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#!/usr/local/bin/perl
use strict;
use warnings;
use feature qw(say);
use Test::More;
use Benchmark qw(cmpthese timethis);
use Data::Dumper qw(Dumper);
my @TESTS = (
[ 0, 1 ],
);
say $_,' > ', get_triples($_) foreach 1..500;
sub get_triples {
my $n = shift;
return $n < 3 ? -1 : join '; ', map { sprintf '(%s)', join ', ', @{$_} }
(
grep { $_->[1] == int $_->[1] } ## Check if all int
map { [ $_, sqrt($n**2-$_**2), $n ] } ## Generate triple
3 .. sqrt($n**2/2) ## Shortest side ($n is hypotenuse)
),(
map { $_->[0]>$_->[1] ? [@{$_}[1,0,2]] : $_ } ## put in numerical order
grep { $_->[1] == int $_->[1] } ## Check all int
map { [ $n, sqrt($_**2-$n**2), $_ ] } ## Generate triple
($n+1) .. ($n**2/2+1) ## Hypotenuse ($n is one of other two sides)
);
}
## Notes:
# Except for $n < 3 there is always a solution
# * If $n is odd then it can always be the short side of a triangle where the
# other sides are ($n^2-1)/2 or ($n^2+1)/2
# * If $n has an odd factor - then we can rewrite $n as $o * $m
# we can use the same trick to get sides of $m($o^2-1)/2 & $m($o^2+1)/2
# * This only leaves us with numbers of the form 2^n - but we know that 4 can be part of a Pyth.Triple
# and so any number of the form 2^($n+2) in a triple of the form ( 3.2^$n, 4.2^$n, 5.2^$n ).
#
# * We have two cases where $n is the hypotenuse and where it is a shorter side.
# * With the former we need to look at the shorter sides these can be 3 -> n-1
# but to avoid dupes we limit our search to sqrt(n^2/2)...
# * The latter is more complex to work out the range
# But we note that the difference between two consecutive squares is ($m+1)^2-($m)^2 = 2m + 1
# So the largest value for the hypotenuse is therefore (n^2+1)/2
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