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Suppose I had a tree, where each edge had a probability assigned to it, and the probabilities all of the edges coming from a node sum up to 1. You could consider these probabilities to come a machine learning model that's trying to recommend items in a hierarchy (say for an online liquor and snacks shop).

One way to choose what to show a customer would be to at each node, traverse the edge with the highest probability, and end when you're at a leaf node. In the example below, that would suggest you recommend the customer "chips."

Example tree

My question is, given such a hierarchy, how would you determine what the 2nd or 3rd best choices are? As in, if you wanted to show the customer 3 items, what would you show? I'm not sure how to formulate that mathematically, and if there is some graph-theory concept that covers this. I'm also not sure this is well defined.

Thank you!

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  • $\begingroup$ I think that you are right, that the problem is that you haven't defined what is meant by "second best choice" (or "$n$th best choice" in general). It may be that once you've defined that, the algorithm for finding it is trivial, or it could be something that requires some thought. It's hard to say in general. $\endgroup$ – Brian Tung Apr 23 at 21:54
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    $\begingroup$ I'm not convinced that your "best choice" algorithm is correct in general. In this tree, if "chips" splits further into 100 brands of chips each with 0.01 probability, then it may be best to recommend "wine" (with probability 0.2) rather than any brand of chips (with probability 0.0063). $\endgroup$ – Misha Lavrov Apr 24 at 5:06
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All items can be identified with nodes with out-degree equal to $0$ in your model. Let us call these nodes item-nodes $i_j$. For each item-node $i_j$, there is a unique path $$(r,...,i_j)$$ from the root node $r$ to $i_j$. By "walking" along such a path and consecutively multyplying the probabilities associated with the path egdes, one can assign a probability $p_{i_j}$to each item-node $i_j$. For example, in the case of "chips", this probability would be $0.7*0.9 = 0.63$. The probabilities $p_{i_j}$ can then be used to rank-order the item-nodes $i_j$.

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  • $\begingroup$ Note that, using this ranking, the most probable item is NOT necessarily the one you get by traversing the most-probable edge each time (which is the OP's example) $\endgroup$ – antkam Apr 24 at 14:55
  • $\begingroup$ As @antkam mentioned, I would ideally not like to adjust the definition of "best" which is already set. $\endgroup$ – Jacqueline Nolis Apr 24 at 22:03
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The solution I ended up going with, and I'm still not 100% sure this is right, is to after selecting the "best" option, going back and removing weight from the edges that were traversed, then deleting the selected option.

In the picture above, I changed the edge from the root to snacks to be (0.7 - 0.9*0.7) to represent the loss of choosing chips. Then I repeated the process to choose the next best option, which in the next case would be cookies (well it's a tie but I just picked the top one).

By repeatedly lowering the weights of the edges used to pick a solution, then removing the solution, I am able to have a set of items picked in a similar way. Again I'm not sure this is the best option!

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