Calculate Nth multiset combination (with repetition) based only on index

Question

How can i calculate the Nth combo based only on it\'s index. There should be (n+k-1)!/(k!(n-1)!) combinations with repetitions.

with n=2, k=5 you get:

0|{0         


        

Answer 1

See the example at: https://en.wikipedia.org/wiki/Combinatorial_number_system#Finding_the_k-combination_for_a_given_number

Just replace the binomial Coefficient with (n+k-1)!/(k!(n-1)!).

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Assuming n=3,k=5, let's say we want to calculate the 19th combination (id=19).

 id=0, {0,0,0,0,0}
 id=1, {0,0,0,0,1}
 id=2, {0,0,0,0,2}
 ...
 id=16, {1,1,1,1,2}
 id=17, {1,1,1,2,2}
 id=18, {1,1,2,2,2}
 id=19, {1,2,2,2,2}
 id=20, {2,2,2,2,2}

The result we're looking for is {1,2,2,2,2}.

Examining our 'T2' triangle: n=3,k=5 points to 21, being the 5th number (top to bottom) of the third diagonal (left to right).

            Indeterminate
                1   1
              1   2   3
            1   3   6   10
          1   4   10  20   35
        1   5   15  35   70   126
      1   6   21  56  126   252  462
    1   7   28  84  210   462  924   1716
 1   8   36  120  330   792  1716  3432  6435

We need to find the largest number in this row (horizontally, not diagonally) that does not exceed our id=19 value. So moving left from 21 we arrive at 6 (this operation is performed by the largest function below). Since 6 is the 2nd number in this row it corresponds to n==2 (or g[2,5] == 6 from the code below).

Now that we've found the 5th number in the combination, we move up a floor in the pyramid, so k-1=4. We also subtract the 6 we encountered below from id, so id=19-6=13. Repeating the entire process we find 5 (n==2 again) to be the largest number less than 13 in this row.

Next: 13-5=8, Largest is 4 in this row (n==2 yet again).

Next: 8-4=4, Largest is 3 in this row (n==2 one more time).

Next: 4-3=1, Largest is 1 in this row (n==1)

So collecting the indices at each stage we get {1,2,2,2,2}

---

The following Mathematica code does the job:

g[n_, k_] := (n + k - 1)!/(k! (n - 1)!)

largest[i_, nn_, kk_] := With[
    {x = g[nn, kk]}, 
    If[x > i, largest[i, nn-1, kk], {nn,x}]
]

id2combo[id_, n_, 0]  := {}
id2combo[id_, n_, k_] := Module[
    {val, offset}, 
    {val, offset} = largest[id, n, k];
    Append[id2combo[id-offset, n, k-1], val]
]

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Update: The order that the combinations were being generated by MBnext_multicombination wasn't matching id2combo, so i don't think they were lexicographic. The function below generates them in the same order as id2combo and matches the order of mathematica's Sort[]function on a list of lists.

void next_combo(unsigned int *ar, unsigned int n, unsigned int k)
{
    unsigned int i, lowest_i;

    for (i=lowest_i=0; i < k; ++i)
        lowest_i = (ar[i] < ar[lowest_i]) ? i : lowest_i;

    ++ar[lowest_i];

    i = (ar[lowest_i] >= n) 
        ? 0           // 0 -> all combinations have been exhausted, reset to first combination.
        : lowest_i+1; // _ -> base incremented. digits to the right of it are now zero.

    for (; i