Problem
In task scheduling, it is common to use a graph representation with a node for each task and a directed edge from task i to task j if i is a precondition for j. This directed graph depicts the precedence constraints in the scheduling problem. Clearly, a schedule is possible if and only if the graph is acyclic; if it isn’t, we’d like to identify the smallest number of constraints that must be dropped so as to make it acyclic.
Given a directed graph G = (V, E), a subset E' ⊆ E is called a feedback arc set if the removal of edges E' renders G acyclic.
FEEDBACK ARC SET (FAS): Given a directed graph G = (V, E) and a budget b, find a feedback arc set of ≤ b edges, if one exists.
(a) Show that FAS is in NP.
FAS can be shown to be NP-complete by a reduction from VERTEX COVER. Given an instance (G, b) of VERTEX COVER, where G is an undirected graph and we want a vertex cover of size ≤ b, we construct a instance (G', b) of FAS as follows. If G = (V, E) has n vertices v1, ..., vn, then make G' = (V', E') a directed graph with 2n vertices w1, w1', ..., wn, wn', and n + 2|E|(directed) edges:
- (wi, wi') for all i = 1, 2, ..., n.
- (wi', wj) and (wj', wi) for every (vi, vj) ∈ E
(b) Show that if G contains a vertex cover of size b, then G' contains a feedback arc set of size b.
(c) Show that if G' contains a feedback arc set of size b, then G contains a vertex cover of size(at most) b. (Hint: given a feedback arc set of size b in G', you may need to first modify it slightly to obtain another one which is of a more convenient form, but is of the same size or smaller. Then, argue that G must contain a vertex cover of the same size as the modified feedback arc set).
Solution
(a) 对于该问题来说, 如果存在一个解, 那么从图G中移除若干条边, 并验证其无环显然是一个需要多项式时间的过程, 因此FAS属于NP.
(b) 首先回顾一下顶点覆盖的概念：
顶点覆盖
对于图G = (V, E), 对于任意点i ∈ V, 我们记Si 为点i所有相连的边的集合.
要求从S1, ..., Sm(m = |V|)中选出b个集合, 使它们的并为E.
这b个集合就被称为图G的一个规模为b的顶点覆盖.
根据题目提示, 设G的一个大小为b的顶点覆盖为C, 对于任意顶点vi ∈ C, 设其在图G' = (V', E')中相对应的顶点为wi和wi'. 初始化一个空的边集E'', 将(wi, wi')加入E''. 对C中的每个顶点都这样处理后, 得到的边集E''就是G’一个规模为b的FAS. 因为对于顶点wi, 去掉边(wi, wi')后, wi就不存在任何出边（出度为0）, 故与其相连的边也不可能位于任何一个环中；而对于顶点wi', 去掉边(wi, wi')后, wi'就不存在任何入边（入度为0）, 故与其相连的边也不可能位于任何环中. 命题得证.
(c) 根据题目提示, 对于图G中的任意一条边(vi, vj), 设它的两个顶点在图G'中对应的顶点为wi, wi', wj, wj', G'中对应的边为(wi, wj'), (wi', wj), (wj, wi'), (wj', wi). 若E''是G'的一个规模为b的FAS, 则这四条边中, 至少有一条边e属于E'', 否则就会构成环（从构成四条边的顶点就可以发现这个事实）. 而边e的两个顶点必然有一个属于{wi, wj}. 设C是一个空集, 如果wi是e的顶点, 则将vi加入C, 否则就将wj加入C. 这样处理完所有图G中的边后, 得到的C即是G的一个规模不超过b的顶点覆盖.
