CombinatorialCase102: Difference between revisions

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This case produces '''Partitions of the set''' <apll>{⍳L}</apll> into exactly <apll>R</apll> parts.  As such, it produces a subset of [[CombinatorialCase101|<apll>101</apll>]], limiting the result to just those rows with <apll>L</apll> subsets.
This case produces '''Partitions of the set <apll>{⍳L}</apll> into exactly <apll>R</apll> parts'''.  As such, it produces a subset of [[CombinatorialCase101|<apll>101</apll>]], limiting the result to just those rows with <apll>L</apll> subsets.


* <apll>L</apll> labeled balls (1), <apll>R</apll> unlabeled boxes (0), at least one ball per box (2)
* <apll>L</apll> labeled balls (1), <apll>R</apll> unlabeled boxes (0), at least one ball per box (2)

Revision as of 20:42, 29 April 2017

This case produces Partitions of the set {⍳L} into exactly R parts. As such, it produces a subset of 101, limiting the result to just those rows with L subsets.

  • L labeled balls (1), R unlabeled boxes (0), at least one ball per box (2)
  • Sensitive to ⎕IO
  • Counted result is an integer scalar
  • Generated result is a nested vector of nested integer vectors.

The count for this function is L SN2 R where L SN2 R calculates the Stirling numbers of the 2nd kind.

For example:

If we have 4 labeled balls (❶❷❸❹) and 2 unlabeled boxes with at least one ball per box, there are 7 (↔ 4 SN2 2) ways to meet these criteria:



       


       



       


       



       



       



       

The diagram above corresponds to the nested array

      ⍪102 1‼4 2
  1 2 3  4
  1 2 4  3
  1 2  3 4
  1 3 4  2
  1 3  2 4
  1 4  2 3
  1  2 3 4

      ⍝ Partitions of {⍳L} into R parts
      ⍝ Labeled balls, unlabeled boxes, ≥1 # Balls per Box
      ⍝ The number to the right in parens
      ⍝    represent the corresponding row from
      ⍝    the table in case 101.

      ⍪102 1‼4 4
  1  2  3  4		        (15)
      ⍪102 1‼4 3
  1 2  3  4			(5)
  1 3  2  4			(8)
  1  2 3  4			(11)
  1 4  2  3			(12)
  1  2 4  3			(13)
  1  2  3 4			(14)
      ⍪102 1‼4 2
  1 2 3  4			(2)
  1 2 4  3			(3)
  1 2  3 4			(4)
  1 3 4  2			(6)
  1 3  2 4			(7)
  1 4  2 3			(9)
  1  2 3 4			(10)
      ⍪102 1‼4 1
  1 2 3 4			(1)
      ⍪102 1‼4 0

In general, this case is related to 101 through the following identities (after sorting the items):

101 1‼L R ↔ ⊃,/102 1‼¨L,¨0..R
102 1‼L R ↔ R {(⍺=≢¨⍵)/⍵} 101 1‼L R

and is related to 112 through the following identities:

102 1‼L R ↔ {(2≢/¯1,(⊂¨⍋¨⍵)⌷¨⍵)/⍵} 112 1‼L R
a←⊃102 1‼L R
b← 110 1‼R R
112 1‼L R ↔ ,⊂[⎕IO+2] a[;b]