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A short proof of konigs matching theorem
- 1. A Short Proof of König's Matching Theorem Romeo Rizzi Journal of Graph Theory 報告者: 陳政謙
- 3. Vertex cover
- 5. Matching
- 7. The relationship between minimum vertex cover and maximum matching • For any graph G = (V, E), n(G) ≤ t(G), where n(G) is the maximum cardinality of a matching of G and t(G) is the minimum cardinality of a vertex cover of G.
- 8. König's theorem • Let G be a bipartite graph. Then n(G) = t(G), where n(G) is the maximum cardinality of a matching of G and t(G) is the minimum cardinality of a vertex cover of G.
- 9. The proof of König's theorem • Let G be a minimal counterexample. … … … n(G) < t(G) If there is a graph H smaller than G, then n(H) = t(H).
- 10. The proof of König's theorem • Then G is connected, is not a circuit, nor a path. If G is not connected G G1 G2 … Gn There exists at least one component Gi which n(Gi) < t(Gi), where 1 ≤ i ≤ n. Then G is not a minimal counterexample.
- 11. The proof of König's theorem • Then G is connected, is not a circuit, nor a path. If G is an odd path … If G is an even path … n(G) = t(G) = 𝑛 −1 2 n(G) = t(G) = 𝑛 2
- 12. The proof of König's theorem • Then G is connected, is not a circuit, nor a path. Bipartite graphs cannot be odd cycle, so this case can be ignored. If G is an even cycle … n(G) = t(G) … … …
- 13. The proof of König's theorem • So, G has a node of degree at least 3. Let u be such a node and v one of its neighbors. … … … u v We consider two cases: n(G - v) < n(G) and n(G - v) = n(G)
- 14. The proof of König's theorem • Case 1: n(G - v) < n(G) By minimality, Gv has a cover W’ with |W’| = n(G - v) < n(G). … … … u v W’ ∪ {v} is a cover of G with cardinality n(G) at most. Hence, t(G) ≤ n(G). The contradiction occurs.
- 15. The proof of König's theorem • Case 2: n(G - v) = n(G) There exists a maximum matching M of G having no edge incident at v.… … … u v f Let W’ be a cover of Gf with |W’| = n(G - v) = n(G). Hence, n(G) = t(G). The contradiction occurs.
- 16. The proof of König's theorem • By the proof, the contradiction occurs in both case 1 and case 2. • So the assumption n(G) < t(G) is not right. • We conclude that n(G) = t(G) holds in bipartite graphs.