TL;DR

Here we are with TASK #1 from the Perl Weekly Challenge #085. Enjoy!

# The challenge

You are given an array of real numbers greater than zero. Write a script to find if there exists a triplet (a,b,c) such that 1 < a+b+c < 2. Print 1 if you succeed otherwise 0.

# The questions

This is one of those annoyingly great challenges! Those where you’re too lazy to find an optimized solution, sort of know that you can do better, do some research but still the answers do not apply. So one question might be… can we solve 3SUM instead?

I guess the answer will be… NO.

OK, fair enough. Legitimate question then:

• should we be wary of corner cases involving the imperfect representation of floating point numbers in computers? Something related to this, for example:
$perl -E 'say 0.1 + 0.2 == 0.3 ? "equal" : "different"' different  • how big is the input list of numbers? In other terms: does it make sense to look for an optimized solution or is it sufficient to give a correct but less scalable one? • (I’m getting tired of this) how should we treat incorrect inputs? # The solution The very boring, hence good level-0, algorithm that comes to mind is: take every possible subset of three items from the input array, calculate the sum and check against the allowed range. Return 1 as soon as you find a match, return 0 otherwise. Generating all the subsets is a challenge by itself, but I think that in this case we should take extra care to generate only the really needed ones. I mean, if we get a list with 0.5 repeated 10000 times, any three of them will do… and we don’t really need to generate all the possible$\binom{10000}{3} = \frac{10000 \cdot 9999 \cdot 9998}{3 \cdot 2} = 166616670000$combinations beforehand, right?!? For this reason, the best here would be to have a way to generate new combinations on the spot, just when they are needed. We need an iterator. Luckily enough, CPAN provides us with Math::Combinatorics: it’s pure-Perl (which makes it easy to install everywhere) and provides us with an iterator-based implementation to get combinations. Yay! This is the solving function:  1 sub triplet_sum { 2 my @R = grep {$_ <= 2.0 } @_; # remove cruft
3    my $combiner = Math::Combinatorics->new(count => 3, data => \@R); 4 while (my ($x, $y,$z) = $combiner->next_combination) { 5$x += $y +$z;
6       return 1 if 1 <= $x &&$x <= 2;
7    }
8    return 0;
9 };


Line 2 collects the input, making sure to keep only the ones that can possibly contribute to a successful sum. This means that everything above 2.0 will get filtered out.

Line 3 creates a Math::Combinatorics object that will be suitable to generate combinations of three items (option count) out of our (filtered) input data @R.

From this point, it’s just a matter of applying our algorithm above:

• take the next triplet (line 4);
• do the sum (line 5);
• check against the target range (line 6).

And until I get an answer to the need for something more efficient… I’ll stick with this solution 😇

The full code, should you be interested into it, is the following:

#!/usr/bin/env perl
use 5.024;
use warnings;
use experimental qw< postderef signatures >;
no warnings qw< experimental::postderef experimental::signatures >;

use FindBin '$Bin'; use lib "$Bin/local/lib/perl5";

use Math::Combinatorics ();

sub triplet_sum {
my @R = grep { $_ <= 2.0 } @_; # remove cruft my$combiner = Math::Combinatorics->new(count => 3, data => \@R);
while (my ($x,$y, $z) =$combiner->next_combination) {
$x +=$y + $z; return 1 if 1 <=$x && \$x <= 2;
}
return 0;
};

my @input = scalar @ARGV ? @ARGV : (0.5, 1.1, 0.3, 0.7);
say triplet_sum(@input);


It assumes that the Math::Combinatorics module is installed in local/lib/perl5 starting from the same position as where the program is saved.