New answers tagged combinatorics
4
This will generate all of them, just like Tuples. Not too hard to redo so as to get one at a time. Just use the correspondence between k-digit numbers base n (n=length of input set) and subsets length k allowing repetitions.
takeWithRepitions[set_, k_] :=
Module[{n = Length[set], rule, vals},
rule = Thread[(Range[n] - 1) -> set];
vals = ...
9
You need to specify assumptions:
In[1]:= FunctionExpand[StirlingS2[n, 10], n > 0 && Mod[n, 1] == 0]
Out[1]= -(1/362880) + 2^(-8 + n)/315 + 1/135 2^(-7 + 2 n) +
1/315 2^(-8 + 3 n) - 3^(-3 + n)/1120 +
1/5 2^(-7 + n) 3^(-3 + n) - 5^(-2 + n)/576 +
1/567 2^(-8 + n) 5^(-2 + n) - 7^(-1 + n)/4320 - 9^(-2 + n)/4480
0
With apologies for the third, yet again independent answer, I recommend calling igraph through RLink using this package (only if you have v9 of Mathematica). Please see the instructions on how to set up the package, then you can do:
graph = AdjacencyGraph[{{1,0},{0,1}}];
res = IGraph["graph.get.isomorphisms.vf2"][graph, graph]
Round[res+1]
This will ...
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