TL;DR

On with TASK #2 from the Perl Weekly Challenge #116. Enjoy!

The challenge

You are given a number $N >= 10.

Write a script to find out if the given number $N is such that sum of squares of all digits is a perfect square. Print 1 if it is otherwise 0.

Example

Input: $N = 34
Ouput: 1 as 3^2 + 4^2 => 9 + 16 => 25 => 5^2

Input: $N = 50
Output: 1 as 5^2 + 0^2 => 25 + 0 => 25 => 5^2

Input: $N = 52
Output: 0 as 5^2 + 2^2 => 25 + 4 => 29

The questions

The first is a meta-question: what’s with all these $N >= 10 this week? Are we’re celebrating something that happened 10 years ago?!?

The number will be an integer, right? Also, I’ll assume that it’s OK to avoid too big integers and assume that we will always remain within reasonable bounds. Whatever this means.

Technically speaking this assumption is totally arbitrary though, as there is no upper limit as to having numbers that match.

Not convinced? Well:

  • take any integer (the bigger, the better!)
  • square it
  • now take a sequence of that square number of any single digit.

Like you chose 3 to start with, so the square is 9 and then we get, let’s say, 777777777 i.e. 7 repeated 9 times.

Each 7 squares to 49, and we have exactly 9 of them… So there you have it, the sum is a perfect square!

The solution

This time I coded the Raku solution first:

#!/usr/bin/env raku
use v6;

sub sum-of-squares (Int $N where * >= 10 --> Int:D) {
   my $M = $N.comb.map(* ** 2).sum;
   my $m = $M.sqrt.Int;
   return $m * $m == $M ?? 1 !! 0;
}

sub MAIN (*@inputs) {
   @inputs = < 34 50 52 > unless @inputs.elems;
   sum-of-squares($_).put for @inputs;
}

I have to admit that I was a bit disappointed in not finding a is-square method in Int. Also, I have to admit that * ** 2 is not an invitation to read 🙄

Then I thought… it’s Raku!

Then I looked in the internet and here we are:

#!/usr/bin/env raku
use v6;
use MONKEY-TYPING;

augment class Int {
   method is-square (--> Bool:D) { self == self.sqrt.Int ** 2 }
}

sub sum-of-squares (Int $N where * >= 10 --> Int:D) {
   $N.comb.map(* ** 2).sum.is-square ?? 1 !! 0;
}

sub MAIN (*@inputs) {
   @inputs = < 34 50 52 > unless @inputs.elems;
   sum-of-squares($_).put for @inputs;
}

It works!!!

Here’s the good, ol’ Perl solution:

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

sub sum_of_squares ($N) {
   my $M = sum map { $_ * $_ } split(m{}mxs, $N);
   my $m = int sqrt $M;
   return $m * $m == $M ? 1 : 0;
}

my @inputs = @ARGV ? @ARGV : qw< 34 50 52 >;
say sum_of_squares($_) for @inputs;

In both cases we split the input as a string into its constituent digits, square them and produce the sum. Then, we check if this is a perfect square, by taking the square root, cutting it to an integer and checking whether squaring it again produces the sum we started with.

Stay safe folks!