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### Re: Bag uniform distribution algorithms

by LanX (Chancellor)
 on Oct 27, 2013 at 21:38 UTC ( #1059935=note: print w/replies, xml ) Need Help??

in reply to Bag uniform distribution algorithms

I'm not sure how "as uniformly distributed as possible" can be qualified ...

Do you have a test-code to check the "quality" of a solution?

But sorting according to a weighting function gives similar results like shown by you.

```  DB<171> %h=( A => 4, B => 2, C => 3, D => 1 )
=> ("A", 4, "B", 2, "C", 3, "D", 1)

DB<172> \$sum=0; \$sum+=\$_ for values %h;
=> ""

DB<173> @list=map  {  my (\$k,\$v)=(\$_,\$h{\$_}); my \$int=\$sum/\$v; map {
+ [ \$k => \$int*(\$_-.5)] } 1..\$v  } keys %h
=> (
["A", "1.25"],
["A", "3.75"],
["A", "6.25"],
["A", "8.75"],
["D", 5],
["C", "1.66666666666667"],
["C", 5],
["C", "8.33333333333333"],
["B", "2.5"],
["B", "7.5"],
)

DB<174> map {\$_->[0]} sort { \$a->[1] <=> \$b->[1] or \$a->[0] cmp \$b->
+[0]} @list
=> ("A", "C", "B", "A", "C", "D", "A", "B", "C", "A")

Changing the weighting function would also allow to repeat the pattern in a way that joined sequences are still equally distributed ( that is A doesn't neighbor A )

An iterator-version shouldn't be too difficult.

Cheers Rolf

( addicted to the Perl Programming Language)

code simplified

##### update

well after second thought it's quite easy to find input where this approach fails ... never mind! :(

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