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My question may be very stupid, and trivial, but looking forward to some suggestions -

As a visualization let's consider a rectangle with four vertices represented by [1,2,3,4]. Now I want to find a triplet of numbers from them like (1, 2, 3).

ly = [1,2,3,4]
triplets = []
for i in ly:
    for j in ly:
        for k in ly:
            triplets.append((i,j,k))

However, I want a combination of those triplets such that (1, 2, 3) is the same as (3, 2, 1), because the angle between these indices is given by 2. I mean I want this triplet occurring only once defined by the middle number. So, in my case, (4, 1, 2) is the same as (2, 1, 4). So I want either (2, 1, 4), or (4, 1, 2), but not both.

Again, to re-paraphrase, I would expect 4 triplets with a distinct number in middle [1,2,3] (or [3,2,1), as I consider [1,2,3] the same as [3,2,1]) then [1,3,2] (or [2,3,1]), then [2,1,3] (or [3,1,2]), and [2,4,3] (or [3,4,2]). So, the first and the last number in the triplet should be distinct.

I hope my question makes sense.

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1 Answer 1

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If I understand what you want correctly, the middle index can be any one of the vertices, and the order of the first and third number doesn't matter, with repeats allowed? So [1,1,1] is a valid triplet.

You can guarantee uniqueness by first iterating over all 4 possible middle indices, and then only iterate over the other 2 indices such that index 1 is less than or equal to index 3:

    ly = [1,2,3,4]
    triplets = []
    for j in ly:
        for i in ly:
            for k in ly:
                if i <= k:
                    triplets.append((i,j,k))
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