Perl Weekly Challenge 63: words and rotations

One way to let me improve my knowledge about Raku (aka Perl 6) is to implement programs in it. Unluckily, I don’t have any production code to implement in Raku yet (sob!). So, why not try solving the Perl Weekly Challenge tasks?

In the following, the assigned tasks for Challenge 63.


The situation here in Italy is slowly coming back to normal: several places have openened the last week and while we are still forced to stay at home as much as possible, we can move a little more freely.
I have visited my mom a couple of other times in the last week, and she came to visit me too.
I’ve worked from office a couple of times, but still I do the main activity from home.

Therefore, Perl Weekly Challenge is not my first thought right now, but at the same time *it is a way to stay in touch with the Perl community and also to get into the same routine that let me think everything is fine**, at least for the time I spend in front of a Perl script!


Me and my wife removed the collar yesterday, and the cat is walking around the house but we don’t allow her to take stairs and jump on the sofa.
She still has the external metal helper on her hibs, we need to wait a couple of week before we can decide to remove it by aoher surgery. So far, she is not walking very well, but at least she is walking.

PWC 63 - Task 1

The first task was about writing a function that allows to match a word into a sentence, given a specific regular expresison. Apparently this was easy, but it took me two trials to get it right.
The requested function can be written as:
sub last_word( $string, $regexp ){
    my $last-word = Nil;
    for $string.split( " " ) {
        $last-word =  $_ if $_ ~~ $regexp;

    return $last-word;
The idea is that the $last-word assumes every single word from the split that matches the regexp. Therefore, at the end of the loop, the $last-word contains either the last word in the sentence or Nil.
Another, sligtly smarter approach is about reversing the words produced by split, so that the first match passes:
sub last_word_shorter( $string, $regexp ){
    for $string.split( " " ).reverse {
        return $_ if $_ ~~ $regexp;
The MAIN is used only for testing purposes:
sub MAIN(){
    say last_word('  hello world',                rx/<[ea]>l/);      # 'hello'
    say last_word("Don't match too much, Chet!",  rx:i/ch.t/);      # 'Chet!'
    say last_word("spaces in regexp won't match", rx:s/ in re / );      #  undef
    say last_word( join(' ', 1..1e6),             rx/ ^ (3.*?) ** 3 / ); # '399933'

    say last_word_shorter('  hello world',                rx/<[ea]>l/);      # 'hello'
    say last_word_shorter("Don't match too much, Chet!",  rx:i/ch.t/);      # 'Chet!'
    say last_word_shorter("spaces in regexp won't match", rx:s/ in re / );      #  undef
    say last_word_shorter( join(' ', 1..1e6),             rx/ ^ (3.*?) ** 3 / ); # '399933'

PWC 63 - Task 2

The second task was about doing a progressive transformation on a string, depending by the iteration number. Essentlially, at every iteration a specific number of chars have to be moved from the beginning to the end, and the cycle repeats until the string has rotated all chars. I decided to implement as follows:
sub rotate( $string where { $string.split( '', :skip-empty ).chars > 1 }, $verbose = False ){
    my $step = 1;
    my @chars  = $string.split( '', :skip-empty );

    while ( $step == 1 || $string ne @chars.join( '' ) ) {
        say "Rotation $step the string is: {@chars}" if $verbose;

        # move the first character to the end
        @chars.push: @chars.shift for 0 ..^ ( $step % $string.chars );

        say "After step $step the string is: { @chars }" if $verbose;


    return $step - 1;
First of all, the function accepts the $string (with the restriction it has to be longer than a single char) and a $verbose flag. Then the function splits the string into an array of characters, and on every iteration pushes to the end the first char of the string depending on the result of $step % $string.chars. If the string returns to its original value, the while stops and the function returns the number of iterations (minus 1).

The article Perl Weekly Challenge 63: words and rotations has been posted by Luca Ferrari on June 1, 2020