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arXiv 2104.08501cs.DS
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基于采样和快速矩阵乘法的近似三角形计数

Approximate Triangle Counting via Sampling and Fast Matrix Multiplication

Jakub Tětek

摘要:

有一个简单的$O(\frac{n^3}{T})$时间算法用于近似三角形计数,其中$T$是图中三角形的数量,$n$是顶点的数量。同时,可以使用时间上的快速矩阵乘法$\tilde{O}(n^ω)$来精确地计数三角形。是否有可能在保持$n^ω$对$n$的依赖性的同时,获得对三角形$T$数量的负依赖性?我们积极回答这个问题,提供了一个算法,运行在时间$O\big(\frac{n^ω}{T^{ω- 2}}\big) \cdot \text{poly}(n^{o(1)}/ε)$。这是最佳的,因为只要$T$的指数独立于$n, T$,它就不能在保持对$n$的依赖性的同时得到改善;这是根据Eden和Rosenbaum的下限得出的[APPROX/RANDOM 2018]。我们的算法改进了现有技术,当$T = ω(1)$和$T = o(n)$。我们还考虑了稀疏图中近似三角形计数的问题,参数化的边缘$m$的数量。最著名的算法在时间上运行$\tilde{O}\big(\frac{m^{3/2}}{T}\big)$ [Eden等人,SIAM Journal on Computing,2017]。还有一个众所周知的算法,用于精确的三角形计数,它在时间上运行$\tilde{O}(m^{2ω/(ω+ 1)})$。我们再次得到了一个算法,保留了$m$的指数,同时在具有大量三角形的图上运行得更快。具体来说,我们的算法在时间上运行$O\Big(\frac{m^{2ω/(ω+1)}}{ T^{2(ω-1)/(ω+1)}}\Big) \cdot \text{poly}(n^{o(1)}/ε)$。这在以下意义上再次是最佳的:如果$T$的指数是恒定的,则在不恶化对$m$的依赖性的情况下不能对其进行改进。当$T = ω(1)$和$T = o(\sqrt{m})$。

英文摘要:

There is a trivial $O(\frac{n^3}{T})$ time algorithm for approximate triangle counting where $T$ is the number of triangles in the graph and $n$ the number of vertices. At the same time, one may count triangles exactly using fast matrix multiplication in time $\tilde{O}(n^ω)$. Is it possible to get a negative dependency on the number of triangles $T$ while retaining the $n^ω$ dependency on $n$? We answer this question positively by providing an algorithm which runs in time $O\big(\frac{n^ω}{T^{ω- 2}}\big) \cdot \text{poly}(n^{o(1)}/ε)$. This is optimal in the sense that as long as the exponent of $T$ is independent of $n, T$, it cannot be improved while retaining the dependency on $n$; this as follows from the lower bound of Eden and Rosenbaum [APPROX/RANDOM 2018]. Our algorithm improves upon the state of the art when $T = ω(1)$ and $T = o(n)$. We also consider the problem of approximate triangle counting in sparse graphs, parameterizing by the number of edges $m$. The best known algorithm runs in time $\tilde{O}\big(\frac{m^{3/2}}{T}\big)$ [Eden et al., SIAM Journal on Computing, 2017]. There is also a well known algorithm for exact triangle counting that runs in time $\tilde{O}(m^{2ω/(ω+ 1)})$. We again get an algorithm that retains the exponent of $m$ while running faster on graphs with larger number of triangles. Specifically, our algorithm runs in time $O\Big(\frac{m^{2ω/(ω+1)}}{ T^{2(ω-1)/(ω+1)}}\Big) \cdot \text{poly}(n^{o(1)}/ε)$. This is again optimal in the sense that if the exponent of $T$ is to be constant, it cannot be improved without worsening the dependency on $m$. This algorithm improves upon the state of the art when $T = ω(1)$ and $T = o(\sqrt{m})$.

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