Tricky algorithm for sorting symbols in an array while preserving relationships via order

Question

The problem

I have multiple groups which specify the relationships of symbols.. for example:

[A B C]

[A D E]

[X Y Z]

What these groups mean is that (for the first group) the symbols, A, B, and C are related to each other. (The second group) The symbols A, D, E are related to each other.. and so forth.

Given all these data, I would need to put all the unique symbols into a 1-dimension array wherein the symbols which are somehow related to each other would be placed closer to each other. Given the example above, the result should be something like:

[B C A D E X Y Z]

or

[X Y Z D E A B C]

In this resulting array, since the symbol A has multiple relationships (namely with B and C in one group and with D and E in another) it's now located between those symbols, somewhat preserving the relationship.

Note that the order is not important. In the result, X Y Z can be placed first or last since those symbols are not related to any other symbols. However, the closeness of the related symbols is what's important.

What I need help in

I need help in determining an algorithm that takes groups of symbol relationships, then outputs the 1-dimension array using the logic above. I'm pulling my hair out on how to do this since with real data, the number of symbols in a relationship group can vary, there is also no limit to the number of relationship groups and a symbol can have relationships with any other symbol.

Further example

To further illustrate the trickiness of my dilemma, IF you add another relationship group to the example above. Let's say:

[C Z]

The result now should be something like:

[X Y Z C B A D E]

Notice that the symbols Z and C are now closer together since their relationship was reinforced by the additional data. All previous relationships are still retained in the result also.

Answer 1

The first thing you need to do is to precisely define the result you want.

You do this by defining how good a result is, so that you know which is the best one. Mathematically you do this by a cost function. In this case one would typically choose the sum of the distances between related elements, the sum of the squares of these distances, or the maximal distance. Then a list with a small value of the cost function is the desired result.

It is not clear whether in this case it is feasible to compute the best solution by some special method (maybe if you choose the maximal distance or the sum of the distances as the cost function).

In any case it should be easy to find a good approximation by standard methods.

A simple greedy approach would be to insert each element in the position where the resulting cost function for the whole list is minimal.

Once you have a good starting point you can try to improve it further by modifying the list towards better solutions, for example by swapping elements or rotating parts of the list (local search, hill climbing, simulated annealing, other).

Answer 2

I think, because with large amounts of data and lack of additional criteria, it's going to be very very difficult to make something that finds the best option. Have you considered doing a greedy algorithm (construct your solution incrementally in a way that gives you something close to the ideal solution)? Here's my idea:

Sort your sets of related symbols by size, and start with the largest one. Keep those all together, because without any other criteria, we might as well say their proximity is the most important since it's the biggest set. Consider every symbol in that first set an "endpoint", an endpoint being a symbol you can rearrange and put at either end of your array without damaging your proximity rule (everything in the first set is an endpoint initially because they can be rearranged in any way). Then go through your list and as soon as one set has one or more symbols in common with the first set, connect them appropriately. The symbols that you connected to each other are no longer considered endpoints, but everything else still is. Even if a bigger set only has one symbol in common, I'm going to guess that's better than smaller sets with more symbols in common, because this way, at least the bigger set stays together as opposed to possibly being split up if it was put in the array later than smaller sets.

I would go on like this, updating the list of endpoints that existed so that you could continue making matches as you went through your set. I would keep track of if I stopped making matches, and in that case, I'd just go to the top of the list and just tack on the next biggest, unmatched set (doesn't matter if there are no more matches to be made, so go with the most valuable/biggest association). Ditch the old endpoints, since they have no matches, and then all the symbols of the set you just tacked on are the new endpoints.

This may not have a good enough runtime, I'm not sure. But hopefully it gives you some ideas.

Edit: Obviously, as part of the algorithm, ditch duplicates (trivial).