NUM / (MATRIX_DIM * 2));
my @Num;
my @Sol=();
sub CheckSolution
{
my $pNum=shift;
my @Sol=();
# INCOMPLETE CODE
#
# Place code here which checks every combination of MATRIX_DIM numbers as rows
# and sees if there are MATRIX_DIM other numbers which can act as columns.
#
# For example, for a 3 x 3 matrix, this code should go through each combination
# of three number (for rows) from the passed array and see if there are three
# other numbers which would work with the selected rows as columns. This should
# be a workable computation for all smaller matrix sizes.
#
return @Sol;
}
while ($Divisor >= 1)
{
@Num=();
foreach (1..int(MAX_NUM / $Divisor))
{
next if (($Divisor * $_) < (MAX_NUM / 10)); #Skip numbers which begin with zero
push @Num, ($Divisor * $_);
}
print 'Checking solution for divisor ',$Divisor,': ',join('-',@Num),"\n";
@Sol=CheckSolution(\@Num);
last if ($#Sol > -1);
$Divisor--;
}
print "\n";
if ($#Sol > -1)
{
print 'Maximum divisor is ',$Divisor,', solution is ',join('-',@Sol),"\n";
}
else
{
print 'No solution found...',"\n";
}
</code><p>
<b>Update:</b> I finished code which will work on any size matrix and posted it [Code to solve the "matrix formation" puzzle|Here].<p>Please note: I use "work" in the loosest terms as solving the 4x4 took 1.5 hours on my notebook.
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