Perl Weekly Challenge 361: numbers

This post presents my solutions to the Perl Weekly Challenge 361.
I keep doing the Perl Weekly Challenge in order to mantain my coding skills in good shape, as well as in order to learn new things, with particular regard to Raku, a language that I love.
This week, I solved the following tasks:

• PWC 361 - Task 1 - Raku
• PWC 361 - Task 2 - Raku

Raku Implementations

PWC 361 - Task 1 - Raku Implementation

The first task was about finding the Zeckendorf sequence for a given integer.

sub MAIN( Int $number is copy
	  where { 0 < $number <= 100 } ) {

    my @fib = 0, 1, 1;
    my @zeckendorf;

    while ( @fib.elems < $number ) {
		@fib.push: @fib[ * - 1 ] + @fib[ * - 2 ];
    }

    for @fib.reverse {
		if ( $_ <= $number ) {
		    @zeckendorf.push: $_;
		    $number -= $_;
		}

        last if $number <= 0;
    }

    @zeckendorf.join( ', ').say;
}

The implementation is quite straighforward: @fib is an array that stores the Fibonacci’s series up to a given number of elements. Then I iterate on the reverse Fibonacci’s series, and keep all the numbers that build up to the given $number, and last I print out the found array. This solution is clearly not optimized.

PWC 361 - Task 2 - Raku Implementation

Given a matrix of integers, representing booleans that provide information if the current row knows the given columns, find out which knows everybody.

sub MAIN() {
    my @party =
            [0, 0, 0, 0, 1, 0],  # 0 knows 4
            [0, 0, 0, 0, 1, 0],  # 1 knows 4
            [0, 0, 0, 0, 1, 0],  # 2 knows 4
            [0, 0, 0, 0, 1, 0],  # 3 knows 4
            [0, 0, 0, 0, 0, 0],  # 4 knows NOBODY
            [0, 0, 0, 0, 1, 0],  # 5 knows 4
    ;

    my %known;
    my $required = @party.elems;

    for 0 ..^ @party.elems -> $current {
		my @who =  @party[ $current ].grep( * ~~ 1, :k );
		$required-- and next unless ( @who );
		%known{ $_ }++ for ( @who );
    }

    my @result = %known.keys.grep( { %known{ $_ } == $required } );
    @result.say and exit if ( @result );
    '-1'.say;
}

The implementation counts how many people the $current one knows, storing for each person’s index the amount of people who know him. In the case someone knows nobody, it means that we need less $required people to satisfy the “knows everyone” concept. Last, I extract from the %known hash the keys (hence, the people) who knows all the others, that means have a counting up to $required.