Weighted variants of triangle detection are an important object of study because of their prominence in fine-grained complexity. We revisit the Node-Weighted Triangle problem, where the goal is to decide if a vertex-weighted graph contains a triangle whose node weights sum to zero. This problem has been the focus of a celebrated line of work, beginning with a subcubic-time algorithm [Vassilevska, Williams; STOC '06], and culminating in algorithms running almost in matrix multiplication time, $O(\textsf{MM}(n) + n^2\cdot 2^{O(\sqrt{\log n})})$ [Czumaj, Lingas; SODA '07], [Vassilevska W., Williams; STOC '09]. This runtime is almost-optimal, since even detecting an unweighted triangle is conjectured to require matrix multiplication time $\textsf{MM}(n)$. However, the superpolylogarithmic $2^{Ω(\sqrt{\log n})}$ overhead persists in a world where near-optimal matrix multiplication is possible (i.e., $\textsf{MM}(n) \leq n^2\text{poly}(\log n)$). In this paper, we present a new algorithm solving Node-Weighted Triangle in $O(\textsf{MM}(n))$ time, closing the gap to unweighted triangle detection completely. Remarkably, our algorithm is much simpler than previous approaches, which use involved recursion schemes and communication protocols.

翻译：三角形检测的加权变体因其在细粒度复杂性中的突出地位而成为重要的研究对象。我们重新审视了节点加权三角形问题，其目标是判断一个顶点加权图是否包含一个节点权值和为零的三角形。这一问题曾是一系列著名工作的焦点，始于一个次立方时间算法 [Vassilevska, Williams; STOC '06]，最终发展到几乎达到矩阵乘法时间的算法，即 $O(\textsf{MM}(n) + n^2\cdot 2^{O(\sqrt{\log n})})$ [Czumaj, Lingas; SODA '07], [Vassilevska W., Williams; STOC '09]。该运行时间几乎是最优的，因为即使检测无权重三角形也被推测需要矩阵乘法时间 $\textsf{MM}(n)$。然而，在可以实现近最优矩阵乘法（即 $\textsf{MM}(n) \leq n^2\text{poly}(\log n)$）的世界中，超多对数级的 $2^{Ω(\sqrt{\log n)}}$ 开销仍然存在。本文提出了一种新算法，可在 $O(\textsf{MM}(n))$ 时间内解决节点加权三角形问题，完全消除了与无权重三角形检测之间的差距。值得注意的是，我们的算法比之前采用复杂递归方案和通信协议的的方法简单得多。