CombinatorialCase001: Difference between revisions
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(Created page with "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) *...") |
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* Generated result is a nested vector of integer vectors. | * Generated result is a nested vector of integer vectors. | ||
The count for this function is <apll>(L+R)PN R</apll> where <apll>L PN R</apll> calculates the number of [https://en.wikipedia.org/wiki/Partition_(number_theory) Partitions] of the number <apll>L</apll> into | The count for this function is <apll>(L+R)PN R</apll> where <apll>L PN R</apll> calculates the number of [https://en.wikipedia.org/wiki/Partition_(number_theory) Partitions] of the number <apll>L</apll> into exactly <apll>R</apll> parts. | ||
For example: | For example: |
Revision as of 14:27, 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