How to partially compare two graphs

Question

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

0 - 1

1 - 2

2 - 3

3 - 0

AND

0 - 1

1 - 2

These two are considered a bad match

0 - 1

1 - 2

2 - 3

3 - 0

AND

0 - 1

1 - 2

2 - 0

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

Answer 1

This is the subgraph isomorphism problem: http://en.wikipedia.org/wiki/Subgraph_isomorphism_problem

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

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

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

Ullmann extends this basic algorithm by narrowing the possibilities:

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

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.

The final algorithm is then:

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