TL;DR

On with TASK #2 from The Weekly Challenge #156. Enjoy!

# The challenge

You are given number, $n > 0. Write a script to find out if the given number is a Weird Number. According to Wikipedia, it is defined as: The sum of the proper divisors (divisors including 1 but not itself) of the number is greater than the number, but no subset of those divisors sums to the number itself. Example 1: Input:$n = 12
Output: 0

Since the proper divisors of 12 are 1, 2, 3, 4, and 6, which sum to 16;
but 2 + 4 + 6 = 12.


Example 2:

Input: $n = 70 Output: 1 As the proper divisors of 70 are 1, 2, 5, 7, 10, 14, and 35; these sum to 74, but no subset of these sums to 70.  # The questions What’s in the challenges this week? Is there anything that calls for lazy, brutey force attacks, or is it just me? # The solution As anticipated, I’m very lazy this week. This totally reflects in the Perl solution, where I summon three different modules to get the job done: #!/usr/bin/env perl use v5.24; use warnings; use experimental 'signatures'; no warnings 'experimental::signatures'; use FindBin '$Bin';
use lib "$Bin/local/lib/perl5"; use Algorithm::Knapsack; use ntheory 'divisors'; use List::Util 'sum'; say$_, ' ', is_weird($_) for (@ARGV ? @ARGV : (12, 70)); sub is_weird ($n) {
my @divs = reverse divisors($n); shift @divs if @divs > 1; return 0 if$n >= sum @divs;
my $ks = Algorithm::Knapsack->new(capacity =>$n, weights => \@divs);
$ks->compute; for my$solution ($ks->solutions) { my$sum = sum @divs[$solution->@*]; return 0 if$sum == $n; } return 1; }  Need the divisors? There you go, ntheory. Need to arrange good approximations? There you go, Algorithm::Knapsack. Need to do sums? There you go, List::Util. Need some glue? There you go, Perl. The Raku translation shows even more lazyness. While some stuff is included just like the batteries, like a handy sum method, looking for proper divisors and using the knapsack algorithm is harder because a very lazy search did not make anything stand out. So I opted for implementing stuff just to minimize the waste of time for looking a pre-made solution. I’m ashamed. Here it is: #!/usr/bin/env raku use v6; sub MAIN (*@args) { my @inputs = @args ?? |@args !! (12, 70); @inputs.map({ put$_, ' ', is-weird($_); }); } sub proper-divisors (Int:D$n) { (1..($n/2)).grep:$n %% * }

sub is-weird (Int:D $n) { my @divs = proper-divisors($n);
return 0 if @divs.sum <= $n; loop { my$sum = @divs.sum;
return 0 if $sum ==$n;
return 1 if $sum <$n;
my $ms = @divs.pop; my$target = $n -$ms;
for (^(2 ** @divs.elems)).reverse -> $k is copy { my$sum = 0;
my $i = 0; while$k > 0 {
$sum += @divs[$i] if $k +& 1; return 0 if$target == $sum; ++$i;
\$k +>= 1;
}
}
}
}


I guess the implementation if is-weird is closer to what I was expected to do. I mean, what I think I might have been expected to do, i.e. code a solution specific to the problem. Whatever, the Perl solution is a solution too.

Anyway, here I’m trying to find out if there’s an arrangement that sums up exactly to the input number, except that I’m trying to bail out early if conditions don’t apply any more. This is obtained by considering the sums with the highest number(s) first, eliminating them on the way. When we end up with a residual list whose sum is lower than the target number, there’s no way to select a subset that sums exactly to that and we can safely return early. Not too much of an optimization but still.

The proper-divisors function finally gave me the occasion to use the whatever variable in a grep. So far I could not manage to get it right, but as you can see I’m still trying.

You might have observed that I’m using the first singular person a lot here, instead of a more inclusive first plural version. This is because I take full responsibility for not thinking too much about the solution to the challenge, and I don’t want to make it appear like it’s somebody else’s fault too.

Let’s hope the best for people in Ukraine, and the rest of the places that are in a war!

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