Can we use BFS algo to identify the furthest vertex from staring vertex v in any graph, in terms of number of edges.
Yes. Let's call the distance from A node to B node the number of edges from A to B.
BFS is finding all the nodes of distance 1, then all nodes of distance 2 and so on. For finding the furthest vertex just retain the last node searched, because he has certain the longest distance.
Suppose we have a N-complete graph G(V, E), which is undirected and weighted. Given K, how could we find the sub-graph G'(V', E'), which is also a complete graph and |V'| = K and the sum of all edges in E' is maximum.
Since the original graph is complete, so you can just sum the weights for all the nodes, and keeps only the maximum K nodes and their edges.
Is this right? If not, you can delete the node with minimal weight sum each time, calculate weights sums again, this is more time consume.
this is an example problem from an old midterm:
Let G = (V, E) be a connected undirected graph where the edges have positive integer edge weights associated with them, and a vertex s ∈ V is the source. Provide an algorithm that for each vertex t ∈ V reports the minimum last edge weight on a non-decreasing path from s to t (∞ if there is no such path). A path v1, v2, . . . vr is non-decreasing if w(v_i, v_i+1) ≤ w(v_i+1, v_i+2) for i = 1, 2, ...r−2.
Am I correct in thinking that the problem wants me to come up with an algorithm that given a graph with a starting vertex, can find the length of the shortest path, that also has edge weights increasing as you go down the path, to every other vertex it can reach?
The question asks to write an algorithm that provides a path from the source vertex s to a (can be any) vertex t where the weights of the path between s and t increase (or stay the same). It then asks to get the minimum weight of the last edge of all these possible paths.
In other words, you need to see which path from s to t (which should be non-decreasing in weights) gives the lowest weight in the last edge.
So I have an exercise that I should prove or disprove that:
1) if e is a minimum weight edge in the connected graph G such that not all edges are necessarily distinct, then every minimum spanning tree of G contains e
2) Same as 1) but now all edge weights are distinct.
Ok so intuitively, I understand that for 1) since not all edge weights are distinct, then it's possible that a vertex has the path with edge e but also another edge e_1 such that if weight(e) = weight (e_1) then there is a spanning tree which does not contain the edge e since the graph is connected. Otherwise if both e_1 and e are in the minimum spanning tree, then there is a cycle
and for 2) since all edge weights are distinct, then of course the minimum spanning tree will contain the edge e since any algorithm will always choose the smaller path.
Any suggestions on how to prove these two though? induction? Not sure how to approach.
Actually in your first proof when you say that if both e and e_1 are in G, then there's a cycle, that's not true, because they're minimal edges, so there doesn't have to be a cycle, and you do need to include them both into the MST, because if |E| > 1 and |V| > 2 then they both have to be there.
Anyways, a counter example for the first one is a complete graph with all edges of the same weight as e, the MST will contain only |V|-1 edges, but you didn't include all the other edges of that same weight, hence you have a contradiction.
As for the second one, if all edges are distinct, then if you remove the minimum edge and want to reconstruct the MST, the only way to go about this is to add a an edge connecting the 2 disjoint sets that were broken up by that minimum-weight edge.
Now suppose that you didn't remove that minimum-weight edge, and added that other edge, now you've created a cycle, and since all edges are distinct the cycle-creating edge will be greater than all of them, hence if you remove any former MST edge from that cycle, it will only increase the size of the MST. Which means that pretty much all former MST edges are critical when all edges have distinct weights.
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.