bimleshsharma has asked for the
wisdom of the Perl Monks concerning the following question:
I have an array having some interger elements. I have to split into 2 subsets. Need to verify that sum of these 2 subsets are equal or not. Subset size is not matter, it can be element count of 3,2 or 1,4 or any but should be in 2 part. For example...
Ex 1:
@a1 = qw(1, 3, 8, 4);
This array can be divided into to subsets (1,3,4)=8 and (8)=8, so these two have equal sum.
Ex 2:
@a1 = qw(1, 6, 2);
This cant be divided into two subsets of equal sum.Because (1,6)!=2 or any combination of a subset is not matching to another subset.
I tried below code but it is working for one set of iteration, actually it should check all possible set of iteration to find possiblities.
my @array= qw(1 3 5 7);
my @array= qw(1 3 5 7);
&test(\@array);
sub test
{
my ($s1,$s2);
my @a=@{$_[0]};
for (my $i=0;$i<=$#a ;$i++) {
($s1,$s2)=0;
for (my $j=0;$j<=$#a ;$j++) {
if ($i == $j) {
$s1=$s1+ $a[$j];
}
else{
$s2+=$a[$j];
}
}
}
print "\n";
}
Re: Divide an array into 2 subsets to verify their sum is equal or not.
by choroba (Bishop) on May 02, 2013 at 07:50 UTC

To sum a list, I used the sum from List::Util. I made the following observation: if the array can be split, then the sum of each part is the half of the sum of the whole array. I used vec to generate binary vectors to be used as Indicator function. Checking a half of the possible vectors is enough, the rest is complementary (i.e. the two subsets are swapped).
#!/usr/bin/perl
use warnings;
use strict;
use List::Util qw(sum);
sub is_divisible {
my $array = shift;
my $sum = sum(@$array) / 2;
for my $bitmask (1 .. 2 ** $#$array  1) {
return 1 if sum(map { $array>[$_] * vec $bitmask, $_, 1} 0 ..
+ $#$array) == $sum;
}
return;
}
my @arrays = (
[qw(1 3 5 7)],
[qw(1 3 8 4)],
[qw(1 6 2)],
[qw(5 5 4 6 2 8 1 9)],
);
for my $array (@arrays) {
print "@$array: ", is_divisible($array) ? 'yes' : 'no', "\n";
}
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Re: Divide an array into 2 subsets to verify their sum is equal or not.
by BrowserUk (Pope) on May 02, 2013 at 09:51 UTC

choroba's Exploreallpossiblecombinations mechanism works okay for smallish sets (max:32 or 64 depending upon your Perl), but will get very slow for anything much larger than 20 or so.
This will very quickly (less than 0.001 of a second) find a solution, if one exists, for sets of 100s or 1000s of elements. :
#! perl slw
use strict;
use Time::HiRes qw[ time ];
use List::Util qw[ sum ];
sub partition {
my $sum = sum @_;
return if $sum & 1;
$sum /= 2;
my @s = sort{ $b <=> $a } @_;
my @a;
my( $t, $n ) = ( 0, 1 );
$t + $s[$n] <= $sum and $t+= $s[$n] and push @a, $n while ++$n < @
+s and $t <= $sum;
@a = delete @s[ @a ];
@s = grep defined, @s;
return unless sum( @a ) == sum( @s );
return \@a, \@s;
}
our $N //= 64;
my( $a, $b ) = partition 1,3,5,7;
print "sum( @{ $a } ) == sum( @{ $b } )" if $a;
my @set = map int( rand 100 ), 1 .. $N;
my $start = time;
( $a, $b ) = partition @set;
printf "Took %f seconds\n", time()  $start;
if( $a ) {
printf "(%u) == sum( @{ $a } ) == sum( @{ $b } )\n", sum @$a;
}
else {
print "No solution existed for the $N element set @set";
}
A few runs:
With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
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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It gives a wrong answer for
2, 12, 4
 [reply] [d/l] 

Corrected. Thanks.
With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
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.
 [reply] 
Re: Divide an array into 2 subsets to verify their sum is equal or not.
by kcott (Chancellor) on May 02, 2013 at 10:35 UTC

G'day bimleshsharma,
Update:
Update2:
Original solution substantially rewritten! It had various problems:
Update3:
Fixed a bug and added some features:
Update4:
Fixed some bugs and changed volume testing.

Display issue with subset output. Duplicate data due to "@a2 ... @a2" not being "@a1 ... @a2".  FIXED.

while loop in check_arrays() had flaws. This has been pretty much rewritten.  FIXED.

Volume testing was decimalbased, now it's octalbased. Previously, volume_power_max=3 [or vpm=3] created arrays of up to 10**3 (1,000) elements; now, the value is 8**3 (512) elements. Decimalbased was a bad choice as neither 1 .. 10 nor 1 .. 100 can be split into two equal portions.

Added a few more tests.
Here's Update4's version of pm_split_equal_sums.pl:
#!/usr/bin/env perl l
use strict;
use warnings;
use List::Util qw{first sum};
use Test::More;
use Time::HiRes qw{time};
use Getopt::Long;
my %opt = (
test_more => 1,
time_hires => 1,
volume_tests => 0,
volume_power_max => 3,
array_limit => 3,
);
GetOptions(map {
join('' => @{[join '' => /(?>^_)([az])/gi]}, $_) . ':i' => \$op
+t{$_}
} keys %opt);
my $test_equal_subsets = [
[1, 3, 8, 4],
[1, 3, 5, 7],
[4, 3, 2, 2, 1],
[4, 3, 2, 2, 2, 2, 1],
[5, 5, 4, 6, 2, 8, 1, 9],
[8, 4, 4, 7, 6, 3],
[1, 1],
[2, 2],
[],
[0],
[0, 0],
[0, 0, 0],
[0, 0, 0, 0],
[ (1) x 100 ],
[ 1 .. 1000 ],
];
my $test_unequal_subsets = [
[1, 6, 2],
[7, 5, 3, 3],
[1, 2 ,3, 7],
[0, 1],
[1, 2],
[1],
[2],
[8, 1, 2, 3],
[ 1 .. 10 ],
[ 1 .. 100 ],
];
if ($opt{volume_tests}) {
for (1 .. $opt{volume_power_max}) {
my @volume = map { (($_), ($_)) } 1 .. 8**$_ / 2;
push @$test_equal_subsets, [@volume];
push @$test_unequal_subsets, [@volume, 8**(2 * $_)];
}
}
if ($opt{test_more}) {
plan tests => scalar @$test_equal_subsets + scalar @$test_unequal_
+subsets;
}
my @expectations = ('Not expecting equal subsets.', 'Expecting equal s
+ubsets.');
my @subsets_data = ([$test_unequal_subsets, 0, 0], [$test_equal_subset
+s, 1, 1]);
for (@subsets_data) {
my ($subsets, $expect_code, $expect_name_index) = @$_;
my $expect_name = $expectations[$expect_name_index];
for (@$subsets) {
my $start = time if $opt{time_hires};
if ($opt{test_more}) {
is(check_arrays($_), $expect_code, $expect_name);
}
else {
check_arrays($_);
}
printf "Took %f seconds\n", time()  $start if $opt{time_hires
+};
}
}
sub check_arrays {
my $full_array = shift;
print 'Checking: (', array_string($full_array), ')';
if (! grep { $_ } @$full_array) {
print "\tSubsets: (", array_string($full_array), ') and ()';
print "\tSubset sum = 0";
return 1;
}
my $full_sum = sum @$full_array;
if ($full_sum % 2) {
print "\tSubsets not equal: sum of starting array is odd ($ful
+l_sum).";
return 0;
}
my $half_sum = $full_sum / 2;
my @sorted_array = sort { $b % 2 <=> $a % 2  $b <=> $a } @$full_
+array;
if (my $big = first { $_ > $half_sum } @sorted_array) {
print "\tSubsets not equal: element ($big) larger than sum of
+rest.";
return 0;
}
my (@a1, @a2);
my $total = 0;
while (@sorted_array) {
push @a1, shift @sorted_array;
$total += $a1[$#a1];
@sorted_array = map { $total + $_ <= $half_sum
? do { push @a1, $_; $total += $_; () } : $_
} @sorted_array;
if ($total == $half_sum) {
(@a2, @sorted_array) = (@a2, @sorted_array);
}
else {
push @a2, pop @a1 if @a1;
}
}
if ($total == $half_sum) {
print "\tSubsets: (", array_string([sort { $a <=> $b } @a1]),
+')';
print "\t and (", array_string([sort { $a <=> $b } @a2]),
+')';
print "\tSubset sum = $half_sum";
return 1;
}
else {
print "\tSubsets not equal: no solution found.";
return 0
}
}
sub array_string {
my $array = shift;
return join(', ' => @$array > 3 * $opt{array_limit}
? ( @$array[0 .. $opt{array_limit}  1],
" ... [snip: @{[@$array  2 * $opt{array_limit}]} elements
+] ...",
@$array[@$array  $opt{array_limit} .. $#$array] )
: @$array);
}
Here's a test run. Note that this uses vpm=8 and final volume test "Took 89.489836 seconds" — you might want to start with a lower value.
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Applies to BrowserUK's solution as well.
Look again :)
#! perl slw
use strict;
use Time::HiRes qw[ time ];
use List::Util qw[ sum ];
sub partition {
my $sum = sum @_;
return if $sum & 1;
$sum /= 2;
my @s = sort{ $b <=> $a } @_;
my @a;
my( $t, $n ) = ( 0, 1 );
$t + $s[$n] <= $sum and $t+= $s[$n] and push @a, $n while ++$n < @
+s and $t <= $sum;
@a = delete @s[ @a ];
@s = grep defined, @s;
return unless sum( @a ) == sum( @s );
return \@a, \@s;
}
my $start = time;
my( $a, $b ) = partition 8, 4, 4, 7, 6, 3;
my @set = map int( rand 100 ), 1 .. $N;
printf "Took %f seconds\n", time()  $start;
if( $a ) {
printf "(%u) == sum( @{ $a } ) == sum( @{ $b } )\n", sum @$a;
}
else {
print "No solution existed for 8, 4, 4, 7, 6, 3";
}
__END__
No solution existed for 8, 4, 4, 7, 6, 3
Took 0.000258 seconds
With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
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.
div  [reply] [d/l] 



> This is an NPcomplete problem http://en.wikipedia.org/wiki/Knapsack_problem
Yes, to be precise a sub class known as "Partition Problem".
See WP article for some efficient algorithms and further links.
I wonder who and why is posting well known scientific problems w/o references ...?
Cheers Rolf
( addicted to the Perl Programming Language)
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A couple of questions:
 Why pass in a reference if the first thing you are going to do is copy the reference array to a local array?
sub check_arrays {
my @full_array = @{shift()};
 Why make a local copy of the array at all, when all the uses (join, sum, sort) of it require you to pass a list?
Ie. Why not my $sum = sum @$ref; etc.
 Isn't resumming your partial array over and over wildly inefficient?
With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
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.
 [reply] [d/l] [select] 

 [reply] 
Re: Divide an array into 2 subsets to verify their sum is equal or not.
by hdb (Monsignor) on May 02, 2013 at 09:47 UTC

use strict;
use warnings;
use List::Util qw/sum/;
# finds one solution
sub findsum {
my ($target, @array) = @_;
while( @array ) {
my $cand = shift @array;
return () if $cand > $target;
return ( $cand ) if $cand==$target;
my @sol = findsum( $target$cand, @array );
return ($cand, @sol) if @sol;
}
return ();
}
my @array = qw(1 3 5 7);
my $total = sum(@array);
die "Odd total $total cannot be split!\n" if $total % 2;
my @sol = findsum( $total/2, sort @array );
if( @sol ) {
print "Solution: ",join( ",", @sol), "\n";
} else {
print "No solution.";
}
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Re: Divide an array into 2 subsets to verify their sum is equal or not.
by davido (Cardinal) on May 03, 2013 at 16:10 UTC

use strict;
use warnings;
use Algorithm::Bucketizer;
use List::Util qw( sum );
use POSIX qw( ceil );
my @arrays = (
[ qw( 1 3 8 4 ) ],
[ qw( 1 6 2 ) ],
[ qw( 1 3 5 7 ) ],
);
foreach my $array ( @arrays ) {
print "( @{$array} ) can ",
can_evenly_distribute( @{$array} ) ? '' : 'not ',
"be evenly distributed.\n";
}
sub can_evenly_distribute {
my @elements = @_;
my $b_size = ceil( sum( @elements ) / 2 );
my $b = Algorithm::Bucketizer>new(
bucketsize => $b_size,
algorithm => 'retry'
);
$b>add_item( $_, $_ ) foreach @elements;
$b>optimize( algorithm => 'random', maxrounds => @elements * 10 );
my @buckets = $b>buckets;
return @buckets == 2
&& sum( $buckets[0]>items ) == sum( $buckets[1]>items );
}
I haven't found any cases where it fails to return the correct answer... but having just said that, someone will probably find one. ;)
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