Using BFS for Weighted Graphs

Question

I was revising single source shortest path algorithms and in the video, the teacher mentions that BFS/DFS can\'t be used directly for finding shorte

Answer 1

Consider a graph like this:

A---(3)-----B
|           |
\-(1)-C--(1)/

The shortest path from A to B is via C (with a total weight of 2). A normal BFS will take the path directly from A to B, marking B as seen, and A to C, marking C as seen.

At the next stage, propagating from C, B is already marked as seen, so the path from C to B will not be considered as a potential shorter path, and the BFS will tell you that the shortest path from A to B has a weight of 3.

You can use Dijkstra's algorithm instead of BFS to find the shortest path on a weighted graph. Functionally, the algorithm is very similar to BFS, and can be written in a similar way to BFS. The only thing that changes is the order in which you consider the nodes.

For example, in the above graph, starting at A, a BFS will process A --> B, then A --> C, and stop there because all nodes have been seen.

On the other hand, Dijkstra's algorithm will operate as follows:

1. Consider A --> C (since this is the lowest-edge weight from A)
2. Consider C --> B (since this is the lowest-weight edge from any node we have reached so far, that we have not yet considered)
3. Consider and ignore A --> B since B has already been seen.

Note that the difference lies simply in the order in which edges are inspected. A BFS will consider all edges from a single node before moving on to other nodes, while Dijkstra's algorithm will always consider the lowest-weight unseen edge, from the set of edges connected to all nodes that have been seen so far. It sounds confusing, but the pseudocode is very simple:

create a heap or priority queue
place the starting node in the heap
dist[2...n] = {∞}
dist[1] = 0
while the heap contains items:
   vertex v = top of heap
   pop top of heap
   for each vertex u connected to v:
       if dist[u] > dist[v] + weight of v-->u:
           dist[u] = dist[v] + weight of edge v-->u
           place u on the heap with weight dist[u]

This GIF from Wikipedia provides a good visualisation of what happens:

Notice that this looks very similar to BFS code, the only real difference is the use of a heap, sorted by distance to the node, instead of a regular queue data structure.

Answer 2

Although this is true, but you could use BFS/DFS in weighted graphs, with a little change in the graph, if your graph's weights are positive integers you can replace an edge with weight n with n edges with weight 1 with n-1 middle nodes. Something like this:

A-(4)-B

will be:

A-(1)-M1-(1)-M2-(1)-M3-(1)-B

And don't consider these middle nodes (like M1,M2,M3 ) in your final BFS/DFS results.

---

This algorithm complexity is O(V * M) and M is the maximum weight of our edges, if we know that in our particular graphs M<log V this algorithm could be considerd, but in general this algorithm may not have a such a good performance.