[论文解读] Linear Sum Assignment with Edition

Sébastien Bougleux, Luc Brun|arXiv (Cornell University)|Mar 14, 2016

Graph Theory and Algorithms参考文献 17被引用 14

一句话总结

本文提出了一种带编辑的线性和分配问题的新公式，将标准分配模型扩展至处理部分分配和ε（ε）元素，通过松弛矩阵表示实现。通过使用满足松弛行和列和约束的(n+1)×(m+1)二值矩阵对ε-分配进行建模，该方法在保持二分图和邻接矩阵等价性的同时，实现了灵活且不完整的匹配。

ABSTRACT

We consider the problem of transforming a set of elements into another by a sequence of elementary edit operations, namely substitutions, removals and insertions of elements. Each possible edit operation is penalized by a non-negative cost and the cost of a transformation is measured by summing the costs of its operations. A solution to this problem consists in defining a transformation having a minimal cost, among all possible transformations. To compute such a solution, the classical approach consists in representing removal and insertion operations by augmenting the two sets so that they get the same size. This allows to express the problem as a linear sum assignment problem (LSAP), which thus finds an optimal bijection (or permutation, perfect matching) between the two augmented sets. While the LSAP is known to be efficiently solvable in polynomial time complexity, for instance with the Hungarian algorithm, useless time and memory are spent to treat the elements which have been added to the initial sets. In this report, we show that the problem can be formalized as an extension of the LSAP which considers only one additional element in each set to represent removal and insertion operations. A solution to the problem is no longer represented as a bijection between the two augmented sets. We show that the considered problem is a binary linear program (BLP) very close to the LSAP. While it can be solved by any BLP solver, we propose an adaptation of the Hungarian algorithm which improves the time and memory complexities previously obtained by the approach based on the LSAP. The importance of the improvement increases as the size of the two sets and their absolute difference increase. Based on the analysis of the problem presented in this report, other classical algorithms can be adapted.

研究动机与目标

• 将线性和分配问题中的部分分配与ε元素形式化。
• 放松标准置换矩阵约束，以允许不完整或缺失的分配。
• 使用(n+1)×(m+1)二值矩阵提供ε-分配的基于矩阵的表示。
• 建立ε-分配与完全二分图Kn+1,m+1的子图之间的等价性。
• 实现对所有元素无需全部匹配的分配问题的高效建模。

提出的方法

• 使用满足松弛行和列和约束的(n+1)×(m+1)二值矩阵X表示ε-分配。
• 在前n行中每行恰好包含一个1（每个u_i最多分配给一个v_j或ε）。
• 在前m列中每列恰好包含一个1（每个v_j最多分配给一个u_i或ε）。
• 固定x_{n+1,m+1} = 1，以表示ε元素至少被分配给一个v_j。
• 使用邻接矩阵X对完全二分图Kn+1,m+1的子图G = ((Uϵ, Vϵ), E)进行建模。
• 确保任何满足约束(15)的矩阵X均可通过(16)中的映射对应到一个有效的ε-分配φ。

实验结果

研究问题

• RQ1 如何将标准线性和分配问题扩展以允许部分分配和ε元素？
• RQ2 在松弛分配框架中，表示ε-分配所需的必要且充分的矩阵约束是什么？
• RQ3 引入一个ε节点如何影响分配矩阵的结构及其二分图表示？
• RQ4 任何满足松弛约束的(n+1)×(m+1)二值矩阵是否都能映射到一个有效的ε-分配？
• RQ5 Kn+1,m+1的子图的邻接矩阵与相应ε-分配之间的关系是什么？

主要发现

• 所有有效ε-分配的集合恰好对应于满足方程(15)中约束的(n+1)×(m+1)二值矩阵的集合。
• X的前n行中每一行恰好包含一个1，确保每个u_i最多被分配给一个v_j或ε。
• X的前m列中每一列恰好包含一个1，确保每个v_j最多被分配给一个u_i或ε。
• X的最后1行和最后1列满足松弛约束：1 ≤ 最后1行的和 ≤ n+1 且 1 ≤ 最后1列的和 ≤ m+1。
• ε始终至少被分配给一个v_j，这是通过x_{n+1,m+1} = 1强制实现的。
• ε-分配与Kn+1,m+1的子图之间存在一一对应关系，其邻接矩阵满足方程(15)。