TL;DR

On with TASK #2 from The Weekly Challenge #143. Enjoy!

The challenge

You are given a positive number, $n.

Write a script to find out if the given number is Stealthy Number.

A positive integer N is stealthy, if there exist positive integers a, b, c, d such that a * b = c * d = N and a + b = c + d + 1.

Example 1

Input: $n = 36
Output: 1

Since 36 = 4 (a) * 9 (b) = 6 (c) * 6 (d) and 4 (a) + 9 (b) = 6 (c) + 6 (d) + 1.

Example 2

Input: $n = 12
Output: 1

Since 2 * 6 = 3 * 4 and 2 + 6 = 3 + 4 + 1

Example 3

Input: $n = 6
Output: 0

Since 2 * 3 = 1 * 6 but 2 + 3 != 1 + 6 + 1

The questions

I only have one meta-questions about this challenge: what’s the matter with all the divisors of integer numbers?

I suspect manwar is manipulating us like pupputs to help solving the Riemann hypothesis. Whatever, happy to help!

The solution

The key insights here were:

• all pairs of divisors come in pairs… by definition. Jokes apart, as soon as we find a divisor, we also have its counterpart to make the pair.
• We’ll have to compare all pairs against each other, possibly in an efficient way. This rings like hash, set, … anything that can check for being there efficiently.
• for each pair of pairs we will have to do two checks, or check an absolute value. In our case, using a hash-like data structure will mean either tracking the sums themselves, or their +1 or -1 versions.

I’ll start with Perl this time.

#!/usr/bin/env perl
use v5.24;
use warnings;
use experimental 'signatures';
no warnings 'experimental::signatures';

sub is_stealthy ($n) {
   my %match;
   for my $k (1 .. sqrt($n)) {
      next if $n % $k;
      my $sum = $k + $n / $k;
      return 1 if $match{$sum - 1} || $match{$sum + 1};
      $match{$sum} = 1;
   }
   return 0;
}

my @inputs = @ARGV ? @ARGV : qw< 36 12 6 >;
say "$_ -> " . is_stealthy($_) for @inputs;

We’re using a plain ol’ hash here. At each iteration, we check if $sum + 1 or $sum - 1 are in, because this is how we can do the required check! This also allows us saving only $sum for future attempts.

I surprised myself in writing this:

return 1 if %match{...

Wow, Raku is getting on me.

So, let’s move to Raku then:

#!/usr/bin/env raku
use v6;
subset PosInt of Int where * > 0;

sub is-stealthy (PosInt:D $n) {
   my $match = SetHash.new;
   for 1 .. $n.sqrt.Int -> $k {
      next unless $n %% $k;
      my Int() $sum = $k + $n / $k;
      return 1 if $match (&) ($sum - 1, $sum + 1);
      $match.set: $sum;
   }
   return 0;
}

sub MAIN (*@args) {
   @args = 36, 12, 6 unless @args.elems;
   "$_ -> {is-stealthy($_)}".put for @args;
}

It’s pretty much the same thing, using a SetHash this time for showing off a bit and also try to be a bit more readable (or, maybe, expressive).

I guess this is it for this challenge… stay safe!