TASK #1 › Pythagorean Triples

Submitted by: Cheok-Yin Fung
You are given a positive integer $N.

Write a script to print all Pythagorean Triples containing $N as a member. Print -1 if it can’t be a member of any.

Triples with the same set of elements are considered the same, i.e. if your script has already printed (3, 4, 5), (4, 3, 5) should not be printed.

The famous Pythagorean theorem states that in a right angle triangle, the length of the two shorter sides and the length of the longest side are related by a2+b2 = c2.

A Pythagorean triple refers to the triple of three integers whose lengths can compose a right-angled triangle.

A2 + B2 = C2

Thing to remember that there’s little difference between A and B here, so you basically need to handle the cases of A and C. The C case is, of course, A2 + B2. Handling the A case requires solving for C2 - B2.

Show Me The Code!

#!/usr/bin/env perl

use feature qw{say state signatures};
use strict;
use warnings;
use utf8;
no warnings qw{ experimental };

use List::Util qw{ uniq };
use JSON;
use Carp;
use Getopt::Long;
my $json = JSON->new->canonical->space_after;

my $n = 5;
GetOptions( 'n=i' => \$n, );

carp 'out of range' if $n <= 0;

my $p = pythagorean_triples($n);
say <<"END";

    INPUT: $n
    OUTPUT: $p

END

sub pythagorean_triples( $n ) {
    my @output;

    push @output, pt_a($n);
    push @output, pt_c($n);

    @output = grep { defined } @output;
    return join ", ", @output if @output;
    return -1;
}

sub pt_a ($n ) {
    my @output;
    my $n2 = $n**2;

    for my $b1 ( 1 .. $n2 ) {
        my $b2 = $b1**2;
        my $c2 = $n2 + $b2;
        my $c  = sqrt $c2;
        next unless int $c == $c;
        my @x = sort { $a <=> $b } map { int $_ } $n, $b1, $c;
        push @output, $json->encode( \@x );
    }
    return uniq @output if @output;
    return undef;
}

sub pt_c ($n ) {
    my @output;
    my $n2 = $n**2;

    for my $a1 ( 1 .. $n2 ) {
        my $a2 = $a1**2;
        for my $b1 ( 1 .. $n2 ) {
            my $b2 = $b1**2;
            my $c2 = $a2 + $b2;
            next if $c2 > $n2;
            if ( $n2 == $c2 ) {
                my @x = sort { $a <=> $b } map { int $_ } $a1, $b1, $n;
                push @output, $json->encode( \@x );
            }
        }
    }
    return uniq @output if @output;
    return undef;
}
 $ ./ch-1.pl

    INPUT: 5
    OUTPUT: [3, 4, 5], [5, 12, 13]


$ ./ch-1.pl -n 1

    INPUT: 1
    OUTPUT: -1


$ ./ch-1.pl -n 15

    INPUT: 15
    OUTPUT: [15, 112, 113], [15, 20, 25], [15, 36, 39], [8, 15, 17], [9, 12, 15]


$ ./ch-1.pl -n 25

    INPUT: 25
    OUTPUT: [15, 20, 25], [25, 312, 313], [25, 60, 65], [7, 24, 25]


$ ./ch-1.pl -n 13

    INPUT: 13
    OUTPUT: [13, 84, 85], [5, 12, 13]


TASK #2 › Binary Tree Diameter

Submitted by: Mohammad S Anwar
You are given binary tree as below:

      1
     / \
   2     5
  / \   / \
 3   4 6   7
          / \
         8  10
        /
       9

Write a script to find the diameter of the given binary tree.

The diameter of a binary tree is the length of the longest path between any two nodes in a tree. It doesn’t have to pass through the root.

I grabbed my Node code, which allows me to build the tree. I use a hash of nodes, so I can individually interact with them outside of the tree. I have is_leaf, so I can start the search only on leaf nodes. I trust I don’t have to prove that, unless you aren’t on a leaf node, you don’t have the longest diameter.

If you’re on a leaf, you’re either at the start or at an end. I test if there are options to move to left, right or parent, and add them to a list. Just as a matter of variable juggling, it’s easier to see “Are there any other nodes to travel to?” and handle that yes/no, than to do that if you’re at a node and the path is longer than 2. At least, I think so.

ETA: Diameter is specifically the count of edges, which is one less than the count of nodes on the path, so Og én, til tre nissen!

Show Me The Code!

#!/usr/bin/env perl

use feature qw{say state signatures};
use strict;
use warnings;
use utf8;
no warnings qw{ experimental };

use List::Util qw{ max };
use JSON;
my $json = JSON->new->space_after->canonical;

my %nodes;
for my $n ( 1 .. 10 ) {
    my $node = Node->new($n);
    $nodes{$n} = $node;
}

$nodes{1}->left( $nodes{2} );
$nodes{1}->right( $nodes{5} );
$nodes{2}->left( $nodes{3} );
$nodes{2}->right( $nodes{4} );
$nodes{5}->left( $nodes{6} );
$nodes{5}->right( $nodes{7} );
$nodes{7}->left( $nodes{8} );
$nodes{7}->right( $nodes{10} );
$nodes{8}->left( $nodes{9} );

my @diameters;
for my $node ( sort values %nodes ) {
    my $v = $node->value();
    my $l = $node->is_leaf();
    push @diameters, btd($node) if $l;
}

my $max = max map { scalar $_->@* } @diameters;
my $done;

@diameters =
    grep {
    my $s1 = join ' ', $_->@*;
    my $s2 = join ' ', reverse $_->@*;
    $done->{$s1}++;
    $done->{$s2}++;
    $done->{$s1} < 2;
    }
    grep { scalar $_->@* == $max }
    sort { scalar $b->@* <=> scalar $a->@* } @diameters;

say join "\n", map { join " ", ( -1 + scalar $_->@* ), ':', $_->@* }

    @diameters;

sub btd ( $node, $path = [] ) {
    my @output;
    my $v = $node->value();
    push $path->@*, $v;

    my @options;
    if ( defined $node->parent() ) {
        my $p  = $node->parent();
        my $pv = $p->value();
        my $is = grep /$pv/, $path->@* ? 1 : 0;
        if ( !grep /$pv/, $path->@* ) {
            push @options, 'parent';
        }
    }
    if ( defined $node->left() ) {
        my $p  = $node->left();
        my $pv = $p->value();
        my $is = grep /$pv/, $path->@* ? 1 : 0;
        if ( !grep /$pv/, $path->@* ) {
            push @options, 'left';
        }
    }
    if ( defined $node->right() ) {
        my $p  = $node->right();
        my $pv = $p->value();
        my $is = grep /$pv/, $path->@* ? 1 : 0;
        if ( !grep /$pv/, $path->@* ) {
            push @options, 'right';
        }
    }

    if (@options) {
        for my $option (@options) {
            if ( $option eq 'parent' ) {
                my $p = $node->parent();
                my $path2->@* = map { int } $path->@*;
                push @output, btd( $p, $path2 );
            }
            if ( $option eq 'left' ) {
                my $p = $node->left();
                my $path2->@* = map { int } $path->@*;
                push @output, btd( $p, $path2 );
            }
            if ( $option eq 'right' ) {
                my $p = $node->right();
                my $path2->@* = map { int } $path->@*;
                push @output, btd( $p, $path2 );
            }
        }
    }
    else {
        push @output, [ map { int } $path->@* ];
    }

    return @output;
}

package Node;

sub new ( $class, $value = 0 ) {
    my $self = {};
    $self->{value}      = $value;
    $self->{left}       = undef;
    $self->{right}      = undef;
    $self->{horizontal} = undef;
    $self->{parent}     = undef;
    return bless $self, $class;
}

sub value ( $self, $value = undef ) {
    if ( defined $value ) {
        $self->{value} = $value;
    }
    else {
        return $self->{value};
    }
}

sub is_root ( $self ) {
    return defined $self->{parent} ? 0 : 1;
}

sub is_leaf ( $self ) {
    return ( !defined $self->{left} && !defined $self->{right} )
        ? 1
        : 0;
}

sub left ( $self, $node = undef ) {
    if ( defined $node ) {
        $self->{left}   = $node;
        $node->{parent} = $self;
    }
    else {
        return $self->{left};
    }
}

sub right ( $self, $node = undef ) {
    if ( defined $node ) {
        $self->{right}  = $node;
        $node->{parent} = $self;
    }
    else {
        return $self->{right};
    }
}

sub horizontal ( $self, $node = undef ) {
    if ( defined $node ) {
        $self->{horizontal} = $node;
        $node->{parent}     = $self;
    }
    else {
        return $self->{horizontal};
    }
}

sub parent ($self ) {
    return $self->{parent};
}
7 : 3 2 1 5 7 8 9
7 : 4 2 1 5 7 8 9

If you have any questions or comments, I would be glad to hear it. Ask me on Twitter or make an issue on my blog repo.