1. Xuding Zhu National Sun Yat-sen University Circular chromatic index

2. G=(V,E): a graph an integer k-coloring of G is such that A a real number A (circular) 0 1 2 0.5 1.5 A 2.5-coloring r-coloring of G is

3. The circular chromatic number of G is { r : G has a circular r-coloring }min The chromatic number of G is = min { k : G has a k-coloring }

4. Why “circular” ? 0=r 3 1 2 4 0r

5. Why “circular” ? 0=r 3 1 2 4 x~y |f(x)-f(y)|_r ≥ 1 The distance between p, p’ in the circle is f is a circular r-coloring if 0r p p’p’

6. Coloring = distribution of resources Circular coloring = distribution of resources of periodic nature

7. A B C D E Downtown Bielefeld AD is a traffic flow AD conflicts with EB AD and CB do not conflict

8. We need to build a traffic light system Each traffic flow is assigned green light phase Conflicting traffic flows get disjoint green light phases

9. Use a graph as a model ABACADAE BABCBDBE CACBCDCE DADBDCDE EAEBECED Each traffic flow is represented by a vertex Conflicting traffic flows are joined by an edge

10. ABACADAE BABCBDBE CACBCDCE DADBDCDE EAEBECED A complete traffic period is represented by a circle Each vertex is assigned a unit length arc of the circle Adjacent vertices are mapped to disjoint arcs Our goal is to minimize the length of the circle

11. If the circle has circumference r, then this is equivalent to a circular r-coloring So the problem is to find

12. A distributed computation problem V: a set of computers D: a set of data files

13. a b c d e If x ~ y, then x and y cannot operate at the same time If x ~ y, then x and y must alternate their turns in operation.

14. Schedule the operating time of the computers efficiently. Efficiency: (number of computers operating on the average)/(total number). The computer time is discrete: time 0, 1, 2, …

15. 1: coloring solution Color the vertices of G with k= colors.

16. a e b d c Color the graph with 3 colors 0 1 0 1 2 At time 0 Operate machines acac 1 bdbd 2 e 3 acac 4 bdbd 5 e The efficiency is 1/3

17. In general, the coloring solution has efficiency

18. 2: Computer scientists solution

21. The efficiency of this scheduling is Better than the coloring solution Theorem [Barbosa and Gafni, 1989] Efficiency of ptimal scheduling = efficiency of optimal scheduling by edge reversal.

22. 3: Circular coloring solution Let r= Let f be a circular r-coloring of G x operates at time k iff for some integer m

23. 0 1 2 3 4 r=4.4 0 1 2 3 4 5 Efficiency =

24. The efficiency of this scheduling is b e a c d For this example, the efficiency is

25. Theorem [Yeh-Zhu] The optimal scheduling has efficiency

26. A.Vince, 1988. star chromatic number About 100 papers published, and publications are accelerating. Quote from Feder, Hell, Mohar (2003): “ The theory of circular colorings of graphs has become an important branch of chromatic graph theory with many exciting results and new techniques. ”

27. Questions concerning Relation between and other graph parameters: 1. Circular chromatic number of special classes of graphs: planar graphs, graphs with forbidden minor, line graphs, distance graphs, … 2. 3. How to calculate 4. Generalization to circular coloring of digraphs

28. Questions concerning Circular chromatic number of line graphs. 2. 3. How to calculate Relation between and other graph parameters: 1. and circular flow number

29. An equivalent definition for p, q: positive integers. A (p, q)-coloring of G is a mapping f: V {0, 1,..., p-1} such that x ~ y

30. An equivalent definition for p, q: positive integers. A (p, q)-coloring of G is a mapping f: V {0, 1,..., p-1} such that x ~ y

31. Circular chromatic index of cubic graphs

32. 8 03 0 6 2 5 1 9 2 7 10 5 8 4 A (11,3)-coloring of L(P)

33. Question: Is it true that every 2-edge connected cubic graph has Problem: Prove that every 2-edge connected cubic planar graph G has Four Color Theorem:

34. Peyman Afshani, Theorem [ Mahsa Ghandehari, Mahya Ghandehari, Hamed Hatami Ruzbeh Tusserkani,Xuding Zhu]

35. Peyman Afshani, Theorem [ Mahsa Ghandehari, Mahya Ghandehari, Hamed Hatami Ruzbeh Tusserkani,Xuding Zhu]

36. Proof of the theorem We need a lemma. For stating this lemma, we need a definition.

37. 1 2 0 0 2 3 0 1 1 0 2 1 0 3 A 4-coloring f 2

38. 1 2 0 0 2 3 0 1 1 0 2 1 0 3 2 We define a digraph induced by this coloring

39. 1 2 0 0 2 3 0 1 1 0 2 1 0 3 A 4-coloring f 0 1 2 3 0 2

40. 1 2 0 0 2 3 0 1 1 0 2 1 0 3 A 4-coloring f 2 0 1 2 3 0

41. 1 2 0 0 2 3 0 1 1 0 2 1 0 3 A 4-coloring f 2 0 1 2 3 0

42. 1 2 0 0 2 3 0 1 1 0 2 1 0 3 A 4-coloring f 2 0 1 2 3 0

43. 1 2 0 0 2 3 0 1 1 0 2 1 0 3 A 4-coloring f 2 0 1 2 3 0

44. 1 2 0 0 2 3 0 1 1 0 21 0 3 A 4-coloring f 2

45. Lemma: If G has a 4-coloring f such that (1): is acyclic; (2): each directed path of contains at most 2 vertices of color 3, then

46. 1 2 0 0 2 3 0 1 1 0 21 0 3 A 4-coloring f 2 is acyclic

47. 1 2 0 0 3 0 1 1 0 21 0 3 A 4-coloring f 2 is acyclic

48. 1 2 0 0 3 0 1 0 21 0 3 A 4-coloring f 2 is acyclic

49. 1 2 0 0 3 0 1 0 21 3 A 4-coloring f 2 is acyclic

50. 1 2 0 0 3 0 1 0 1 3 A 4-coloring f 2 is acyclic

51. 1 2 0 0 3 0 0 1 3 A 4-coloring f 2 is acyclic

52. 1 2 0 0 0 1 3 A 4-coloring f is acyclic

53. 1 2 0 0 0 3 A 4-coloring f is acyclic

54. 1 2 0 0 2 3 0 1 1 0 21 0 3 A 4-coloring f 2 is acyclic Each directed path contains at most two vertices of color 3.

55. 1 2 0 0 2 3 0 1 1 0 21 0 3 A 4-coloring f 2 is acyclic How to get a (11, 3)-coloring

56. 1 2 0 0 2 3 0 1 1 0 21 0 3 A 4-coloring f 2 is acyclic How to get a (11, 3)-coloring Multiply each color by 3.

57. 1 2 0 0 2 3 0 1 1 0 21 0 3 A 4-coloring f 2 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 3 6

58. 3 0 0 6 9 0 3 3 0 0 6 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 3 6

59. 3 0 0 6 9 0 3 3 0 0 6 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 3 6 But this is not a (11,3)-coloring

60. 3 0 0 6 9 0 3 3 0 0 6 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 3 6 But this is not a (11,3)-coloring 1 1 1

61. 3 0 0 6 9 0 3 3 0 0 6 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 3 6 But this is not a (11,3)-coloring 1 1 1

62. 3 0 0 6 9 0 4 4 0 0 6 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 4 6 But this is not a (11,3)-coloring 1 1 1

63. 3 0 0 6 9 0 4 4 0 0 6 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 4 6 But this is not a (11,3)-coloring 1 1 1

64. 3 0 0 7 9 0 4 4 0 0 7 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 4 7 But this is not a (11,3)-coloring 1 1 1

65. 3 0 0 7 9 0 4 4 0 0 7 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 4 7 But this is not a (11,3)-coloring 1 1 1

66. 3 0 0 7 10 0 4 4 0 0 7 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 4 7 But this is not a (11,3)-coloring 1 1 1

67. 3 0 0 7 10 0 4 4 0 0 7 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 4 7 But this is not a (11,3)-coloring 1 1 1

68. 3 0 0 7 10 0 4 4 0 0 7 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 4 7 But this is not a (11,3)-coloring 1 1 2

69. 3 0 0 7 10 0 4 4 0 0 7 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 4 7 But this is not a (11,3)-coloring 1 1 2

70. 3 0 0 7 10 0 5 5 0 0 7 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 4 7 But this is not a (11,3)-coloring 1 1 2

71. 3 0 0 7 10 0 5 5 0 0 7 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 4 7 But this is not a (11,3)-coloring 1 1 2

72. 3 0 0 8 10 0 5 5 0 0 7 is acyclic How to get a (11, 3)-coloring Multiply each color by 3. 96 4 8 But this is not a (11,3)-coloring 1 1 2

73. 3 0 0 8 10 0 5 5 0 0 7 How to get a (11, 3)-coloring Multiply each color by 3. 96 4 8 But this is not a (11,3)-coloring 1 1 2 Each vertex changed color at most twice

74. 3 0 0 8 10 0 5 5 0 0 7 How to get a (11, 3)-coloring Multiply each color by 3. 96 4 8 But this is not a (11,3)-coloring 1 1 2 Each vertex changed color at most twice The deleted edges are OK

75. Exercise: Find and prove a generalization of this lemma. Lemma: If G has a 4-coloring f such that (1): is acyclic; (2): each directed path of contains at most 2 vertices of color 3, then

76. G cubic, has a perfect matching, girth at least 4 First we prove: 1: is acyclic; We shall construct a 4-edge coloring c of G so that 2: each directed path of contains at most two edges of color 3.

77. The M-edges will be colored by color 0 G has a perfect matching M (the brown edges)

78. Delete the M-edges, we have a 2-factor

79. Color each even cycle by colors 1, 2 1 2 1 2 1 2 1 2

80. Color each odd cycle by colors 1, 2, except one edge is colored by color 3. 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 3

81. Such a 4-edge coloring of G is called a VALID edge coloring of G (w.r.t. the matching M). 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 3

82. Given a valid 4-edge coloring c of G, is 1: acyclic ? 2: every directed path of contains at mots 2 edges of color 3 ? It depends. We shall modify the coloring c so that 1 and 2 hold.

83. 1 and 2 can be combined into a single requirement 1&2: Every directed walk of contains at most two edges of color 3. Only odd cycles of G-M contains edges of color 3. 3 1 2 1 2 1 2 Red edges are colored by color 0

84. 3 1 2 1 2 1 2 The line graph L(G) around the cycle is

85. 3 1 2 1 2 1 2 0 0 0 0 0 0 0 Consider directed walks of starting and ending with edges of color 3. Which M-edges will be contained in such a walk ?

86. 3 1 2 1 2 1 2 0 0 0 0 0 0 0

87. 3 1 2 1 2 1 2 0 0 0 0 0 0 0

88. 3 1 2 1 2 1 2 0 0 0 0 0 0 0 For each odd cycle, at most 4 M-edges incident to it can be used in a directed walk which contains 3 edges of color 3.

89. 3 1 2 1 2 1 2 0 0 0 0 0 0 0 For each odd cycle, at most 4 M-edges incident to it can be used in a directed walk which contains 3 edges of color 3.

90. 3 1 2 1 2 1 2 0 0 0 0 0 0 0 For each odd cycle, at most 4 M-edges incident to it can be used in a directed walk which contains 3 edges of color 3. Completely blocked M-edges One direction is blocked

91. 3 1 2 1 2 1 2 0 0 0 0 0 0 0 not 3 This M-edge becomes completely blocked

92. 3 1 2 1 2 1 2 0 0 0 0 0 0 0 By considering the coloring of the other odd cycles, more M-edges incident to C become completely blocked.

93. 3 1 2 1 2 1 2 0 0 0 0 0 0 0 Input of C Output of C Lemma: There is a valid 4-edge coloring c such that no odd cycle of G-M has both input and output. This implies that no directed walk contains 3 edges of color 3, because each odd cycle has only one edge of color 3.

94. 3 1 2 1 2 1 2 0 0 0 0 0 0 0 3 3

95. number of not completely blocked M-edges which are chords of some odd cycles C of G-M number of not completely blocked M-edges connecting two distinct odd cycles. 3 1 3 1

96. Lemma: If c is a valid 4-edge coloring of G such that is minimum, then no odd cycles C has both inputs and outputs.

97. Assume c is a valid 4-edge coloring of G. is an odd cycle which contains both inputs and outputs. We shall find another valid coloring c ’ such that

98. First we consider the case that C is chordless. Uncolor the edges of C. Each M-edge incident to C has at least one direction blocked, because it is incident to another colored cycle.

99. Each M-edge has at least one blocked direction (under the condition that C is uncolored). For simplicity, for each M-edge incident to C, we draw exactly one direction blocked. By assumption, at least one edge is an input and one edge is an output.

100. As C is odd, there are two consecutive M-edges that have a common blocked direction. The next two consecutive M-edges have opposite blocked directions.

101. Color C as above. 1 2 3

102. 1 2 3 We have more blocked directions.

103. 1 2 3 At most one M-edge incident to C is not completely blocked Originally at least two M-edges incident to C are not completely blocked. M-edges not incident to C remain the same status

104. Next we consider the case C has a chord. Details omitted

105. Up to now we have proved G cubic, has a perfect matching, girth at least 4 Corollary: If G is 2-edge connected, has maximum degree at most 3, and girth at least 4, then

106. Proof. If G is cubic, then G has a perfect matching. Otherwise, G G G’G’ G ’ is cubic, either 2-edge connected, or contains one cut edge. In any case G ’ has a perfect matching

107. Up to now we have proved G cubic, has a perfect matching, girth at least 4 Next we omit the condition that G has girth at least 4, cubic, has a perfect matching.

108. We prove it by induction on the number of vertices that G has maximum degree 3, and does not contain as subgraphs We may assume G is 2-edge connected.

109. G has maximum degree 3, 2-edge connected, and If G has girth at least 4, then we are done !

110. If not H1, H2, then find a (11, 3)-coloring by induction. a bc a b c c a b

111. c c c c+3 c+6

112. If the reduced graph is H1 or H2, then the We can find a (7,2)-edge coloring.

115. Circular chromatic indexes of graphs of large girth G has large girth G is simple

116. Theorem [Thomas Kraiser, Daniel Kral, Riste Skrekovski] If G is cubic and has large girth, then depends on the girth.

117. Theorem [Thomas Kraiser, Daniel Kral, Riste Skrekovski, Xuding Zhu] For any graph G of large girth, depends on the girth.

118. Conjecture 1: For any graph G of maximum of maximum degree k, either or True for k=1, 2, 3. If it is true, is it sharp ?

119. For any k, find a k-regular graph G with For k=2, the 5-cycle. For k=3, the Petersen graph. For k =4, ???

120. Is there any relation between and for cubic graphs other than In particular, is it true that cubic graphs of large girth have

121. The flower snarks J3 J5 J7

122. Theorem [Steffen] (I conjecture equality holds) Question: Is it true that 3 ?

123. We see GAPS among circular chromatic indexes of graphs. (5/2, 3), (7/3, 5/2), (9/4, 7/3), … (11/3, 4), Conjecture: Among the rational numbers which are the circular chromatic indexes of graphs, there is no strictly increasing bounded infinite sequence.

124. Real numbers Circular chromatic indexes of graphs This conjecture implies, in particular, that there are a lot of gaps.

125. We have many density results: For each rational 2 < r < 4, there is a planar G with For each rational 2 4), there is a Kn-minor free G with For each rational 2 < r < 8/3, there is a K4-minor free G with

126. What makes line graphs so different ? (circular chromatic indexes have gaps). Feeling: What makes it different from other classes of graphs is that for line graphs.

127. Conjecture: Among the rational numbers which are the circular chromatic numbers of graphs of bounded Delta-chi ratio, there is no strictly increasing bounded infinite sequence. The Delta-chi ratio of G: