BFS algorithm in grids/mazes

[Panphobia]

Ok so I know the concept of a BFS, but the thing is I still do not know exactly how to implement it. I am just asking how would I lets say with a BFS, in either java or python, doesnt matter really, get the shortest path from A-B with this grid/maze and the # are walls

code:

A......... ##....##.. .####...## .......... .......### B......... ####...### .......... #...#.###. ..........

If you could explain it in laymens terms, or a code example would be great!

[DemonWasp]

A Breadth-First Search works like this:

code:

nodes-to-search = {node A} while nodes-to-search is not empty, do:     get closest from nodes-to-search     if closest is B, then         return the path stored in closest     else         add neighbours(closest) to nodes-to-search (ignoring nodes which have already been searched or are duplicates

There's some complication to computing the path, but it works like this:
- when you add neighbours(closest), assign "closest" as the "parent" of each neighbour
- then the "parent" of each node points to the node it was discovered from
- by following one parent to the next, you can get from B to A
- reverse that path to get from A to B

[Panphobia]

So node-to-search is a set?

[mirhagk]

Hey DemonWasp, I think you're confusing dijisktras with BFS. Breadth First Search is nearly identical to Depth First Search, the difference being which node you check next. With DFS you check the last node you discovered whereas with BFS you check the first one you discovered.

Essentially you get a queue (which is something that you add items to the end, but get items from the front, so the items you get are the first items you get out. It's FIFO unlike stacks which are LIFO).

Every time you discover a new node you add it to the queue, and each time in the while loop you grab the first thing from the queue, and check if it's the end, and if not you add new neighbours to the queue.

[mirhagk]

Python:

def DFS(x,y,Map):     if (Map[x][y]=="exit"): #check if we're at the exit         return [(x,y)] #if so then we solved it so return this spot     if (Map[x][y]!="path"): #if it's not a path, we can't try this spot         return []     Map[x][y]="explored" #make this spot explored so we don't try again     for i in [[x-1,y],[x+1,y],[x,y-1],[x,y+1]]: #new spots to try             result = DFS(i[0],i[1],Map) #recursively call itself             if len(result)>0: #if the result had at least one element, it found a correct path, otherwise it failed                 result.append((x,y)) #if it found a correct path then return the path plus this spot                 return result     return [] #return the empty list since we couldn't find any paths from here from collections import deque def BFS(x,y,Map):     queue = deque( [(x,y,None)]) #create queue     while len(queue)>0: #make sure there are nodes to check left         node = queue.popleft() #grab the first node         x = node[0] #get x and y         y = node[1]         if Map[x][y] == "exit": #check if it's an exit             return GetPathFromNodes(node) #if it is then return the path         if (Map[x][y]!="path"): #if it's not a path, we can't try this spot             continue         Map[x][y]="explored" #make this spot explored so we don't try again         for i in [[x-1,y],[x+1,y],[x,y-1],[x,y+1]]: #new spots to try             queue.append((i[0],i[1],node))#create the new spot, with node as the parent     return []             def GetPathFromNodes(node):     path = []     while(node != None):         path.append((node[0],node[1]))         node = node[2]     return path         def GetMap():     return [         ["wall","wall","wall","wall","wall","wall","wall","wall"],         ["wall","path","path","path","path","path","path","wall"],         ["wall","wall","wall","path","wall","path","path","wall"],         ["wall","path","path","path","wall","wall","path","wall"],         ["wall","path","wall","path","path","path","path","wall"],         ["wall","path","wall","wall","wall","wall","path","wall"],         ["wall","path","path","path","path","path","path","wall"],         ["wall","wall","wall","exit","wall","wall","wall","wall"]             ] def DrawMap(Map,path):     for x in range(0,len(Map)):         for y in range(0,len(Map[x])):             if ((x,y) in path):                 assert Map[x][y] in ("path","exit")                 print("-",end="")             elif (Map[x][y]=="wall"):                 print("#",end="")             elif (Map[x][y]=="exit"):                 print(".",end="")             else:                 print(' ',end="")         print() print("unsolved") DrawMap(GetMap(),[]) print("\n") print("solved with DFS") print("path is ",len(DFS(1,1,GetMap()))," spots long") DrawMap(GetMap(),DFS(1,1,GetMap())) print("\n") print("solved with BFS") print("path is ",len(BFS(1,1,GetMap()))," spots long") DrawMap(GetMap(),BFS(1,1,GetMap()))

As you can see from this updated example (from the one I sent you). BFS is very similar to DFS, with only the order you check nodes changed.

The example map has been changed to show that DFS and BFS often find very different solutions, and that BFS allows finds a path that has the shortest number of nodes, whereas DFS does not.

You could also create many optimizations from this, but you need to make sure you understand the algorithm before you do so.

[Panphobia]

so this is a BFS right? not a DFS because your method tells me its DFS O:

[Panphobia]

never mind lol, i will try to convert the bfs to java

[mirhagk]

The code includes both a BFS and a DFS for comparison. The 2nd method is BFS, the first is DFS

[Panphobia]

This is my current code edited from yours, how does it not work

code:

import collections def BFS(x,y,Map):     queue = collections.deque( [(x,y,None)]) #create queue     while len(queue)>0: #make sure there are nodes to check left         node = queue.popleft() #grab the first node         x = node[0] #get x and y         y = node[1]         if Map[x][y] == "B": #check if it's an exit             return GetPathFromNodes(node) #if it is then return the path         if (Map[x][y]!="*"): #if it's not a path, we can't try this spot             continue         Map[x][y]="explored" #make this spot explored so we don't try again         for i in [[x-1,y],[x+1,y],[x,y-1],[x,y+1]]: #new spots to try             queue.append((i[0],i[1],node))#create the new spot, with node as the parent     return []             def GetPathFromNodes(node):     path = []     while(node != None):         path.append((node[0],node[1]))         node = node[2]     return path         def GetMap():     return [         ["#","#","#","#","#","#","#","#"],         ["#","A","*","*","*","*","*","#"],         ["#","#","#","*","#","*","*","#"],         ["#","*","*","*","#","#","*","#"],         ["#","*","#","*","*","*","*","#"],         ["#","*","#","#","#","#","*","#"],         ["#","*","*","*","*","*","*","#"],         ["#","#","#","B","#","#","#","#"]             ] def DrawMap(Map,path):     for x in range(0,len(Map)):         for y in range(0,len(Map[x])):             if ((x,y) in path):                 assert Map[x][y] in ("*","B")                 print("-")             elif (Map[x][y]=="#"):                 print()             elif (Map[x][y]=="B"):                 print(".")             else:                 print(' ')         print() print("solved with BFS") print("path is ",len(BFS(1,1,GetMap()))," spots long") DrawMap(GetMap(),BFS(1,1,GetMap()))

[mirhagk]

Change A to * in your map. The starting point should be determined by the caller, not the map. If the starting point isn't * then it stops searching instantly.

You've also really messed up the DrawMap part. It should look like

code:

if ((x,y) in path):                 assert Map[x][y] in ("*","B")                 print("-",end="")             elif (Map[x][y]=="#"):                 print("#",end="")             elif (Map[x][y]=="B"):                 print(".",end="")             else:                 print(' ',end="")

[Panphobia]

Yea but its an error at print(end=) and i had to take it out

[DemonWasp]

mirhagk @ Mon Nov 19, 2012 9:11 am wrote:

...confusing dijisktras with BFS...

Nope. Dijkstra's algorithm is completely different. For one, it uses edge weights, which BFS ignores. The pseudocode I wrote out is almost exactly identical to what's displayed on Wikipedia: http://en.wikipedia.org/wiki/Breadth-first_search

[Panphobia]

I got it to work, now the problem is, it takes like 5 seconds to execute, is there a quicker algorithm because dwite it needs to be less than 2.5 seconds

[mirhagk]

Oh yeah there are lots better algorithms. Both DFS and BFS are brute force solutions. Depending on your problem Djikstras or A* could be very beneficial. Basically if you can come up with an estimate of how close something is to the answer you can use A*.

In general most of the problems require semi-unique solutions, because brute forcing is very easy in some languages (imperative languages especially), and they are meant to require at least some thought.

@DemonWasp Sorry I was confused by your variable names, nodes-to-search and closest. It seemed like you were doing some sort of sorting of the queue, which I now see you aren't.

Your pseudo-code isn't really specific on what node to grab next, which is the only difference between many of these graph searching algorithms (In fact djikstras is just that pseudo code where you grab the node that has the least sum of the edge nodes, and A* is just the same except you add a heuristic to the calcuation etc etc). The pseudo code is really more suited to a generic method for searching.

[Panphobia]

hmmmm i just searched dijkstra on wikipedia, and the pseudocode is making me very sad lol. Could you code up a quick example using dijkstra or A*, and also is Lee's Algorithm good for this situation?