After mulling on this one for a while, in particular thinking about how to track when a letter was due to get output again and how to maintain a distribution, I think I have something good. Compute a rolling score for each element, where it accumulates based upon probability of output and gets dinged when its element gets selected; maximum score always gets chosen.
```use strict;
use warnings;
use List::Util 'sum';
use Data::Dump qw[ pp ];

sub gen {
my \$href = shift;
my %score = %\$href;
my \$norm = sum values %score;
my @series;

for (1 .. \$norm) {
my (\$max, \$elem) = 0;
for (keys %score) {
(\$max, \$elem) = (\$score{\$_}, \$_) if \$score{\$_} >= \$max;
\$score{\$_} += \$href->{\$_};
}
push @series, \$elem;
\$score{\$elem} -= \$norm;
}

return \@series;
}

pp gen( { A => 4, B => 2, C => 3, D => 1 } );
pp gen( { A => 5, B => 4, C => 3, D => 2, E => 1 } );
outputs
```["A", "C", "B", "A", "D", "C", "A", "B", "C", "A"]
["A" .. "D", "A", "B", "E", "A", "C", "B", "A", "D", "C", "B", "A"]

It also naturally extends to infinite series and non-integer element counts. The infinite series case is why it's necessary to have a >= in the score comparison; you could alternatively initialize \$max to something negative since the sum of all scores is necessarily zero at all times. N*M complexity, where N is the length of desired series and M is number of distinct elements.

This approach also lends itself to designing a metric for how smooth a series is:

```use strict;
use warnings;
use List::Util 'sum';

sub measure {
my @series = @_;
my %count;
\$count{\$_}++ for @series;
my \$norm = sum values %count;
\$_ /= \$norm for values %count;
my %score = %count;

my \$metric = 0;
for my \$elem (@series) {
\$score{\$elem} -= 1;
for (keys %score) {
\$metric += \$score{\$_}**2;
\$score{\$_} += \$count{\$_};
}
}

return \$metric/@series;
}

print measure("A", "C", "B", "A", "D", "C", "A", "B", "C", "A"), "\n";
print measure("A", "C", "B", "A", "C", "D", "A", "B", "C", "A"), "\n";
print measure("A", "A", "A", "A", "B", "B", "C", "C", "C", "D"), "\n";
outputs
```0.37
0.37
3.45

#11929 First ask yourself `How would I do this without a computer?' Then have the computer do it the same way.

In reply to Re: Bag uniform distribution algorithms by kennethk
in thread Bag uniform distribution algorithms by davido

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