Algorithm to partition graph into complete subgraphs

Question

I need an algorithm to partition the vertices of an undirected graph into one or more subgraphs, such that each subgraph is a complete graph (every vertex adjacent to every other vertex). Each vertex needs to be in exactly one of the subgraphs.

Here's an example:

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

Here's an image that better describes the problem:

The input list is sorted in decreasing order by distance, that's why I connect A-B-C instead of B-D.

I thought this might be called "strongly connected components", and have already tried the following solutions:

• Finding a Strongly Connected Components in unDirected Graphs: it's looking for something different

• Finding all cycles in undirected graphs: it gives me many cycles and not the best, it doesn't care about the input order.

• An algorithm to create clusters from data pairs in python: it connects all the components, just because there is a path between them (A-B-C-D-E).

• Kosaraju's algorithm: it works only with a directed graph.

Answer 1

Here's a class that implements the segmentation into complete subgraphs. It's by no means optimized and can likely be improved significantly, but it's a starting point

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

... and here's a working example:

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']]

Answer 2

You can find all paths, and then group by connectivity:

from itertools import groupby as gb
d = [('A', 'B'), ('B', 'C'), ('A', 'C'), ('B', 'D'), ('D', 'E')]
def connect_num(node):
    return [sum(a == node for a, _ in d), sum(b == node for _, b in d)]

def group_paths(data):
   new_d = sorted([[i, connect_num(i)] for i in data], key=lambda x:max(x[1]))
   return [[k for k, _ in b] for _, b in gb(new_d, key=lambda x:max(x[1]))]

def get_paths(start, c = [], seen = []):
   new_vals = [a for a, _ in d if a not in seen+c]
   if (vals:=[b for a, b in d if a == start]):
      for i in vals:
         yield from get_paths(i, c=c+vals, seen=seen)
   else:
      yield c
      for i in new_vals:
         yield from get_paths(i, c = [i], seen=c+seen)

result = sorted(map(set, get_paths(d[0][0])), key=len, reverse=True)
new_result = [a for i, a in enumerate(result) if not any(all(k in c for k in a) for c in result[:i])]
final_result = [group_paths(i) for i in new_result]

Output:

#final_result[0]
[['E', 'D'], ['A', 'C', 'B']]

Answer 3

Another attempt:

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)

Prints:

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

---

For input

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

Prints:

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

---

For input:

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

Prints:

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

Answer 4

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'}]

WARNING: This relies on the fact that dict keeps insertion order in python 3.7+. In older versions you have to use matrix = DefaultOrderedDict(dict) https://stackoverflow.com/a/35968897/9392216: