找到"好"邻居的算法 - 图着色？

Question

我有一群人,他们每个人都有朋友名单和敌人名单.我想把它们排成一行(桌子上没有圆圈),所以没有敌人,只有朋友彼此相邻.

输入示例:https://gist.github.com/solars/53a132e34688cc5f396c

我想我需要使用图形着色来解决这个问题,但我不确定如何 - 我想我必须省略朋友(或敌人)列表以使其更容易并映射到图表.

有谁知道如何解决这些问题,并告诉我,如果我走在正确的道路上？

代码示例或在线示例也不错,我不介意编程语言,我通常使用Ruby,Java,Python,Javascript

非常感谢你的帮助!

Answer 1

评论中已经提到，这个问题相当于旅行商问题。我想详细说明一下：

每个人相当于一个顶点，顶点之间的边代表可以坐在一起的人。现在，找到可能的座位安排相当于在图中找到哈密顿路径。

所以这个问题是NPC。最幼稚的解决方案是尝试所有可能的排列，从而产生O(n!)运行时间。有很多众所周知的方法，其性能比O(n!)网络上的免费方法更好。我想提一下Held-Karp，它在 python 中运行O(n^2*2^n)并且代码非常简单：

#graph[i] contains all possible neighbors of the i-th person
def held_karp(graph):
    n = len(graph)#number of persons

    #remember the set of already seated persons (as bitmask) and the last person in the line
    #thus a configuration consists of the set of seated persons and the last person in the line
    #start with every possible person:
    possible=set([(2**i, i) for i in xrange(n)])

    #remember the predecessor configuration for every possible configuration:
    preds=dict([((2**i, i), (0,-1)) for i in xrange(n)])

    #there are maximal n persons in the line - every iterations adds a person
    for _ in xrange(n-1):
        next_possible=set()
        #iterate through all possible configurations
        for seated, last in possible:
            for neighbor in graph[last]:
                bit_mask=2**neighbor
                if (bit_mask&seated)==0: #this possible neighbor is not yet seated!
                    next_config=(seated|bit_mask, neighbor)#add neighbor to the bit mask of seated
                    next_possible.add(next_config)
                    preds[next_config]=(seated, last)
        possible=next_possible

    #now reconstruct the line
    if not possible:
      return []#it is not possible for all to be seated

    line=[]
    config=possible.pop() #any configuration in possible has n person seated and is good enough!
    while config[1]!=-1:
        line.insert(0, config[1])
        config=preds[config]#go a step back

    return line

免责声明：此代码未经正确测试，但我希望您能了解其要点。