As you know finding maximum independent set is NP. Is there an algorithm to find out whether a given graph has an Independent set of at least k vertices? Note that we don't want to find it. We just want to find out if such thing exists.
Quoting Wikipedia:
In the independent set decision problem, the input is an undirected graph and a number k, and the output is a Boolean value: true if the graph contains an independent set of size k, and false otherwise.
This problem is NP-complete. Your problem asks the same question, just differently phrased, because a graph has an independent set of at least k vertices if and only if it has an independent set of exactly k vertices.
That means your problem is NP-complete too.
There is a nice lower bound on the size of maximum independent set ("graph independence number" denoted by alpha),
where d(v) is the degree of the vertex v and the sum is over all the vertices of a simple graph G. It was discovered independently by Wei and Caro (I found it in "Lower bounds on the independence number in terms of the degrees" by JR Griggs) and there are other lower bounds for different types of graphs.
What this means is that a maximum independent set will have at least k vertices with k given by this lower bound.
Related
I was reading about the Maximum Weight Independent Set problem which is:
Input: An undirected graph G = (V, E)and a non-negative weight Wv for
each vertex v ∈ V
Output: An independent set S ∈ V of G with the maximum-possible sum
∑Vw of vertex weights
and that same source (not the SO post) mentions that the problem can be solved by 4 recursive calls with a divide & conquer approach.
I googled but couldn't find such an algorithm. Does anyone have an idea how would this be solved by divide & conquer? I do understand that the running time is much worse than the Dynamic Programming I am just curious on the approach
I understand the part in the manuscript in such a way that only line graphs are taken into consideration. In that case, I believe the footnote to mean the following.
If a larger line graph is taken as input and split (say at an edge which in incident with the nodes a and b), there are four cases to consider to have a proper combination step.
You would have to solve the "left" branch by solving for the cases
a is included in the maximum independent set
a is not included in the maximum independent set
and the same goes for the "right" branch. In total there are four possible ways to combine the results, out of which a maximal one is to be returned.
I have a complete graph with n vertex. The MinimumVertexCover of the complete graph is n-1.Can I take less then n-1 as a MinimumVertexCover? If the answer is No please Explain it with the proof?
The answer is No.
I think the easiest way to prove it is by Gallai’s result. Recall that an independent set of vertices is one in which two vertices share an edge. Now Gallai’s theorem says that:
α(G)+β(G)=n
where,
α(G)=
size of largest independent set,
β(G)=size of minimum vertex cover, and
n=
number of vertices in G.
Now clearly α(Kn)=1
since a set of two vertices in Kn is not independent. It follows therefore that β(G)=n−1, as required.
In Singapore, this year's (2016) NOI (National Olympiad in Informatics) included the following problem "ROCKCLIMBING" (I was unable to solve it during the contest.) :
Abridged Problem Statement
Given a DAG with N <= 500 vertices, find the maximum number of vertices in a subset of the original vertices such that there is no path from 1 vertex in the set to another vertex in the same set, directly or indirectly.
Solution
The solution was to use transitive closure algorithm, and then to form a bipartite graph by duplicating each vertex i to form i' such that if vertex j can be reached from vertex i directly or indirectly in the original graph, then there is a directed edge from i to j' in the new graph.
However, during the solution presentation, the presenters did not explain how or why N - MCBM (MCBM being the Maximum Cardinality Bipartite Matching) of the new bipartite graph is also the maximum size of the set of vertices that cannot reach each other directly or indirectly in the original DAG.
I looked up other problems related to DAGs and bipartite graphs, such as the Minimum Path Cover problem on DAGs, but I could not find anything that explains this.
Does anyone know a way in which to prove this equality?
The problem statement can be found here: ROCKCLIMBING
Thank you in advance.
There are two things going on here:
A set is independent if and only if its complement is a vertex cover (see wikipedia). This means that the size of a max independent set is equal to the size of a minimum vertex cover.
Konig's theorem proves that
In any bipartite graph, the number of edges in a maximum matching equals the number of vertices in a minimum vertex cover.
Therefore to find the size of the max independent set we first compute the size MCBM of the max matching, and then compute its complement which equals N-MCBM.
An alternative viewpoint is as follows:
If we use A<B to mean we can climb from A to B, we have defined a partially ordered set
There is a result called Dilworth's theorem that says the maximum number of incomparable elements is equal to the minimum number of chains
The proof shows how to construct the minimum number of chains by constructing a maximum matching in your bipartite graph.
I am trying to find the maximal set for an undirected graph and here is the algorithm that i am using to do so:
1) Select the node with minimum number of edges
2) Eliminate all it's neighbors
3) From the rest of the nodes, select the node with minimum number of edges
4) Repeat the steps until the whole graph is covered
Can someone tell me if this is right? If not, then why is this method wrong to calculate the maximal independent set in a graph?
What you have described will pick a maximal independent set. We can see this as follows:
This produces an independent set. By contradiction, suppose that it didn't. Then there would have to be two nodes connected by edges that were added into the set you produced. Take whichever one of them was picked first (call it u, let the other be v) Then when it was added to the set, you would have removed all of its neighboring nodes from the set, including node v. Then v wouldn't have been added to the set, giving a contradiction.
This produces a maximal independent set. By contradiction, suppose that it didn't. This means that there is some node v that can be added to the independent set produced by your algorithm, but was not added. Since this node wasn't added, it must have been removed from the graph by the algorithm. This means that it must have been adjacent to some node added to the set already. But this is impossible, because it would mean that the node v cannot be added to the produced independent set without making the result not an independent set. We have a contradiction.
Hope this helps!
There is not one definite maximal independent set in any graph; take for example the cycle over 3 nodes, each of the nodes forms a maximal independent set. Your algorithm will give you one of the maximal independent sets of the graph, without guaranteeing that it has maximum cardinality.On the other hand, finding the maximum independent set in a graph is NP-complete (since that problem is complementary to that of finding a maximum clique), so there probably isn't an efficient algorithm.
After your clarify situation in comments, your solutions is right.
Even better, according to Corollary 3 from this paper http://courses.engr.illinois.edu/cs598csc/sp2011/Lectures/lecture_7.pdf
your get good aproximation for subset order.
Greedy gives a 1 / (d + 1) -approximation for (unweighted) MIS in graphs of degree at most d
Some months ago there was a nice question regarding a "1:n matching problem" and there seems to be no poly-time algorithm.
I would like to add constraints to find a maximum matching for the 1:n matching problem with a polynomial algorithm. I would like to say: "For vertex A1 choose either {B1,B2,B5} or {B2,B3} if the vertices are not already taken from another A-vertex" i.e. I would not allow all possible combinations.
This could be expressed if we introduce helper vertices H for each choice and substitute edges with trees => we get a problem similar to the ordinary bipartite matching. Every vertex of A or B can have only one edge in the matching. The edges to or from vertices in H are either all in the matching or none of them is present in the matching. Imagine the following tri-partite graph:
Now define h_ij="tree rooted that contains H_ij" to express the matching easily:
Then in the example M={h12,h22} would be one 'maximum' matching, although not all vertices from B are involved
The set {h12,h23} is not a matching because then B3 would have be choosen twice.
Would this problem then be solvable in polynomial time? If yes, is there a polytime solution for the weighted (w(h_ij)) variant? If no, could you argue or even proof it for a "simple-man" like me or suggest other constraints to solve the 1:n matching problem?
E.g. could the graph transformed to a general graph which then could be solved with the weighted matching for general graphs? Or could branchings or even matching forests help here?
PS: not a homework ;-)
There is a difference between maximal and maximum. I have assumed you meant maximum for the below writeup.
You don't seem to have defined your problem very clearly, but if I have understood your intent correctly, It seems like your problem is NP complete (and 'equivalent' to Set Packing).
We can assume that the allowed sets sizes is the same (k) for all A_i to find a [1:k] matching, as any other set size can be ignored. To find max k, we just run the algorithm for [1:k] for k = 1,2,3.. etc.
So your problem is (I think...):
Given m set families F_i = {S_1i, .., S_n(i)i} (|F_i| = size of F_i = n(i), need not be same as |F_j|), each set of size k, you have to find one set from each family (say S_i) such that
S_i and S_j are disjoint for any i neq j.
number of S_i's is maximum.
We can show that it is NP-Complete for k=3 in two steps:
The NP-Complete problem Set Packing can be reduced it. This shows that it is NP-Hard.
Your problem is in NP and can be reduced to Set Packing. This and 1) implies your problem is NP-Complete. It also helps you leverage any approximation/randomized algorithms already existing for Set-Packing.
Set Packing is the problem:
Given n sets S_1, S_2, ..., S_n, find the maximum number of pairwise disjoint sets among these.
This problem remains NP-Complete even if |S_1| = |S_2| = ... = |S_n| = 3 and is called the 3-Set packing problem.
We will use this to show that your problem is NP-Hard, by providing an easy reduction from 3-Set packing to your problem.
Given S_1, S_2, .., S_n just form the families
F_i = {S_i}.
Now if your problem had a polynomial time solution, then we get a set of Sets {S_1, S_2, ..., S_r} such that
S_i and S_j are disjoint
Number of S_i is maximum.
This easy reduction gives us a solution to the 3-set Packing problem and thus your problem is NP-Hard.
To see that this problem is in NP, we reduce it to Set-Packing as follows:
Given F_i = {S_1i, S_2i, ..., S_ni}
we consider the sets T_ji = S_ji U {i} (i.e. we add an id of the family into the set itself) and run them through the Set-Packing algorithm. I will leave it to you to see why a solution to Set-Packing gives a solution to your problem.
For a maximal solution, all you need is a greedy algorithm. Just keep picking up sets till you can pick no more. This would be polynomial time.