Perl Weekly Challenge 44: only 100 and 200 bucks

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 44.

PWC 44 - Task 1

Given the 123456789 string, it is required to find all possible algebric sums that made up the result of 100. My solution composes a List of Seqs that compose all the available combinations of digits and operators. That is obtained applying the zip operator (Z) to the lists @operators and @digits and than performing a hyperoperator with the cross string concatenation X~. Having obtained all possible combinations of digits and operators, I then loop against the obtained strings and convert every single digit in an integer number, since .Int keeps track of the sign. Then it does suffice to sum every result and see if I come to the final result number of 100:
for [X~] map { .Slip },  (@digits Z @operators) {

    my $expression = .Str.subst( ' ', '', :g );

    my $sum = 0;
    for $expression ~~ m:g/ (<[- + ]>\d+)/ {
        $sum += .Int;

    say "Expression [$expression] = $sum" if $sum == 100;

PWC 44 - Task 2

This has puzzled me for a long time before I did come to a very simple solution. The problem was to find the shortest way to double or increment a number (1) to reach the final result of 200.
Initially I started considering 200 increments, than trying to group them in order to reduce them by doubling a subset (e.g., 50 icnrements are like 25 doubled). No way to get a right approach.
Then it comes to my mind a simple solution: since the doubling has to be the shortest operation, check from the end result how many times I can reduce it by a factor of 2, and if not possible, place an increment in the operation list and try again after having subtracted one value. Therefore, quite simply:
# $current-val = 200 initially
while ( $current-val > $initial-value ) {

  @moves.unshift: '%s to get %d $'
        .sprintf( $current-val %% 2 ?? 'double' !! 'add one buck', $current-val );
  $current-val = $current-val %% 2 ?? $current-val / 2 !! $current-val - 1;

After that, it does suffice to print out the @moves array and count how many elements it has in order to get the requested result.
In a previous implementation I resolved the problem with a given control flow, but I did not liked the repetition of the descriptive strings:
while ( $current-val > $initial-value ) {
 given  $current-val %% 2 {
     when .so {
         @moves.unshift: 'double to get %d $'.sprintf( $current-val );
         $current-val /= 2;
     when ! .so {
         @moves.unshift: 'add one buck to get %d $'.sprintf( $current-val );
         $current-val -= 1;
However, as you can see, while the given approach is the most readable (to me), the ?? !! one is the most compact.

The article Perl Weekly Challenge 44: only 100 and 200 bucks has been posted by Luca Ferrari on January 20, 2020