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哈密顿图同构的证明
姜咏江
一个图是不是著名的哈密顿图?采用图的01表示方法可以做到。将当前顶点用1表示,与其相邻的顶点用0表示,并将它们放在一张表中,这就是图的01表示法。
1. 证明过程要证明下面的图1和图2同构,只要将其中的一个图的0和1表达的表列出,并复制一个,然后清除表头的标注。表1是哈密顿图,表2是对图2的操作表。我们在表2中用排列图2顶点的方法来证明图2与图1同构。
表1 哈密顿图
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | x14 | 15 | 16 | 17 | 18 | 19 | 20 |
1 | 0 |
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| 0 |
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| 0 |
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0 | 1 | 0 |
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| 0 |
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| 0 | 1 | 0 |
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| 0 |
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| 0 | 1 | 0 |
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| 0 |
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0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 |
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| 0 | 1 | 0 |
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| 0 |
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0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 |
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| 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 |
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| 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 |
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| 0 |
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| 0 | 1 | 0 |
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| 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 |
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| 0 | 1 | 0 |
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| 0 |
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| 0 | 1 | 0 |
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| 0 |
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| 0 | 1 | 0 |
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| 0 |
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| 0 | 1 |
操作:
(1)在图2中选择一个圈。这里选的顶点是qrihg。
(2)按照q的邻点,添写p。
(3)按照r的邻点,添写s。
(4)按照i的邻点,添写j。
(5)按照h的邻点,添写a。
(6)按照g的邻点,添写f。
表2 选择圈排列
q | r | i | h | g | f |
| p |
| s |
| j |
| a |
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1 | 0 |
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| 0 |
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| 0 |
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0 | 1 | 0 |
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| 0 |
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| 0 | 1 | 0 |
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| 0 |
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| 0 | 1 | 0 |
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| 0 |
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0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 | 1 |
表2中有黄色底纹的中间位置应是图2边上两点关联的顶点。例如,f、p共同关联的顶点是o,这叫添写共同关联顶点。
(6)添写已经确定顶点的共同关联顶点(见表3的黄色位置)。
(7)依据图2顶点的相邻关系,添写剩余的顶点位置(见表4的绿色位置)。
表3 选择关联顶点
q | r | i | h | g | f | o | p | t | s | k | j | b | a |
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1 | 0 |
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0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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| 0 | 1 | 0 |
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表4 图2的0和1表示
q | r | i | h | g | f | o | p | t | s | k | j | b | a | e | d | n | m | l | c |
1 | 0 |
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0 | 1 | 0 |
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| 0 | 1 |
当所有的顶点在表中添写完成之后,需要认真核对各顶点之间的相邻关系是否与图2一致。若完全一致,则证明图1与图2同构。图1与图2的顶点与顶点,边与边的一一对应关系,可由表1和表4立即得到。
2. 总结本例的原哈密顿图同构才用先找“圈”来确定一些相互关联的顶点位置,然后再依据多点共同关联的顶点,再确定那个顶点的位置,逐步扩充安排顶点的位置,可一次性得到结果。
在用图的01表示法中,可能产生顶点的不同排列顺序,但这并不影响对图同构的证明。产生这种情况的原因,是因为图定义本身没有图形位置的概念。
2016-04-08
参考:http://blog.sciencenet.cn/blog-340399-968548.html
http://blog.sciencenet.cn/blog-340399-958453.html
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