use strict;
use warnings;
@_ = (2..10,'J','Q','K','A');
comb('', 4, 2..4,6..10,'J','Q','K','A',@_,@_,@_);
sub comb {
my ($str, $depth, @items) = @_;
if (!$depth--) { interpret($str); return; }
comb("$str $items[$_]", $depth, @items[($_+1)..$#items]) for (0..$
+#items)
}
sub interpret {
my @cards = split / /, "5$_[0]";
# Do whatever with @cards;
}
Actually though, the possible set can't contain any 5's except the original 5, nor can it contain any 10, J, Q, K - since all hands involving those will have at least 2 points. This cuts the number of combinations significantly.
@_ = (2..4,6..9,'A');
comb('', 4, @_,@_,@_,@_);
Nor can you have more than one of any card.
comb('', 4, 2..4,6..9,'A');
It should be fairly easy to figure out whether there are straights or 15's in the permutations? of this set. Code to follow...
use strict;
use warnings;
comb('', 4, 1..4,6..9);
sub comb {
my ($str, $depth, @items) = @_;
if (!$depth--) { interpret($str); return; }
comb("$str $items[$_]", $depth, @items[($_+1)..$#items]) for (0..$
+#items)
}
sub interpret {
my @cards = split / /, "5$_[0]";
@cards = sort {$b <=> $a} @cards;
my $flag = 1;
for (0..3) { $flag = 0 if $cards[$_+1] != ($cards[$_]-1); }
return if $flag;
for (perm(@cards)) {
return if sum(split //) == 15;
}
print join " ", @cards;
}
BEGIN {
my @c_out;
sub perm {
@c_out = ();
permute('', $_, @_) for (0..$#_);
return @c_out;
}
sub permute {
my ($str, $depth, @chars) = @_;
if (!$depth--) {
push @c_out, $str.$_ for @chars;
}
else {
permute($str.$chars[$_], $depth, @chars[($_+1)..($#chars)]
+)
for (0..$#chars);
}
}
}
sub sum { my $n; $n += $_ for @_; return $n; }
Very rough code, but upshot is that there are no hands with a 5 that don't have at least 2 points. I could actually have done this whole thing on paper, since there's very few sets of cards that have to be tested for 15's and straights.
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