Perl Weekly puzzle week 12

week12.pl

#!/usr/bin/perl6
use v6;

use Test;

# 12.1
my @primes = lazy (2,3,*+2 ... ∞).grep: *.is-prime;

# from wikipedia: https://en.wikipedia.org/wiki/Euclid_number
# Write a script that finds the smallest Euclid Number that is not prime

for ^∞ -> $n { 
    my $i = ( [*] @primes[0..$n] ) + 1;
    if ! $i.is-prime {
        say "12.1) (first $n primes + 1) -> $i is not prime";
        last;
    }
};

# Write a script that finds the common directory path, given a collection of paths and directory separator. For example, if the following paths are supplied.
# 
# /a/b/c/d
# /a/b/cd
# /a/b/cc
# /a/b/c/d/e
# and the path separator is /. Your script should return /a/b as common directory path.
# 

my @input=("/a/b/c/d", "/a/b/cd", "/a/b/cc", "/a/b/c/d/e");
my $output="/a/b";
my $sep="/";

is shortestCDP(@input,$sep),$output,"Test of shortestCDP";
is shortestCDP(("/a/b/c/d", "/a/b/cd", "/a/cc", "/a/b/c/d/e"),$sep),"/a","Test of shortestCDP";
is shortestCDP(("/a/b/c/d", "/a/b/c", "/a/b/c/d/e"),$sep),"/a/b/c","Test of shortestCDP";

done-testing;

say "12.2) shortest path is " ~  shortestCDP(@input,$sep); 


sub shortestCDP(@input, $sep) {
    my $l=@input[0].chars;
    my @dirs;
    for @input -> $i {
        my $d=[$i.split($sep, :skip-empty)];
        $l=min($l, $d.elems);
        @dirs.push($d);
    }
    my @found;
    my $i=0;
    while $i < $l {
        my $alleq=True;
        for @dirs -> $d {
            next if $d[$i] eq @dirs[0][$i];
            $alleq=False;
            last;;
       }
        $i++;
        next if $alleq==False;
        @found.push(@dirs[0][$i-1]);
    }
    return $sep ~ @found.join($sep);
}