It is known that any two trees on the same $n$ leaves can be displayed by a network with $n-2$ reticulations, and there are two trees that cannot be displayed by a network with fewer reticulations. But how many reticulations are needed to display multiple trees? For any set of $t$ trees on $n$ leaves, there is a trivial network with $(t - 1)n$ reticulations that displays them. To do better, we have to exploit common structure of the trees to embed non-trivial subtrees of different trees into the same part of the network. In this paper, we show that for $t \in o(\sqrt{\lg n})$, there is a set of $t$ trees with virtually no common structure that could be exploited. More precisely, we show for any $t\in o(\sqrt{\lg n})$, there are $t$ trees such that any network displaying them has $(t-1)n - o(n)$ reticulations. For $t \in o(\lg n)$, we obtain a slightly weaker bound. We also prove that already for $t = c\lg n$, for any constant $c > 0$, there is a set of $t$ trees that cannot be displayed by a network with $o(n \lg n)$ reticulations, matching up to constant factors the known upper bound of $O(n \lg n)$ reticulations sufficient to display \emph{all} trees with $n$ leaves. These results are based on simple counting arguments and extend to unrooted networks and trees.

翻译：已知任意两棵具有相同 $n$ 个叶子的树可通过一个具有 $n-2$ 个网状结构的网络来展示，且存在两棵树无法用更少网状结构的网络展示。但展示多棵树需要多少网状结构？对于任意 $t$ 棵具有 $n$ 个叶子的树集合，存在一个具有 $(t - 1)n$ 个网状结构的平凡网络来展示它们。为改进此结果，必须利用树之间的共同结构，将不同树的非平凡子树嵌入网络的同一部分。本文证明，当 $t \in o(\sqrt{\lg n})$ 时，存在一个 $t$ 棵树的集合，其几乎不存在可被利用的共同结构。更精确地说，我们证明对于任意 $t\in o(\sqrt{\lg n})$，存在 $t$ 棵树，使得任何展示它们的网络均需 $(t-1)n - o(n)$ 个网状结构。对于 $t \in o(\lg n)$，我们得到一个稍弱的界。我们还证明，即使对于 $t = c\lg n$（其中 $c > 0$ 为任意常数），也存在一个 $t$ 棵树的集合，无法用 $o(n \lg n)$ 个网状结构的网络展示，这与已知展示所有具有 $n$ 个叶子的树所需 $O(n \lg n)$ 个网状结构的上界在常数因子内匹配。这些结果基于简单的计数论证，并可推广至无根网络与无根树。