How to loop trough all unique pair combinations of N items (pairing algorithm)

[AndyRR]

I have a list of items A,B,C,D,E,F.. and I'd like to loop through all possible unique pair combinations (note: AB and BA are the same pair). I know that the possible pair combinatios increase by (N-1)!! (!!=double factorial):
2 items : 1!! = 1=1
4 items : 3!! = 1*3=3
6 items : 5!! = 1*3*5=15
8 items : 7!! = 1*3*5*7=105
10 items : 9!! = 1*3*5*7*9=945

However I can't figure out an algorithm how to loop through all unique pair combinations once in Labview. I have N items, where N is even and reasonably low so that the number of pair combination is not too high. This is how it would look like for for 6 items: 15 different unique pair combinations. One row of index pairs is what I'd like to get for each loop index.
[0,1],[2,3],[4,5] (= AB,CD,EF)
[0,1],[2,4],[3,5] (= AB,CE,DF)
[0,1],[2,5],[3,4] (= AB,CF,DE)
[0,2],[1,3],[4,5] (= AC,BD,EF)
[0,2],[1,4],[3,5] (= AC,BE,DF)
[0,2],[1,5],[3,4] (= AC,BF,DE)
[0,3],[1,2],[4,5] (= AD,BC,EF)
[0,3],[1,4],[2,5] (= AD,BE,CF)
[0,3],[1,5],[2,4] (= AD,BF,CE)
[0,4],[1,2],[4,5] (= AE,BC,DF)
[0,4],[1,3],[2,5] (= AE,BD,CF)
[0,4],[1,5],[2,3] (= AE,BF,CD)
[0,5],[1,2],[3,4] (= AF,BC,DE)
[0,5],[1,3],[2,4] (= AF,BD,CE)
[0,5],[1,4],[2,3] (= AF,BE,CD)

I have score for each possible pair and I'm trying to get a group of pairs out of the N items with the highest total group score.

Thanks in advance

[GerdW]

Hi Andy,

use autoindexing:

An U8 array is used here instead of an Array of strings...

[NeilR]

Gerd I'm afraid that isn't correct. Your algorithm just gives an array increasing like this ...

1                        (1)

1+2+3                 (6)

1+2+3+4+5         (15)

1+2+3+4+5+6+7  (28)  etc

rather than in a double factorial manner.

[GerdW]

Hi Neil,

the OP wanted to have all "unique combinations" of array elements (without permutations as "AB" is the same as "BA").

That's what my snippet provides…

[NeilR]

Gerd, I think you'll find that your snippet gives "unique pairs" rather than "unique pair combinations".

Having said that, I think that the unique pairs can subsequently be used to generate the necessary combinations ... it's just a case of working out how !

[GerdW]

Hi Neil,

ok, I missed the word "combinations".

Atleast the OP now already has a "1 row list" of all possible combinations. Now we only need to pick some elements from it.

But well, this is a nice job of learning recursion and implementing it in LabVIEW. 😄

[AndyRR]

Thanks for the replys so far. What I actually want are the unique combinations of the pairs. Greg, out of your list the unique pair combinations would be:
1st unique pair combination: [0,1] and [2,3]
2nd unique pair combination: [0,2] and [1,3]
3rd unique pair combination: [0,3] and [1,2]
The question remains how to get these.

[jcarmody]

This was an interesting conversation a few years ago - http://forums.ni.com/t5/LabVIEW/How-can-I-create-list-of-permutations/td-p/1256988