Once More into the Breech!

TASK #1 › JortSort

Submitted by: Mohammad S Anwar
You are given a list of numbers.

Write a script to implement JortSort. It should return true/false depending if the given list of numbers are already sorted.

First, let me say that I love Jenn Schiffer. Sarah Cooper might be the Internet’s own comedian, but Jenn is WebDev’s own comedian. She’s the greatest!

Here, she’s created a new sorting algorithm. No. “jortSort isn’t a sorting algorithm. It’s a sorting toolset.”.

It’s actually more a boolean check: Is this sorted (Y/N)?

And the source is on the page!

var jortSort = function( array ) {

  // sort the array
  var originalArray = array.slice(0);
  array.sort( function(a,b){return a - b} );

  // compare to see if it was originally sorted
  for (var i = 0; i < originalArray.length; ++i) {
    if (originalArray[i] !== array[i]) return false;

  return true;

JortSort does the work of sorting, but simply tells you if you did it right or not. It steadfastly avoids being helpful. It, to me, is like replacing Stack Overflow with a small JS function. I’ve seen many SO users I’d love to see replced with small JS functions.

Again, I love Jenn Schiffer.


Show Me The Code!

#!/usr/bin/env perl

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

# JortSort - https://jort.technology/ - https://github.com/jennschiffer/jortsort

my @examples;
push @examples, [ 1 .. 5 ];
push @examples, [ 1, 3, 2, 4, 5 ];
push @examples, [ 1 .. 20 ];
push @examples, [ sort { rand 1 <=> rand 1 } 1 .. 20 ];

for my $input (@examples) {
    my $o = jortsort( $input->@* );
    my $i = join ',',$input->@*;
    say <<"END";
    Input: \@n = ($i)
    Output: $o

# basically? It's sorted already, or go back and try again.
sub jortsort (@array ) {
    my @copy = sort { $a <=> $b } @array;
    for my $i ( 0 .. -1 + scalar @array ) {
        return 0 if $copy[$i] ne $array[$i];
    return 1;
 $ ./ch-1.pl
    Input: @n = (1,2,3,4,5)
    Output: 1

    Input: @n = (1,3,2,4,5)
    Output: 0

    Input: @n = (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20)
    Output: 1

    Input: @n = (15,18,3,8,11,17,12,7,19,20,16,2,10,9,1,4,13,5,14,6)
    Output: 0

TASK #2 › Long Primes

Submitted by: Mohammad S Anwar
Write a script to generate first 5 Long Primes.

A prime number (p) is called Long Prime if (1/p) has an infinite decimal expansion repeating every (p-1) digits.

And this is the serious one. Let’s think about Long Primes!

So, for a number n, we must check:

  • if n is prime (so 4 is out)
  • if 1/n is cyclic (so 2 is out)
  • if the cycle for 1/n is as long as n-1 (so 3 is out)

The Wikipedia article gives us the first few, with a link to the page on the Online Encyclopedia of Integer Sequences (OEIS) for this, so we can check our work.

The lowest Long Prime is 7. Taken to 200 digits, that’s 0.14285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714285714, and we can clearly see the six-digit repetitions (142857). And it’s the six-digit repetitions we’re looking for: 3 has a one-digit repetition, which is also a 2-digit repetition, but the one-digit rep comes first, so it isn’t good enough to be a Long Prime.

Problem is, 1/7 == 0.142857142857143, using default Perl variables, which covers us for 7, but when we get longer, we need to deal with bigger floats.

We need Big Floats.

use Math::BigFloat;


if (1) {
    my $b1    = Math::BigFloat->new(1);
    my $b7    = Math::BigFloat->new(7);
    my $ratio = $b1->bdiv($b7);
    say 1 / 7;
    say $ratio;


accuracy works on either the variable or class level to determine how many significant digits to show, so we can show we have enough digits to show the pattern.

From there, we need two to make a pattern: my ($f1,$f2) = $alt =~ m{(\d{$i})}gmx. I know that this is the kind of thing that makes Perl haters bring out line noise comparisons, so:

  • We don’t need to check every entry to see if there’s repetition, but just the first two. We get a list context and fill two scalars with my ($f1,$f2)=...
  • We’re just doing matching, and everything that $alt =~ m{regex} matches gets sent to ($f1,$f2)
  • and m{(\d{$i})}gmx … is like an Onion: it has layers. At center, it’s a number made of $i digits, so \d{$i}. We want to get the output, not just a boolean showing it exists, so we put in parentheses, like (\d{$i}). As I said, that’s a match, so m{(\d{$i})}, and we want it to be global, so we add m{(\d{$i})}g. I also add mx because that’s best practice for regular expressions, even though it’s single line and I don’t need extra whitespace for comments. Maybe I should do that more…

Anyway, now, we have $f1 and $f2. If $f2 isn’t defined, then we know it doesn’t repeat, and we we can return 0. Beyond that, there are two other failure points: If $f1 and $f2 match and the cycle isn’t n-1 characters long, and if $f1 and $f2 don’t match as the cycle is n-1 characters long.

If I was doing this with any other language, I might be thinking about how to expand the number of significant digits we’re seeing.

Show Me The Code!

#!/usr/bin/env perl

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

# https://en.wikipedia.org/wiki/Full_reptend_prime

#  The first five long primes are: 7, 17, 19, 23, 29

use Math::BigFloat;
my @long_primes;
for my $i ( 1 .. 50 ) {
    next unless is_long_prime($i);
    push @long_primes, $i;
say 'The first five Long Primes are: ', join ', ', @long_primes;

sub is_long_prime ($n ) {
    return 0 unless is_prime($n);
    my $bign = Math::BigFloat->new($n);
    my $big1 = Math::BigFloat->new(1);
    my $big  = $big1->bdiv($bign);
    $big =~ s/0+$//mix;
    my $alt = $big;
    $alt =~ s/^0\.//;

    my $l = $n - 1;
    for my $i ( 1 .. $l ) {
        my ( $f1, $f2 ) = $alt =~ m{(\d{$i})}gmx;
        return 0 if !defined $f2;
        return 0 if $f1 == $f2 && $i < $l;
        return 0 if $f1 != $f2 && $i == $l;

    return 1;

sub is_prime ( $n ) {
    my @factors = factor($n);
    return scalar @factors == 1 ? 1 : 0;

sub factor ( $n ) {
    my @factors;
    for my $i ( 1 .. $n - 1 ) {
        push @factors, $i if $n % $i == 0;
    return @factors;
$ ./ch-2.pl
The first five Long Primes are: 7, 17, 19, 23, 29, 47

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.