Using the elements of a subset in a Do loop

Question

I would like to use the elements of Subsets in a Do loop. I have:

w = Subsets[{3, 4, 5, 6}, {2}]

which yields:

{{3, 4}, {3, 5}, {3, 6}, {4, 5}, {4, 6}, {5, 6}}

Now I want to use these ordered pairs in a Do loop. For example:

Do[ma[i]+mb[j],{i,...},{j,...}]

where {i,j} should be only these ordered pairs.

Answer 1

 w = Subsets[{3, 4, 5, 6}, {2}]

{{3, 4}, {3, 5}, {3, 6}, {4, 5}, {4, 6}, {5, 6}}

This will run only over this ordered pairs. Is this what you need? -

Do[Print[k], {k, w}];

In Mathematica you can also use other constructs similar to Do. For example

Table[Print[k], {k, w}];

Or even better, use functional programming:

Print /@ w;

All these give same result. Here /@ is a shortcut notation for Map which in full form would be Map[Print, w]. The major difference with Do and Table - there is no running index specification.

If you need to run a computation that separates the indexes (as you mentioned in the comment), it is still easy:

ma[#1] + mb[#2] & @@@ w

where @@@ is is a flavor of Apply . These will work too:

ma[#[[1]]] + mb[#[[2]]] & /@ w

Table[ma[k[[1]]] + mb[k[[2]]], {k, w}]

but why would you stray away from elegance? ;-) These [[...]] mean Part.

Answer 2

I suppose that the most likely reason to use subsets in a Do loop rather than just mapping (/@) a function is to conserve memory. Therefore I would like to point out that the third argument of Subsets specifies which subset(s) within the sequence to return. If one were attempting to find all subsets that sum to a certain value without first storing all subsets in memory one might use it like this:

SeedRandom[1]

set = RandomInteger[123, 50];

n = 3;

Do[
 If[Total@# == 37, Print@#] & @@ Subsets[set, {n}, {i}],
 {i, Binomial[Length@set, n]}
]

{14,0,23} {0,4,33} {3,24,10} {3,4,30} {3,1,33} {23,4,10} {15,4,18} {1,26,10}

(This is a brute force approach to that particular problem and there are better ways, but it serves as an illustration.)

In practice this is painfully slow. To make it practical you would process the subsets in blocks. Here in blocks of 10,000 subsets each:

SeedRandom[1]
set = RandomInteger[2000, 250];
n = 3;
Do[
  If[Total@# == 137, Print@#] & /@ Quiet@Subsets[set, {n}, {i, i + 10000}],
  {i, 1, Binomial[Length@set, n], 10000}
]

{28,16,93}

{28,106,3}