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Choosing a tree \(t_2\) that's a line (that is, \(N\) nodes connected as a chain\) always minimizes \(\text{similarity}(S(t_1), S(t_2))\).
Proof: Let \(t_\text{line}\) be a line of size-\(N\), and \(t_x\) be any tree of size \(N\). We'll show for any \(t_1\) and any \(t_x\), \(\text{similarity}(S(t_1), S(t_\text{line})) \le \text{similarity}(S(t_1), S(t_x))\)\). Let’s also not consider subtrees of size \(1\) (because all trees of size \(n\) have \(n\) subtrees of size \(1\)). There are two cases:
*Case 1*: Consider \(t_x\) such that \(\text{diameter}(t_x) \le \text{diameter}(t_1)\). We can choose any two nodes from \(t_x\) and get the path between them. This will be a subtree of \(t_1\). However, \(\text{similarity}(S(t_1), S(t_\text{line}))\) is at most \(\binom{n}{2}\), because \(S(t_\text{line})\) itself has only \(\binom{n}{2}\) elements.
*Case 2*: Consider \(t_x\) such that \(\text{diameter}(t_x) > \text{diameter}(t_1)\). For any pair of nodes \((u, v) \in t_x\), either the path from \(u\) to \(v\) is a subtree of \(t_1\), or there is a corresponding pair in \(t_\text{line}\) that is not a subtree of \(t_1\).
Therefore, \(\text{similarity}(S(t_1), S(t_\text{line})) \le \text{similarity}(S(t_1), S(t_x))\). To compute the former, simply calculate the diameter of \(t_1\) using two DFS's and use combinatorics. In particular, each length \(i\) line (of \(i + 1\) nodes) appears in \(S(t_\text{line})\) exactly \(N - i\) times, and all of these sub-lines are isomorphic to a subtree of \(t_1\). The answer will be:
\[ \sum_{i=0}^{\text{diameter}(t_\text{line})} N - i \]
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