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我正在尝试实施 Krager 的 Min。python中的cut算法解决以下问题。这个问题来自斯坦福的 edx 课程，“算法：设计与分析，第 1 部分”

该文件包含一个简单无向图的邻接表表示。有 200 个顶点标记为 1 到 200。文件中的第一列表示顶点标签，特定行（除第一列之外的其他条目）告诉所有与该顶点相邻的顶点。例如，第 6 行看起来像：“6 155 56 52 120 ......”。这只是意味着标签为 6 的顶点与标签为 155、56、52、120、......等的顶点相邻（即共享一条边）

运行最小割问题的随机收缩算法，并在上图中使用它来计算最小割。

它涉及以下邻接列表：https ://pastebin.pl/view/39b087f8

这是我在 python 中编写的代码——我知道实现很幼稚，但我只想在优化之前把它弄好。

import random
import copy
with open('Contraction hwk.txt') as f:  #Please input the correct file path
    proto = f.readlines()
proto2 = [(x.replace('\t',' ').strip('\n')).split() for x in proto]
adjl = [[int(x) for x in i] for i in proto2]

solnset = []
for x in range(1000):#Repeated a 100 times cause the algo is not always correct
    wadjl = copy.deepcopy(adjl)
    nodes = list(range(1,len(adjl)+1))
    while len(nodes) > 2:
        randnodeval = random.choice(nodes)  #Select a random node
        randnode = wadjl[randnodeval-1] #Assign the random node to a var --> randnode
        randedgeval = random.choice(randnode[1:]) # Picks another random node(edge node) that is adjacent to the initial this will contract with the original random node
        if randedgeval == randnodeval:
            continue
        for node in wadjl[randedgeval-1][1:]:#This loop will go to all nodes adjacent to the edge node and replace the value of the edge node with the random node
            if node == randnodeval:#If the node is the same as the random node it removes the node as an adjacent node (Deletion of edge node from random node)
                modnode = wadjl[node-1][1:]
                edgevalnode = modnode.index(randedgeval)
                wadjl[node-1].pop(edgevalnode+1)
                continue
            modnode = wadjl[node-1][1:]
            edgevalnode = modnode.index(randedgeval)
            wadjl[node-1][edgevalnode+1] = randnodeval 
        randnodeidx = wadjl[randedgeval-1][1:].index(randnodeval)#This yeilds the index of the random node in the adjaceny list of the edge node 
        wadjl[randedgeval-1].pop(randnodeidx+1)#The random node is removed from that list(Deletion of random node from edgenode)
        randnode.extend(wadjl[randedgeval-1][1:])#The adjacency list of the edge node is merged with that of the random node
        del wadjl[randedgeval-1][1:]#The edge node is deleted
        try:#repeates/edges to itself are removed from the random node
            repeats = [i for i,x in enumerate(randnode) if x == randnodeval]
            for repeate in repeats:
                randnode.pop(repeat)
        except:
            pass
        #Edge node is removed from the nodes list
        noderemove = nodes.index(randedgeval) 
        nodes.pop(noderemove)
    for entry in wadjl:
        if len(entry) > 1:
            minc = len(entry) - 1 #edges in min cut case
            solnset.append(minc) #append solution to solution set
            break
# print(solnset)
print(min(solnset))

根据一些互联网搜索，答案似乎是 17。我得到的答案是 20，而且我不认为我的算法实现是正确的，因为解决方案集太多样化了。我相信如果实施得当，解决方案应该经常出现在这个集合中。

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1 回答 1

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您发布的代码不会随机均匀地选择一条边，除非收缩图恰好具有均匀度（几乎总是不是这种情况）。通过选择一个随机节点，然后选择该节点的一个随机邻居，它会超重具有少数邻居的节点，这导致最佳切割边缘比它们应该更频繁地收缩，从而产生更大的最终切割。

理论上，Karger 的算法也需要 Θ(n choose 2) 次迭代才能以恒定概率成功，这里 n 选择 2 是 200 (200 - 1) / 2 = 19,900，但这张图并不接近最坏的情况，如 100大多数时候，迭代似乎绰绰有余。

这是我的实现：

import fileinput
import random


def find(parents, i):
    r = i
    while r in parents:
        r = parents[r]
    while i in parents:
        p = parents[i]
        parents[i] = r
        i = p
    return i


def unite(parents, i, j):
    parents[i] = j


def karger(n, edges):
    edges = list(edges)
    random.shuffle(edges)
    parents = {}
    for i, j in edges:
        if n <= 2:
            break
        i = find(parents, i)
        j = find(parents, j)
        if i == j:
            continue
        unite(parents, i, j)
        n -= 1
    return sum(find(parents, i) != find(parents, j) for (i, j) in edges)


def main():
    lines = list(fileinput.input())
    n = len(lines)
    edges = set()
    for line in lines:
        fields = iter(map(int, line.split()))
        u = next(fields)
        edges.update((min(u, v), max(u, v)) for v in fields)
    print(min(karger(n, edges) for k in range(1000)))


if __name__ == "__main__":
    main()
于 2020-09-07T13:53:51.447 回答