# Combinations of X elements taken Y at a time with each combination not sharing more than Z element with another

Good evening,

As the title explains, I am looking for an algorithm (I'd implement it in Python) to generate all of the combinations for a given basket of X stocks taken Y at a time (order is not relevant and no repetitions allowed) and not sharing more than Z element between one and another.

I have added the extra restriction because my base case is a universe of 50 stocks and combinations of (i.e basket) 10 stocks, which would generate 10 272 278 170 possible baskets. Since I would have to run further analysis on these baskets, it would be computationally impossible.

As such, I have thought about applying a further restriction which would drastically reduce the number of baskets. In practice, for the same concrete example given above (universe of 50 and baskets of 10) I would like to only generate those baskets of 10, that share a maximum of 5 stocks with any single other basket.

I have tried to come up with a generic algorithm to do so (universe of X, basket of Y and minimum different components of Z) but I haven't managed to come up with a proper solution. Maybe this type of restricted combination does have an actual name in specific literature but my google search have been fruitless.

Hope someone has faced and solved that problem successfully or can come up with a great solution for it!

Thanks!

• Out of curiosity, what are you using this list of combinations for? What are the "stocks" and "baskets"? Feb 12 '21 at 22:12
• They are actually financial stocks, and baskets are structured products on combinations of them (the size of which can vary but in the type of product I'm looking at, the standard is usually 10). The other part of my code analyses some market data to generate buy/sell signals but I needed to run it on a large universe of sufficiently different baskets to spot some interesting opportunities (as I don't know beforehand which ones could prove interesting and the brute force method is computationally impossible as I mentioned)! Feb 15 '21 at 8:51
• Interesting! Well, in case it is of interest, I found a larger system of 3333 baskets with no two having five or more in common. It also samples all of the 50 stocks roughly evenly, which each stock appearing in 500 to 800 baskets. pastebin.com/0VpzYiMc Feb 16 '21 at 14:22
• Great, thanks Mike! That will allow me to run it on a much wider scope and potentially spot more interesting combination, much appreciated! Out of curiosity did you come up with those yourself or was it extracted from some paper ? It's quite the interesting branch of maths but I'm afraid I am very limited when it comes to this topic. Feb 17 '21 at 11:39
• I found it myself. I started $45$ baskets, given by breaking the 50 stocks into 10 groups and picking all ways to choose two of those groups. Then, I looked at all ways to selected $5$ of those groups and take two stocks from each selected group to make a basket of $10$. I looked at these one of a time by brute force, and added them if they created no conflict with previously added baskets. Furthermore, is a way to extend my collection by $3125$ more (nearly doubling it!) by using a "quadratic residue code." Feb 17 '21 at 20:22

What you are looking for in the literature is known as a constant weight binary code (Wikipedia). Specifically, if $$A(n,d,w)$$ is the maximum number of binary vectors, each with $$w$$ ones and $$n-w$$ zeroes, such that the Hamming distance between any two codewords is $$d$$, then the largest possible size of your set of sets would be $$A(X,2(Y-Z) ,Y)$$. In the special case you mentioned, you want $$A(50,10,10)$$.

Not much is known about the exact optimal values of $$A(n,d,w)$$. The main ways to find constant weight binary codes usually involve a lot of computational effort. This table lists the values of $$A(n,10,w)$$ for $$n\ge 32$$ and $$w\le 16$$, but the highest relevant value for your problem is $$A(32,10,10)\ge 500$$. This means that even when you restrict yourself to only using the first $$32$$ elements of your $$50$$ element set, you can get $$500$$ sets. If you click on that link, they actually provide the list of codewords.

The question you need to consider is; since there is little hope of finding the maximal collection of subsets of a size of $$50$$, each with size $$10$$ and at most $$5$$ elements in common, how many subsets do you realistically need? Is $$500$$ enough?

Solution for $$315$$ subsets of $$\{1,\dots,29\}$$ of size $$5$$ where each subset has at most two in common with each other:

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• Thank you Mike for the answer. I have been reading about it and it's much more complex than I first assumed it was. That being said, the highest relevant value A(32,10,10) that you point out and its lower bound would already suffice for my actual application. I am however unclear as to how I could leverage the explicit list of 500 codewords provided and implement it in my specific case. Would you mind further explaining that to me ? Thank you very much anyway for the quick answer! Feb 11 '21 at 22:34
• Happy to help! You need to convert each codeword from hexadecimal to binary. Each result should be a binary vector with 32 bits and exactly 10 ones, which naturally translates into subsets of $\{1,2,\dots,32\}$. Feb 11 '21 at 23:13
• Great Mike, with that I'll be able to implement that case! I was also contemplating looking at a basket of 8 elements with at most 4 elements in common, which would be A(X,8,8) as far as I understand. I am not sure how to process the solution in the link though. Same question for a basket of 5 elements with at most 2 elements in common (X,6,5) and the solution in the link Feb 12 '21 at 8:32
• For the A(28,8,8) file, the codewords all smushed together instead of on separate lines. So you first need to split that long word into sections of length 7, then convert each hex block to binary to get words of length 28. Feb 12 '21 at 14:26
• For the second file, the first several rows represent permutations. Those permutations generate a subgroup, and you need to apply all the permutations in that subgroup to the short list of codewords provided to get the entire list. At least, I am pretty sure... Feb 12 '21 at 15:02