Re: Bag uniform distribution algorithms

[kcott]

G'day Dave,

By alternatively splicing elements from either end of the distribution, I get your original 'qw( A C B A D C A B C A )'.

$ perl -Mstrict -Mwarnings -E '
    my %bag = ( A => 4, B => 2, C => 3, D => 1 );
    my @distribution;

    for my $key (sort { $bag{$b} <=> $bag{$a} } keys %bag) {
        my $base_offset = int(@distribution / ($bag{$key} + 1)); 
        my $offset = $base_offset;

        for (1 .. $bag{$key}) {
            next unless $_ % 2;
            splice @distribution, $offset, 0, $key;

            if ($_ < $bag{$key}) {
                splice @distribution, -$offset, 0, $key;
            }

            $offset += $base_offset + 1;
        }   
    }   

    say "@distribution";
'
A C B A D C A B C A

If I round up the $base_offset value, the middle two elements are reversed but everything else remains the same. I don't see this as being more or less uniform but maybe it's more correct.

$ perl -Mstrict -Mwarnings -E '
    my %bag = ( A => 4, B => 2, C => 3, D => 1 );
    my @distribution;

    for my $key (sort { $bag{$b} <=> $bag{$a} } keys %bag) {
        my $base_offset = int(@distribution / ($bag{$key} + 1) + 0.5);
+ 
        my $offset = $base_offset;

        for (1 .. $bag{$key}) {
            next unless $_ % 2;
            splice @distribution, $offset, 0, $key;

            if ($_ < $bag{$key}) {
                splice @distribution, -$offset, 0, $key;
            }

            $offset += $base_offset + 1;
        }   
    }   

    say "@distribution";
'
A C B A C D A B C A

I haven't done extensive testing on this solution. Beyond your sample input, I tried:

my %bag = ( A => 4, B => 2, C => 3, D => 1, Z => 1 );

gives: A C B A C D Z A B C A; and

my %bag = ( A => 4, B => 2, C => 3, D => 1, Y => 5, Z => 1 );

gives: Y A C Y B A C Z D Y A B Y C A Y; and

my %bag = ( A => 4, B => 2, C => 3, D => 1, X => 6, Y => 5, Z => 1 );

gives: X Y A X C Y B A X Y Z D C X A B Y C X A Y X; and

my %bag = ( A => 4, B => 2, C => 3, D => 1, W => 4, X => 6, Y => 5, Z 
+=> 1 );

gives: X Y A W X C Y B A W X Y Z D C X W A B Y C X W A Y X

The above 4 tests all included rounding; without rounding, the results are:

A C B A Z D C A B C A
Y A C Y B A C D Z Y A B Y C A Y
X Y A C X Y B A C X D Z Y X A B Y X C A Y X
X Y A W C X Y B A W C X D Z Y X W A B Y X C W A Y X