We show that any one-round algorithm that computes a minimum spanning tree (MST) in the unicast congested clique must use a link bandwidth of $\Omega(\log^3 n)$ bits in the worst case. Consequently, computing an MST under the standard assumption of $O(\log n)$-size messages requires at least $2$ rounds. This is the first round complexity lower bound in the unicast congested clique for a problem where the output size is small, i.e., $O(n\log n)$ bits. Our lower bound holds as long as every edge of the MST is output by an incident node. To the best of our knowledge, all prior lower bounds for the unicast congested clique either considered problems with large output sizes (e.g., triangle enumeration) or required every node to learn the entire output.

翻译：我们证明，在单播拥塞团簇中，任何计算最小生成树（MST）的单轮算法在最坏情况下必须使用 Ω(log³ n) 比特的链路带宽。因此，在标准假设的 O(log n) 大小消息下，计算 MST 至少需要 2 轮。这是针对输出规模较小（即 O(n log n) 比特）的问题，在单播拥塞团簇中得出的首个轮复杂度下界。该下界成立的前提是 MST 的每条边都由其关联节点输出。据我们所知，先前所有针对单播拥塞团簇的下界要么考虑输出规模较大的问题（如三角形枚举），要么要求每个节点学习整个输出。