我知道这个问题以前已经被问过了,但是我在任何帖子中都没有找到答案。有人能否建议一个算法来枚举图中的所有哈密顿路径?
一点背景:我正在解决一个问题,在这个问题中,我必须枚举每个哈密顿路径,进行一些分析,并返回结果。为此,我需要能够枚举所有可能的哈密顿路径。
谢谢。
我知道这个问题以前已经被问过了,但是我在任何帖子中都没有找到答案。有人能否建议一个算法来枚举图中的所有哈密顿路径?
一点背景:我正在解决一个问题,在这个问题中,我必须枚举每个哈密顿路径,进行一些分析,并返回结果。为此,我需要能够枚举所有可能的哈密顿路径。
谢谢。
我的Java代码:(绝对基于递归方法)
算法:
+从一个点开始连接到它可以看到的另一个点(形成一条路径)。
+移除路径并在最新的点递归地查找新路径,直到连接图中的所有点。
+如果无法从最新点形成哈密尔顿路径,则删除路径并回溯到初始图。
public class HamiltonPath {
public static void main(String[] args){
HamiltonPath obj = new HamiltonPath();
int[][]x = {{0,1,0,1,0}, //Represent the graphs in the adjacent matrix forms
{1,0,0,0,1},
{0,0,0,1,0},
{1,0,1,0,1},
{0,1,0,1,0}};
int[][]y = {{0,1,0,0,0,1},
{1,0,1,0,0,1},
{0,1,0,1,1,0},
{0,0,1,0,0,0},
{0,0,1,0,0,1},
{1,1,0,0,1,0}};
int[][]z = {{0,1,1,0,0,1},
{1,0,1,0,0,0},
{1,1,0,1,0,1},
{0,0,1,0,1,0},
{0,0,0,1,0,1},
{1,0,1,0,1,0}};
obj.allHamiltonPath(y); //list all Hamiltonian paths of graph
//obj.HamiltonPath(z,1); //list all Hamiltonian paths start at point 1
}
static int len;
static int[]path;
static int count = 0;
public void allHamiltonPath(int[][]x){ //List all possible Hamilton path in the graph
len = x.length;
path = new int[len];
int i;
for(i = 0;i<len;i++){ //Go through column(of matrix)
path[0]=i+1;
findHamiltonpath(x,0,i,0);
}
}
public void HamiltonPath(int[][]x, int start){ //List all possible Hamilton path with fixed starting point
len = x.length;
path = new int[len];
int i;
for(i = start-1;i<start;i++){ //Go through row(with given column)
path[0]=i+1;
findHamiltonpath(x,0,i,0);
}
}
private void findHamiltonpath(int[][]M,int x,int y,int l){
int i;
for(i=x;i<len;i++){ //Go through row
if(M[i][y]!=0){ //2 point connect
if(detect(path,i+1))// if detect a point that already in the path => duplicate
continue;
l++; //Increase path length due to 1 new point is connected
path[l]=i+1; //correspond to the array that start at 0, graph that start at point 1
if(l==len-1){//Except initial point already count =>success connect all point
count++;
if (count ==1)
System.out.println("Hamilton path of graph: ");
display(path);
l--;
continue;
}
M[i][y]=M[y][i]=0; //remove the path that has been get and
findHamiltonpath(M,0,i,l); //recursively start to find new path at new end point
l--; // reduce path length due to the failure to find new path
M[i][y] = M[y][i]=1; //and tranform back to the inital form of adjacent matrix(graph)
}
}path[l+1]=0; //disconnect two point correspond the failure to find the..
} //possible hamilton path at new point(ignore newest point try another one)
public void display(int[]x){
System.out.print(count+" : ");
for(int i:x){
System.out.print(i+" ");
}
System.out.println();
}
private boolean detect(int[]x,int target){ //Detect duplicate point in Halmilton path
boolean t=false;
for(int i:x){
if(i==target){
t = true;
break;
}
}
return t;
}
}
建议使用BFS/DFS算法,但不要停留在第一个解决方案上。BFS/DFS的主要用途(在这种情况下)是找到所有的解决方案,你需要给它设置一个条件,以便在找到第一个解决方案后停止。
Python3的解决方案:
def hamiltonians(G, vis = []):
if not vis:
for n in G:
for p in hamiltonians(G, [n]):
yield p
else:
dests = set(G[vis[-1]]) - set(vis)
if not dests and len(vis) == len(G):
yield vis
for n in dests:
for p in hamiltonians(G, vis + [n]):
yield p
G = {'a' : 'bc', 'b' : 'ad', 'c' : 'b', 'd' : 'ac'}
print(list(hamiltonians(G)))
http://puzzledraccoon.wordpress.com/2012/06/07/how-to-cool-a-data-center/