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**k-Factor Factor: a spanning subgraph of graph G**

k-Factor: a spanning k-regular subgraph 1-Factor of K6

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k-Factor 2-Factor of K7

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**Odd Component Odd component: a component of odd order**

o(H): the number of odd components of H Odd component o(H)=4 Even component

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Tutte’s Condition A graph G has 1-factor if and only if o(G-S)<=|S| for every SV(G)

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**Proof of Tutte’s Condition ()**

Suppose G has 1-factor. Consider a set SV(G). Every odd component of G-S has a vertex matched to one vertex of S. o(G-S)<=|S| since these vertices of S must be distinct. S Odd component Even component

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**Proof of Tutte’s Condition (⇐)**

n(G) is even. 2. Let G’=G+e. G’ satisfies Tutte’s condition. Let S=. Then o(G)<= ||. Then o(G’-S) <= |S| for every SV(G). <= o(G-S) e e Odd component S Even component

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**Proof of Tutte’s Condition (⇐)**

3. Let U be the set of vertices in G that have degree n(G)-1. 4. Two cases are discussed: (Case 1) G-U consists of disjoint complete graphs and (Case 2) G-U is not a disjoint union of cliques (complete graph).

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**Proof of Tutte’s Condition (⇐)**

5. Case 1: G-U consists of disjoint complete graphs. 6. The vertices in each component of G-U can be paired in any way, with one extra in the odd components. We can match the leftover vertices to vertices of U since each vertex of U is adjacent to all of G-U. 8. The remaining vertices in U can be matched because U is a clique and n(G) is even. 7. Let S=U. o(G-U)<=|U|. G has 1-factor.

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**Proof of Tutte’s Condition (⇐)**

9. Case 2: G-U is not a disjoint union of cliques. 10. It suffices to show if G satisfies Tutte’s condition and G has no 1-factor, there exists an edge e such that G+e has no 1-factor. 11. Suppose G satisfies Tutte’s condition and has no 1-factor. There exists an edge e such that G+e has no 1-factor. By 1, G+e satisfies Tutte’s condition. Kn(G) has no 1-factor by repeating the same argument. It is a contradiction since n(G) is even. Now, let’s show 10.

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**Proof of Tutte’s Condition (⇐)**

12. Suppose that G satisfies Tutte’s condition, G has no 1-factor, and adding any missing edge to G yields a graph with 1-factor. 13. G-U has two vertices x,z at distance 2 with a common neighbor y not in U. 14. G-U has another vertex w not adjacent to y since y is not in U. 15. G+xz has 1-factor and G+yw has 1-factor. 16. Let M1 be 1-factor in G+xz, and let M2 be 1-factor in G+yw. 17. xzM1 and ywM2 since G has no 1-factor. xzM1-M2 and ywM2 -M1. xzF and ywF, where F= M1M2.

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**Proof of Tutte’s Condition (⇐)**

18. Each vertex of G has degree 1 in each of M1 and M2. Every vertex of G has degree 0 or 2 in F. The components of F are even cycles and isolated vertices. 19. Let C be the cycle of F containing xz. 20. If C does not also contain yw, then the desired 1-factor consists of the edges of M2 from C and all of M1 not in C. F has 1-factor avoiding xz and yw. G has 1-factor. A contradiction. : in M1 – M2 : in M2 – M1 : in both (hence not in F) C

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**Proof of Tutte’s Condition (⇐)**

21. Suppose C contains both yw and xz. 22. Starting from y along yw, we use edge of M1 to avoid using yw. When we reach {x,z}, we use zy if we arrive at z; otherwise, we use xy. In the remainder of C we use the edge of M2. C+{xy,yz} has 1-factor avoiding xz and yw. We have 1-factor of G by combing with M1 or M2 outside V(C). Another contradiction.

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Theorem 3.3.9 Every regular graph with positive even degree has a 2-factor. Let G be a 2k-regular graph with vertices v1,…,vn. Every component in G is Eulerian with some Eulerian circuit C by Theorem 2. For each component, define a bipartite graph H with vertices u1,…,un and w1,…,wn by adding edge (ui, wj) if vj immediately follows vi somewhere on C. 8 1 5 10 6 2 9 7 4 3

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Theorem 3.3.9 3. H is k-regular because C enters and exists each vertex k times. 4. H has a 1-factor M by Corollary 5. The 1-factor in H can be transformed into a 2-regular spanning subgraph of G.

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Join Join: The join of simple graphs G and H, written GH, is the graph obtained from the disjoint union G+H by adding the edges {xy: xV(G), yV(H)}

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Berge-Tutte Formula The largest number of vertices saturated by a matching in G is minSV(G){n(G)-d(S)}, where d(S)=o(G-S)-|S|. Given SV(G), at most |S| edges can match vertices of S to vertices in odd components of G-S, so every matching has at least o(G-S)-|S|=d(S) unsaturated vertices. Every matching has at least maxSV(G){d(S)} unsaturated vertices. Every matching has at most minSV(G){n(G)-d(S)} saturated vertices. The largest number of vertices saturated by a matching in G is equal to minSV(G){n(G)-d(S)}. or less than

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**Berge-Tutte Formula 2. Let d=maxSV(G){d(S)}. We need to show**

there exists a matching in G that has exactly d unsaturated vertices. 3. Let S=. We have d>=0. 4. Let G’=GKd. 5. If G’ has a perfect matching, then we obtain a matching in G having at most d unsaturated vertices from a perfect matching in G’, because deleting the d added vertices eliminates edges that saturate at most d vertices of G. It suffices to show G’ satisfies Tutte’s Condition.

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**Berge-Tutte Formula 6. o(G’-S’)<=|S’| for S’=. n(G’) is even.**

7. o(G’-S’)<=|S’| if S’ is nonempty but does not contain V(Kd). 8. o(G’-S’)<=|S’| if S’ contains V(Kd). n(G’) is even. d has the same parity as n(G). d(S) = o(G-S)-|S| has the same parity as n(G) for each S. o(G-S) |S| d(S) n(G) odd odd odd even even odd even even even even odd odd odd odd even even G’-S’ has only one component. Let S=S’-V(Kd). G’-S’=G-S. o(G’-S’)= o(G-S)= d(S)+|S|<=|S|+d=|S’|.

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