[Purdue-pm] Perl Weekly Challenge

[Mark Senn]

This week's Perl Weekly Challenge problem was at
https://perlweeklychallenge.org/blog/perl-weekly-challenge-023/
It includes recaps of some previous problems by several people.



My blog for this week's forward difference task is at
    https://engineering.purdue.edu/~mark/pwc-023-1.pdf
The Perl 6 code:
    #
    # Perl Weekly Challenge - 023
    # Task #1
    #
    # See
    #     engineering.purdue.edu/~mark/blog/pwc-023-1.pdf
    # for more information.
    #

    # Run using Perl v6.d.
    use v6.d;

    # Get command line arguments.
    my $order = shift @*ARGS;
    my @x = @*ARGS;

    my $i = 1;
    while  @x.elems > 1 && $i <= $order  {
        # <<->> is a hyperoperator
        # if @x = (10,20,30,40), then
        @x = @x[1..*] <<->> @x[0..^*-1];
        # gives (20-10,30-20,40-30) = (10,20,30)

        say "order {$i++}: {@x}";
    }



My blog for this week's factoring numbers task is at
    https://engineering.purdue.edu/~mark/pwc-023-2.pdf
The Perl 6 code:
    #
    # Perl Weekly Challenge - 023
    # Task #2
    #
    # See
    #     engineering.purdue.edu/~mark/blog/pwc-023-2.pdf
    # for more information.
    #

    # Run using Perl v6.d.
    use v6.d;

    my $n = 228;    # number to factor

    # for 2, 3, 5, 7, ..., if the number is prime enter the loop
    for  (2, 3, *+2 ... *).grep({.is-prime}) -> $p  {

        while  $n %% $p  {    # $n evenly divisible by the current prime?
            say $p;           #     print the prime
            $n div= $p;       #     divide $n by the prime
        }

        # Is $p > $n.sqrt?
        # But, I suspect $p ** 2 > $n (= $p * $p > $n) , is faster to compute.
        if ($p ** 2 > $n)  {
            $n.say;
            last;
        }

    }



-mark