Perl Weekly Challenge 48: Survivors and Palindrome Dates

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 48.

PWC 48 - Task 1

The first task required to find out a survivor in a kind of Philosopher Dining Problem: there are 50 people sitting in a circle, one of them has a sword and kills the next one passing the sword to the next-to-the-just-killed one. This goes on until a single one survives.
I decided to implement the people as a list where a True value indicates that that person has the sword, a False that the person does not have a sword and a Nil that the person has been killed.
The only problem I found, is to rotate over the array, so I defined a specific method to find the next alive person given a specific index:
sub next-alive( @people, $current-person ) {
    my $next = $current-person;

    loop {
        $next = $next >= @people.elems ?? $next % @people.elems !! $next;
        return $next if @people[ $next ].defined;
I did not find any more elegant approach to loop over the array.
Having that, it becomes quite simple to iterate over the people and loop until only one remains:
while ( @people.grep( *.defined ) > 1 ) {
    # find out who has the sword
    my $killer      = @people.first: *.so, :k;
    # then find out the next person to kill
    my $killed      = next-alive( @people, $killer );
    @people[ $killed ] = Nil;  # killed!
    @people[ $killer ] = False; # pass the sword
    # now get the next person that will hold the sword
    my $next-killer = next-alive( @people, $killed );
    @people[ $next-killer ] = True; # the next killer
While there is only one not-Nil value in the array, I select the first people with the sword and then the next one alive near the killer, and finally the next one to the killed person. Then it is just a metter of changing the values.
The final result is that running the program produces:
% perl6 ch-1.p6
The person who survives is 37

PWC 48 - Task 2

The second task was simpler to me: produce palindrome dates.
I decided to implement this using Date and placing the formatter property, then I looped one day at a time and print out the date if its stringyfied version is the same as its reversed (i.e., flip) stringyfied value:
    my $current-date = :year( $year-start ),
                                 formatter => { sprintf( "%02d%02d%04d", .month, .day, .year ) } );
    my $end-date     = :year( $year-end ), :day(31), :month(12) );

    for 1 .. $end-date - $current-date {
        $current-date += 1;  # add one day at a time
        # print the date if its string representation is the same as
        # the flipped string representation
        "Palindrome: $current-date".say if $current-date.Str ~~ $current-date.Str.flip;
However, after some thoughts (night was useful!), I realized that such a brute force way was too much expensive. Knowing that the date must be palindrome, it means that given an year we already know what the month and the day will be, and therefore the only thing to do is to check if such a date is good or not. I then re-implemeted the program as follows:
    for $year-start .. $year-end {
        $_ ~~ / ^ $<day>=\d ** 2 $<month>=\d ** 2 $ /;
        my $month = $/<month>.flip;
        my $day   = $/<day>.flip;
        next if  $month > 12 || $month == 0;
        next if $day > 31 || $day == 0;
        "Palindrome date %02d%02d%04d".sprintf( $month, $day, $_ ).say if try :year( $_),
                                                                                    :month( $month),
                                                                                    :day( $day ) );
First, I extract the parts that will become the month and the day from the year string. Then I flip them because I need them reversed to make a palindrome date. It is simple enough to exclude obvious out of range values, like a month greater than 12 or a day greater than 31. This however does not suffice to give me a valida te, so I try to construct a Date object, and if I succeed I then have a valid palindrome date.
Please note the use of try within the if: Date will throw an exception if the construction fails, so I don’t want my loop to break if I’m trying a stupid date with out of range values.

The article Perl Weekly Challenge 48: Survivors and Palindrome Dates has been posted by Luca Ferrari on February 17, 2020