in reply to decomposing binary matrices
I think this does the job, but it needs testing on some more demanding sample data.
c:\test>600418.pl
This input
0 0 1 1 0
1 0 1 1 1
0 1 0 1 0
1 1 0 0 1
0 1 1 1 0
Inverted looks like this
0 : 01010
1 : 00111
2 : 11001
3 : 11101
4 : 01010
This subset can be removed from the main group:
0 : 01010
4 : 01010
c:\test>600418.pl
This input
0 0 0 1 1 0 0 0
0 1 0 1 1 1 1 0
1 0 1 0 1 0 1 1
1 1 1 0 0 1 1 1
1 0 1 1 1 0 0 1
Inverted looks like this
0 : 00111
1 : 01010
2 : 00111
3 : 11001
4 : 11101
5 : 01010
6 : 01110
7 : 00111
This subset can be removed from the main group:
1 : 01010
5 : 01010
This subset can be removed from the main group:
0 : 00111
2 : 00111
7 : 00111
Relating the subset numbering back to the pre-inversion set is left as an exercise.
#! perl -slw
use strict;
=comment
my @grid = (
[ 0, 0, 1, 1, 0 ],
[ 1, 0, 1, 1, 1 ],
[ 0, 1, 0, 1, 0 ],
[ 1, 1, 0, 0, 1 ],
[ 0, 1, 1, 1, 0 ],
);
=cut
my @grid = (
# 0 1 2 3 4 5 6 7
[ 0, 0, 0, 1, 1, 0, 0, 0, ],
[ 0, 1, 0, 1, 1, 1, 1, 0, ],
[ 1, 0, 1, 0, 1, 0, 1, 1, ],
[ 1, 1, 1, 0, 0, 1, 1, 1, ],
[ 1, 0, 1, 1, 1, 0, 0, 1, ],
);
print "This input\n";
print "\t@$_" for @grid;
my @inverted;
for my $i ( 0 .. $#grid ) {
$inverted[ $_ ] .= $grid[ $i ][ $_ ] for 0 .. $#{ $grid[ $i ] };
}
print "\nInverted looks like this\n";
print "\t$_ : $inverted[ $_ ]" for 0 .. $#inverted;
my @bitCounts;
push @{ $bitCounts[ $inverted[ $_ ] =~ tr[1][1] ] }, $_ for 0 .. $#inv
+erted;
#print @{ $_||[] } for @bitCounts;
for my $c ( 2 .. $#bitCounts ) {
#print( "skipping $c elements < ${ \ scalar @{ $bitCounts[ $c ] }
+} bits" ),
next unless @{ $bitCounts[ $c ] } >= $c;
my @set = @{ $bitCounts[ $c ] };
for my $offset ( 0 .. @set - $c ) {
my $matches = grep{
$inverted[ $set[ $offset ] ] eq $_
} @inverted[ @set[ $offset .. $#set ] ];
#print "set:@set matches:$matches";
last unless $c <= $matches;
print "\nThis subset can be removed from the main group:\n";
print "\t$_ : $inverted[ $_ ]" for grep{
$inverted[ $set[ $offset ] ] eq $inverted[ $_ ]
} @set[ $offset .. $#set ];
}
}
Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.
Re^2: decomposing binary matrices
by hv (Prior) on Feb 16, 2007 at 15:16 UTC
|
Thanks. First, I should note that there must be at least as many values as variables, since each variable must take a distinct value within the set of possible values.
Second, variables that are part of an n-element submatrix need not have n bits set. The sparsest counterexample is:
my @grid = (
# 0 1 2 3 4 5
[ 0, 0, 0, 1, 1, 0, ],
[ 0, 0, 0, 1, 0, 1, ],
[ 0, 0, 0, 0, 1, 1, ],
[ 1, 1, 0, 0, 0, 0, ],
[ 1, 0, 1, 0, 0, 0, ],
[ 0, 1, 1, 0, 0, 0, ],
);
.. which can decompose into two 3-element submatrices.
The least sparse version of that is: my @grid = (
# 0 1 2 3 4 5
[ 0, 0, 0, 1, 1, 1, ],
[ 0, 0, 0, 1, 1, 1, ],
[ 0, 0, 0, 1, 1, 0, ],
[ 1, 1, 1, 1, 1, 1, ],
[ 1, 1, 1, 1, 1, 1, ],
[ 1, 1, 0, 1, 1, 1, ],
);
.. which can decompose the same way.
Update: swapped 2 bits in the last row of the sparse matrix, so it actually represents what I'm saying
Hugo | [reply] [Watch: Dir/Any] [d/l] [select] |
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Hm. Sorry to have wasted your time. I'd picked up on this bit of the OP
... since A and E are restricted to only two values between them, they must consume those two values;
... and hung my hat on it, but that obviously doesn't apply in the same way to the two examples above.
Question: Would this example also decompose into the (same?) two groups as the above?
my @grid = (
[ 0, 1, 0, 1, 0, 0, ],
[ 0, 0, 0, 1, 1, 0, ],
[ 0, 1, 0, 0, 1, 0, ],
[ 1, 0, 0, 0, 0, 1, ],
[ 1, 0, 1, 0, 0, 0, ],
[ 1, 0, 0, 0, 0, 1, ],
);
Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.
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Yes: labelling the columns A..F and the rows 1..6, we know variables {B, D, E} must consume values {1, 2, 3} between them, and likewise {A, C, F} must consume {4, 5, 6}. In this example, however, the latter can be further decomposed: C can only be 5, so {A, F] are left with {4, 6}.
Hugo
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