如何部分比较两个图 [英] How to partially compare two graphs

问题描述

例如，这两个图形被认为是完美的部分匹配

For example, these two graphs is considered to be a perfect partial match:

0 - 1

1 - 2

2 - 3

3 - 0

和

0 - 1

1 - 2

这两个被认为是一个糟糕的比赛

These two are considered a bad match

0 - 1

1 - 2

2 - 3

3 - 0

和

0 - 1

1 - 2

2 - 0

的数字不具有匹配的，只要这些节点之间的关系可以完美匹配

The numbers don't have to match, as long as the relation between those nodes can perfectly match.

推荐答案

这是子图同构问题：<一href="http://en.wikipedia.org/wiki/Subgraph_isomorphism_problem">http://en.wikipedia.org/wiki/Subgraph_isomorphism_problem

目前由于乌尔曼在文章中提到的一种算法。

There is one algorithm mentioned in the article due to Ullmann.

乌尔曼的算法是一个深度优先搜索的扩展。深度优先搜索将像这样的工作：

Ullmann's algorithm is an extension of a depth-first search. A depth-first search would work like this:

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_isomporhism(graph,subgraph):
  assignments=[]
  if search(graph,subgraph,assignments):
    return assignments
  return None

有关的基本算法， possible_assignments [I] =范围（0，graph.n_vertices）。也就是说，所有的顶点是一种可能性。

For the basic algorithm, possible_assignments[i] = range(0,graph.n_vertices). That is, all the vertices are a possibility.

乌尔曼通过缩小的可能性扩展了这个基本算法：

Ullmann extends this basic algorithm by narrowing the possibilities:

def update_possible_assignements(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

的想法是，如果节点的子图i的可能可能匹配图形的节点j，然后为每一个节点x即相邻节点i的子图，它有可能找到一个节点y表示相邻到节点j的曲线图。这个过程有助于多于威力第一是显而易见的，因为每次我们消除一个可能的分配，这可能会导致其他可能的分配来消除，因为它们是相互依存

The idea is that if node i of the subgraph could possibly match node j of the graph, then for every node x that is adjacent to node i in the subgraph, it has to be possible to find a node y that is adjacent to node j in the graph. This process helps more than might first be obvious, because each time we eliminate a possible assignment, this may cause other possible assignments to be eliminated, since they are interdependent.

最后的算法则：

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_isomporhism(graph,subgraph):
  assignments=[]
  possible_assignments = [[True]*graph.n_vertices for i in range(subgraph.n_vertices)]
  if search(graph,subgraph,asignments,possible_assignments):
    return assignments
  return None

这篇关于如何部分比较两个图的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！