Uniform Cost Graph Search opening too many nodes - search

Question

I'm working on an assignment from an archived AI course from 2014.
The parameter "problem" refers to an object that has different cost functions chosen at run (sometimes it is 1 cost per move; sometimes moves are more expensive depending on which side of the pacman board the moves are done on).
As written below, I get the right behavior but I open more search nodes than expected (about 2x what the assignment expects).
If I turn the cost variable to negative, I get the right behavior on the 1-unit cost case AND get a really low number of nodes. But the behavior is opposite for the cases of higher costs for a given side of the board.
So basically question is: Does it seem like I'm opening any nodes unnecessarily in the context of a uniform cost search?
def uniformCostSearch(problem):
"""Search the node of least total cost first."""
def UCS(problem, start):
q = util.PriorityQueue()
for node in problem.getSuccessors(start): ## Comes a tuple ((x,y), 'North', 1)
q.push([node], node[2]) ##Push nodes onto queue one a time (so they are paths)
while not q.isEmpty():
pathToCheck = q.pop() ##Pops off the lowest priorty path on the queue?
#if pathToCheck in q.heap:
# continue
lastNode = pathToCheck[-1][0] ## Gets the coordinates of that last node in that path
if problem.isGoalState(lastNode): ##Checks if those coordinates are goal
return pathToCheck ## If goal, returns that path that was checked
else: ## Else, need to get successors of that node and put them on queue
for successor in problem.getSuccessors(lastNode): ##Gets those successors the popped path's last node and iterates over them
nodesVisited = [edge[0] for edge in pathToCheck] ##Looks at all the edges (the node plus its next legal move and cost) and grabs just the coords visited (nodes)
if successor[0] not in nodesVisited: ## ##Checks to see if those visited were in THIS particular path (to avoid cyclces)
newPath = pathToCheck + [successor] ## If ONE successor was not in path, adds it to the growing path (will do the next one in next part of loop)
cost = problem.getCostOfActions([x[1] for x in newPath])
q.update(newPath, cost) #Pushes that validated new path onto the back of the queue for retrieval later
return None
start = problem.getStartState()#Starting point of stack
successorList = UCS(problem, start)
directions = []
for i in range(len(successorList)):
directions += successorList[i]
return directions[1::3]

Answer 1

I figured it out.
Basically, while I'm checking that I don't revisit nodes within a given path, I'm not checking if I'm visiting nodes in others paths on the queue. I can check that by adding a nodesVisited list that just appends all nodes ever visited and checking THAT for duplicate visits.