电大离散数学形成性考核作业5答案图论部分.doc

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电大离散数学作业5

离散数学图论部分形成性考核书面作业
本课程形成性考核书面作业共3次，内容主要分别是集合论部分、图论部分、数理逻辑部分的综合练习，基本上是按照考试的题型（除单项选择题外）安排练习题目，目的是通过综合性书面作业，使同学自己检验学习成果，找出掌握的薄弱知识点，重点复习，争取尽快掌握。本次形考书面作业是第二次作业，大家要认真及时地完成图论部分的综合练习作业。
要求：将此作业用A4纸打印出来，手工书写答题，字迹工整，解答题要有解答过程，要求2010年12月5日前完成并上交任课教师（不收电子稿）。并在05任务界面下方点击“保存”和“交卷”按钮，以便教师评分。

一、填空题
1．已知图G中有1个1度结点，2个2度结点，3个3度结点，4个4度结点，则G的边数是 15 ．
2．设给定图G(如右由图所示)，则图G的点割集是
{f} ．
3．设G是一个图，结点集合为V，边集合为E，则
G的结点 度数之和 等于边数的两倍．
4．无向图G存在欧拉回路，当且仅当G连通且 等于出度 ．
5．设G=<V，E>是具有n个结点的简单图，若在G中每一对结点度数之和大于等于 n-1 ，则在G中存在一条汉密尔顿路．
6．若图G=<V, E>中具有一条汉密尔顿回路，则对于结点集V的每个非空子集S，在G中删除S中的所有结点得到的连通分支数为W，则S中结点数|S|与W满足的关系式为 W(G-V1) £½V1½ ．
7．设完全图K有n个结点(n³2)，m条边，当 n为奇数 时，K中存在欧拉回路．
8．结点数v与边数e满足 e=v-1 关系的无向连通图就是树．
9．设图G是有6个结点的连通图，结点的总度数为18，则可从G中删去
4 条边后使之变成树．
10．设正则5叉树的树叶数为17，则分支数为i = 5 ．

二、判断说明题（判断下列各题，并说明理由．）
1．如果图G是无向图，且其结点度数均为偶数，则图G存在一条欧拉回路．．
(1) 不正确，缺了一个条件，图G应该是连通图，可以找出一个反例，比如图G是一个有孤立结点的图。

2．如下图所示的图G存在一条欧拉回路．

(2) 不正确，图中有奇数度结点，所以不存在是欧拉回路。

3．如下图所示的图G不是欧拉图而是汉密尔顿图．

G

解：正确
因为图中结点a，b，d，f的度数都为奇数，所以不是欧拉图。
如果我们沿着(a,d,g,f,e,b,c,a)，这样除起点和终点是a外，我们经过每个点一次仅一次，所以存在一条汉密尔顿回路，是汉密尔顿图

4．设G是一个有7个结点16条边的连通图，则G为平面图．
解：(1) 错误
假设图G是连通的平面图，根据定理，结点数v，边数为e，应满足e小于等于3v-6，但现在16小于等于3*7-6，显示不成立。所以假设错误。

5．设G是一个连通平面图，且有6个结点11条边，则G有7个面．
(2) 正确
根据欧拉定理，有v-e+r=2，边数v=11，结点数e=6，代入公式求出面数r=7

三、计算题
1．设G=<V，E>，V={ v1，v2，v3，v4，v5}，E={ (v1,v3)，(v2,v3)，(v2,v4)，(v3,v4)，(v3,v5)，(v4,v5) }，试
(1) 给出G的图形表示； (2) 写出其邻接矩阵；
(3) 求出每个结点的度数； (4) 画出其补图的图形．
解：(1)
o
o
o
o
v1
o
v5
v2
v3
v4

(2) 邻接矩阵为

(3) v1结点度数为1，v2结点度数为2，v3结点度数为3，v4结点度数为2，v5结点度数为2

(4) 补图图形为
o
o
o
o
v1
o
v5
v2
v3
v4

2．图G=<V, E>，其中V={ a, b, c, d, e}，E={ (a, b), (a, c), (a, e), (b, d), (b, e), (c, e), (c, d), (d, e) }，对应边的权值依次为2、1、2、3、6、1、4及5，试
（1）画出G的图形； （2）写出G的邻接矩阵；
（3）求出G权最小的生成树及其权值．
（1）G的图形如下：

（2）写出G的邻接矩阵

（3）G权最小的生成树及其权值

3．已知带权图G如右图所示．
(1) 求图G的最小生成树； (2)计算该生成树的权值．

解：(1) 最小生成树为
1
2
3
5
7

(2) 该生成树的权值为(1+2+3+5+7)=18

4．设有一组权为2, 3, 5, 7, 17, 31，试画出相应的最优二叉树，计算该最优二叉树的权．
3
5
2
5
10
7
17
31
17
34
65

权为 2*5+3*5+5*4+7*3+17*2+31=131

四、证明题
1．设G是一个n阶无向简单图，n是大于等于3的奇数．证明图G与它的补图中的奇数度顶点个数相等．
证明：设，．则是由n阶无向完全图的边删去E所得到的．所以对于任意结点，u在G和中的度数之和等于u在中的度数．由于n是大于等于3的奇数，从而的每个结点都是偶数度的（度），于是若在G中是奇数度结点，则它在中也是奇数度结点．故图G与它的补图中的奇数度结点个数相等．

2．设连通图G有k个奇数度的结点，证明在图G中至少要添加条边才能使其成为欧拉图．
证明：由定理3.1.2，任何图中度数为奇数的结点必是偶数，可知k是偶数．
又根据定理4.1.1的推论，图G是欧拉图的充分必要条件是图G不含奇数度结点．因此只要在每对奇数度结点之间各加一条边，使图G的所有结点的度数变为偶数，成为欧拉图．
故最少要加条边到图G才能使其成为欧拉图．

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