平面图(new)_46002149

例如右图. 注: 平面图 画法不唯一.
v1 v2 v4 v3 v5 v2 v4
v1 v3 v5
下面是两个 v1 重要的非平面图 v:2 K5和K3,3 v4 1 2 3 v5 f e d b c d v2 v4 a v3 v5 f e
对偶图的性质

1. G*是连通的. 2. G*是平面图. 3. 若G是平面连通图, 则(G*)*=G. 4. m(G*)=m(G), n(G*)=r(G), d(v*)=d(f). 5. 设C是平面图G的一个圈, S*是G*中与C的各边ei对应的 G*的边集合 , 则S*是G*的一个割集. 证明: ∵ C把G的域分成两部分, ∴ E(G*)-S*把G*的点分 成不连通的两部分.
Proposition The dual G* of a plane graph G is itself a planar graph. In fact, there is a natural embedding of G* in the plane. We place each vertex f* in the corresponding face f of G, and then draw each edge e* in such a way that it crosses the corresponding edge e of G exactly once (and crosses no other edge of G). It is intuitively clear that we can always draw the dual as a plane graph in this way, but we do not prove this fact
2.对偶图(偶图)的定义:

给定平面图G=<V,E>, 可以定义另一个图 G*=<V*,E*>,如下: 1. 对于G的每个面f, 都有G*的顶点f*与之对应. 2. 对于G的每条边e, 都有G*的边e*与之对应. 3. G*中的点f*与g*被边e*相连 G中的面f和g被e分隔. 图G*称为G的对偶图.
Euler公式的推广形式
定理：对任意p(p≥1)个连通分支的平面图G，
有 n-m+r=p+1。 推论：对任意平面图G，有 n-m+r≥2。



Corollary 1. Let G be a simple planar graph on at least three vertices. Then m ≤ 3n − 6. Furthermore, m = 3n − 6 if and only if every planar embedding of G is a triangulation. Proof: It clearly suffices to prove the corollary for connected graphs. Let G be a simple connected planar graph with n ≥ 3 Consider any planar embedding G of G. Because G is simple and connected, on at least three vertices, d(f) ≥ 3 for all f ∈ F( G). Therefore, by Euler’s Formula (1)
Proposition
Let G be a planar graph, and let f be a face in some planar embedding of G. Then G admits a planar embedding whose outer face has the same boundary as f.
Proposition
The dual of any plane graph is
connected. Proof Let G be a plane graph and G* a plane dual of G. Consider any two vertices of G*. There is a curve in the plane connecting them which avoids all vertices of G. The sequence of faces and edges of G traversed by this curve corresponds in G* to a walk connecting the two vertices.
n
(i 2)( f
i3
i1
fi 2 ) 0
证明：G的边被分成三类：在C上、C的内部（称内弦）、 C的外部（称外弦） 以m1表示内弦的数目，则C内部的面数为m1 +1
n
故

i3
f i 1 m1 1
因为每条内弦是C内部两个面的边界，而C上的边恰 好是C内部的一个面的边界
n if 2 m n 2 f 1 所以 i1 1 i1 n i3 i3

2=n-m+r≤6-9+4=1 矛盾。
Corollary All planar embeddings of a connected planar graph have the same number of faces.
Proof
极大平面图(maximal planar)

定义: 设G为一个n≥3的简单平面图, 若在G中任意不 相邻的两个顶点之间再加一条边所得图为非平面图, 则称G为极大平面图.

定理1. （1）K5是非平面图.


证明: （1）(反证法) intC1 若有可能, 设G是对应K5的一个 平面图, 用v1, v2, v3, v4, v5表示G的顶点. intC2 intC3 ∵ G是完全图, ∴ vi vjE(G), ij. 因此圈C= v1v2 v3 v1是一条平面Jordon曲 线, 而点v4必然在intC内或extC内. 假设 v4intC, 则v1 v4, v2 v4, v3 v4把intC分成三个区域：int C1, int C2, int C3. v5必然在四个区域extC, int C1, int C2, int C3之一中.若 v5extC, ∵ v4intC, 从Jordan曲线定理知, 边v4v5必然 在某点和C相交. 矛盾. 若v5intCi, 如v5intC1, ∵v3extC1, ∴ 边v5v3必然在某点和C1相交. 矛盾. （2）的证明类似。

defined by π(s) = p if and only if the points z, s, and p are collinear, is called a stereographic projection from z
Theorem

A graph G is embeddable on the plane if and only if it is embeddable on the sphere.
（2）K3,3是非平面图.
二、对偶图(dual)
1. 平面图的面、边界及面的次数 设G是个平面图, 图中边围 成的区域,其内部不含有结点, 也不含有边,称这样区域为G 的一个面(域)(face). 面的边界(boundary):围成一个面f的所有 边构成的回路,称之为这个f面的边界. 此回路中的边 数,称之为面f的度数,记作d(f). 注1：在计算面的度时，割边被计算两次。 注2：每个平面恰有一个无界的面----外部分(outer face). F(G): 平面G中面的集合 . r(G)=|F(G)|----面的个数.

Consider a sphere S resting on a plane P, and denote by z the point that is diametrically opposite the point of contact of S and P. The mapping
π : S \{z} → P,

v3*

v5

v1* F1 F2 v * 2
F3
Note:
isomorphic plane graphs may well have nonisomorphic duals.
Theorem If G is a plane graph, then
定理：设G是n阶平面图，G含有H圈C，以fi1, fi2分别表示在C 的内部和外部的、度为i的面数，则
t (n 2) m t2
推论：Ｋ5是非平面图．

推论2. 若G是简单连通平面图, 则≤5.
证:
当n=1, 2时, 结论显然成立. d ( v ) 2m 由推论1，有 若n≥3,
vV
n d( v ) 2m 6n 12.
∴
vV
≤5.
推论3. K3,3是非平面图.

equivalently, m ≤ 3n − 6 (2) Equality holds in (2) if and only if it holds in (1), that is, if and only if d(f) = 3 for each f ∈ F( G).

推论1’：设平面图G是简单图，且每个面的边界数至少 是t，则有
Proof:
Jordon曲线: 一条连续的, 自身不相交的, 起点 和终点相重合的曲线. 设J是平面上的一条Jordon曲线, 平面的剩下部 分被分成两个不相交的开集, 称为J的内部和外 部, 分别记为intJ和extJ, 并且用IntJ和ExtJ表 示它们的闭包. 显然IntJExtJ=J. Jordon曲线定理(The Jordan Curve Theorem): 连接intJ的点和extJ的点的任何连线必在某点 和J相交.