Question

我尝试使用Mathematica在下图中找到一个汉密尔顿循环：

g = Graph[{1 \[UndirectedEdge] 2, 1 \[UndirectedEdge] 6, 
   2 \[UndirectedEdge] 4, 1 \[UndirectedEdge] 3, 
   3 \[UndirectedEdge] 4, 4 \[UndirectedEdge] 5, 
   5 \[UndirectedEdge] 6, 3 \[UndirectedEdge] 9, 
   4 \[UndirectedEdge] 7, 7 \[UndirectedEdge] 8, 
   8 \[UndirectedEdge] 11, 5 \[UndirectedEdge] 8, 
   5 \[UndirectedEdge] 12, 6 \[UndirectedEdge] 12, 
   9 \[UndirectedEdge] 10, 10 \[UndirectedEdge] 11, 
   11 \[UndirectedEdge] 12, 9 \[UndirectedEdge] 15, 
   15 \[UndirectedEdge] 16, 16 \[UndirectedEdge] 17, 
   16 \[UndirectedEdge] 19, 19 \[UndirectedEdge] 20, 
   20 \[UndirectedEdge] 23, 20 \[UndirectedEdge] 17, 
   17 \[UndirectedEdge] 18, 15 \[UndirectedEdge] 21, 
   18 \[UndirectedEdge] 24, 21 \[UndirectedEdge] 22, 
   22 \[UndirectedEdge] 23, 23 \[UndirectedEdge] 24, 
   3 \[UndirectedEdge] 15, 14 \[UndirectedEdge] 15, 
   14 \[UndirectedEdge] 12, 22 \[UndirectedEdge] 15, 
   12 \[UndirectedEdge] 18, 11 \[UndirectedEdge] 13, 
   13 \[UndirectedEdge] 14, 13 \[UndirectedEdge] 10, 
   1 \[UndirectedEdge] 10, 12 \[UndirectedEdge] 24}, 
  VertexLabels -> "Name"]

然后

EdgeList[g, 4 \[UndirectedEdge] _]

显示4与2,3,5,7连接

但代码

First[FindHamiltonianCycle[g]]

返回错误，即

{1 \[UndirectedEdge] 2, 2 \[UndirectedEdge] 4, 4 \[UndirectedEdge] 8, 
 8 \[UndirectedEdge] 9, 9 \[UndirectedEdge] 6, 6 \[UndirectedEdge] 3, 
 3 \[UndirectedEdge] 11, 11 \[UndirectedEdge] 21, 
 21 \[UndirectedEdge] 19, 19 \[UndirectedEdge] 15, 
 15 \[UndirectedEdge] 14, 14 \[UndirectedEdge] 16, 
 16 \[UndirectedEdge] 17, 17 \[UndirectedEdge] 18, 
 18 \[UndirectedEdge] 22, 22 \[UndirectedEdge] 20, 
 20 \[UndirectedEdge] 13, 13 \[UndirectedEdge] 23, 
 23 \[UndirectedEdge] 24, 24 \[UndirectedEdge] 10, 
 10 \[UndirectedEdge] 12, 12 \[UndirectedEdge] 7, 
 7 \[UndirectedEdge] 5, 5 \[UndirectedEdge] 1}

为什么mathematica从4到8的边缘，因为它不存在？算法中是否有错误或我做错了什么？

我顺便使用Mathematica 8。

Answer 1

我在Windows 7上使用8.0.4，但我不明白这一点。它看起来像是在8.0.4中修复的错误

g = Graph[{UndirectedEdge[1, 2], UndirectedEdge[1, 6], UndirectedEdge[2, 4], 
UndirectedEdge[1, 3], UndirectedEdge[3, 4], UndirectedEdge[4, 5], 
UndirectedEdge[5, 6], UndirectedEdge[3, 9], UndirectedEdge[4, 7], 
UndirectedEdge[7, 8], UndirectedEdge[8, 11], UndirectedEdge[5, 8], 
UndirectedEdge[5, 12], UndirectedEdge[6, 12], UndirectedEdge[9, 10], 
UndirectedEdge[10, 11], UndirectedEdge[11, 12], UndirectedEdge[9, 15], 
UndirectedEdge[15, 16], UndirectedEdge[16, 17], UndirectedEdge[16, 19], 
UndirectedEdge[19, 20], UndirectedEdge[20, 23], UndirectedEdge[20, 17], 
UndirectedEdge[17, 18], UndirectedEdge[15, 21], UndirectedEdge[18, 24], 
UndirectedEdge[21, 22], UndirectedEdge[22, 23], UndirectedEdge[23, 24], 
UndirectedEdge[3, 15], UndirectedEdge[14, 15], UndirectedEdge[14, 12], 
UndirectedEdge[22, 15], UndirectedEdge[12, 18], UndirectedEdge[11, 13], 
UndirectedEdge[13, 14], UndirectedEdge[13, 10], UndirectedEdge[1, 10], 
UndirectedEdge[12, 24]}, VertexLabels -> "Name"];

r = First[FindHamiltonianCycle[g]]

给出

{UndirectedEdge[1, 2], UndirectedEdge[2, 4], UndirectedEdge[4, 7], 
  UndirectedEdge[7, 8], UndirectedEdge[8, 5], UndirectedEdge[5, 6], 
  UndirectedEdge[6, 12], UndirectedEdge[12, 24], UndirectedEdge[24, 18], 
  UndirectedEdge[18, 17], UndirectedEdge[17, 16], UndirectedEdge[16, 19], 
  UndirectedEdge[19, 20], UndirectedEdge[20, 23], UndirectedEdge[23, 22], 
  UndirectedEdge[22, 21], UndirectedEdge[21, 15], UndirectedEdge[15, 14], 
  UndirectedEdge[14, 13], UndirectedEdge[13, 11], UndirectedEdge[11, 10], 
  UndirectedEdge[10, 9], UndirectedEdge[9, 3], UndirectedEdge[3, 1]}

您可以看到边缘4<->8不存在。验证

In[7]:= Cases[r, UndirectedEdge[4, 8]]
Out[7]= {}

用mathematica显示汉密尔顿循环

1 个答案: