Hi all

In a recent post, someone asked how to find if two numeric ranges overlap. I couldn't resist to try a solution based on bit vectors. That is, create 2 bit vectors with the bits inside the range set to "1", and see if the & operation on them give something different to a vector of '0's. Something like:

use strict;
use warnings;

my ($bv1,$bv2,$bv3) = ('') x 3;
vec ($bv1,$_,1)=1 for (11..19)  # First range
vec ($bv2,$_,1)=1 for (15..25)  # Second range
vec ($bv3,19,1)=0                      # The "0" vector -- should have
+ the length of the shorter.

print unpack ("b*",$bv1),"\n";
print unpack ("b*",$bv2),"\n";
print unpack ("b*",$bv3),"\n",'-'x60,"\n";
my $olap = $bv1 & $bv2;
print unpack ("b*",$olap),"\n\n";

print +(($bv1 & $bv2) eq $bv3 ? "don't overlap\n" : "overlap\n");
[download]

This works as expected:

1  000000000001111111110000
2  00000000000000011111111111000000
0s 000000000000000000000000
------------------------------------------------------------
&  000000000000000111110000
[download]

but I find a bit tricky to have to build the '0' vector for comparisons and the fact that the bit vectors 000 and 0000000000000000 are not equals

Is there a better way to see if a bit vector has all its bits set to '0'?

Thanks in advance