CombinatorialCase001: Difference between revisions
From NARS2000
Jump to navigationJump to search
mNo edit summary |
No edit summary |
||
Line 6: | Line 6: | ||
* 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>(M+N)PN N</apll> where <apll>M PN N</apll> calculates the number of [https://en.wikipedia.org/wiki/Partition_(number_theory) Partitions] of the number <apll>M</apll> into exactly <apll>N</apll> parts. | The count for this function is <apll>(M+N)PN N</apll> where <apll>M PN N</apll> calculates the number of [https://en.wikipedia.org/wiki/Partition_(number_theory)#Restricted_part_size_or_number_of_parts Partitions] of the number <apll>M</apll> into exactly <apll>N</apll> parts. | ||
For example: | For example: |
Latest revision as of 18:49, 19 June 2017
This case produces the Partitions of the number M into at most N parts.
- M unlabeled balls (0), N 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 (M+N)PN N where M PN N calculates the number of Partitions of the number M into exactly N 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:
|
|
|
|
|
|
|
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 M into at most N parts ⍝ Unlabeled balls & boxes, any # Balls per Box ⍪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, (M+N)PN N ↔ +/M PN¨0..N.
Because partitions of M into N non-negative parts (001) is the same as partitions of M+N into N positive parts (002), these cases are related by the following identity:
001 1‼M N ↔ (⊂[⎕IO+1] ¯1+002 1‼(M+N) N)~¨0