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Here is a very procedural approach. Could be shortened somewhat; not sure it would give a speed gain or otherwise be worth the trouble. ll = {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, …

Could use integer linear programming via Maximize, for example. The constraint I use is overkill, all we really need to enforce is that values be nonnegative. vals = {140, 280, 420, 560, 700}; goal = …

One way to get a single solution faster is to set it up as a feasibility problem and use FindMinimum. This in turn uses fast machine precision linear programming under the hood. I recast as a 0-1 ILP …

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 …

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 …

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 …

Could do this as a polynomial algebra problem with matrix values used as exponents, with a distinct variable for each, summed in rows and product taken of row sums. Use a distinct variable to record t …

This might not be any better than a brute force check of all tuples, but it can be recast as an integer linear programming problem, mostly with 0-1 variables. The idea is to use a sum of the form -1*a …

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 …

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] …

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 …

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], { …

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 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 …

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[ …