Set puzzle
The trick here is to allocate three different bits for each enum, with the result that the cards of a matching set OR together to produce a 4-digit octal number that contains only the digits 1, 2, 4, or 7. This OR is done by funny looking [+|]
, which is the reduction form of +|
, which is the numeric bitwise OR. (Because Perl 6 stole the bare |
operator for composing junctions instead.)
enum Color (red => 0o1000, green => 0o2000, purple => 0o4000);
enum Count (one => 0o100, two => 0o200, three => 0o400);
enum Shape (oval => 0o10, squiggle => 0o20, diamond => 0o40);
enum Style (solid => 0o1, open => 0o2, striped => 0o4);
my @deck = Color.enums X Count.enums X Shape.enums X Style.enums;
sub MAIN($DRAW = 9, $GOAL = $DRAW div 2) {
sub show-cards(@c) { { printf "%9s%7s%10s%9s\n", @c[$_;*]».key } for ^@c }
my @combinations = [^$DRAW].combinations(3);
my @draw;
repeat until (my @sets) == $GOAL {
@draw = @deck.pick($DRAW);
my @bits = @draw.map: { [+] @^enums».value }
@sets = gather for @combinations -> @c {
take @draw[@c].item when /^ <[1247]>+ $/ given ( [+|] @bits[@c] ).base(8);
}
}
say "Drew $DRAW cards:";
show-cards @draw;
for @sets.kv -> $i, @cards {
say "\nSet {$i+1}:";
show-cards @cards;
}
}
Output:
Drew 9 cards:
purple two diamond open
red two squiggle striped
purple three squiggle open
purple two squiggle striped
red three oval striped
red one diamond striped
purple two oval solid
green three diamond solid
red two squiggle open
Set 1:
purple two diamond open
purple two squiggle striped
purple two oval solid
Set 2:
purple two diamond open
red one diamond striped
green three diamond solid
Set 3:
red two squiggle striped
red three oval striped
red one diamond striped
Set 4:
purple three squiggle open
red three oval striped
green three diamond solid