Given an undirected and connected graph G(V,E) with weight function w:E->R, an edge e(u,v) belongs to E. Which algorithm that runs in linear run-time complexity can assure if there's a minimal spanning tree that contains the edge e?
Perform depth-first search starting from u and ignoring any edge with weight equal to weight of given edge or heavier. If DFS completes without visiting v, this means there is no cycle where given edge is the heaviest edge, therefore given edge is contained in some minimal spanning tree.
This is asking you to implement Prim's Minimum Spanning Tree twice.
The first time run the algorithm as normal. Make a note of the weight of the mst.
The second time, instead of starting with an empty tree as Prim's algorithm normally does, start with the nodes u and v already on the tree -- which is the same as having edge e(u,v). Then continue building the mst.
at the end if the second mst is the same weight as the first mst, then edge e(u,v) can be in a mst. If not, there is a lighter mst.
By the way, 2*O(n) is still O(n)
Related
In a weighted undirected graph, I need to modify Kruskal's algorithm to find the MST conditional to the fact that it includes a given edge 'e' in O(m log n) time. How can I do that?
Consider kruskal's algorithm:
sort all edges in increasing order of edge weight
for every edge from u->v with weight w in the list:
if u and v are not connected, take this edge
This algorithm at any step does not consider the graph to be "individual" nodes; rather, they are all components that can be connected by any arbitrary node within the component. Thus, if we want to force some edges to be taken in the MST, we can simply connect them before running kruskal's, and remove the edge(s) forced from the edge list provided to kruskal's.
As when we traverse the undirected connected graph using DFS and mark out only edges which we travel during DFS, we end up with a DFS tree which is basically a tree and traversing a tree requires O(v) complexity where v is the number of vertices, then why it stated that complexity is O(v + e)?
I know it's a noob question but I am confused.
The DFS tree is the tree made of the edges you traversed. You indeed only traverse O(V) edges. But in order to traverse an edge, you first examine the edge to check if it will lead to a vertex you already encountered. This means you examine more edges than you traverse. Indeed, you examine O(E) edges. So the total work is O(V+E).
Note: Because your graph is connected, you are sure that E > V. In that case the complexity can be rewritten O(E).
you find all the nodes of graph through edges so the time complexity depends upon no. of edges as well that's why the O(e) is also included. If you consider complete graph TC will be O(V^2).
Consider two different graphs:
A graph having more edges than vertices, for instance a connected graph with a high minimum degree.
When the DFS algorithm visits a vertex, it will have to follow each of the edges that connect this vertex, repeating a DFS traversal from the neighboring vertices. Of course, if that neighbor was already visited, the DFS algorithm will backtrack, but at least we can state that the edge had to be processed.
This procedure will process all edges of the graph. So in this case we can say the algorithm is O(e)
A graph having fewer edges than vertices, often a disconnected graph (in the extreme case there are no edges).
When the DFS algorithm has visited all vertices that it can reach from a traversal that started in vertex A, it must still iterate over the remaining vertices, to find vertices that might not have been visited. These unvisted vertices do not belong to the same connected component). Another DFS traversal must start from there.
This way all vertices are processed. So in this case the algorithm has a O(v) time complexity
So in general, the algorithm has a O(max(e, v)) time complexity. You could also say that the algorithm must visit all edges and all vertices, and so the algorithm has a O(e+v) time complexity. Both are equivalent.
Suppose we are given a graph G and a spanning tree T of it. How can we check that if this spanning tree is a MST or not?
I suggest that for each edge i that is not in T, all edges j on the unique path of T from one end of i to the other end, we must have w_i >= w_j
You want to check that for each edge (u, v) not in the MST, the path from u to v in the MST has no edge with weight larger than that of (u, v). For a single vertex in the tree, you can use a single BFS or DFS to find the largest-weight edge on the path to all other nodes, so this gives an O(n2) algorithm (n times O(n) search). You can probably do better by not starting from scratch for each vertex. That said, it may be more efficient to just calculate an MST and see if the sum of all edge weights is the same. As mentioned by #Niklas in the comments, there are linear time methods for verifying an MST, but they seem to be significantly more involved.
I have the following problem on my homework:
Give an O(n+m) algorithm to find that whether an edge e would be a part of the MST of a graph
(We are allowed to get help from others on this assignment, so this isn't cheating.)
I think that I could do a BFS and find if this edge is a edge between two layers and if so whether this edge was the smallest across those layers. But what could I say when this edge is not a tree edge of the BFS tree?
As a hint, if an edge is not the heaviest edge in any cycle that contains it, there is some MST that contains that edge. To see this, consider any MST. If the MST already contains the edge, great! We're done. If not, then add the edge into the MST. This creates a cycle in the graph. Now, find the heaviest edge in that cycle and remove it from the graph. Everything is now still connected (because if two nodes used to be connected by a path that went across that edge, now they can be connected by just going around the cycle the other way). Moreover, since the cost of the edge was deleted wasn't any smaller than the cost of the edge in question (because the edge isn't the heaviest edge in the cycle), the cost of this tree can't be any greater than before. Since we started with an MST, we must therefore end with an MST.
Using this property, see if you can find whether the edge is the heaviest edge on any cycle in linear time.
We will solve this using MST cycle property, which says that, "For any cycle C in the graph, if the weight of an edge e of C is larger than the weights of all other edges of C, then this edge cannot belong to an MST."
Now, run the following O(E+V) algorithm to test if the edge E connecting vertices u and v will be a part of some MST or not.
Step 1
Run dfs from one of the end-points(either u or v) of the edge E considering only those edges that have weight less than that of E.
Step 2
Case 1
If at the end of this dfs, the vertices u and v get connected, then edge E cannot be a part of some MST. This is because in this case there definitely exists a cycle in the graph with the edge E having the maximum weight and it cannot be a part of the MST(from the cycle property).
Case 2
But if at the end of the dfs u and v stay disconnected, then edge E must be the part of some MST as in this case E is never the maximum weight edge of the cycles that it is a part of.
Find if there are any paths that are cheaper than the current one (u,v) that creates a cycle to u and v. If yes, then (u,v) is not on the mst. Otherwise it is. This can be proved by the cut property and the cycle property.
Does the opposite of Kruskal's algorithm for minimum spanning tree work for it? I mean, choosing the max weight (edge) every step?
Any other idea to find maximum spanning tree?
Yes, it does.
One method for computing the maximum weight spanning tree of a network G –
due to Kruskal – can be summarized as follows.
Sort the edges of G into decreasing order by weight. Let T be the set of edges comprising the maximum weight spanning tree. Set T = ∅.
Add the first edge to T.
Add the next edge to T if and only if it does not form a cycle in T. If
there are no remaining edges exit and report G to be disconnected.
If T has n−1 edges (where n is the number of vertices in G) stop and
output T . Otherwise go to step 3.
Source: https://web.archive.org/web/20141114045919/http://www.stats.ox.ac.uk/~konis/Rcourse/exercise1.pdf.
From Maximum Spanning Tree at Wolfram MathWorld:
"A maximum spanning tree is a spanning tree of a weighted graph having maximum weight. It can be computed by negating the weights for each edge and applying Kruskal's algorithm (Pemmaraju and Skiena, 2003, p. 336)."
If you invert the weight on every edge and minimize, do you get the maximum spanning tree? If that works you can use the same algorithm. Zero weights will be a problem, of course.
Although this thread is too old, I have another approach for finding the maximum spanning tree (MST) in a graph G=(V,E)
We can apply some sort Prim's algorithm for finding the MST. For that I have to define Cut Property for the maximum weighted edge.
Cut property: Let say at any point we have a set S which contains the vertices that are in MST( for now assume it is calculated somehow ). Now consider the set S/V ( vertices not in MST ):
Claim: The edge from S to S/V which has the maximum weight will always be in every MST.
Proof: Let's say that at a point when we are adding the vertices to our set S the maximum weighted edge from S to S/V is e=(u,v) where u is in S and v is in S/V. Now consider an MST which does not contain e. Add the edge e to the MST. It will create a cycle in the original MST. Traverse the cycle and find the vertices u' in S and v' in S/V such that u' is the last vertex in S after which we enter S/V and v' is the first vertex in S/V on the path in cycle from u to v.
Remove the edge e'=(u',v') and the resultant graph is still connected but the weight of e is greater than e' [ as e is the maximum weighted edge from S to S/V at this point] so this results in an MST which has sum of weights greater than original MST. So this is a contradiction. This means that edge e must be in every MST.
Algorithm to find MST:
Start from S={s} //s is the start vertex
while S does not contain all vertices
do
{
for each vertex s in S
add a vertex v from S/V such that weight of edge e=(s,v) is maximum
}
end while
Implementation:
we can implement using Max Heap/Priority Queue where the key is the maximum weight of the edge from a vertex in S to a vertex in S/V and value is the vertex itself. Adding a vertex in S is equal to Extract_Max from the Heap and at every Extract_Max change the key of the vertices adjacent to the vertex just added.
So it takes m Change_Key operations and n Extract_Max operations.
Extract_Min and Change_Key both can be implemented in O(log n). n is the number of vertices.
So This takes O(m log n) time. m is the number of edges in the graph.
Let me provide an improvement algorithm:
first construct an arbitrary tree (using BFS or DFS)
then pick an edge outside the tree, add to the tree, it will form a cycle, drop the smallest weight edge in the cycle.
continue doing this util all the rest edges are considered
Thus, we'll get the maximum spanning tree.
This tree satisfies any edge outside the tree, if added will form a cycle and the edge outside <= any edge weights in the cycle
In fact, this is a necessary and sufficient condition for a spanning tree to be maximum spanning tree.
Pf.
Necessary: It's obvious that this is necessary, or we could swap edge to make a tree with a larger sum of edge weights.
Sufficient: Suppose tree T1 satisfies this condition, and T2 is the maximum spanning tree.
Then for the edges T1 ∪ T2, there're T1-only edges, T2-only edges, T1 ∩ T2 edges, if we add a T1-only edge(x1, xk) to T2, we know it will form a cycle, and we claim, in this cycle there must exist one T2-only edge that has the same edge weights as (x1, xk). Then we can exchange these edges will produce a tree with one more edge in common with T2 and has the same sum of edge weights, repeating doing this we'll get T2. so T1 is also a maximum spanning tree.
Prove the claim:
suppose it's not true, in the cycle we must have a T2-only edge since T1 is a tree. If none of the T2-only edges has a value equal to that of (x1, xk), then each of T2-only edges makes a loop with tree T1, then T1 has a loop leads to a contradiction.
This algorithm taken from UTD professor R. Chandrasekaran's notes. You can refer here: Single Commodity Multi-terminal Flows
Negate the weight of original graph and compute minimum spanning tree on the negated graph will give the right answer. Here is why: For the same spanning tree in both graphs, the weighted sum of one graph is the negation of the other. So the minimum spanning tree of the negated graph should give the maximum spanning tree of the original one.
Only reversing the sorting order, and choosing a heavy edge in a vertex cut does not guarantee a Maximum Spanning Forest (Kruskal's algorithm generates forest, not tree). In case all edges have negative weights, the Max Spanning Forest obtained from reverse of kruskal, would still be a negative weight path. However the ideal answer is a forest of disconnected vertices. i.e. a forest of |V| singleton trees, or |V| components having total weight of 0 (not the least negative).
Change the weight in a reserved order(You can achieve this by taking a negative weight value and add a large number, whose purpose is to ensure non-negative) Then run your family geedy-based algorithm on the minimum spanning tree.