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I was studying disjoint set data structure. I understand how rank helps us make shallow trees and path compression decrease tree height.

I can find the below code in most articles or study materials for path compression.

int find(subset[] subsets, int I) {
    if (subsets[i].parent != I) {
        subsets[i].parent= find(subsets, subsets[i].parent);
        return subsets[i].parent;
    }
}

I am wondering what about ranks. When we will do path compression the rank for the root also change but we did not update that. Could you please explain to me if I am missing something?

I have checked with the online tool so it looks like if we don't update the rank then it would not work as expected. I think it's more about the approximation of path compression. I am thinking in the worst-case scenario it can possible to create a dense tree.

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Vimit Dhawan
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    The wikipedia page https://en.wikipedia.org/wiki/Disjoint-set_data_structure gives an overview of the code and a proof of efficiency of what you seem to be asking about. – Paul Hankin Nov 03 '22 at 08:31
  • Thanks @PaulHankin It's difficult to read a Wikipedia page but let me read for this one – Vimit Dhawan Nov 03 '22 at 08:35

1 Answers1

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No, you don't update the ranks in path compression. In fact the reason that the word "rank" is used instead of "height", is because the rank doesn't accurately reflect the height, because it isn't updated in path compression.

Rank is a worst-case height that is accurate enough to provide the promised complexities. Whenever I write a union-find structure, though, I use the size of the subtree instead of rank. It works just as well and is also useful for other things.

Matt Timmermans
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