CombinatorialCase001: Difference between revisions
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This case produces the Partitions of the number L into at most R parts. | This case produces the '''Partitions of the number''' L into at most R parts. | ||
* <apll>L</apll> unlabeled balls (0), <apll>R</apll> unlabeled boxes (0), any # of balls per box (1) | * <apll>L</apll> unlabeled balls (0), <apll>R</apll> unlabeled boxes (0), any # of balls per box (1) | ||
Revision as of 17:37, 29 April 2017
This case produces the Partitions of the number L into at most R parts.
- L unlabeled balls (0), R unlabeled boxes (0), any # of balls per box (1)
- Not ⎕IO-sensitive
- Counted result is an integer scalar
- Generated result is a nested vector of integer vectors.
The count for this function is (L+R)PN R where L PN R calculates the number of Partitions of the number L into exactly R parts.
For example:
If we have 6 unlabeled balls (●●●●●●) and 3 unlabeled boxes with any # of balls per box, there are 7 (↔ (6+3)PN 3) ways to meet these criteria:
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The diagram above corresponds to the nested array
⍪001 1‼6 3
6
5 1
4 2
4 1 1
3 3
3 2 1
2 2 2
⍝ Partitions of L into at most R parts
⍝ Unlabeled balls & boxes, any #bpb
⍪001 1‼5 5
5
4 1
3 2
3 1 1
2 2 1
2 1 1 1
1 1 1 1 1
⍪001 1‼5 4
5
4 1
3 2
3 1 1
2 2 1
2 1 1 1
⍪001 1‼5 3
5
4 1
3 2
3 1 1
2 2 1
⍪001 1‼5 2
5
4 1
3 2
⍪001 1‼5 1
5
Identities
As shown in Wikipedia, (L+R)PN R ↔ +/L PN¨0..R.
Because partitions of L into R non-negative parts (001) is the same as partitions of L+R into R positive parts (002), these cases are related by the following identity:
001 1‼L R ↔ (⊂[⎕IO+1] ¯1+002 1‼(L+R) R)~¨0
