将图形分割为完全子图的算法

Question

我需要一个算法来划分一个无向图的顶点到一个或多个子图，这样每个子图是一个完整的图（每个顶点相邻的其他每个顶点）。每个顶点必须正好位于其中一个子图中。你知道吗

举个例子：

input = [
    (A, B),
    (B, C),
    (A, C),
    (B, D),
    (D, E),
]
output = myalgo(input)  # [(A, B, C), (D, E)]

下面的图片可以更好地描述问题：

输入列表按距离按降序排序，这就是为什么我连接A-B-C而不是B-D

我认为这可能被称为“强连接组件”，并且已经尝试了以下解决方案：

• Finding a Strongly Connected Components in unDirected Graphs：它在寻找不同的东西

• Finding all cycles in undirected graphs：它给了我许多周期，但不是最好的，它不关心输入顺序。

• An algorithm to create clusters from data pairs in python：它连接所有组件，仅仅因为它们之间有一条路径（a-B-C-D-E）。

• Kosaraju's algorithm：它只适用于有向图。

Answer 1

另一次尝试：

lst = [
    ('A', 'B'),
    ('B', 'C'),
    ('A', 'C'),
    ('B', 'D'),
    ('D', 'E'),
]

d = {}
for i, j in lst:
    d.setdefault(i, []).append(j)
    d.setdefault(j, []).append(i)

from itertools import combinations

rv, seen_segments, seen_vertices = [], set(), set()
for k, v in d.items():
    if len(v) == 1:
        segment = set((k, v[0])).difference(seen_vertices)
        seen_vertices.update(segment)
        rv.append([tuple(segment), ])
    else:
        graph = []
        for i, j in combinations([k] + v, 2):
            if not j in d[i]:
                break
            else:
                graph.append(tuple(sorted((i, j))))
        else:
            if graph:
                graph = [segment for segment in graph if segment not in seen_segments]
                seen_segments.update(graph)
                if graph:
                    rv.append(graph)

from pprint import pprint
pprint(rv)

印刷品：

[[('A', 'B'), ('A', 'C'), ('B', 'C')], [('D', 'E')]]

---

用于输入

lst = [
    ('A', 'B'),
    ('B', 'C'),
]

印刷品：

[[('A', 'B')], [('C',)]]

---

输入：

lst = [
    ('A', 'B'),
    ('B', 'C'),
    ('C', 'D'),
]

印刷品：

[[('B', 'A')], [('D', 'C')]]

Answer 2

from collections import defaultdict


def create_adjacency_matrix(connections):
    matrix = defaultdict(dict)
    for a, b in connections:
        matrix[a][b] = 1
        matrix[b][a] = 1
    return matrix


def is_connected_to_all(vertex, group, matrix):
    for v in group:
        if vertex != v and vertex not in matrix[v]:
            return False
    return True


def group_vertexes(input):
    matrix = create_adjacency_matrix(input)
    groups = []
    current_group = set()
    for vertex in matrix.keys():
        if is_connected_to_all(vertex, current_group, matrix):
            current_group.add(vertex)
        else:
            groups.append(current_group)
            current_group = {vertex}
    groups.append(current_group)
    return groups

input = [("A", "B"), ("B", "C"), ("A", "C"), ("B", "E"), ("E", "D")]
print(group_vertexes(input))
# [{'C', 'B', 'A'}, {'D', 'E'}]

警告：这取决于dict在python3.7+中保持插入顺序的事实。在旧版本中，必须使用matrix = DefaultOrderedDict(dict)https://stackoverflow.com/a/35968897/9392216：

Answer 3

下面是一个类，它实现了对完整子图的分段。它绝不是优化的，可能会得到显著的改进，但这是一个起点

class SCCManager:
    def __init__(self, edges):
        self.clusters = []
        self.edges = edges

    def clusters_in(self, conn):
        first, second = conn
        f_clust = None
        s_clust = None
        for i, clust in enumerate(self.clusters):
            if first in clust:
                f_clust = i
            if second in clust:
                s_clust = i
            # break early if both already found
            if f_clust and s_clust:
                break
        return (f_clust, s_clust)

    def all_connected(self, cluster, vertex):
        for v in cluster:
            connected = (v, vertex) in self.edges or (vertex, v) in self.edges
            # break early if any element is not connected to the candidate
            if not connected:
                return False
        return True

    def get_scc(self):
        for edge in self.edges:
            c_first, c_second = self.clusters_in(edge)

            # case 1: none of the vertices are in an existing cluster
            # -> create new cluster containing the vertices
            if c_first == c_second == None:
                self.clusters.append([edge[0], edge[1]])
                continue

            # case 2: first is in a cluster, second isn't
            # -> add to cluster if eligible
            if c_first != None and c_second == None:
                # check if the second is connected to all cluster components
                okay = self.all_connected(self.clusters[c_first], edge[1])
                # add to cluster if eligible
                if okay:
                    self.clusters[c_first].append(edge[1])
                continue

            # case 3: other way round
            if c_first == None and c_second != None:
                okay = self.all_connected(self.clusters[c_second], edge[0])
                if okay:
                    self.clusters[c_second].append(edge[0])
                continue

            # case 4: both are in different clusters
            # -> merge clusters if allowed
            if c_first != c_second:
                # check if clusters can be merged
                for v in self.clusters[c_first]:
                    merge = self.all_connected(self.clusters[c_second], v)
                    # break if any elements are not connected
                    if not merge:
                        break
                # merge if allowed
                if merge:
                    self.clusters[c_first].extend(self.clusters[c_second])
                    self.clusters.remove(self.clusters[c_second])

            # case 5: both are in the same cluster
            # won't happen if input is sane, but doesn't require an action either way


        return self.clusters

。。。下面是一个有效的例子：

inp = [
    ('A', 'B'),
    ('B', 'C'),
    ('A', 'C'),
    ('B', 'D'),
    ('D', 'E'),
    ('C', 'E')
]

test = SCCManager(inp)
print(test.get_scc())

[['A', 'B', 'C'], ['D', 'E']]