CombinatorialCase012: Difference between revisions

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This case produces '''Compositions''' of the number <apll>L</apll> into <apll>R</apll> parts.  A composition is a way of representing a number as the sum of all positive integers, in this case it’s a way of representing <apll>L</apll> as the sum of <apll>R</apll> positive integers.  It can also be thought of as a partition of <apll>L</apll> into <apll>R</apll> ordered parts.
This case produces '''Compositions of the number <apll>L</apll> into <apll>R</apll> parts'''.  A composition is a way of representing a number as the sum of all positive integers, in this case it’s a way of representing <apll>L</apll> as the sum of <apll>R</apll> positive integers.  It can also be thought of as a partition of <apll>L</apll> into <apll>R</apll> ordered parts.


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

Revision as of 20:45, 29 April 2017

This case produces Compositions of the number L into R parts. A composition is a way of representing a number as the sum of all positive integers, in this case it’s a way of representing L as the sum of R positive integers. It can also be thought of as a partition of L into R ordered parts.

  • L unlabeled balls (0), R labeled boxes (1), at least one ball per box (2)
  • Not ⎕IO-sensitive
  • Counted result is an integer scalar
  • Generated result is an integer matrix.

The count for this function is (L-R)!L-1.

For example:

If we have 5 unlabeled balls (●●●●●) and 3 labeled boxes (123) with at least one ball per box, there are 6 (↔ (5-3)!5-1) ways to meet these criteria:

 
 
 
 


1 2 3
 
 
 

 

1 2 3
 
 


 
 
1 2 3
 

 
 
 

1 2 3
 

 

 
 
1 2 3


 
 
 
 
1 2 3

The diagram above corresponds to

      012 1‼5 3
1 1 3
1 2 2
1 3 1
2 1 2
2 2 1
3 1 1
      ⍝ Compositions of L into R parts
      ⍝ Unlabeled balls, labeled boxes, ≥1 # Balls per Box
      012 1‼5 5
1 1 1 1 1
      012 1‼5 4
1 1 1 2
1 1 2 1
1 2 1 1
2 1 1 1
      012 1‼5 3
1 1 3
1 2 2
1 3 1
2 1 2
2 2 1
3 1 1
      012 1‼5 2
1 4
2 3
3 2
4 1
      012 1‼5 1
5

In general, because the counts of both compositions (012) and combinations (010) is a binomial coefficient, there might be a mapping between the two, and indeed there is, as seen by the following identities:

010 1‼L R ↔ +\0 ¯1↓012 1‼⍠1 R L+1
012 1‼L R ↔ ¯2-\(010 1‼⍠1 R L-1),L

where ‼⍠1 uses the Variant operator to evaluate in origin 1.