This is not a homework. I am trying to do exercises from a textbook to understand MST (minimum spanning tree).
Suppose I have a cycle C in a weighted undirected graph G. As I understand, the following is correct:
The heaviest edge in C belongs to no MST of G. That is, there is no MST of G, which contains that edge.
The lightest edge in C belongs to some MST of G. That is, there is an MST of G, which contains that edge.
Now I wonder if the followings claims are correct too.
The lightest edge in C belongs to all MST of G. That is, there is no MST of G, which does not contain that edge.
Any edge in C except the heaviest one belongs to some MST. That is, for each edge in C except the heaviest one there is an MST, which contains that edge.
Could you prove the last claim?
Even for the first claim if there are multiple edges which are lightest, all need not be included in the MST.
The first one of your claims is always true. The lightest edge is on the MST for any graph.
The second one is not always true. It is always true if the entire graph is a
cycle and thus every node has 2 edges incident to it. However, in the general case,
an edge (u,v) of weight k is never on MST whenever there is a path between the nodes u and v
connecting them at a total weight less than k.
I don't think your claims are valid. The problem is that you are only considering a cycle in a larger graph.
Consider for example a graph G consisting of 6 nodes in a cycle (with random weights >1). Your claims might hold for that graph but now add 1 node in the center of the graph and connect it with 6 links of cost 1. The MST of your entire graph now will consist of only those 6 edges (which form a star).
If you now look at your claims, you'll see:
The lightest edge in your cycle does not belong to the MST (=star)
None of the edges in the cycle are in the MST
Related
Let G = (V , E) be a weighted undirected connected graph that contains a cycle, and let e be the maximum-weight edge among all edges in the cycle. I need to prove that there exists a minimum spanning tree of G which does NOT include e.
The idea is intuitively clear and I can show it on a cycle, consisting of 3 nodes. But I do not know how to show that formally for any cycle.
Assume that exists MST with e. Removing e from it, splits tree in two parts. Expecially, it splits cycle nodes into two non empty parts, call them A and B. Since these nodes form a cycle there is at least one more edge between A and B nodes, call it f. Than MST-e+f is a spanning tree with weight less than MST. That means it is not possible to have MST with e.
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.
Let's say there's Graph G such that it all its edges have weights that correspond to distinct integers. So no two edge has the same weight.
Let E be all the edges of G. Let emax be an edge in E with the maximum weight.
Another property of Graph G is that every edge e belongs to some cycle in G.
I have to prove that no minimum spanning tree of G contains the edge emax.
I can see why this is true, since all edges are distinct and every edge belongs to a cycle, the minimum spanning tree algorithm can simply choose the edge with lower weight in the cycle that contains emax.
But I'm not sure how to concretely prove it.
This is related to the Cycle Property of the Minimum Spanning Tree, which is basically saying that given a cycle in a graph the edge with the greatest weight does not belong in the MST (easily proven by contradiction in the link above). Thus since the edge emax belongs to a cycle it must not be in the MST.
Proof by contradiction works here. Suppose you have a minimum spanning tree including the maximum edge. If you remove that edge you have two components no longer connected from each other. Every vertex is in one component or the other. There is a cycle including the maximum edge. Start on a vertex at one side of the maximum edge and move along the cycle. Because you will eventually cycle round to the other side of the maximum edge in the other component you will find - before then - an edge which has one vertex in one of the disconnected components and one vertex in another of the disconnected components. Since the components are disconnected that edge is not in the minimum spanning tree. By adding it to the tree you reconnect the components and create a minimum spanning tree with smaller weight than you started with - so you original minimum spanning tree was not minimum.
Does a MST contain the maximum weight edge?
Sometimes, Yes.
It depends on the type of graph. If the edge with maximum weight is the only bridge that connects the components of a graph, then that edge must also be present in the MST.
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.
If we have an (arbitrary) connected undirected graph G, whose edges have distinct weights,
does every MST of G contains the minimum weighted edge?
is there an MST of G that does not contain the maximum weighted edge?
Also, I'm more thankful if someone can give a hint of the key things one must keep in mind when dealing with such MST questions.
This is a homework problem. Thanks.
is there an MST of G that does not contain the maximum weighted edge?
There may be, but there doesn't have to be. Consider a 4-vertex graph as follows:
[A]--{2}--[B]
| |
| |
{1} {3}
| |
| |
[C]-{50}--[D]
The minimum spanning tree consists of the edge set {CA, AB, BD}. The maximum edge weight is 50, along {CD}, but it's not part of the MST. But if G were already equal to its own MST, then obviously it would contain its own maximum edge.
does every MST of G contains the minimum weighted edge?
Yes. MSTs have a cut property. A cut is simply a partition of the vertices of the graph into two disjoint sets. For any cut you can make, if the weight of an edge in that cut is smaller than the weights of the other edges in the cut, then this edge belongs to all MSTs in the graph. Because you guaranteed that the edge weights are distinct, you have also guaranteed that there is an edge which is smaller than all other edges.
Also, I'm more thankful if someone can give a hint of the key things one must keep in mind when dealing with such MST questions.
Your best bet is to reason about things using the properties of MSTs in general, and to try to construct specific counterexamples which you think will prove your case. I gave an instance of each line of reasoning above. Because of the cut and cycle properties, you can always determine exactly which edges are in an MST, so you can systematically test each edge to determine whether or not it's in the MST.
Does every MST of G contains the minimum weighted edge?
Yes. Lets assume we have a MST which does not contain the min weight edge. Now the inclusion of this edge to the MST will result in a cycle. Now there will always be another edge in the cycle which can be removed to remove the cycle and still maintain the graph(MST) connected.
Is there an MST of G that does not contain the maximum weighted edge?
Depends on the graph. If the graph itself is a tree then we need to include all of its n-1 edges in the MST, so the max weight edge cannot be excluded. Also if the max weight edge is a cut-edge so that its exclusion will never result in connectivity, then the max weight edge cannot be excluded. But if the max weight edge is a part of a cycle then it is possible to exclude from the MST.
For your first question the answer is no, and kruskal's algorithm proves it. It will always select the minimum cost edge.
For the second question the answer is yes, and it's trivial to find an example graph:
1 - 2 (cost 10)
2 - 3 (cost 100)
3 - 1 (cost 1000)
The third edge will never be selected as it introduces a cycle. So basically, if the edge with the maximum cost would create a cycle if inserted in the MST, it won't be inserted.
I see you too are studying for CSC263 through the 2009 test? (Same here!)
Another way to see that the minimum is always in the MST is to look simply at this minimum edge (call it e):
e
v1 ---------------- v2
(Assume this has connections to other verticies). Now, for e NOT to be included in the final MST means at one point we have, without loss of generality, v1 in the MST but not v2. However, the only way to add v2 without adding e would be to say that the addition of v1 didn't add e to the queue (because by definition, e would be at the top of the queue because it has lowest priority) but this contradicts the MST construction theorem.
So essentially, it is impossible to have an edge with minimum weight not get to the queue which means that any MST constructed would have it.