论文标题
计数置换中的4个模式等同于计数图中的4循环
Counting 4-Patterns in Permutations Is Equivalent to Counting 4-Cycles in Graphs
论文作者
论文摘要
如果存在$π$的$π$，则排列$σ$出现在排列$π$中。自然的问题是检查$σ$是否出现在$π$中，如果是这样，则计算出事件的数量。我们知道，对于任何固定的长度〜k，我们可以检查给定的长度k是否出现在n中的长度为n的置换中，但是能够计算$ f（k）\ cdot n^{o（k/\ log k）$ time中的所有此类事件都会反驳指出时间的时间假设（ETH）。这激发了对固定小长度k的不同模式的计数发生复杂性的系统研究。我们为k = 4调查了这个问题。最近，Even-Zohar和Leng [Arxiv 2019]确定了两种类型的4个模式。对于第一种类型，他们设计了$é（n）$ time算法，而第二算法则可以提供$é（n^{1.5}）$ time算法。这提出了一个问题，第二类的排列是否本质上比第一类更难。 我们在计数第二类的4个pattern和在稀疏的无向图中计数4循环之间建立了联系。通过设计双向减少，我们表明这两个问题的复杂性都是相同的，这取决于多毛体因子。这使我们能够为为什么要计算两种类型的4个模式的复杂性存在差异。特别是，即使对于在M边缘上图中检测4周期的更简单问题，最著名的算法在$ O（m^{4/3}）$时间中起作用。我们的降低意味着一个$ o（n^{4/3- \ varepsilon}）$ time算法用于计数出现，这意味着要计数（因此也检测到）4个循环的兴奋突破。在另一个方向上，通过插入用于计数4循环的最快已知算法，我们获得了一种用于计数$ O（n^{1.48}）$时间的任何4个模式的算法。
Permutation $σ$ appears in permutation $π$ if there exists a subsequence of $π$ that is order-isomorphic to $σ$. The natural question is to check if $σ$ appears in $π$, and if so count the number of occurrences. We know that for any fixed length~k, we can check if a given pattern of length k appears in a permutation of length n in time linear in n, but being able to count all such occurrences in $f(k)\cdot n^{o(k/\log k)}$ time would refute the exponential time hypothesis (ETH). This motivates a systematic study of the complexity of counting occurrences for different patterns of fixed small length k. We investigate this question for k=4. Very recently, Even-Zohar and Leng [arXiv 2019] identified two types of 4-patterns. For the first type they designed an $Õ(n)$ time algorithm, while for the second they were able to provide an $Õ(n^{1.5})$ time algorithm. This brings up the question whether the permutations of the second type are inherently harder than the first type. We establish a connection between counting 4-patterns of the second type and counting 4-cycles in a sparse undirected graph. By designing two-way reductions we show that the complexities of both problems are the same, up to polylogarithmic factors. This allows us to provide a reasonable argument for why there is a difference in the complexities for counting 4-patterns of the two types. In particular, even for the simpler problem of detecting a 4-cycle in a graph on m edges, the best known algorithm works in $O(m^{4/3})$ time. Our reductions imply that an $O(n^{4/3-\varepsilon})$ time algorithm for counting occurrences would imply an exciting breakthrough for counting (and hence also detecting) 4-cycles. In the other direction, by plugging in the fastest known algorithm for counting 4-cycles, we obtain an algorithm for counting occurrences of any 4-pattern in $O(n^{1.48})$ time.
