Prove that Dense Subgraph is NP Complete by Generalisation
Last Updated : 17 Apr, 2024
Prerequisites: NP-Completeness, NP Class, Dense Subgraph
Problem: Given graph G = (V, E) and two integers a and b. A set of a number of vertices of G such that there are at least b edges between them is known as the Dense Subgraph of graph G.
Explanation: To prove the Dense Subgraph problem as NP-completeness by generalization, we are going to prove that it is a generalization of the known NP-complete problem. In this case, we are going to take Clique as the known problem which is already known to be NP-complete, and explained in Proof of Clique Is an NP-Complete and we need to show the reduction from Clique → Dense Subgraph.
Clique is a subset of vertices of an undirected graph such that every two distinct vertices in the clique are adjacent.
Proof:
1. Input Conversion: We need to convert the input from Clique to the input of the Dense Subgraph.
Clique Input: An undirected graph G(V, E) and integer k.
Dense Subgraph Input : An undirected graph G'(V, E) and two integers a and b.
We are going to transform the input from Clique for Dense Subgraph such that
- G' = G(V, E)
- a = k
- b = (k * (k - 1))/2
This conversion is going to take O(1) time so it's polynomial in nature.
2. Output Conversion: We need to convert the solution from Dense Graph to the solution for the Clique problem.
Solution of Dense Graph will result in a set a which would be a Clique of size k as k = a. So direct output from Dense Graph can be used by Clique. Since no conversion is required so it's again polynomial in nature.
3. Correctness: We have restricted the range of input value b such that (k¦2) with value as (k * (k - 1))/2.
Now we are looking for a subgraph having k vertices and are connected by at least (k * (k - 1))/2 edges.
- Since in a complete graph, n vertices can have at most (n * (n - 1))/2 edges between them so we can say that we need to find a subgraph of k vertices that have exactly (k * (k - 1))/2 edges which means output graph should have an edge between each pair of vertices which is nothing but Clique of k vertices.
- Similarly, a Clique of k vertices on a graph G(V, E) must have (k * (k - 1))/2 edges which is nothing but the Dense-Subgraph of graph G(V, E)
Red Edges and Vertices denotes a Dense-Subgraph with a = 4 and b = 5 So, this means Dense-Subgraph has a solution↔ Clique has a solution.
The complete reduction takes polynomial time and Clique is NP complete so Dense Subgraph is also NP complete.
Conclusion:
Hence we can conclude that Dense-Subgraph is NP Complete
For more details, please refer: Design and Analysis of Algorithms.
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