1. Introduction of graph theory
Edges and Cycles
Introduction of graph theory

2. Line Graphs and Edge-coloring
Definition : Line graph
L(G) is the simple graph whose vertices are the edges of G, with ef  E (L (G )) when e and f have a common endpoint in G . d e f g h d e f h g

3. Some question between G and L(G)
Eulerian circuits
Spanning cycles
Matching
(G)
Independent set ’(L(G))
Menger’s Theorem
Edge coloring
Vertex coloring

4. Edge-Coloring
Edge of K2n :
solution

5. k -edge-coloring of G Definition
Is a labeling : E(G)  S |S| = k. the labels are colors
Color class : edges of one color
Proper : incident edges have different labels
K-edge-colorable : has a proper k-edge-coloring
Edge-chromatic number : ’(G)
another name : chromatic index

6. Multiplicity m of xy :
There are m edges with endpoints x, y. we write  (xy). And  (G) for the maximum of the edge multiplicities in G.

7. ’(G)  2(G)-1
Always assigning the current edge the least-indexed color different from those already appearing on edges incident to it. Since no edge is incident to more than 2((G)-1) other edges, this never uses more than 2(G)-1 colors.

8. König Theorem If G is bipartite, then ’(G) = (G).
By Corollary , every regular bipartite graph H has a 1-factory, and H has (H)-edge-coloring. We can add vertices and edges to make every bipartite graph G become (G)-regular bipartite graph H.

9. A decomposition of a regular graph G into 1-factors is a 1-factorization of G. a graph with a 1-factorization is 1-factorable.
Odd cycle is not 1-factorable.

10. The Petersen graph is 4-edge-chromatic.
The Petersen graph is 3-regular; 3-edge-colorability requires a 1-factorization.
Deleting a perfect matching leaves a 2-factor, all components are cycles.
To show that every 2-factor is isomorphic to 2C5 .
Consider the drawing consisting of two C5 and a matching (the cross edges) between them.

11. Every cycle uses an even number of cross edges, so a 2-factor H has an even number m of cross edges.
m = 0 H = 2C5
m = 2 violates the 2-factor requirement.
m = 4 form a 2P5 whose only completion to a 2-factor in H is 2C5

12. Vizing, Gupta Theorem if G is a simple graph, then ’(G)  (G)+1.
Let f be a proper (G)+1-edge-coloring of a subgraph G’ of G. if G’  G, then some edge uv is uncolored by f. after possibly recoloring some edges, we extend the coloring to include uv ;call this an augmentation.

13. Vizing, Gupta Theorem (cont.)
Every vertex has some color not appearing on its incident edges. Let a0 be the missing color at u. we generate a list of neighbor of u, begin with v0 = v.
Having selected uvi-1 with color ai-1 ,let ai be a color missing at vi-1 .if ai is missing at u, then we use ai on uvi-1 and shift color aj from uvj to uvj-1 for 1 j  i-1,we call this downshifting from i.

14. Vizing, Gupta Theorem (cont.)
downshifting from I
v0 missing a1,v1 missing a2 … vi-1 missing ai , if ai is also missing at u.
completing augmention. ai
ai-1 a2 a1
vi-1 v2 v1 u v0 …

15. Vizing, Gupta Theorem (cont.)
The list of selected colors eventually repeats( or augmentation by downshifting)
Let l be the smallest index such that a color missing at vl is in the list a1,…,al;let this color be ak.
ak missing at vl is also missing at vk -1 and appears on uvk .

16. Vizing, Gupta Theorem (cont.)
If a0 does not appear at vl , we downshift from vl and use color a0 on uvl . Compeletion a0 al u vl … ak a1
ak-1 vk v0
Vk-1

17. Vizing, Gupta Theorem (cont.)
else, let P be the maximal alternating path of edges colored a0 and ak that begins at vl along color a0.
Case1 : if P reaches vk , we downshift from vk , and switch colors on P. ak
vk-1 vk v2 v1 v0 vl a0

18. Vizing, Gupta Theorem (cont.)
Case2 : if P reaches vk-1 , we downshift from vk-1 , give a0 to uvk-1 and switch colors on P. ak
vk-1 vk v2 v1 v0 vl a0
ak-1 ak
vk-1 vk v2 v1 v0 vl a0

19. Vizing, Gupta Theorem (cont.)
Case3 : if P does not reaches vk-1 or vk, we downshift from vl , give a0 to uvl and switch colors on P. ak
vk-1 vk v2 v1 v0 vl a0 al

20. A simple graph G is Class 1 if ’(G) = (G)
A simple graph G is Class 1 if ’(G) = (G). It is Class 2 if ’(G) = (G)+1.
A graph to be class 1 that is conjectured to be sufficient when (G) > (3/10)n(G)
Overfull conjecture states that if (G) > n(G)/3
A simple graph G is Class 1  G has no overfull subgraph.

21. When G has multiple edges, ’(G)  3/2 (G) and ’(G)  (G) + (G).
These bounds follow from that of
’(G)  max{(G), maxP1/2(d(x)+ (xy)+ (yz)+d(z))}
where P = {x,y,z  V(G): yN(x)N(z)}

22. Shannon Theorem If G is a graph, then ’(G)  3/2 (G)
Let k = ’(G), and assume k  3/2 (G). Let G’ be a minimal subgraph of G with ’(G’) = k. since k  (G’)+ (G’), we obtain (G’)1/2 (G). Let e with endpoints x, y be an edge with multiplicity (G’)
F be a proper k-1-edge-coloring of G’-e. in G’-e both x and y are missing (k-1)-((G)-1) colors, and no color is missing at both.

23. Shannon Theorem (cont.)
2(k-(G))+((G)/2)-1  2(k-(G))+(G’)-1  k-1.
Then k  (3/2)(G)

24. Conjecture..
If ’(G)  (G) + 2, then ’(G) =
max HGe(H)/n(H)/2

25. Introduction of graph theory
Edges and Cycles
Introduction of graph theory

26. Hamiltonian Cycles
Hamiltonian Graph is a graph with a spanning cycle, also called a Hamiltonian cycle.

27. Necessary Conditions
Bipartite graphs : Km,n is Hamiltonian only if m=n.
Proposition : If G has a Hamiltonian cycle, then for each nonempty set SV, the graph G−S has at most |S| components. S

28. Let c(H) denote the number of components of a graph H.
The necessary condition is that c(G-S)  |S| for all   S  V.
Petersen graph is non-Hamiltonian graph satisfying the condition. Because that 2C5 is the only 2-factor of the Petersen graph, it has no spanning cycle.

29. example Bipartite , fails the necessary condition of proposition 7.2.3
The necessary condition is not sufficient

30. A graph G is t-tough if |S|  tc(G-S) for every cutset S  V
A graph G is t-tough if |S|  tc(G-S) for every cutset S  V. the toughness of G is the maximum t such that G is t-tough
By proposition 7.2.3, spanning cycle require toughness at least 1.for some years it was thought that toughness 2 would be sufficient.
Some value of toughness suffices remains open.

31. Sufficient Conditions
(G)  n(G)/2 suffices. And no smaller minimum degree is sufficient.
Consisting of cliques of (n+1)/2 and (n+1)/2 sharing a vertex has minimum degree (n-1)/2 but is not Hamiltonian.
Biclique with partite sets of sizes (n-1)/2 and (n+1)/2

32. Dirac Theorem
If G is a simple graph with at least three vertices and (G)  n(G)/2 , then G is Hamiltonian.
Uses contradiction. If there is a non-Hamiltonian graph satisfying the hypotheses, then adding edges cannot reduce the minimum degree. Thus we may restrict our attention to maximal non-Hamiltonian graph with minimum degree at least n/2.

33. Dirac Theorem (cont)
When uv in G, the maximality of G implies that G has a spanning path v1,…,vn from u = v1 to v = vn, because every spanning cycle in G+uv contains uv.
We can avoid using the edge uv, this will build a spanning cycle in G.

34. Dirac Theorem (cont)
If u’s neighbor follows a neighbor of v on the path: v u vi
Vi+1
To prove that such a cycle exists, we show that there is a common index in the sets S and T,
S = { i : uvi+1}, T = { i : vvi}
|ST| + |ST| = |S| + |T| = d(u) + d(v)  n
Neighbor S nor T contains the index n, thus |ST| < n and |ST|  1.

35. Observed that this argument, we can weaken the requirement of minimum degree n/2 to require only that d(u)+d(v)  n whenever uv and G+uv was Hamiltonian and thereby provided a spanning u,v-path.
Lemma let G be a simple graph. If u, v are distinct non-adjacent vertices of G with d(u) + d(v)  n(G), then G is Hamiltonian if and only if G+uv is Hamiltonian.

36. (Hamiltonian) closure
Denote C(G), is the graph with vertex set V(G) obtained from G by iteratively adding edges joining pairs of nonadjacent vertices whose degree sum is at least n, until no such pair remains.

37. Bondy-Chvátal Theorem
A simple n-vertex graph is Hamiltonian if and only if its closure is Hamiltonian.
Lemma. The closure of G is well-define
Let e1,…,er and f1,…,fs be sequences of edges added in forming C(G), the first yielding G1 and the second G2. let uv must be added.
if f1, being initially addable to G, must belong to G1. similarly if f1, … ,fi-1  E(G1), then fi becomes addable to G1 and therefore belongs to G1.

38. Chvátal Theorem
Let G be a simple graph with vertex degrees d1… dn, where n  3. if i < n/2 implies that di > i or dn-i  n-i, then G is Hamiltonian.
Because G is Hamiltonian if and only if C(G) is, thus we can consider the case where C(G)=G. called G is closed.
We prove that Chvátal Theorem implies that G = Kn.
We prove the contrapositive; if G is a closed n-vertex graph that is not a complete graph, then we construct a value of i less than n/2 for which at least I vertices have degree at most I and at least n-i vertices have degree less than n-i.

39. Chvátal Theorem
G  Kn, we can choose among the pairs of nonadjacent vertices a pair u, v with maximum degree sum.
G is closed, uv implies that d(u)+d(v) < n.
If d(u)  d(v) we have d(u) < n/2. let i = d(u).
we need to find I vertices with degree at most i.
Every vertex of V-{v} that is not adjacent to v has degree at most d(u)=i, there are n-1-d(v) such vertices.
d(u)+d(v)  n-1 => n-1-d(v)  i.

40. Chvátal Theorem We also need n-i vertices with degree less than n-i.
Every vertex of V-{u} that is not adjacent to u has degree at most d(v), and we have d(v) < n-d(u) = n-i.
There are n-1-d(u) such vertices. Since d(u)  d(v), we can add u itself to the set of vertices with degree at most d(v). We obtain n-i vertices with degree less than n-i.

41. Example Non-Hamiltonian graphs with “large” vertex degrees is :
dj = i for j  i
dj = n-i-1 for i+1  j  n-i
dj = n-1 for j > n-i. i
n-i-1
n-1 dn d1 di
dn-i

42. Let G be a simple graph realizing this degree sequence (if exists).
G must be :
(Ki(bar)+Kn-2i)vKi
This graph is not Hamiltonian.
Kbi Ki
Kn-2i

43. Spanning path A Hamiltonian path is a spanning path.
Remark : A graph G has a spanning path if and only if the graph GvK1 has a spanning cycle.
Theorem : Let G be a simple graph with vertex degrees d1… dn. if i < (n+1)/2 implies (di  i or dn+1-i  n-i), then G has a spanning cycle.

44. Proof
Let G’ = GvK1, let n’ = n+1 and let d’1,… ,d’n’ be the degree sequence of G’. It suffices to show G’ satisfies Chvátal Theorem.
We have d’n’ = n, d’j = dj+1 for j < n’. For i < (n+1)/2 =n’/2, the hypothesis on G yields
d’i = di +1  i+1 > i or d’n’-i = dn+1-i +1  n-i+1 = n’-i.
G’ satisfies Chvátal Theorem.

45. Remark :
Every regular simple graph G with vertex degrees at least n(G)/3 is Hamiltonian.
It may possible to lower the degree condition further when connectivity is high.

46. Chvátal Erdős Theorem If (G)  (G), then G has a Hamiltonian cycle.
Suppose that (G)  (G). Let k = (G), and let C be a longest cycle in G. Since (G)  (G), and every graph with (G)  2 has a cycle of length at least (G) + 1, C has at least k + 1 vertices.
Let H be a component of G – V(C). The cycle C has at least k vertices with edges to H;((G) = k)
Let u1,…, uk be k vertices of C with edges to H, in clockwise order.

47. Chvátal Erdős Theorem
Let ai be the vertex immediately following ui on C. if any two of these vertices are adjacent, then we construct a longer cycle by using aiaj. H ui uj ai aj
And no ai has a neighbor in H. hence {a1,…,ak}plus a vertex of H forms an independent set of size k+1
This implies that C is a Hamiltonian cycle.

48. Planarity, Colorings, and Cycles
Introduction of graph theory

49. Tait’s Theorem
A theorem relating face-coloring and edge-coloring of plane graphs, and used this in an approach to the Four Color Theorem.
Definition : A proper face-coloring of a 2- edge-connected plane graph is an assignment of colors to its faces so that faces having a common edge in their boundaries have distinct colors.

50. Tait’s Theorem
We restrict our attention to face-coloring on 2-edge-connecting graph, because when a plane graph has a cut edge, it’s dual graph has a loop, and is thus uncolorable.
To prove Four Color Theorem it suffices to prove that all triangulations are 4-colorable. (adding edge to graph)
Equivalently, we could show that all duals of triangulations are 4-face-colorable.

51. Tait’s Theorem
The dual G* of a plane triangulation G is a 3-regular, 2-edge-connected plane graph.
A simple 2-edge-connected 3-regular plane graph is 3-edge-colorable if and only if it is 4-face-colorable.

52. Tait’s Theorem Pf: Let G be such a graph.
First, suppose that G is 4-face-colorable3-edge-coloring.
Let the four colors be denoted by : c1 = 00, c2 = 01
c3 = 10, c4 = 11. Color E(G) by: 10 01 11

53. Tait’s Theorem
Because G is 2-edge-connected, each edge bounds two distinct faces.
The color 00 never occurs as a sum. we check the proper property.
At vertex v the faces bordering the three incident edges must have distinct colors {ci, cj, ck}
if color 00 is not in this set, the sum of any two is the third.
If ck = 00, then ci and cj appear on two of the edges, and the third is ci+ cj, which is not in {ci, cj, ck}.

54. Tait’s Theorem
For the converse, G has a proper 3-edge-coloring using color a, b, c. 00 01 11 10 a b c

55. Tait’s Theorem
Let Ea, Eb, Ec be the edge sets having the three colors, we construct a 4-face-coloring.
G is 3-regular, each color appears at every vertex, and the union of any two of Ea, Eb, Ec is 2-regular and makes it a union of disjoint cycle.
Let H1 = EaEb, H2 = EbEc, to each face of G, assign the color whose ith coordinate (i{1, 2}) is the parity of the number of cycles in Hi that contain it.(0 for even, 1 for odd)

56. Tait’s Theorem 00 01 11 10 a b c 00 01 11 10 a b c H1 H2

57. This is a proper 4-face-coloring.
Faces F, F’ sharing an edge e are distinct faces, G is 2-edge-connected. E belongs to a cycle C in at least one of H1, H2 (or both).
One of F, F’ is inside C and the other is outside. All other cycles in H1, H2 fail to separate F and F’.
If e has color a, c or b, the parity of the number of cycles containing F and F’ is different in H1, in H2 or in both, respectively.

58. A proper 3-edge-coloring of a 3-regular graph is called a Tait coloring.
Lemma7.3.3: if G is a 3-regular graph with edge-connectivity 2, then G has subgraphs G1, G2 and vertices u1,v1V(G1) and u2,v2V(G2) such that u1v1 , also u2v2 and a ladder of some length joining G1, G2 at u1,v1,u2,v2. u1 u2 v1 v2 G1 G2

59. Pf: if G has an edge cut of size 2, and these 2 edge are incident, then the third edge is a cut edge. Hence the four endpoints in our minimum edge cut xy, uv are distinct.
If xy and uv, then we find it.
Else , we extend the ladder until we obtain a nonadjacent pair at the base of the ladder.

60. Theorem: All 2-edge-connected 3-regular simple planar graphs are 3-edge-colorable if and only if all 3-connected 3-regular simple planar graphs are 3-edge-colorable.
pf: it suffices to show that 3-colorability for 3-connected implies it for 2-connected.
Using induction on n(G)
Basis step n(G) = 4: the only 2-connected 3 regular planar is K4 .

61. Induction step (n(G)>4): since (G) = ’(G) when G is 3-regular, we may restrict our attention to 3-regular graph with edge-connectivity 2.
Lemma7.3.3 give us a decomposition of G into G1, G2 and a ladder joining them.
Both G1+u1v1and G2+u2v2 are 2-edge-connected and 3-regular. By the induction hypothesis, they are 3-edge-colorable.

62. Let fi be a proper 3-edge-coloring of Gi+uivi.
Let f1(u1v1) = 1, and f2(u2v2) chosen from{1, 2} to have the same parity as the length of the ladder.
Return to G…see figure. u1 u2 v1 v2 G1 G2 3 1 2

63. Thus the Four Color theorem reduces to finding Tait colorings of 3-edge-connected 3-regular planar graphs. The statement of their existence was known as Tait’s conjecture and is equivalent to the Four Color Theorem.

64. Grinberg’s Theorem
Every Hamiltonian 3-regular graph has a Tait coloring.
Tait believed that this completed a proof of the Four Color theorem, because he assumed that every 3-connected 3-regular planar graph is Hamiltonian.
Until 1964 was an explicit counterexample found.

65. Grinberg’s Theorem
If G is a loopless plane graph having a Hamiltonian cycle C, and G has fi’ faces of length i inside C and fi’’ faces of length i outside C, then i(i-2)(fi’- fi’’)=0
Pf: we want to show that i(i-2)fi’ = i(i-2)fi’’.
Outside could be inside, inside could be outside.
We need only show that i(i-2)fi’ is constant.
When there are no inside edges, the sum is n-2. using this as the basis step.
Suppose that i(i-2)fi’ = n-2 when k edges inside C.

66. Grinberg’s Theorem
we can obtain any graph with k+1 edges inside C by adding an edge.
The added edge cuts a face of some length r into two faces of lengths s and t, s+t = r+2.
Since (s-2)+(t-2) = (r-2).
By the induction hypothesis, the sum is n-2.

67. Grinberg’s condition and the Tutte graph. x-y Hamiltonian path.9 4 5 x y H G

68. H’ has five 5-faces, three 4-faces, one 9-faces
H’ has five 5-faces, three 4-faces, one 9-faces. Grinberg’s condition becomes 2a4+ 3a5+7a9 = 0, where ai = fi’- fi’’.
Since the unbounded face is always outside, the equation reduces mod 3 to 2a4  7 mod 3.
The possibility of a4 are +3,+1,-1,-3. the correct choice is -1. means there are 2 4-faces lie outside the Hamiltonian.
However, 2 4-faces has 2-degree vertex. Can lies outside the Hamiltonian.

69. Faster way…
Subdividing one edge incident to each vertex of degree 2.
Resulting graph has seven 5-faces, one 4-faces, and one 11-faces.
Equation becomes 2(1) = 9-3a5, no solution.

70. High connectivity makes it harder to avoid spanning cycles.
Tutte proved that every 4-connected planar graph is Hamiltonian.
Barnette conjectured that every planar 3-connected 3-regular bipartite graph is Hamiltonian.