枚舉*所有*哈密爾頓路徑

Question

我知道這已被問過，但我沒有在任何帖子中找到答案。有人可以建議我一個枚舉圖中所有哈密頓路徑的算法嗎？

有點背景：我正在研究一個問題，我必須列舉每個哈密爾頓路徑，做一些分析並返回結果。爲此，我需要列舉所有可能的哈密爾頓路徑。

謝謝。

Answer 1

按照建議使用BFS/DFS，但不要停在第一個解決方案。 BFS/DFS的主要用途（在這種情況下）將是找到所有的解決方案，你需要給它一個條件停止在第一個。

Answer 2

我的Java代碼：（絕對基於遞歸方法）

算法：

+開始在1點連接到另一點它可以看到（以形成路徑）。

+刪除路徑並遞歸查找最新點的新路徑，直到連接圖的所有點。

+除去路徑和回溯到初始圖表如果斜面形成從最新點

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; 
} 

}

Answer 3

深度優先窮舉搜索給你的答案漢密爾頓路徑。我剛剛完成了一個Java實現了這個問題（包括代碼）寫了起來：

http://puzzledraccoon.wordpress.com/2012/06/07/how-to-cool-a-data-center/

Answer 4

解決方案在Python3：

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)))