The full version of this question is quoted below:
Let G be a connected graph with n vertices, m edges with distinct edge
weights. Let T be a tree of G with n vertices and n-1 edges (i.e. a
spanning tree), and define a bottleneck edge of T to be the edge of T
with the smallest weight. The max-bottleneck tree is a spanning tree
of G if there is no spanning tree with larger bottleneck edge. Prove
or provide a counter example for the following statement:
Every max-spanning tree of G is a max bottleneck tree of G
I think since the graph has unique edge weights, then every spanning tree of G is also unique. Then there is only one maximum spanning tree of G, and if I can prove that this tree is also a max bottle neck tree, then that would prove this statement to be true, but only if it's true for all graphs that have unique edge weights.
I've tried looking for counter examples to prove this false but so far it looks like every graph I draw with unique edge weights winds up have the max spanning tree also be a max bottleneck tree. I think I can use that to prove that this statement is true, but I am not sure how to word it.
Negate all the edge weights in the graph. Then the problems get changed to Minimum Spanning Tree and Minimum Bottleneck Spanning Tree respectively.
Now every Minimum Spanning Tree is also a Minimum Bottleneck Spanning Tree. Proof by Cut Property.
http://flashing-thoughts.blogspot.in/2010/06/everything-about-bottleneck-spanning.html
Related
Having an undirected graph with v vertices and e edges with each edges having a non-negative weight of at most w, the task is to find(if possible!) a subset of edges that connects all vertices with minimum cost. Also this graph has a root node which in the sub-graph should only be directly connected to exactly k other vertices(so it most have exactly k edges connected to it in the sub-graph). The algorithm should work in O(e⋅log(v)⋅log(w)) time.
So after some thinking I thought that we need to find a minimum spanning tree since it gives us the Minimum-cost subgraph which takes O(e.log(v)) time.However for insuring that the root node has exactly k edges in the resulting tree I couldn't come up with any working idea.
I think that maybe changing the weights of k edges that are connect to the root may enforce their selection,since in the Kruskal's algorithm we first sort the edges, if we pick exactly k edges that are already connected to the edge root in the graph and change their weights to some minimum amount then in the tree construction phase, chances are that this k edges make it to the MST. There are still problems here:
K edges should be picked so that it would be possible for them to make it to an MST.
The origin problem is from the exercise of Introduction of Algorithm.
23.1-5 Let e be a maximum-weight edge on some cycle of connected graph G=(V, E). Prove that there is a minimum spanning tree of G'=(V, E - {e}) that is also a minimum spanning tree of G. That is, there is a minimum spanning tree of G that does not include e.
The question is that: I think the proposition that all the minimum spanning tree of G do not include e is right. The e is the only one maximum-weight edge on some cycle. Is it ?
Update: 2016-10-28 20:21
Add the restriction that e is the only one maximum-weight edge on some cycle.
One test case is when there are nodes labeled 0..n-1 and there are links only between node i and node (i + 1) mod n (that is, a ring). In this case the minimum spanning tree is created by leaving out just one of the links. If e is the unique maximum weight edge it is not in the unique spanning tree, which is all the other links. If there is more than one edge of maximum weight then there are as many different minimum spanning trees as there are edges of maximum weight, each one of them leaving out a different edge of maximum weight and keeping the other ones in.
Consider the case when there is just one edge of maximum weight. Supposing somebody hands you a minimum spanning tree that uses this edge. Delete it from the tree, giving you two disconnected components. Now try adding each of the other edges in the cycle, one at a time. If the edge doesn't connect the two components, delete it again. If any of the edges connect the two components, you have a spanning tree of smaller weight than before, so it can't have been a minimum spanning tree. Can it be the case that none of the edges connect the two components? Adding an edge that doesn't connect the two components doesn't increase the set of nodes reachable from either component, so if no single edge connected the two components, adding all of them at the same time won't. But we know that adding all of these edges adds a path that connects the two nodes connected by the previous maximum weight edge, so one of the edges must connect the components. So our original so-called minimum spanning tree wasn't, and an edge which is of unique maximum weight in a cycle can't be part of a minimum spanning tree.
Your guess is correct：
all the minimum spanning tree of G do not include e is right.
First we need to prove:
e is not a light edge crossing any cut of G.
Let C be any cut that cuts e, since e is in a cycle, so e is not a light edge for any of those cuts, and all the other cuts won't have the edge e crossing it, we won't have that the edge is light for any of those cuts either.
Then we need to prove:
if e is not a light edge crossing any cut of G, then all the minimum spanning tree of G do not include e.
Which is exactly the inverse proposition of 23.1-3.
Could someone explain to be what a minimum leaf spanning tree is? I am confused to what exactly is a leaf in a spanning tree. I understand a spanning tree contains simple paths with no cycles and it spans all vertices in a graph G, but what is a minimum leaf one?
A leaf is a vertex of degree one in a tree. The degree of a vertex is equal to the number of edges that contain the vertex. A minimum leaf spanning tree is a problem that given a graph G = (V, E) and an integer i, is there a spanning tree T in G that contains at most i leaves?
By definition, a leaf here means a vertex with degree 1. So a minimum leaf spanning tree is a spanning tree with the minimum number of vertices with degree 1.
Given a connected undirected graph, the problem of finding the spanning tree with the minimum max degree has been well-studied (M. F¨urer, B. Raghvachari, "Approximating the minimum degree spanning tree to within one from the optimal degree", ACM-SIAM Symposium on Discrete Algorithms (SODA), 1992). The problem is NP-hard and an approximation algorithm has been described in the reference.
I am interested in the following problem - given a connected undirected graph G = (V1,V2,E), find the spanning tree with the maximum min degree over all internal nodes (non-leaf nodes). Can someone please tell me if this problem has been studied; is it NP-hard or does there exist a polynomial-time algorithm for solving it? Also, the graph can be considered to be bipartite for convenience.
As noted in Evgeny Kluev's comment, the leaves of a (finite) tree have degree 1. (Else, cycles would exist and the structure would not be a tree.)
If instead you mean to find a spanning tree with a node of maximum degree, from among all possible spanning trees on a connected undirected graph G, then just form a spanning tree whose root R is a node M of G with maximal degree among all the nodes of G, and all neighbors of M are children of R.
It looks like exact cover by 3-sets can be reduced to this problem. Represent the 3-sets by vertices of degree 4, each with 3 edges connecting it to 3 nodes representing its elements in the original problem instance. The additional 4th edge connects all the "3-set" nodes to a single vertex V.
This graph is biparite - every edge is between a "3-set" node and an "element" node (or V). Now this graph has a spanning tree of max min degree = 4 if and only if the original problem has a solution.
Obviously there need to be enough of the 3-sets so that the node V doesn't lower the max min degree of the tree, but this limit doesn't change the NP-hardness of the problem.
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.