|WikiProject Mathematics||(Rated Start-class, Mid-priority)|
|WikiProject Computer science||(Rated Start-class, High-importance)|
Breadth first recursion now redirects here.
- 1 Request for major editing:
- 2 What type of graph does this apply to?
- 3 Pseudocode
- 4 Dijsktra's Algorithm vs. BFS
- 5 Usage in 2D grids for computer games
- 6 C++ implementation
- 7 Connectedness, search and traversal
- 8 Complexity
- 9 Python Implementation
- 10 Pseudo code
- 11 Merger_proposal
- 12 Breadth-first search implementation at AfD
- 13 Completeness
- 14 Example with connected components
- 15 C# implementation appears to be incorrect
- 16 Example explosion
- 17 Testing Bipartiteness
- 18 White, Grey and Black
Request for major editing:
I have been looking at this and also depth-first search and the preliminary explanation has the same meaning. It says that it starts at the root node "or a node", and explores each neighboring node of the "root" fully without making a cycle until it finds the goal. Something needs to be done about this. I'm just a student trying to learn this so I don't think I can really help. Maybe I'll ask my professor to fix this.
- No that's not a mistake. Both algorithms have this property. The property you state does not uniquely identify the algorithm. The algorithms use completely different data structures and result in different traversal orders.
"Since all nodes discovered so far have to be saved, the space complexity of breadth-first search is O(|V| + |E|)." I don't see it. Where does the |E| come from?
What type of graph does this apply to?
It appears to be assumed that the graphs being dealt with have a root node, but this is not stated explicitly. Presumably any node will do if the graph is undirected, but what happens for a directed graph? If the article deals only with undirected graphs then it might be helpful if this was made clear at the beginning. Autoplunger (talk) 20:05, 30 January 2011 (UTC)
The pseudocode looks funny, and a little too close to an actual implementation. I have taken liberty to modify it a little.
You did a poor job in modifying it, language is unclear, look at psuedocode on page for topological sort for good reference. —Preceding unsigned comment added by 126.96.36.199 (talk) 17:21, 22 November 2010 (UTC)
The pseudocode doesn't just look funny, it's wrong. "parent" is never used and nothing is searched for. It's a breadth-first traversal with an unnecessary extra variable.
To clarify: the marking of 'distance' (another unneeded variable) as INFINITE is apparently a check to make sure there are no cycles. But it relies upon there being a collection of unique graph nodes. This is just terrible pseudocode. — Preceding unsigned comment added by 188.8.131.52 (talk) 21:35, 3 July 2016 (UTC)
Dijsktra's Algorithm vs. BFS
When BFS is used with a priority queue, it is a total algorithm to find the shortest path between any two nodes in a weighted graph. Clearly, therefore, finding the shortest path between two nodes in a weighted graph is an application of BFS.
Somebody deleted this application because they felt that this should be an application of Dijsktra's Algorithm instead. What is the logic here? Clearly it is an application of both DA and BFS. DA is an optimization of the backtracing step of BFS, so you don't have to use back-pointers to remember how you got to the destination. The distinction between BFS and DA therefore has little to do with whether or not the graph is weighted. Finally, BFS is much easier to remember than DA, so BFS is undoubtedly the more popular algorithm to find shortest paths in any graph (weighted or unweighted). Therefore this application should be listed.
- Hi, I think I was the one who deleted that passage. Why I did it? BFS simply does not find a shortest path in a weighted graph! It finds the path that has the fewest number of edges, but if edges have edge weights, BFS will find an arbitrary expensive path of n edges instead of the cheap path that has n+1 edges. Therefore BFS is not an application for finding a shortest path in a weighted graph simply because it computes absolute nonsense if run on a weighted graph. BFS will always find the path that has the fewest number of nodes which just happens to be the shortest path if all weights are the same. You certainly can modify BFS to use a priority queue instead of a normal queue so that it then really finds a shortest path. But then DFS is the same as BFS just with a stack instead of a queue. So you would have to add the part about shortest paths in the DFS article, too. All three algorithms differ (mainly) only by just the datastructure used to save the nodes in, but their output is absolutely different. --Regnaron 21:14, 26 September 2006 (UTC)
The original formulation of BFS did not restrict the queue to be a FIFO queue. If you use BFS with a FIFO queue on a weighted graph then you're effectively discarding the edge weights. That's why you get a different result from the algorithm with a priority queue versus FIFO queue: you're changing the input. DFS is completely different because you get a result that you won't get merely by changing the inputs. BFS has been used with a priority queue in scheduling simulators long before Dijkstra's algorithm got its name. It is not limited to unweighted graphs!
- If you use BFS with a FIFO queue on a weighted graph then you're effectively discarding the edge weights. That's why you get a different result from the algorithm with a priority queue versus FIFO queue: you're changing the input. And exactly that result you get is wrong. It is not a shortest path but the path with fewest nodes. If you use it with a priority queue, the result is correct, but then, where is the difference to Dijkstras algorithm? Find the vertex that is first in the queue, update its value and never look at it again. Just the same as Dijkstra. --Regnaron 05:43, 4 October 2006 (UTC)
Your question, "Where is the difference to Dijkstra's Algorithm?" gets to the heart of the matter. BFS with PQ is easy to remember, is immediately clear to anybody who has learned BFS, and is the oldest and most popular algorithm for finding all shortest paths in a weighted graph. By contrast, Dijkstra's Algorithm is obscure, known only in elite circles, difficult to memorize, and named after a professor. The differences between the algorithms have nothing to do with the problem they are solving, and everything to do with name-dropping elitism and obfuscation.
What you seem to be suggesting is that "Dijkstra's Algorithm solves the problem, therefore BFS with PQ does not". BFS and PQ long predate Dijkstra's Algorithm: they were used in the 1940s back in Alan Turing's days when lattice theory, AI, and scheduling systems were first developed for computers. So your statement should be turned around. What problem is Dijkstra's Algorithm solving that BFS and PQ did not already solve? If anything needs to be deleted, it's the Dijkstra's Algorithm entry, not the BFS entry.
BFS is what we call a blind search algorithm, cause it uses no more information than links between nodes. On the other hand are informed search algorithms, such as Best-first search. This last algorithm effectively uses a priority queue to handle the visiting nodes. So modifying BFS with such queue results in a different class of algorithm: BestFS. By the way this class of algorithm is still not optimal. You need to move forward to A* search algorithm (and in the process finding an admissible heuristic) to solve the "shortest" path in a weighted graph. So I'm with Regnaron in this matter. —Preceding unsigned comment added by 184.108.40.206 (talk) 01:46, 16 June 2008 (UTC)
Usage in 2D grids for computer games
It is worth mentioning that when BFS is used in that manner, the neighbour list should be created such that north, east, south and west get priority over north-east, south-east, south-west and north-west. The reason for this is that BFS tends to start searching in a diagonal manner rather than adjacent, and the path found will not be the correct one. BFS should first search adjacent nodes, then diagonal nodes.
I think a better explanation or a reference is required here. As I understand it the resulting path will be zig-zagged if diagonal moves are prioritized over moves to adjacent grid point. The resulting path should still be correct.
- This section is mostly unintelligible and describes an obscure application in a very special field. There are thousands of equally obscure applications which are only interesting to their practitioners and add no additional insights to BFS. Suggest to remove it. --Mbetter 07:59, 17 August 2007 (UTC)
- I agree this is not helpful and is incorrect as it is currently worded. As Mbetter points out the topic is not especially important to BFS. I am removing it. Add it as a bullet to the applications section if you really feel it is worth mentioning. Jludwig 18:22, 2 October 2007 (UTC)
In the sample implementation, why do we use an std::vector<int> to keep track of key-value pairs? The method we currently use wastes more space, runs slower (when attempting to recover path data later on), makes reading the code less clear, and is less versatile (e.g., in expanding the function for a situation where graph size is unknown beforehand) than the much more elegant solution for such an associative array—an std::map<int,int>. I will change the code accordingly if no reasonable objections appear within a week or so. --220.127.116.11 00:41, 10 August 2007 (UTC)
- This has been changed accordingly. --18.104.22.168 03:16, 18 August 2007 (UTC)
Connectedness, search and traversal
The article is vague about stating the problem that BFS is supposed to solve. Is the goal to search for a path to a given vertex, or to traverse all vertices? The distinction is important for instance if the graph is not connected. (The link to "graph search algorithms" actually leads to a page about traversal!) --Mbetter 08:10, 17 August 2007 (UTC)
The sections on complexity are slightly inconsistent and worded incorrectly. I am changing them to correct the reason for the time and space complexity and to reflect the common notation. Jludwig 18:26, 2 October 2007 (UTC)
I think the Python implementation should store visited nodes in a set rather than a list, because testing membership in a set is O(1) rather than O(n) with the list. —Preceding unsigned comment added by 22.214.171.124 (talk) 08:30, 16 June 2008 (UTC)
I believe that the Python implementation is still not correct because pop(0) on a list is linear time. In the worst case, the length of the queue is O(|V|), which makes the complexity of this implementation O(|V|^2). Also, checking to see whether the child is already in the queue is currently O(n) and needs to be O(1).
I will wait a few days to see if anyone contradicts me; otherwise I will go ahead and fix it (or at least make a note of the problem).--AllenDowney
It seems to be wrong. "Otherwise enqueue any successors (the direct child nodes) that have not yet been examined." But successors might have been already enqueued earlier (and not examined yet). So we should color them grey when enqueue, and enqueue only white nodes. Poemich (talk) 11:41, 3 May 2009 (UTC) kij —Preceding unsigned comment added by 126.96.36.199 (talk) 08:58, 4 August 2009 (UTC)
Merging the Implementation in to the main document would make more sense. And avoid deletion.
- No. There's a good reason we don't clutter up the main document with large numbers of implementations. See Wikipedia talk:WikiProject Computer science#Source code written by editors. —David Eppstein (talk) 13:40, 11 May 2009 (UTC)
See Wikipedia:Articles for deletion/Breadth-first search implementation. As usual for AfD discussions, please leave a detailed comment explaining your opinion rather than just a keep or delete without an explanation. —David Eppstein (talk) 21:59, 12 May 2009 (UTC)
<harry_ftw> afaik, there's no such thing as a "breath first search"
- I tried a Google search for that, hoping to find something about underwater rescue techniques, but sadly it's all just spelling mistakes. —David Eppstein (talk) 06:47, 18 May 2009 (UTC)
Example with connected components
C# implementation appears to be incorrect
When checking whether to enqueue a new node, the posted version appears to be check if the node is already in the queue by comparing it with every other node in the queue (an O(n) operation). This implementation doesn't have the same characteristics or behavior as the Java implementation, or the description in Cormen, Rivest. Could someone who knows C# fix this? 188.8.131.52 (talk) 20:14, 26 April 2010 (UTC)
- There was a lot of weirdness in that implementation, beyond what you noted. I fixed that specific problem, and also changed it so that it's essentially the same as the Java implementation. -Xodarap00 (talk) 21:58, 9 May 2010 (UTC)
We only need one implementation. As it is, both languages conflate language features with explanation of the algorithm. We should either have only a pseudocode implementation, or maybe Python or some other reads-mostly-like-pseudocode language if it can be made sufficiently close to pseudocode. The fact that
java.util.LinkedList is in the example suggests to me that the example says more about Java than about BFS. —Ben FrantzDale (talk) 12:53, 11 May 2010 (UTC)
Doesn't this algorithm for testing bipartiteness assume that every node is reachable from the start node? I don't think it will work correctly for graphs that aren't connected. — Preceding unsigned comment added by 184.108.40.206 (talk) 20:00, 5 April 2013 (UTC)
I concur. The pseudo-code that's given is specific to BFS for a tree graph. It doesn't work correctly for graphs with multiple components, and doesn't agree with the text. Frankly, this article is a hot mess. I'll edit this when I get home and can refer to Cormen et al. 220.127.116.11 (talk) 09:37, 29 May 2013 (UTC)
White, Grey and Black
The use of a colour attribute which is WHITE, GREY or BLACK, which I assume originated with the Corman et al. algorithms book, is useful for proving things about the algorithm. But it isn't used for anything in this article and it isn't needed for correctness of the algorithm. The pseudocode is still valid if "if adj.color == WHITE" is replaced by "if adj.distance == INFINITY" and then all statements involving colour are removed. So at the moment it only serves to complicate something simple. Who will object if I make this simplification? McKay (talk) 04:35, 28 July 2015 (UTC)