Suppose you had n indistinguishable balls and k distinguishable boxes. Enumerate the ways of distributing the balls into boxes. Some boxes may be empty.
We can represent each distribution in the form of n stars and k − 1 vertical lines. The stars represent balls, and the vertical lines divide the balls into boxes. For example, here are the possible distributions for n = 3, k = 3:
***|| | 3+0+0 |
**|*| | 2+1+0 |
**||* | 2+0+1 |
*|**| | 1+2+0 |
*|*|* | 1+1+1 |
*||** | 1+0+2 |
|***| | 0+3+0 |
|**|* | 0+2+1 |
|*|** | 0+1+2 |
||*** | 0+0+3 |
This visualization is known in combinatorics circles as stars and bars.
From this visualization, we see that what we are doing is taking n + k − 1 slots, and in each slot placing a star or a bar, subject to the constraint that there be n stars and k − 1 bars. Another way of looking at this is that we are choosing a subset of size k − 1 from a set of size n + k − 1 (the subset specifying where the bars go).
Now we can fire up our subset-generating machine.
function Distributions(n, k, f) { Subsets(n + k - 1, k - 1, function(s) { s.push(n + k); f(s.map(function(v, i) { return v - (s[i-1]||0) - 1; })); s.pop(); }); }
We ask to generate subsets of size k − 1 from a set of size n + k − 1. For each such subset, we draw an artificial bar at the end (slot n + k), then calculate the number of stars between the bars. The number of stars between two bars is the distance between the two bars, minus 1 because the bar takes up space, too.
Another solution is to reduce this to a problem we already know how to solve: enumerating integer compositions. After distributing the balls into boxes, we go around like Santa Claus and give each box one extra ball, which produces a composition. Conversely, for any composition, remove one ball from each box, and you get a distribution.
function Distributions(n, k, f) { Compositions(n + k, k, function(s) { f(s.map(function(v) { return v - 1; })); }); }
We added k extra balls, so we need to generate compositions of n + k. When we get each composition, we take one ball away from each box and call that the distribution.
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