TL;DR

On with TASK #2 from The Weekly Challenge #164. Enjoy!

# The challenge

Write a script to find the first 8 Happy Numbers in base 10. For more information, please check out Wikipedia.

Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1.

Those numbers for which this process end in 1 are happy numbers, while those numbers that do not end in 1 are unhappy numbers.

Example

19 is Happy Number in base 10, as shown:

19 => 1^2 + 9^2
=> 1   + 81
=> 82 => 8^2 + 2^2
=> 64  + 4
=> 68 => 6^2 + 8^2
=> 36  + 64
=> 100 => 1^2 + 0^2 + 0^2
=> 1 + 0 + 0
=> 1


# The questions

You know, the usual suspectsâŠ

• âfirst 8 happy numbers** implies an ordering based on the numbersâ values, right?

• can I skip reading the Wikipedia page and assume the explanation and the example are sufficient?

# The solution

This challenge seems to scream about using some sort of cache. I a number is happy, so are all the âintermediatesâ that separate that number from becoming 1. On the other hand, if we find a cycle without a 1 inside, we know for sure that all elements in that cycle are not happy.

So, to start with, we keep a couple variables with our past knowledge about previous calls to the function, as state variables to keep this knowledge. Both will be hash references because thatâs how we can track flags easily in Perl. The $is_happy is initialized to contain 1, of course, as itâs the quintessential happy number. We can, of course, put these caches at work immediately. sub is_happy ($n) {
state $is_happy = { 1 => 1 }; state$is_not_happy = {};
return 1 if $is_happy->{$n};
return 0 if $is_not_happy->{$n};
...


If we get past this point, the number is unknown and we have to investigate. We will track our effort as a round: if we end up with a 1 the whole round will be added to the $is_happy hash, otherwise all elements will be added to the $is_not_happy cache:

   ...
my %round;
while (! $round{$n}) {
$round{$n} = 1;
$n = sum map {$_ * $_ } split m{}mxs,$n;
if ($n == 1) {$is_happy->{$_} = 1 for keys %round; return 1; } }$is_not_happy->{$_} for keys %round; return 0; }  And this is really it. At this point, we just have to iterate until we find enough happy numbers, ending with the following complete program: #!/usr/bin/env perl use v5.24; use warnings; use experimental 'signatures'; no warnings 'experimental::signatures'; use List::Util 'sum'; my$wanted = shift // 8;
my $n = 1; my @happy; while (@happy <$wanted) {
push @happy, $n if is_happy($n);
++$n; } say join ', ', @happy; sub is_happy ($n) {
state $is_happy = { 1 => 1 }; state$is_not_happy = {};
return 1 if $is_happy->{$n};
return 0 if $is_not_happy->{$n};
my %round;
while (! $round{$n}) {
$round{$n} = 1;
$n = sum map {$_ * $_ } split m{}mxs,$n;
if ($n == 1) {$is_happy->{$_} = 1 for keys %round; return 1; } }$is_not_happy->{$_} for keys %round; return 0; }  I played a bit with the idea to provide an iterator, but laziness won eventually. The Raku counterpart allows us to show off a bit with hyperstuff and Unicode operators. At the end of the day, though, itâs the same algorithm with a shinier look. #!/usr/bin/env raku use v6; sub MAIN (Int:D$wanted = 8) {
my $n = 1; my @happy; while @happy <$wanted {
@happy.push: $n if is-happy($n);
++$n; } @happy.join(', ').put; } sub is-happy (Int:D$n is copy) {
state $is-happy = SetHash.new(1); return True if$n â $is-happy; state$is-not-happy = SetHash.new;
return False if $n â$is-not-happy;

my $round = SetHash.new; while$n â $round {$round.set($n);$n = $n.combÂ»ÂČ.sum; if$n == 1 {
$is-happy âȘ=$round;
return 1;
}
}
$is-not-happy âȘ=$round;
return 0;
}


One objection that might be moved to the caching approach is that it might explode. Well, this should not be the case, because beyond a certain point weâre sure that the next number will be lower.

As an example, consider:

$9 \rightarrow 9^2 = 81 > 9 \\ 99 \rightarrow 2 \cdot 9^2 = 162 > 999 \\ 999 \rightarrow 3 \cdot 9^3 = 243 < 999$

Iâm not going to calculate what this exact threshold is (maybe the Wikipedia page has something on it), but you can be sure that whatever number beyond 243 will go down for sure, so at the end of the day we should be pretty safe with our cache.

WellâŠ then at least everyting between 200 and 243 goes down as well because itâs maxed by 239, and 199 goes down as well. Then I stop!

Stay safe!

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