合1性质与多重集PQ-树：序计数的困难性
Consecutive Ones Property and PQ-Trees for Multisets: Hardness of Counting Their Orderings
摘要：
一个二元矩阵满足连续一性质（COP），如果它的列可以被置换，使得所得矩阵的每行中的一是连续的。等价地，集合F = {Q_1，.，Q_m}满足COP，如果R中的符号可以被排列，使得每个集合Q_i的元素连续出现，作为R的符号排列的连续段。我们认为COP版本的多集，并证明计数其解决方案是困难的（#P-完全）。我们证明完整性的结果也计数的PQ-树，这是通常用于测试的COP集的边界，从而表明，多项式算法是不可能存在的，当处理多集。我们使用的组合方法的基础上简约减少从哈密顿路径问题，表明我们的问题的决策版本，因此是NP完全的。
英文摘要：
A binary matrix satisfies the consecutive ones property (COP) if its columns can be permuted such that the ones in each row of the resulting matrix are consecutive. Equivalently, a family of sets F = {Q_1,..,Q_m}, where Q_i is subset of R for some universe R, satisfies the COP if the symbols in R can be permuted such that the elements of each set Q_i occur consecutively, as a contiguous segment of the permutation of R's symbols. We consider the COP version on multisets and prove that counting its solutions is difficult (#P-complete). We prove completeness results also for counting the frontiers of PQ-trees, which are typically used for testing the COP on sets, thus showing that a polynomial algorithm is unlikely to exist when dealing with multisets. We use a combinatorial approach based on parsimonious reductions from the Hamiltonian path problem, showing that the decisional version of our problems is therefore NP-complete.
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