### np-complete

#### Reduction from Maximum independent set to Dominating set to prove the Dominating set is NP-complete

```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.```