Finding the path with the maximum minimal weight

Question

I'm trying to work out an algorithm for finding a path across a directed graph. It's not a conventional path and I can't find any references to anything like this being done already.

I want to find the path which has the maximum minimum weight.

I.e. If there are two paths with weights 10->1->10 and 2->2->2 then the second path is considered better than the first because the minimum weight (2) is greater than the minimum weight of the first (1).

If anyone can work out a way to do this, or just point me in the direction of some reference material it would be incredibly useful :)

EDIT:: It seems I forgot to mention that I'm trying to get from a specific vertex to another specific vertex. Quite important point there :/

EDIT2:: As someone below pointed out, I should highlight that edge weights are non negative.

Answer 1

I am copying this answer and adding also adding my proof of correctness for the algorithm:

I would use some variant of Dijkstra's. I took the pseudo code below directly from Wikipedia and only changed 5 small things:

1. Renamed dist to width (from line 3 on)
2. Initialized each width to -infinity (line 3)
3. Initialized the width of the source to infinity (line 8)
4. Set the finish criterion to -infinity (line 14)
5. Modified the update function and sign (line 20 + 21)

1  function Dijkstra(Graph, source):
2      for each vertex v in Graph:                                 // Initializations
3          width[v] := -infinity  ;                                // Unknown width function from 
4                                                                  // source to v
5          previous[v] := undefined ;                              // Previous node in optimal path
6      end for                                                     // from source
7      
8      width[source] := infinity ;                                 // Width from source to source
9      Q := the set of all nodes in Graph ;                        // All nodes in the graph are
10                                                                 // unoptimized – thus are in Q
11      while Q is not empty:                                      // The main loop
12          u := vertex in Q with largest width in width[] ;       // Source node in first case
13          remove u from Q ;
14          if width[u] = -infinity:
15              break ;                                            // all remaining vertices are
16          end if                                                 // inaccessible from source
17          
18          for each neighbor v of u:                              // where v has not yet been 
19                                                                 // removed from Q.
20              alt := max(width[v], min(width[u], width_between(u, v))) ;
21              if alt > width[v]:                                 // Relax (u,v,a)
22                  width[v] := alt ;
23                  previous[v] := u ;
24                  decrease-key v in Q;                           // Reorder v in the Queue
25              end if
26          end for
27      end while
28      return width;
29  endfunction

Some (handwaving) explanation why this works: you start with the source. From there, you have infinite capacity to itself. Now you check all neighbors of the source. Assume the edges don't all have the same capacity (in your example, say (s, a) = 300). Then, there is no better way to reach b then via (s, b), so you know the best case capacity of b. You continue going to the best neighbors of the known set of vertices, until you reach all vertices.

Proof of correctness of algorithm:

At any point in the algorithm, there will be 2 sets of vertices A and B. The vertices in A will be the vertices to which the correct maximum minimum capacity path has been found. And set B has vertices to which we haven't found the answer.

Inductive Hypothesis: At any step, all vertices in set A have the correct values of maximum minimum capacity path to them. ie., all previous iterations are correct.

Correctness of base case: When the set A has the vertex S only. Then the value to S is infinity, which is correct.

In current iteration, we set

val[W] = max(val[W], min(val[V], width_between(V-W)))

Inductive step: Suppose, W is the vertex in set B with the largest val[W]. And W is dequeued from the queue and W has been set the answer val[W].

Now, we need to show that every other S-W path has a width <= val[W]. This will be always true because all other ways of reaching W will go through some other vertex (call it X) in the set B.

And for all other vertices X in set B, val[X] <= val[W]

Thus any other path to W will be constrained by val[X], which is never greater than val[W].

Thus the current estimate of val[W] is optimum and hence algorithm computes the correct values for all the vertices.

Answer 2

I am not sure that Prim will work here. Take this counterexample:

V = {1, 2, 3, 4}

E = {(1, 2), (2, 3), (1, 4), (4, 2)}

weight function w:
  w((1,2)) = .1, 
  w((2,3)) = .3
  w((1,4)) = .2
  w((4,2)) = .25

If you apply Prim to find the maxmin path from 1 to 3, starting from 1 will select the 1 --> 2 --> 3 path, while the max-min distance is attained for the path that goes through 4.

Answer 3

You can still use Dijkstra's!

Instead of using +, use the min() operator.
In addition, you'll want to orient the heap/priority_queue so that the biggest things are on top.

Something like this should work: (i've probably missed some implementation details)

let pq = priority queue of <node, minimum edge>, sorted by min. edge descending
push (start, infinity) on queue
mark start as visited
while !queue.empty:
   current = pq.top()
   pq.pop()
   for all neighbors of current.node:
      if neighbor has not been visited
          pq.decrease_key(neighbor, min(current.weight, edge.weight))

It is guaranteed that whenever you get to a node you followed an optimal path (since you find all possibilities in decreasing order, and you can never improve your path by adding an edge)

The time bounds are the same as Dijkstra's - O(Vlog(E)).

EDIT: oh wait, this is basically what you posted. LOL.

Answer 4

Use either Prim's or Kruskal's algorithm. Just modify them so they stop when they find out that the vertices you ask about are connected.

EDIT: You ask for maximum minimum, but your example looks like you want minimum maximum. In case of maximum minimum Kruskal's algorithm won't work.

EDIT: The example is okay, my mistake. Only Prim's algorithm will work then.

Answer 5

Ok, answering my own question here just to try and get a bit of feedback I had on the tentative solution I worked out before posting here:

Each node stores a "path fragment", this is the entire path to itself so far.

0) set current vertex to the starting vertex
1) Generate all path fragments from this vertex and add them to a priority queue
2) Take the fragment off the top off the priority queue, and set the current vertex to the ending vertex of that path
3) If the current vertex is the target vertex, then return the path
4) goto 1

I'm not sure this will find the best path though, I think the exit condition in step three is a little ambitious. I can't think of a better exit condition though, since this algorithm doesn't close vertices (a vertex can be referenced in as many path fragments as it likes) you can't just wait until all vertices are closed (like Dijkstra's for example)

Answer 6

This can be solved using a BFS style algorithm, however you need two variations:

• Instead of marking each node as "visited", you mark it with the minimum weight along the path you took to reach it.

For example, if I and J are neighbors, I has value w1, and the weight of the edge between them is w2, then J=min(w1, w2).

• If you reach a marked node with value w1, you might need to remark and process it again, if assigning a new value w2 (and w2>w1). This is required to make sure you get the maximum of all minimums.

For example, if I and J are neighbors, I has value w1, J has value w2, and the weight of the edge between them is w3, then if min(w2, w3) > w1 you must remark J and process all it's neighbors again.

Answer 7

You could also use the "binary search on the answer" paradigm. That is, do a binary search on the weights, testing for each weight w whether you can find a path in the graph using only edges of weight greater than w.

The largest w for which you can (found through binary search) gives the answer. Note that you only need to check if a path exists, so just an O(|E|) breadth-first/depth-first search, not a shortest-path. So it's O(|E|*log(max W)) in all, comparable to the Dijkstra/Kruskal/Prim's O(|E|log |V|) (and I can't immediately see a proof of those, too).