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京都大学 情報学研究科 数理工学専攻 2018年8月実施 アルゴリズム基礎

Author

祭音Myyura

Description

Let be a connected simple undirected graph with a set of vertices and a set of edges, let be a spanning tree of rooted at a vertex , and let be a numbering on , where we assume that the following conditions (a) and (b) hold.

(a) For each edge , vertex is either an ancestor or a descendant of in .

(b) For each vertex and the parent of in , .

Let denote the set of leaves in . For each vertex , let denote the set of neighbors of in , and let denote the set consisting of vertex and the descendants of in . Define a function

such that

Answer the following questions.

(i) Prove that no leaf is a cut-vertex in .

(ii) Prove that a necessary and sufficient condition for the root to be a cut-vertex in is that has at least two children in .

(iii) Prove that a necessary and sufficient condition for a vertex to be a cut-vertex in is that has a child in such that

Kai

(1)

Let , i.e., is a leaf of .

Sinc is a spanning tree of , is connected.

After removing the vertex from , the remaining graph is still connected and spans all the vertices .

Note that is a subgraph of , hence is connected, which implies that is not a cut-vertex.

(2)

(Necessity) Assume that is a cut-vertex.

If has only one child in , i.e., is a leaf vertex of , then by the same argument of (1), we know that is still connected.

This contradicts the assumption taht is a cut-vertex.

(Sufficiency) Assume that has at least two children in . Let two of them be and .

Consider the two subtrees and . Since and are different children of the root , the vertices in and are not in an ancestor-descendant relationship with each other.

By condition (a), there cannot be any edge between a vertex in and a vertex in .

After deleting , the two subtrees and cannot be connected to each other. Hence is disconnected, which implies that is a cut-vertex.

(3)

(Necessity) Assume that is a cut-vertex.

We prove that there must exist a child of such that by contradiction.

Suppose that every child of satisfies . Then for each child of , there exist some vertex and some neighbor such that .

Since , the vertex is a descendant of . By condition (b), every descendant of has numbering greater than .

Therefore, the vertex of is not in .

By condition (a), since is a descendant of , and y has numbering smaller than , the vertex must be a proper ancestor of .

Thus every child subtree of has an edge to some proper ancestor of , i.e., after deleting , every child subtree can still connect to the part above through an edge to a proper ancestor of .

Therefore, remains connected, which contradicts the assumption that is a cut-vertex.

(Sufficiency) Assume that has a child satisfying .

Note that is a neighbor of , i.e., , when computing , the vertex can be reached from by one graph edge.

Hence , together with the assumption we have

This means that the subtree cannot reach any proper ancestor of through a graph edge. The smallest numbered vertex reachable from is exactly .

Equivalently, has no edge to any proper ancestor of .

After deleting , the subtree is disconnected from the part of the graph above . Therefore is disconnected.

Hence is a cut-vertex.