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This result is stated in: Kocsis, L., & Szepesvári, C. (2006, September). Bandit based monte-carlo planning. In European conference on machine learning (pp. 282-293). Springer, Berlin, Heidelberg. It is the content of Theorems 5 & 6. Sadly there is no detailed proof for theorem 5. The main reference for this is: Auer, P., Cesa-Bianchi, N., & ...


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I'm assuming you have parent pointers, you can probably avoid them by maintaining a couple stacks though. Find the extremal two nodes $\ell \leq p \leq q \leq u$ and their common ancestor $a$ in $O(h)$ time. If nodes $p$ and $q$ are the same, return $S = \{p\}$. While $p$ is the left child of its parent and its parent is not the common ancestor, let $p := ...


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Yes, this is sort of common subexpression elimination. Simple approach for binary trees: introduce nodes 1 for a ... 6 for f. Traverse your trees. When you have new pair like (1, 2) or (4, 5) introduce for them new nodes like 7 and 8, when you hit old pair, add edge from it. You will need unordered map from pair of nodes to node, representing this pair: (...


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