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Take all subsets of length 10, then for each one find all splits into two sets of five such that the first of the ten is in the first part of the split. In[29]:= Timing[ msets = Subsets[Range[12], { …

This has [quadratic, he said] actually perhaps cubic complexity in a worst case (okay, now I'm just confused. More below).. Not the fastest of the lot, but it seems reasonable, or at least not entirel …

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 …

Depending on whether you care about permutations or not, here are some ways to go about it. One is to solve a system of equations via Reduce and count the solutions. vars = Array[a, 6]; eqn = Total[ …

Could set this up as a 1-0 integer linear programming problem. Module[{vars = Array[a, 10]}, vars*Range[10] /. Solve[Flatten@{vars.Range[10] == 28, Total[vars] == 4, Map[0 <= # <= 1 &, v …

Not fastest, but gives an idea of how Solve might be used with tolerable efficiency. In[181]:= Timing[ soln = Solve[{x^2 + y^2 - z^2 == 0, z >= y - 1 >= x - 1 >= 0, 0 <= x + y + z <= 1000}, {x …

This is not really very efficient but here goes. We can create a rational function that is effectively a generating function in three variables, one to force 8 factors, one to force a sum e1ual to 24, …

Edited for correctness: I use the variant for directed graphs from here. I take your graph as above, extract edges, rename so vertices are integers from 1 to #vertices. After finishing we revert to t …

This might get you started. It generates (I think) all graphs of given number of vertices and degree, with the following caveats: They are undirected. They do not have self edges. Edges only have we …

Could use integer linear programming. In Mathematica this can be done with Reduce[]. One way (probably not the best) to set this up is shown below. It uses an array of 0-1 variables, where a 1 in posi …

Could do this procedurally by constructing iterator lists. indices[n_, max_] := Module[ {jj = Array[j, n], starts, ends}, starts = Prepend[Most[jj] + 1, 1]; ends = max - Range[n - 1, 0, -1]; …

Not really what you want, but you can find the closest hit using FindMinimum, as below. n = 6; target = 10; SeedRandom[1111]; vals = RandomReal[1, n]; coeffs = Array[c, 6]; c1 = Map[# >= 0 &, coeff …

One way is to place the first element in each of the subsets formed from s-1 of the remaining elements, find complements, and recursively subdivvy those complements. subsetSubsets[set_, s_ /; s <= 0] …

I will use David Carraher's example to illustrate a way using integer linear programming. x = Range[7]; a[1] = {1, 4, 7}; a[2] = {1, 4}; a[3] = {4, 5, 7}; a[4] = {3, 5, 6}; a[5] = {2, 3, 6, 7}; a[6] …

A simple recursion handles this. fullGrayCode[1, n_Integer] /; n >= 1 := Transpose[{Range[0, n - 1]}] fullGrayCode[m_Integer, n_Integer] /; m >= 2 && n >= 1 := Module[ {gcm1 = fullGrayCode[m - 1, n …