这是子图同构问题：http :
//en.wikipedia.org/wiki/Subgraph_isomorphism_problem

由于Ullmann，本文中提到了一种算法。

乌尔曼算法是深度优先搜索的扩展。深度优先搜索将像这样工作：

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，如果子图中的节点i可能与图中的节点j相匹配，那么对于子图中与节点i相邻的每个节点x，必须找到与节点j相邻的节点y。在图中。这个过程的帮助比最初显而易见的要多，因为每次我们消除一个可能的分配时，这可能会导致其他可能的分配被消除，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