I have to solve this problem and it has been bugging me for hours and I can't seem to find a valid solution that satisfies the time complexity required.
For any edge e in any graph G, let G\e denote the graph obtained by deleting e from G.
(a)Suppose we are given an edge-weighted directed graph G in which the shortest path σ from vertex s to vertex t passes through every vertex of G. Describe an algorithm to compute the shortest-path distance from s to t in G\e, for every edge e of G, in O(VlogV) time. Your algorithm should output a set of E shortest-path distances,one for each edge of the input graph. You may assume that all edge weights are non-negative.[Hint: If we delete an edge of the original shortest path, how do the old and new shortest paths overlap?
(b) Describe an algorithm to solve the replacement paths problem for arbitrary undirected graphs in O(V log V ) time.
a) Consider the shortest path P between vertices s and t. Since P is a shortest path, there is no edge between any two vertices u and v in P in which the length of shortest path between u and v is bigger than 1 in the induced graph P. Since every vertex in G is present in P, So every edge of G is present in the induced graph of P. So we conclude that every edge in G is between two adjacent vertices in P (not the induced graph of P). The only thing you should check for each edge e which connects vertices u and v in G, is that there is another edge which connects u and v in G. If so, the shortest path doesn't change in G/e, otherwise s and t will lie on different components.
First of all I want to mention that the complexity can't only depend on V since the output should contain E values, therefore I guess it should be O(E + Vlog(V)).
I think the following idea should work for problem b) and thus for a) as well. Let σ be the shortest s-t path.
For every edge e in G/E(σ), the shortest path in G/e is still σ.
If we remove an edge e in σ, then the shortest path in the remaining graph would like this: it would start going along σ, then would continue outside (might be from the very first vertex s, then go back to σ (might be at the very last vertex t). Let's then iterate over all edges e=(u,v) that either go from G/σ to σ, or from one vertex of σ, that is closer to s, to another vertex of σ, that is closer to t, as edges of potential candidate paths for some shortest s-t path in G/e' for some e'. Suppose that w is the last vertex in the shortest s-u path U that belongs to σ. Then U+e+σ[v..t] is a potential candidate for a shortest s-t path in graphs G/e' for all e' in σ[w..v]. In order to efficiently store this candidate, we can use a Segment Tree data structure with min operation. This would give us O(E log(E)) total complexity for all updates after we consider all edges. It might be possible to make it O(Vlog(V)) if we look closer at the structure of the updates, but I am not sure at the moment, might be not. Precomputing vertices w for each shortest s-u path above can be done in a single Dijkstra run, if we already have σ.
So in the end the solution runs in O(Elog(E)) time. I am not sure how to make it O(E + Vlog(V)), it is possible to make Dijkstra to run in this time by using Fibonacci heap, but for the updates for candidates I don't really see a faster solution at the moment.
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Let V be the set of vertices in a graph. I understand that given a graph of |V| vertices with no negative cycles, a shortest path will always have |V|-1 edges. I still don't quite understand why checking every edge |V|-1 times guarantees that Bellman Ford's algorithm will produce the shortest path. Can someone help me understand this better?
In the first phase (checking every edge once), you consider all potential shortest paths with only one edge. So if the shortest path has only one edge, you will have found it after the first phase. (But you don't yet know that you've found the shortest path.)
After the second phase of checking all edges, you will have considered all potential paths of two edges, since you consider all possible extensions by one edge of the paths you already considered. So if the shortest path has at most two edges, you will have found it after the second phase.
And so on… If the shortest path has at most |V|−1 edges (which it does), you will have found it after |V|−1 phases.
This is done in the relaxation step. In this step, vertice distances are updated incrementally. Intuitively, these updates are propagated throughout the graph: When you find a shorter path to one node, then its neighbours might also benefit from this shortcut. In the next iteration, they get updated too, and so on.
But this updating cannot continue indefinitely. The maximum number of times it can occur is precisely |V| - 1. You can visualise it as a path of updates propagating from one vertex, throughout all other vertices, back to the source. The source itself of course does not find a shortcut to itself.
In my opinion, it's necessary to do some math to understand it fully (although the other answers already give some intuition). Let's prove the following statement by induction:
Let s be the source vertex and u be any other vertex in the directed graph. After i iterations of Bellman-ford algorithm:
if d[u] is not infinity, it is equal to the length of some path from s to u.
if there is a path from s to u with at most i edges, then d[u] is at most the length of the shortest path from s to u with at most i edges.
Proof
The base case is i=0 (before any iteration). At this moment, d[s] = 0 and d[u] = ∞ for all u. It's obvious that (1) and (2) are valid for this case.
Let's consider that it's valid for i (inductive hypothesis).
At the i+1 iteration, consider a moment when a vertex v distance is updated by d[v] := d[u] + w[u][v]. By the inductive hypothesis, d[u] is the length of some path from s to u. Then d[u] + w[u][v] is the length of the path from source to v that follows the path from source to u and then goes to v. So, (1) is valid for the (i+1)-th iteration and by induction is valid for all natural i.
Now, consider a shortest path P (there may be more than one) from s to v with at most i+1 edges. Let u be the last vertex before v on this path. Then, the part of the path from s to u is a shortest path from s to u with at most i edges. Indeed, suppose that it's not. Then there would exist some strictly shorter path from s to u with at most i edges, which could be appended to the edge uv to obtain a path with at most i edges that is strictly shorter than P, which is contradiction. By inductive hypothesis, d[u] after i iterations is at most the length of the shortest path from s to u. Therefore, d[u] + w[u][v] is at most the length of P. In the (i+1)-th iteration, d[v] gets compared with d[u] + w[u][v], and is set equal to it if d[u] + w[u][v] is smaller. Therefore, after i+1 iterations, d[v] is at most the length of P, which is a shortest path from s to v that uses at most i+1 edges. So, (2) is valid for the (i+1)-th iteration and by induction is valid for all natural i.
To finish, notice that if the directed graph has |V| vertices and no negative cycles, a shortest path from a source s to a vertex u has at most |V| - 1 edges. So, by the previous result, Bellman-Ford indeed find all shortest paths from s.
Notice that The Bellman-Ford Algorithm is based on a recursive relationship, which states that:
d(i,v) is the shortest distance from the source to each node v with at most i edges between them.
As i grows to i+1, you allow one more edge, and you update the path length only if it improves.
Since we know that the shortest path can have at most n-1 edges, there is no point to grow i further, i.e, the shortest path has to be:
d(n-1,v).
I have been trying to prove this for a decent amount of time, but nothing is coming to my mind. Anybody have some tips? I'd be happy to see an explanation.
I also found this question on StackOverflow, that says this:
If the direct u->v traffic doesn't knock out any links in v->u, then
the v->u problem is equivalent of finding the maximum flow from v->u
as if the u->v flow doesn't happen.
He describes how it can be solved, but still there is no answer to the question that the author asked.
So, the problem is:
Given a directed graph, at each vertex the number of incoming and outgoing edges is the same. Let there be k edge-disjoint paths from b to a in this graph.
How can I prove that it also contains k edge-disjoint paths from a to b?
Thanks in advance!
We can try to argue about the general case where the graph is a multi-graph (i.e. can have multi-edges and loops).
Note: Following the convention that two copies of an edge count towards in and degree twice. And a loop count towards both in and out degree once. Also assuming when you say paths, you mean simple paths.
Using induction on the number of vertices in the graph G.
Base case: G has only vertices a and b.
Now as there are k edge-disjoint paths from a to b, all of them are simply k copies of the edge a->b. Thus to have in and out degrees same for both vertices there have to be k copies of b->a, and thus the claim holds.
Induction G has n >=1 vertices apart from a and b.
Let the nth vertex be u. Let in-degree of u, same as its out-degree be d. Let the d vertices with edges "going into" u be s1,s2,..,sd and similarly the d vertices with edges going out from u be t1,t2,..,td (note all these vertices may not be unique). Just pair these vertices up. Say si with ti (1<=i<=d). Now just "short-circuit" the vertex u, i.e. rather than having the edge si->u->ti, just have si->ti. Let the new graph be G'. It is trivial to see in and out degree of vertices are still the same in G' (as it was in G). And it is not hard to argue that the new graph still has k disjoint paths from a to b. Additionally G' has one less vertex. So apply inductive hypothesis, and claim holds for G'. Lastly not hard to check again, un-short circuiting u still keeps the claim intact for G.
Problem : A directed graph G with n vertices and a special vertex u is provided. We call a vertex v ‘interesting’ if there is a path from v to a vertex w such that there is a cycle containing the vertices w and u. Write an O(n) time algorithm which takes G (the whole graph) and the node u as input and returns all the interesting vertices.
Ineffiecient Algorithm : My idea initially was to consider the node u and compute all the cycles that contain u. (This itself seems like traversing through the nodes using DFS and then forward-tracking as well when you encounter a visited node) Now from each vertex on these cycles we can compute the number of nodes on the graph that do not belong to the cycle(s) but is connected with each particular vertex w not equal to u on a cycle. Add all these values to get the desired answer. This isn't an O(n) algorithm.
There are two cases:
If there are no cycles containing u, then no vertex can be "interesting".
If there are any cycles containing u, then a vertex v is "interesting" if and only if there's a path from v to u. (We don't need to worry about the w in the problem description, because if a cycle contains two vertices u and w, then any path that ends at u can be extended to end at w and vice versa.)
So, the most efficient algorithm would be as follows:
Using DFS, determine if u is in a cycle. (We don't need to find all cycles; we just need to determine whether there are any.)
Using DFS in the "reverse" direction, find all vertices from which u is reachable.
DFS requires O(|V| + |E|) time, which is greater than O(n) = O(|V|) unless |E| is in O(n); but then, there's no way to even read in the entire graph definition in less than |E| time, so this is unavoidable. Whoever gave you this question must not have really thought this through.
Given a weighted undirected graph G and two nodes U,V to get the shortest path. How can i get the shortest path from U to V that uses a even number of edges (if possible to get it) ?
I've found some articles on the web speaking that a modification on the original graph is necessary. But i can't understand how to do it.
There is some good material to study on this problem ?
You'll need to build an intermediate graph and run Dijkstra's on that graph.
Given a graph G = (V, E), create a new graph G' = (V', E'), with V' a new set of vertices v_even and v_odd for every vertex v in V and E' the set of vertices as follows:
If (u, v) is an edge in G, then (u_odd, v_even) and (u_even, v_odd) are edges in G', with the same weight.
Obviously, the new graph has twice as many edges and vertices as the original graph.
Now, if you wanted to find the shortest path between s and t in G, simply run Dijkstra's on G' to find the shortest path between s_even and t_even.
The running time is still O(|V| log |E|).
Make a copy of your graph with the same weights and call it G'.
Connect every vertex of G to the corresponding vertex in G' and set the weight of the new edges to zero.
Delete copy of u from G' and delete v from G.
Now, the set of edges you added between G and G' constitute a matching M. Take that matching and find a minimum augmenting path for that matching.
Such a path must have u as one of its end points and copy of v as the other endpoint, because they are the only uncovered vertices. If you merge back the copies and get rid of the added edges, the path you found corresponds to an even path, because it starts at one copy and ends at another. Also, every even path corresponds to an augmenting path (by same argument), therefore the minimum of one is also the minimum of the other. This is explained in here.
What about running Dijkstra where each node has two values. One is odd (coming from even value) and other is even value.
Here is an excise
Suppose we are given the minimum spanning tree T of a given graph G
(with n vertices and m edges) and a new edge e = (u, v) of weight w
that we will add to G. Give an efficient algorithm to find the minimum
spanning tree of the graph G + e. Your algorithm should run in O(n)
time to receive full credit.
I have this idea:
In the MST, just find out the path between u and v. Then find the edge (along the path) with maximum weight; if the maximum weight is bigger than w, then remove that edge from the MST and add the new edge to the MST.
The tricky part is how to do this in O(n) time and it is also I get stuck.
The question is that how the MST is stored. In normal Prim's algorithm, the MST is stored as a parent array, i.e., each element is the parent of the according vertex.
So suppose the excise give me a parent array indicating the MST, how can I release the above algorithm in O(n)?
First, how can I identify the path between u and v from the parent array? I can have two ancestor arrays for u and v, then check on the common ancestor, then I can get the path, although in backwards. I think for this part, to find the common ancestor, at least I have to do it in O(n^2), right?
Then, we have the path. But we still need to find the weight of each edge along the path. Since I suppose the graph will use adjacency-list for Prim's algorithm, we have to do O(m) (m is the number of edges) to locate each weight of the edge.
...
So I don't see it is possible to do the algorithm in O(n). Am I wrong?
The idea you have is right. Note that, finding the path between u and v is O(n). I'll assume you have a parent array identifying the MST. tracking the path (for max edge) from u to v or u to root vertex should take only O(n). If you reach root vertex, just track the path from v to u or root vertex.
Now that you have the path from u -> u1 ... -> max_path_vert1 -> max_path_vert2 -> ... -> v, remove the edge max_path_vert1->max_path_vert2 (assuming this is greater than the added edge) and reverse the parents for u->...->max_path_vert1 and mark parent[u] = v.
Edit: More explanation for clarity
Note that, in MST there will be exactly one path between any pair of vertices. So, if you can trace from u->y and v->y, you have only traced through atmost n vertices. If you traced more than n vertices that means you visited a vertex twice, which will not happen in an MST. Ok, now hopefully you're convinced it's O(n) to track from u->y and v->y. Once you have these paths, you have established a path from u->v. Do you see how? I'm assuming this is an undirected graph, since finding MST for directed graph is a different concept in itself. For undirected graph, when you have a path from x->y you have a path from y-x. So, u->y->v exist. You don't even need to trace back from y->v, since weights for v->y will be same as that of y->v. Just find the edge with the maximum weight when you trace from u->y and v->y.
Now for finding edge weights in O(1); how are you storing your current weights? Adjacency list or adjacency matrix? For O(1) access, store it the way parent vertex array is stored. So, weight[v] = weight(v, parent[v]). So, you'll have O(1) access. Hope this helps.
Well - your solution is correct.
But regarding implementation, I dont see why you are using G instead of T to find the path between u and v. Using any search traversal in T for the path between u and v, will give you O(n). - That is, you can assume that v is the root and performs a Depth-First Search algorithm [in this case, you will have to assume all neighbors of v as children] - and stop the DFS once you find u - then, the nodes in the stack corresponds to the path between u and v.
It is easy afterward to find the cost of each edge in the path (O(n)), and it is easy as well to delete/add edges. In total O(n).
Does that help somehow ?
Or maybe you are getting O(n^2) - according to my understanding - because you access the children of a vertex v in T in O(n) -- Here, you have to present your data structure as a mapped array so that the cost is reduced to O(1). [for instace, {a,b,c,u,w}(vertices) -> {0,1,2,3,4}(indices of vertices).