use Perl

And actually I made a horrible error in my code: the way I wrote it the traversal order is actually undefined, because the order of iteration of hyper operators is undefined. So, although I'd say that having a concurrent traversal ordering would be a nice addition, the last line of my sub should probably be something like: ".() for @actions;"

Nice work, but still something lacking in the "show" function: both your version and SF's have too much repetition. What we really want to to say that there's a list of three actions to perform in one of three orders, with only one of the blocks moving. something like:

sub show(%tree, Traversal $method) {
    return unless %tree;
    my @actions = (
        { show %tree<RIGHT>, $method },
        { show %tree<LEFT>, $method },
    );
    @actions.=splice( $method.value, 0 ) <== { take %tree<VALUE> };
    @actions>>.();
}

Perhaps losing a bit of readability ... but I think we can remove even the last bit of repetition here:

sub show(%tree, Traversal $method) {
    return unless %tree;
    my @actions = <RIGHT LEFT>.map: -> $way { show %tree<$way>, $method }.assuming( $^way );
    @actions.=splice( $method.value, 0 ) <== { take %tree<VALUE> };
    @actions>>.();
}

I agree, though I think it might need a special-case for zero-arity so that could could write an arbitrary reduction:

    'sum' => { @stack = [+] @stack }, ...

hmm, can that be written as assignment-operator?

    'sum' => { @stack [+]= () },

Thanks for the subset suggestion. I'd tried putting "where" clauses inline, but that was very clunky. Subsets definitely solve that.

I'm not sure, however, I like your version of the consumption logic any better than mine: using a modifier-op ("shift" in both the conditional and consequent confuses me (I like conditionals to be pure). I would tend to agree, though, that my form was equally confusing. It might be best to be explicit:

        my $bowls_this_frame = @scores[0] eq "X" ?? 1 !! 2;
        @scores.shift for ^$bowls_this_frame


Advanced Challenge 3...

A couple of different approaches here: one is
to maintain a list of excluded candidates, and then ignore them during
recounts. The other is to actually delete them from the ballot and then
recount. Both solutions work: the first uses junctions; the other uses the
"is rw" trait on a for loop. So both are of some interest.

First: the junctional approach...

The final challenge was blackjack. The instructions said that we
could choose to implement that aces always equal 11, but that seems
insufficiently interesting: I wanted to use junctions!

To start with, a simple definition of a deck of cards. The challenge
required humanly named cards: so I started with a list of pairs, and
then crossed this list against the list of suites. Finally, "@cards =
@deck.pick( @deck.elems )" results in a random shuffle.

event 6 asks us to generate the prime numbers from 1 to 200:


% ./perl6 -e 'my @primes; for 2 .. 200 -> $n { @primes.push($n) unless $n % any(@primes)==0 }; .say for @primes'


and event 7 asks for a random scheduling of a round-robin tournament with 6 teams:

I couldn't stop with just the "beginners" challenges. "Expert" event 1 asks use to take a phone number, and convert it to a valid word using standard mapping of digits to letters -- http://www.microsoft.com/technet/scriptcenter/funzone/games/games08/aevent1.mspx

The actual challenge uses a file that contains the list of words: I decided to hard code the array:

Having played with the "pairs" challenge, I looked through the other beginners events. Most dealt with files and databases: I didn't want to tackle that right now. So I skipped to challenge-10: interpret a bowling game that is input as an array.

For this one, my thought was to exploit "multi" subs: there're only so many patterns to each frame: I can just code them explicitly. Lets start with my solution:


my @scores = 2,5,7,"/",8,1,"X",9,"/",5,3,7,0,4,5,"X",2,0;

Having thought it through, my "diagonal cross" should almost certainly be prefix, not infix -- I can't think what a diagonal cross with a different array would mean. So:

say [+] %eq% @x

if I can figure out how to implement it!

So Patrick suggested on rakudo.org that playing with Microsoft's "Winter Scripting Games" would be a good way to get started with Rakudo.

I liked that idea, so I went to the first challenge (for beginners) -- simply count the number of pairs in a list. For their test data (7,5,7,7,13), the correct answer is to be "3", because "7" is paired 3 different ways.