Reduce Independent Set to Vertex Cover
Convert an independent-set threshold into a vertex-cover threshold by complementing the chosen vertex set.
Problem
Given an undirected graph and integer , construct in polynomial time an integer (on the same graph, or an explicitly described isomorphic copy) such that has an independent set of size at least if and only if has a vertex cover of size at most . Prove both directions.
Hint 1
If is independent, every edge must touch .
Hint 2
Relate the two sizes by .
Independent Set → Vertex Cover
Full solution
Keep the same graph and set . The map is clearly polynomial.
Claim: is an independent set if and only if is a vertex cover. Indeed, is independent exactly when no edge has both endpoints in , which is exactly the statement that every edge has at least one endpoint in .
Therefore has an independent set of size if and only if it has a vertex cover of size . (If an independent set larger than exists one may discard vertices; dually, a smaller cover may be padded only when the language uses “at most,” which is the standard decision form here.)
Vertex Cover is in NP via a candidate cover of size . With Independent Set NP-complete, the reduction yields NP-completeness of Vertex Cover.
Checkpoint
This uses set complement on vertices, not graph complement on edges. The parameter must change: .
Common mistakes
- Leaving , which reverses the intended meaning.
- Confusing Vertex Cover with Edge Cover or Dominating Set.
- Claiming the identity without checking edges that touch both and .
Provenance note
This is an independently written presentation of the standard IS/VC duality taught in the course graph-problems lectures and aligned with Sipser / Garey–Johnson. It is not transcribed from a single exam prompt.
Sources and verification
- Merged summary, pp. 55-57
- Sipser 3e, sec. 7.5
- Garey-Johnson, GT1 and GT20
- Catalog mapping:
data/exercises.json; statements and solutions are independently written.
Open in the visualization gallery: Independent Set → Vertex Cover