CombinatorialCase000/100: Difference between revisions
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(Created page with "<apll>L</apll> Pigeons into <apll>R</apll> holes Cases '''000''' and '''100''' are both trivial. * <apll>L</apll> unlabeled (0) or labeled (1) balls into <apll>R</apll> unla...") |
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* <apll>L</apll> unlabeled (0) or labeled (1) balls into <apll>R</apll> unlabeled boxes (0), at most one per box (0) | * <apll>L</apll> unlabeled (0) or labeled (1) balls into <apll>R</apll> unlabeled boxes (0), at most one per box (0) | ||
* Not ⎕IO-sensitive | * Not <apll>⎕IO</apll>-sensitive | ||
* Counted result is a Boolean scalar | * Counted result is a Boolean scalar | ||
* Generated result is a Boolean matrix. | * Generated result is a Boolean matrix. | ||
Revision as of 13:21, 29 April 2017
L Pigeons into R holes
Cases 000 and 100 are both trivial.
- L unlabeled (0) or labeled (1) balls into R unlabeled boxes (0), at most one per box (0)
- Not ⎕IO-sensitive
- Counted result is a Boolean scalar
- Generated result is a Boolean matrix.
The count for this function is L≤R.
If L>R, then there is no answer, or more accurately, the result is an empty matrix of shape 0 R. If L≤R, then the result is a one-row matrix with L leading 1s and the rest 0s. Combining these two cases yields a result of ((L≤R) R)⍴R↑L⍴1.
For example:
If we have 3 unlabeled balls (●●●) and 5 unlabeled boxes with at most one ball per box, there is only 1 (↔ 3≤5) way to meet these criteria:
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as well as
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The diagram above corresponds to
000 1‼3 5
1 1 1 0 0
⍝ L pigeons into R holes
⍝ Unlabeled balls & boxes, ≤1 #bpb
000 1‼4 5
1 1 1 1 0