Perl Weekly Challenge 94: I’M (nearly) BACK!

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 94.

I’M (nearly) BACK

My last PWC contribution was number 76.
Yes, I was away from keyboard for almost four months due to an eye surgery, trabeculectomy, that has done good results on the eye pressure (the main aim), but quite bad results on the eyesight.
Therefore, I’m not really happy with that, and I’m trying to figure out how to deal with my new (downgraded) life.
As a way to force my brain to stay focused on something different, I decided to start over the Perl Weekly Challenge.
This is also why my blog activity has dropped in the last quarter of 2020.{:target=”_blank”}.

PWC 94 - Task 1

The first task was about getting out anagrams by a set of words, i.e., different words written in different ways but with the very same letters. I wrote a get-anagrams function that accepts the array of string and iterate on every string of the array. For every string found in the array, from the beginning to the end, I compute all the possible permutations of the letters that compose such string, and then try to see if there is another word that matches the resulting anagram. If there is a word, such word is added to a @results temporary array, that in turn is materialized into the global @results array within MAIN.
Since the string array is iterated from the very beginning to the end, there is no possibility there is a repetition of the same anagrams.

sub get-anagrams( @S where { @S.elems > 1 } ) {
    my @results;

    # cycle thru the words using ordering
    # so that the same anagram is not
    # backward referenced
    for 0 ..^ @S -> $outer-index {

        # use a temporary array to store these anagrams
        # and initialize it with the current word
        my @anagrams = @S[ $outer-index ];
        for @anagrams[ 0 ].comb.permutations {
            my $current = .join;
            # push this word if it is not the same as the current one,
            # it is within @S and is not already into @results
            @anagrams.push: $current if ( @S[ $outer-index .. *-1 ].grep( $current )
                                          && ! @anagrams.grep( $current )
                                          && ! @results.List.flat.grep( $current ) );

        # if there is more than one word (the first one is the current)
        # then there is at least an anagram, so push to the final results
        @results.push: [ @anagrams ] if ( @anagrams.elems > 1 );

    return @results;

It is quite simple then to produce a MAIN:

multi sub MAIN( *@S ) {
    my @results = @S.elems > 1 ??  get-anagrams( @S ) !! @S;
    say "Valid anagrams: { @results.join(' - ') }";

multi sub MAIN(){
    my Str @S = <opt bat saw tab pot top was>;
    my @results = get-anagrams( @S );
    say "Valid anagrams: { @results.join(' - ') }";

The special case for @S with a single string begin returned as the anagram of itself is a special case requested by the task itself.
Invoking the program produces the result as follows:

% raku ch-1.p6        
Valid anagrams: opt pot top - bat tab - saw was

There is room for improvements:
  • ensure that all searched words have the very same length (it is not clear to me if this should be a requirement of @S or must be ensured at run-time);
  • grep against different cases (upper or lower case).

PWC 94 - Task 2

The second task was not too clear to me: it is required to print out a tree structure as a linked list.
My doubts are about:
  • do we have to produce a list and then print the list or just print the tree as a list?
  • is the tree to be printed left to right when dealing with nodes?
    I assume no list has to be created and that the tree must be printed from left to right when inspecting the tree.
    First of all I created a tree-like structure with very hand Node class:

class Node {
    has Int $.value  is rw;
    has Node $.left  is rw;
    has Node $.right is rw;

    method me-recursive() {
        my $str =;
        $str ~= ' -> ' ~ $!  if $!left;
        $str ~= ' -> ' ~ $! if $!right;

    method me() { $!value; }


All the magic happens in the me-recursive method:
  • it does print the current node value;
  • appends the arrow and the value of the $!left node, if any;
  • appends the arrow and the value of the $!right node, if any.

Since every call to me-recursive proceed in recursion on the left node first, the tree is printed from left to right until no children have been found, than the traversing starts over the right direction.
The end result is the one asked from the challenge:

% raku ch-2.p6        
1 -> 2 -> 4 -> 5 -> 6 -> 7 -> 3

Again, I’m not sure I’ve fully understood this challenge.

The article Perl Weekly Challenge 94: I'M (nearly) BACK! has been posted by Luca Ferrari on January 4, 2021