Shortest distance algorithm Python

I wanted to create a simple breadth first search algorithm, which returns the shortest path.

An actor information dictionary maps and actor to the list of movies the actor appears in:

actor_info = { "act1" : ["movieC", "movieA"], "act2" : ["movieA", "movieB"], "act3" :["movieA", "movieB"], "act4" : ["movieC", "movieD"], "act5" : ["movieD", "movieB"], "act6" : ["movieE"], "act7" : ["movieG", "movieE"], "act8" : ["movieD", "movieF"], "KevinBacon" : ["movieF"], "act10" : ["movieG"], "act11" : ["movieG"] }

The inverse of this maps movies to the list of actors appearing in them:

movie_info = {'movieB': ['act2', 'act3', 'act5'], 'movieC': ['act1', 'act4'], 'movieA': ['act1', 'act2', 'act3'], 'movieF': ['KevinBacon', 'act8'], 'movieG': ['act7', 'act10', 'act11'], 'movieD': ['act8', 'act4', 'act5'], 'movieE': ['act6', 'act7']}

so for a call

shortest_dictance("act1", "Kevin Bacon", actor_info, movie_info)

I should get 3 since act1 appears in movieC with Act4 who appears in movieD with Act8 who appears in movie F with KevinBacon. So the shortest distance is 3.

So far I have this:

def shotest_distance(actA, actB, actor_info, movie_info):'''Return the number of movies required to connect actA and actB.    If theres no connection return -1.'''    # So we keep 2 lists of actors:    #   1.The actors that we have already investigated.    #   2.The actors that need to be investigated because we have found a     #      connection beginning at actA. This list must be     #      ordered, since we want to investigate actors in the order we     #      discover them.    #      -- Each time we put an actor in this list, we also store    #         her distance from actA.    investigated = []    to_investigate = [actA]    distance = 0    while actB not in to_investigate and to_investigate!= []:        for actor in to_investigate:            to_investigated.remove(actA)            investigated.append(act)            for movie in actor_info[actor]:                for co_star in movie_info[movie]:                    if co_star not in (investigated and to_investigate):                        to_investigate.append(co_star) .... .... return d    

I can't figure the appropriate way to keep track of the distances discovered each of iteration of the code. Also the code seems to be very ineffecient time wise.