Merge pull request #5543 from jacoby/master

Challenge 148

3 files changed, 68 insertions, 0 deletions

diff --git a/challenge-148/dave-jacoby/blog.txt b/challenge-148/dave-jacoby/blog.txt
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+++ b/challenge-148/dave-jacoby/blog.txt
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+https://jacoby.github.io/2022/01/19/its-triplets-weekly-challenge-148.html
diff --git a/challenge-148/dave-jacoby/perl/ch-1.pl b/challenge-148/dave-jacoby/perl/ch-1.pl
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+++ b/challenge-148/dave-jacoby/perl/ch-1.pl
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+#!/usr/bin/env perl
+
+use strict;
+use warnings;
+use feature qw{ say postderef signatures state };
+no warnings qw{ experimental };
+
+use Lingua::EN::Numbers qw( num2en );
+
+# You COULD try to make up a way to do this, but this wheel has been
+# invented already and is sufficiently round.
+
+my @numbers;
+for my $i ( 0 .. 100 ) {
+    my $e = num2en $i;
+    next if $e =~ /e/mx;
+    push @numbers, $i;
+}
+say join ', ', @numbers;
diff --git a/challenge-148/dave-jacoby/perl/ch-2.pl b/challenge-148/dave-jacoby/perl/ch-2.pl
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+#!/usr/bin/env perl
+
+use strict;
+use warnings;
+use feature qw{ say postderef signatures state };
+no warnings qw{ experimental };
+
+use Algorithm::Permute;
+
+my @triplets;
+
+my $i = 0;
+while ( scalar @triplets < 5 ) {
+    for my $j ( 1 .. $i ) {
+        for my $k ( 1 .. $j ) {
+            my $p = Algorithm::Permute->new( [ $i, $j, $k ] );
+            while ( my @res = $p->next ) {
+                my $t = test_cardano(@res);
+                if ( $t == 1 ) {
+                    push @triplets, \@res;
+                }
+            }
+        }
+    }
+    $i++;
+    last if $i > 1000;
+}
+
+for my $ct (@triplets) {
+    my ( $a, $b, $c ) = $ct->@*;
+    print <<"END";
+    A: $a\tB: $b\tC: $c
+END
+}
+
+sub test_cardano ( $a, $b, $c ) {
+    my $sqrtc = sqrt $c;
+
+    # not necessary for the first five
+    if ( $a > $b * $sqrtc ) {
+        return cuberoot( $a + $b * $sqrtc ) + cuberoot( $a - $b * $sqrtc );
+    }
+
+    return cuberoot( $a + $b * $sqrtc ) +
+        -1 * cuberoot( abs( $a - $b * $sqrtc ) );
+}
+
+sub cuberoot ($n ) { return $n**( 1 / 3 ) }