### Re: Find combination of numbers whose sum equals X

by johngg (Canon)
 on Nov 20, 2020 at 17:38 UTC Need Help??

The word "combination" in the title brought Algorithm::Combinatorics to mind. I'm not sure if duplicate sums, e.g. the several 5+5+10+80 combinations in the third example, should all be shown but I have eliminated them. This code

```use strict;
use warnings;

use feature qw{ say };

use Algorithm::Combinatorics qw{ combinations };
use List::Util               qw{ sum };

my @tests = (
{
target => 100,
values => [ 1, 99, 2, 40, 50, 100, 60, 90, 3, 5, 95, 100 ],
},
{
target => 10,
values => [ 1, 3, 2, 4 ],
},
{
target => 100,
values => [ 5, 5, 5, 5, 10, 15, 80, 99 ],
},
);

foreach my \$rhTest ( @tests )
{
say
qq{\nFind sums from },
join( q{, }, @{ \$rhTest->{ values } } ),
qq{ making \$rhTest->{ target }};

say for do {
my %seen;
grep { ! \$seen{ \$_ } ++ }
grep { \$_ == \$rhTest->{ target } }
@{ \$rhTest->{ values } };
};

for my \$sumsOf ( 2 .. scalar @{ \$rhTest->{ values } } )
{
my \$combIter = combinations( \$rhTest->{ values }, \$sumsOf );
my %seen;

while ( my \$raComb = \$combIter->next() )
{
next if \$seen{ join q{+}, sort { \$a <=> \$b } @{ \$raComb }
+} ++;
say join q{+}, @{ \$raComb }
if \$rhTest->{ target } == sum @{ \$raComb };
}
}
}

produces

```
Find sums from 1, 99, 2, 40, 50, 100, 60, 90, 3, 5, 95, 100 making 100
100
1+99
40+60
5+95
2+3+95
2+90+3+5
2+40+50+3+5

Find sums from 1, 3, 2, 4 making 10
1+3+2+4

Find sums from 5, 5, 5, 5, 10, 15, 80, 99 making 100
5+15+80
5+5+10+80
5+5+5+5+80

Cheers,

JohnGG

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