I know of the reduction from the Vertex cover to Dominating set.
However, I was seeing if I could get a reduction from the maximum independent set problem straight to the Dominating set problem in order to prove the latter NP-complete.
Does anyone know if this has been done? I can't find anything online.
I was hoping to find something along the lines of a proof like:
If there is a dominating set of size k -> there is a maximum independent set of size k.
AND
If there is a maximum independent set of size k -> then there is a dominating set of size k.
Yes you can get a reduction from the maximum independent set problem straight to the Dominating set problem -- but not that straight, you need to construct another graph in the following manner. We then can prove that if the original graph has an independent set of size k iff the new graph has a dominating set of some size related to k. The construction is polynomial.
Given a graph G = (V, E) we can construct another graph G' = (V', E') where for each edge e_k = (v_i, v_j) in E, we add a vertex v_{e_k} and two edges (v_i, v_{e_k}) and (v_{e_k}, v_j).
We can prove G has an independent set of size k iff G' has a dominating set of size |V|-k.
(=>) Suppose I is a size-k independent set of G, then V-I must be a size-(|V|-k) dominating set of G'. Since there is no pair of connected vertex in I, then each vertex in I is connected to some vertex in V-I. Moreover, every new added vertex are also connected to some vertices in V-I.
(<=) Suppose D is a size-(|V|-k) independent set of G', then we can safely assume that all vertices in D is in V (since if D contains an added vertex we can replace it by one of its adjacent vertex in V and still have a dominating set of the same size).
We claim V-D is a size-k independent set in G and prove it by contradiction: suppose V-D is not independent and contains a pair of vertices v_i and v_j and the edge e_k = (v_i, v_j) is in E. Then in G' the added vertex v_{e_k} need to be dominated by either v_i or v_j, that is at least one of v_i and v_j is in D. Contradiction. Therefore V-D is a size-k independent set in G.
Combining the two directions you get what you want.
Related
Can we convert a directed weighted graph in such a way that each of its path from a specified source to destination is of equal cost?
The cost of each of the path should be equal to the maximum cost path in original graph. How to convert any directed weighted graph to such type of graph? Is it possible to convert every directed weighted graph into such type of graph?
Source and destination of graph is predefined.
It is possible to convert graph in such a way.
Note that if (resulting) graph has property than all paths between given
vertices (s and d) have same cost, than that property holds for each pair of vertices that lay on any path between s and d. That is seen by checking costs between s (or d) and any inner vertex x. With that we can say that each vertex x has cost from s.
To create vertex costs:
set cost to s to 0,
pass graph in topological order and set cost to vertex as max predecessor costs + 1.
To create graph with required property change edge cost in a way that edge a -> b have cost cost_of_Vertex_b - cost_of_vertex_a.
To get predefined cost, scale all costs by a factor.
Given a graph G,which are the sufficient and necessary conditions , so that this graph has a unique minimum spanning tree?In addition , how can I proove these conditions?
So far , I had found that those conditions are :
1)For every partition of V(G) into two subsets, the minimum weight edge with one endpoint in each subset is unique.
2)The maximum-weight edge in any cycle of G is unique.
But I am not sure if this is correct.Even in case this is correct,I cannot prove its correctness.
Actually, there is a necessary and sufficient condition for unique MST. In the book A First Course In Graph Theory, it is given as an exercise:
Exercise 4.30
Let G be a connected weighted graph and T a minimal spanning tree of G. Show that T is a unique minimal spanning tree of G if and only if the weight of each edge e of G that is not in T exceeds the weight of every other edge on the cycle in T+e.
I write my proof here.
This is false because at least the first condition is not necessary. The proof is by counterexample (source).
Take G to be any tree where all edge weights are 1. Then G has a
unique MST (itself), but any partition with more than one edge
crossing it has several minimum weight edges.
EDIT:
In response to your modified question...
There is a well-known sufficient (but not necessary) condition for the uniqueness of a MST:
If the weight of each edge in a connected graph is distinct, then the graph contains exactly one (unique) minimum spanning tree.
The proof is as follows (source):
For the sake of contradiction, suppose there are two different MSTs of
G, say T1 and T2. Let e = v-w be the min weight edge of G that is in
one of T1 or T2, but not both. Let's suppose e is in T1. Adding e to
T2 creates a cycle C. There is at least one edge, say f, in C that is
not in T1 (otherwise T1 would be cyclic). By our choice of e, w(e) ≤
w(f). Since all of the edge weights are distinct, w(e) < w(f). Now,
replacing f with e in T2 yields a new spanning tree with weight less
than that of T2 (contradicting the minimality of T2).
However, regarding "sufficient and necessary" conditions for the uniqueness of a MST, I do not believe any are known to exist.
Show that given a graph G and a number k, there is some way to transform it to a graph H such that G has an independent set of size of size at least k if and only if H has a clique of size at least k.
An independent set is a group of nodes where for any pair of nodes in the set, there is not an edge between those nodes. A clique is a group of nodes where for any pair of nodes in the set, there is an edge between those nodes. Therefore, an independent set in a graph G is a clique in the complement of G and vice-versa.
Given this, a simple transformation would be given G and k to produce Gc (the complement of G) and k. Then, G has an independent set of size k if and only if Gc has a clique of size k.
Hope this helps!
A friend presented me with a conjecture that seems to be true but neither of us can come up with a proof. Here's the problem:
Given a connected, bipartite graph with disjoint non-empty vertex sets U and V, such that |U|<|V|, all vertices are in either U or V, and there are no edges connecting two vertices within the same set, then there exists at least one edge which connects vertices a∈U and b∈V such that degree(a)>degree(b)
It's trivial to prove that there is at least one vertex in U with degree higher than one in V, but to prove that a pair exists with an edge connecting them is stumping us.
For any edge e=(a,b) with a∈U and b∈V, let w(e)=1/deg(b)-1/deg(a). For any vertex x, the sum of 1/deg(x) over all edges incident with x equals 1, because there are deg(x) such edges. Hence, the sum of w(e) over all edges e equals |V|-|U|. Since |V|-|U|>0, w(e)>0 for som edge e=(a,b), which means that deg(a)>deg(b).
Prove it by contradiction, i.e. suppose that deg(a) ≤ deg(b) ∀(a,b)∈E, where E is the edgeset of the graph (with the convention that the first element is in U and the second in V).
For F⊆E, designate by V(F) the subset of V which is reachable through edgeset F, that is:
V(F) = { b | (a,b)∈F }
Now build an edgeset F as follows:
F = empty set
For a ∈ U:
add any edge (a,b)∈E to F
Keep adding arbitrary edges (a,b)∈E to F until |V(F)| = |U|
The set V(F) obtained is connected to all nodes in U, hence by our assumption we must have
∑a∈U deg(a) ≤ ∑b∈V(F) deg(b)
However, since |U|=|V(F)| and |U|<|V| we know that there must be at least one "unreached" node v∈V\V(F), and since the graph is connected, deg(v)>0, so we obtain
∑a∈U deg(a) < ∑b∈V deg(b)
which is impossible; this should be an equality for a bipartite graph.
here is the question. I am wondering if there is a clear and efficient proof:
Vertex Cover: input undirected G, integer k > 0. Is there a subset of
vertices S, |S|<=k, that covers all edges?
Dominating Set: input undirected G, integer k > 0. Is there a subset of
vertices S, |S|<= k, that dominates all vertices?
A vertex covers it's incident edges and dominates it's neighbors and itself.
Assuming that VC is NPC, prove that DS is NPC.
There is a quite nice and well known reduction:
Given an instance (G,k) of Vertex Cover build an instance of Dominating Set (H,k), where for H you take G, remove all isolated vertices, and for every edge (u,v) add an additional vertex x connected to u and v.
First realize that a Vertex Cover of G is a Dominating Set of H: it's a DS of G (after removing isolated vertices), and the new vertices are also dominated. So if G has a VC smaller k, then H has a DS smaller k.
For the converse, consider D, a Dominating Set of H.
Notice that if one of the new vertices is in D, we can replace it with one of it's two neighbors and still get an Dominating Set: it's only neighbors are are the two original vertices and they are also connected - everything x can possible dominate is also dominated by u or v.
So we can assume that D contains only vertices from G. Now for every edge (u,v) in G the new vertex x is dominated by D, so either u or v is in D. But this means D is a Vertex Cover of G.
And there we have it: G has a Vertex Cover smaller k if and only if H has a Dominating Set smaller k.
Taken from :
CMSC 651 Advanced Algorithms , Lecturer Samir Khuller
I think that second problem is not NP.
Let's try the following algorithm.
1. Get the original Graph
2. Run any algorithm which checks if a graph is connected or not.
3. mark all used edges of step 2
4. if the graph is connected then return the set of marked edges otherwise there is no such a set.
If I understood correctly your problem then it is not NP Complete.