例如,这两个图被认为是完美的部分匹配:
0 - 1
1 - 2
2 - 3
3 - 0
和
0 - 1
1 - 2
这两者被认为是一个糟糕的匹配
0 - 1
1 - 2
2 - 3
3 - 0
和
0 - 1
1 - 2
2 - 0
这些数字不必匹配,只要这些节点之间的关系可以完全匹配即可。
这是子图同构问题: http://en.wikipedia.org/wiki/Subgraph_isomorphism_problem
由于 Ullmann,文章中提到了一种算法。
Ullmann算法是深度优先搜索的扩展。 深度优先搜索的工作方式如下:
def search(graph,subgraph,assignments):
i=len(assignments)
# Make sure that every edge between assigned vertices in the subgraph is also an
# edge in the graph.
for edge in subgraph.edges:
if edge.first<i and edge.second<i:
if not graph.has_edge(assignments[edge.first],assignments[edge.second]):
return False
# If all the vertices in the subgraph are assigned, then we are done.
if i==subgraph.n_vertices:
return True
# Otherwise, go through all the possible assignments for the next vertex of
# the subgraph and try it.
for j in possible_assignments[i]:
if j not in assignments:
assignments.append(j)
if search(graph,subgraph,assignments):
# This worked, so we've found an isomorphism.
return True
assignments.pop()
def find_isomorphism(graph,subgraph):
assignments=[]
if search(graph,subgraph,assignments):
return assignments
return None
对于基本算法,possible_assignments[i] = range(0,graph.n_vertices)
. 也就是说,所有顶点都是可能的。
Ullmann 通过缩小可能性来扩展此基本算法:
def update_possible_assignments(graph,subgraph,possible_assignments):
any_changes=True
while any_changes:
any_changes = False
for i in range(0,len(subgraph.n_vertices)):
for j in possible_assignments[i]:
for x in subgraph.adjacencies(i):
match=False
for y in range(0,len(graph.n_vertices)):
if y in possible_assignments[x] and graph.has_edge(j,y):
match=True
if not match:
possible_assignments[i].remove(j)
any_changes = True
这个想法是,如果子图的节点 i 可能与图的节点 j 匹配,那么对于子图中与节点 i 相邻的每个节点 x,必须能够找到与图中节点 j 相邻的节点 y。 这个过程比最初显而易见的更有帮助,因为每次我们消除可能的赋值时,这都可能导致其他可能的赋值被消除,因为它们是相互依赖的。
最终的算法是:
def search(graph,subgraph,assignments,possible_assignments):
update_possible_assignments(graph,subgraph,possible_assignments)
i=len(assignments)
# Make sure that every edge between assigned vertices in the subgraph is also an
# edge in the graph.
for edge in subgraph.edges:
if edge.first<i and edge.second<i:
if not graph.has_edge(assignments[edge.first],assignments[edge.second]):
return False
# If all the vertices in the subgraph are assigned, then we are done.
if i==subgraph.n_vertices:
return True
for j in possible_assignments[i]:
if j not in assignments:
assignments.append(j)
# Create a new set of possible assignments, where graph node j is the only
# possibility for the assignment of subgraph node i.
new_possible_assignments = deep_copy(possible_assignments)
new_possible_assignments[i] = [j]
if search(graph,subgraph,assignments,new_possible_assignments):
return True
assignments.pop()
possible_assignments[i].remove(j)
update_possible_assignments(graph,subgraph,possible_assignments)
def find_isomorphism(graph,subgraph):
assignments=[]
possible_assignments = [[True]*graph.n_vertices for i in range(subgraph.n_vertices)]
if search(graph,subgraph,assignments,possible_assignments):
return assignments
return None