NP Problems: Weighted Vertex Cover vs Weighted Independent Set

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SUMMARY

The discussion centers on the relationship between weighted independent sets and weighted vertex covers in graph theory. It establishes that if G = (V, E) is a graph with vertices assigned positive integer weights, then an independent set S with maximum weight corresponds to a vertex cover V - S with minimum weight. The proof leverages the contradiction method, demonstrating that if S is not maximal, a lighter vertex cover T can be found, contradicting the assumption of S's maximality.

PREREQUISITES
  • Understanding of graph theory concepts, specifically independent sets and vertex covers.
  • Familiarity with the properties of weighted graphs.
  • Knowledge of proof techniques, particularly proof by contradiction.
  • Basic mathematical notation and operations involving sets and weights.
NEXT STEPS
  • Study the properties of NP problems and their classifications.
  • Explore algorithms for finding maximum independent sets in weighted graphs.
  • Learn about the relationship between vertex covers and independent sets in graph theory.
  • Investigate the implications of the theorem in practical applications, such as network design and resource allocation.
USEFUL FOR

Researchers, computer scientists, and mathematicians interested in graph theory, particularly those focusing on optimization problems and NP-completeness.

JohnDoe2013
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There is a Theorem that states that if G = (V, E) is a graph, then S is an independent set $\Leftarrow\Rightarrow$ V - S is a vertex cover.

Suppose the vertices have positive integer weights. Does it follow from the theorem that:

S is an independent set with maximum weight $\Leftarrow\Rightarrow$ V - S is a vertex cover with minimum weight?
 
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I think so. If $S\subseteq V$, let $w(S)$ denote the sum of weights of vertices is $S$, and let $W=w(V)$. Suppose that $S$ is an independent set with maximum weight, but $V-S$ is not minimal, i.e., there exists a $T\subseteq V$ such that $T$ is a vertex cover and $w(T)<w(V-S)=W-w(S)$. Then $w(S)<W-w(T)=w(V-T)$ and $V-T$ is an independent set, a contradiction with maximality of $S$. The other direction is shown similarly.
 

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