12.1 Chromatic Number

Transcription

1 12.1 Chromatic Number

2 History of Graph Colorings Theorem The Four Color Theorem: Any planar graph can be colored in at most four colors. But, it wasn t always the Four Color Theorem...

3 Four Color Conjecture 1852 Francis Guthrie

4 Four Color Conjecture 1852 Francis Guthrie Augustus DeMorgan

5 Four Color Conjecture 1852 Francis Guthrie Augustus DeMorgan 1879 Arthur Cayley credit to DeMorgan

6 Four Color Conjecture 1852 Francis Guthrie Augustus DeMorgan 1879 Arthur Cayley credit to DeMorgan 1879 Alfred Kempe proof 1880 Peter Guthrie Tait proof

7 Four Color Conjecture 1852 Francis Guthrie Augustus DeMorgan 1879 Arthur Cayley credit to DeMorgan 1879 Alfred Kempe proof 1880 Peter Guthrie Tait proof 1890 Percy Heawood disproves Kempe, proves Five Color Theorem

8 Four Color Conjecture 1891 Julius Petersen disproves Tait The Petersen graph is an example of a snark.

9 Four Color Conjecture 1891 Julius Petersen disproves Tait The Petersen graph is an example of a snark. Definition A snark is a connected, bridgeless cubic graph with chromatic index 4. Chromatic index is analogous to chromatic number but deals with edge colorings.

10 Four Color Conjecture 1940 s Danilo Blanu sa new snarks

11 Four Color Theorem 1960 s and 1970 s Heinrich Heesch computer work

12 Four Color Theorem 1960 s and 1970 s Heinrich Heesch computer work 1976 Kenneth Appel and Wolfgang Haken proof requiring various computer applications

13 Four Color Theorem 1960 s and 1970 s Heinrich Heesch computer work 1976 Kenneth Appel and Wolfgang Haken proof requiring various computer applications 1980 s Ulrich Schmidt finds error

14 Four Color Theorem 1960 s and 1970 s Heinrich Heesch computer work 1976 Kenneth Appel and Wolfgang Haken proof requiring various computer applications 1980 s Ulrich Schmidt finds error 1989 Appel and Hakem publish book addressing errors

15 Four Color Theorem 1960 s and 1970 s Heinrich Heesch computer work 1976 Kenneth Appel and Wolfgang Haken proof requiring various computer applications 1980 s Ulrich Schmidt finds error 1989 Appel and Hakem publish book addressing errors 2005 Benjamin Watson and Georges Gonthier proof with only one necessary program

16 Chromatic Number χ(g) Definition A coloring of a graph G is an assignment of colors to the vertices of G.

17 Chromatic Number χ(g) Definition A coloring of a graph G is an assignment of colors to the vertices of G. Example Three colorings on C 4.

18 Proper Colorings Definition A proper coloring exists when adjacent vertices are colored differently. Example

19 Proper Colorings Definition A proper coloring exists when adjacent vertices are colored differently. Example Definition The smallest integer k such that a proper coloring of G exists is called the chromatic number and is denoted by χ(g).

20 Chromatic Number What is the chromatic number for... C 4?

21 Chromatic Number What is the chromatic number for... C 4? C 5?

22 Chromatic Number What is the chromatic number for... C 4? C 5? C 2n?

23 Chromatic Number What is the chromatic number for... C 4? C 5? C 2n? C 2n+1?

24 Chromatic Number What is the chromatic number for... C 4? C 5? C 2n? C 2n+1? K n?

25 Chromatic Number What is the chromatic number for... C 4? C 5? C 2n? C 2n+1? K n? Bipartite graph?

26 Chromatic Number What is the chromatic number for... C 4? C 5? C 2n? C 2n+1? K n? Bipartite graph? Tree?

27 Chromatic Number Wheel W 2n?

28 Chromatic Number Wheel W 2n? Wheel W 2n+1?

29 Example Example What is χ(g) for the given graph? G

30 Example Example Find χ(g) for the given graph. G

31 Greedy Algorithm for Vertex Coloring Greedy Algorithm Greedy Algorithms work by making the decision that seems most promising at any moment; it never reconsiders this decision, whatever situation may arise later

32 Greedy Algorithm for Vertex Coloring Greedy Algorithm Greedy Algorithms work by making the decision that seems most promising at any moment; it never reconsiders this decision, whatever situation may arise later The Greedy Algorithm Given a graph G(V, E) with vertex set V = {v 1,..., v n } and adjacency list A vj, construct a function c : V Z + such that if e = {v i, v j } then c(v 1 ) c(v j ). 1 Set c(v j ) = 0 for all 1 j n 2 Set c(v 1 ) = 1 3 For 2 j n, choose a color k > 0 for each vertex v j that differs from its neighbors by c(v j ) = min{k Z + k > 0, c(w) k w A vj }

33 Chromatic Number Example Find the chromatic number for the following graph using the Greedy Algorithm.

34 Solution

35 Solution which corresponds to

36 Example Example Find the chromatic number for the following graph using the Greedy Algorithm.

37 Solution

38 Theorems Theorem Brook s Theorem(1941) If G is connected then if G is not an odd cycle nor complete then χ(g) (G).

39 Theorems Theorem Brook s Theorem(1941) If G is connected then if G is not an odd cycle nor complete then χ(g) (G). Definition A graph is k-colorable if a proper coloring exists on k colors.

40 Theorems Theorem Brook s Theorem(1941) If G is connected then if G is not an odd cycle nor complete then χ(g) (G). Definition A graph is k-colorable if a proper coloring exists on k colors. Theorem Vizing s Theorem (1964): Every undirected graph G can be colored using a number of colors that is at most one larger than (G).

41 Theorems Theorem Brook s Theorem(1941) If G is connected then if G is not an odd cycle nor complete then χ(g) (G). Definition A graph is k-colorable if a proper coloring exists on k colors. Theorem Vizing s Theorem (1964): Every undirected graph G can be colored using a number of colors that is at most one larger than (G). Theorem Konig (1936) A graph is 2-colorable iff it has no circuits of odd length.

42 Application Example Suppose the Dean wants to schedule committee meetings and wants to schedule them in as few time slots as possible. How many would be needed if a meeting cannot take place at the same time as another meeting if the committees have any common members. Curriculum Committee: A,B,F Athletics: D,E,F Academic Affairs: C,D,G Chairs Meeting: A,E,G Student Senate: B,D,E

43 Solution We have to decide which are the vertices and which are the edges.

44 Solution We have to decide which are the vertices and which are the edges. The way that makes the most sense is to have the committees be the vertices and an edge is incident to vertices if there is a member on both committees. Curriculum Athletics Student Senate Academic Affairs Chairs Meeting

45 Solution We have to decide which are the vertices and which are the edges. The way that makes the most sense is to have the committees be the vertices and an edge is incident to vertices if there is a member on both committees. Curriculum Athletics Student Senate Academic Affairs Chairs Meeting χ(g) = 4

46 Chromatic Polynomials Definition P G (x) is the chromatic polynomial for the graph G. It counts the number of ways to color G in at most x colors.

47 Chromatic Polynomials Definition P G (x) is the chromatic polynomial for the graph G. It counts the number of ways to color G in at most x colors. Example Find P K3 (x) and find out how many colorings there are on 1, 3 and 5 colors.

48 Solution x x 1 x 2 P K3 (x) = x(x 1)(x 2) = x 3 3x 2 + 2x

49 Solution x x 1 x 2 P K3 (x) = x(x 1)(x 2) = x 3 3x 2 + 2x For 1 color, there are

50 Solution x x 1 x 2 P K3 (x) = x(x 1)(x 2) = x 3 3x 2 + 2x For 1 color, there are no proper colorings For 3 colors, there are

51 Solution x x 1 x 2 P K3 (x) = x(x 1)(x 2) = x 3 3x 2 + 2x For 1 color, there are no proper colorings For 3 colors, there are 6 proper colorings For 5 colors, there are

52 Solution x x 1 x 2 P K3 (x) = x(x 1)(x 2) = x 3 3x 2 + 2x For 1 color, there are no proper colorings For 3 colors, there are 6 proper colorings For 5 colors, there are 60 proper colorings

53 Properties of Chromatic Polynomials 1 degree is n

54 Properties of Chromatic Polynomials 1 degree is n 2 lead coefficient is 1

55 Properties of Chromatic Polynomials 1 degree is n 2 lead coefficient is 1 3 constant term is 0

56 Properties of Chromatic Polynomials 1 degree is n 2 lead coefficient is 1 3 constant term is 0 4 x is always a factor

57 Properties of Chromatic Polynomials 1 degree is n 2 lead coefficient is 1 3 constant term is 0 4 x is always a factor 5 Either P G (x) = x n or the sum of the coefficients is 0

58 Properties of Chromatic Polynomials 1 degree is n 2 lead coefficient is 1 3 constant term is 0 4 x is always a factor 5 Either P G (x) = x n or the sum of the coefficients is 0 6 χ(g) = k is the smallest integer such that P G (k) 0

59 Properties of Chromatic Polynomials 1 degree is n 2 lead coefficient is 1 3 constant term is 0 4 x is always a factor 5 Either P G (x) = x n or the sum of the coefficients is 0 6 χ(g) = k is the smallest integer such that P G (k) 0 7 the absolute value of the coefficient of the x n 1 term is the size of the graph

60 Example Example Find P K4 (x) and P K5 (x).

61 Solution for K 4 K 4

62 Solution for K 4 K 4 K 4 x x 1 x 2 x 3 P K4 (x) = x(x 1)(x 2)(x 3)

63 Solution for K 5

64 Solution for K 5 x x 1 x 2 x 3 P K5 (x) = x(x 1)(x 2)(x 3)(x 4) x 4

65 Solution for K 5 x x 1 x 2 x 3 P K5 (x) = x(x 1)(x 2)(x 3)(x 4) Pattern? x 4

66 Solution for K 5 x x 1 x 2 x 3 x 4 P K5 (x) = x(x 1)(x 2)(x 3)(x 4) Pattern? P Kn (x) = x(x 1)(x 2) (x (n 1))

67 Example Example Find P G (x) for the following graph.

68 Example Example Find P G (x) for the following graph. P G (x) = x(x 1) 5

69 Example Example Find P G (x) for the following graph. P G (x) = x(x 1) 5 Chromatic Polynomial for a Tree P Tn (x) = x(x 1) n 1

70 Example Example Find P C4 (x) a b d c

71 Solution There are two different cases to consider: Case 1: x choices for a b and d get the same color since they are nonadjacnet, so x 1 choices x 1 choices for c, since it is adjacent to vertices with the same color But, remember... we are looking for number of proper colorings, not just chromatic number - we already know that.

72 Solution There are two different cases to consider: Case 1: x choices for a b and d get the same color since they are nonadjacnet, so x 1 choices x 1 choices for c, since it is adjacent to vertices with the same color But, remember... we are looking for number of proper colorings, not just chromatic number - we already know that. Case 2: x choices for a x 1 choices for b We force d to be a different color than b, so there are x 2 choices x 2 choices for c In math, or means +, so we have P C4 (x) = x(x 1)(x 2) 2 + x(x 1) 2 = x 4 4x 3 + 6x 2 3x

73 Example Example Find P G (x) for the following graph.

74 Solution x x 1 x 2 x 2 x 3 P G (x) = x(x 1)(x 2) 2 (x 3)

75 Two Piece Theorem Theorem Two Piece Theorem Let the vertex set of G be partitioned into two disjoint sets w 1 and w 2 and let G 1 and G 2 be the subgraphs generated by w 1 and w 2, respectively. Suppose that in G 1, no edge joins a vertex of w 1 to a vertex of w 2. Then P G (x) = P G1 (x)p G2 (x)

76 Two Piece Theorem Theorem Two Piece Theorem Let the vertex set of G be partitioned into two disjoint sets w 1 and w 2 and let G 1 and G 2 be the subgraphs generated by w 1 and w 2, respectively. Suppose that in G 1, no edge joins a vertex of w 1 to a vertex of w 2. Then Example Find P G (x) for the given graph. P G (x) = P G1 (x)p G2 (x)

77 Solution Example Find P G (x) for the given graph. P G1 (x) = x(x 1)(x 2) = x 3 3x 2 + 2x P G2 (x) = x(x 1)(x 2) 2 = x 4 5x 3 + 8x 2 4x P G (x) = (x 3 3x 2 + 2x)(x 4 5x 3 + 8x 2 4x)

78 Fundamental Reduction Theorem Theorem The Fundamental Reduction Theorem P G (x) = P G α (x) P G α (x) where G α = G α is a deletion and G α is a contraction.

79 Example Example Find P C4 (x) using the Fundamental Reduction Theorem. a α b a b a,b d c d c d c

80 Example Example Find P C4 (x) using the Fundamental Reduction Theorem. a α b a b a,b d c d c d c P G α (x) = x(x 1) 3 = x 4 3x 3 + 3x 2 x P G α (x) = x(x 1)(x 2) = x3 3x 2 + 2x P G (x) = x 4 3x 3 + 3x 2 x (x 3 3x 2 + 2x) = x 4 4x 3 + 6x 2 3x

81 Example Example Use the Fundamental Reduction Theorem to find the chromatic polynomial for the given graph. G

82 Solution Multiple choices of α would work here. G α

83 Solution Multiple choices of α would work here. G α So now we have... G α P G α (x) = x(x 1) 3 = x 4 3x 3 + 3x 2 x

84 Solution (cont.) G α P G α (x) = x(x 1)2 = x 3 2x 2 + x

85 Solution (cont.) G α P G α (x) = x(x 1)2 = x 3 2x 2 + x Putting this together, we have P G (x) = x 4 3x 3 + 3x 2 x (x 3 2x 2 + x) = x 4 4x 3 + 5x 2 2x

86 Principle of Inclusion-Exclusion Theorem Suppose we have a set A, A = N. If the elements of A have the properties a 1, a 2,..., a p then the number of objects having none of these properties is given by N(a 1 a p) = N i N(a i )+ i j N(a i a j )±...+( 1) p N(a 1 a 2 a p )

87 Proof If x A has none of a 1,..., a p then it is counted once on the left hand side once on the right hand side in the N term. If x has a certain set of properties then x is not counted on the left hand side and the following happens on the right hand side:

88 Proof If x A has none of a 1,..., a p then it is counted once on the left hand side once on the right hand side in the N term. If x has a certain set of properties then x is not counted on the left hand side and the following happens on the right hand side: Suppose x has r p properties. ( ) r x is counted once in the N term: 0 x is counted r times in the ( ) r N(a i ) term: = r 1 ( ) r x is counted times in the N(a 2 i a j ) term

89 Proof If x A has none of a 1,..., a p then it is counted once on the left hand side once on the right hand side in the N term. If x has a certain set of properties then x is not counted on the left hand side and the following happens on the right hand side: Suppose x has r p properties. ( ) r x is counted once in the N term: 0 x is counted r times in the ( ) r N(a i ) term: = r 1 ( ) r x is counted times in the N(a 2 i a j ) term Altogether, x is counted ( ) ( ) ( ) ( ) r r r + ±... + ( 1) r r r }{{} binomial expansion for (x 1) r And, for x = 1, we get that this sum is (1 1) r = 0

90 Example Example How many integers between 1 and 1000 are not divisible by 3, 5 or 7?

91 Solution For our properties, we have a 1 is the property the integer is divisible by 3 a 2 is the property the integer is divisible by 5 a 3 is the property the integer is divisible by 7

92 Solution For our properties, we have a 1 is the property the integer is divisible by 3 a 2 is the property the integer is divisible by 5 a 3 is the property the integer is divisible by 7 N(a 1 ) = = 333 N(a 2 ) = = 200 N(a 3 ) = = 142

93 Solution For our properties, we have a 1 is the property the integer is divisible by 3 a 2 is the property the integer is divisible by 5 a 3 is the property the integer is divisible by 7 N(a 1 ) = = 333 N(a 2 ) = = 200 N(a 3 ) = = 142 N(a 1 a 2 ) = = 66 N(a 1 a 3 ) = = 47 N(a 2 a 3 ) = = 28

94 Solution For our properties, we have a 1 is the property the integer is divisible by 3 a 2 is the property the integer is divisible by 5 a 3 is the property the integer is divisible by 7 N(a 1 ) = = 333 N(a 2 ) = = 200 N(a 3 ) = = 142 N(a 1 a 2 ) = = 66 N(a 1 a 3 ) = = 47 N(a 2 a 3 ) = = 28 N(a 1 a 2 a 3 ) = = 9

95 Solution For our properties, we have a 1 is the property the integer is divisible by 3 a 2 is the property the integer is divisible by 5 a 3 is the property the integer is divisible by 7 N(a 1 ) = = 333 N(a 2 ) = = 200 N(a 3 ) = = 142 N(a 1 a 2 ) = = 66 N(a 1 a 3 ) = = 47 N(a 2 a 3 ) = = 28 N(a 1 a 2 a 3 ) = = 9 So, by the Principle of Inclusion-Exclusion, we have N(a 1a 2a 3) = 1000 ( ) + ( ) 9 = 457

96 Inclusion-Exclusion and the Chromatic Polynomial Example Find P C4 (x) using the Principle of Inclusion-Exclusion. e 4 e 1 e 3 e 2

97 Solution Let a i be the property that edge e i has both incident vertices colored with the same color. So, what we will want is none of the a i to be true. That is, we are looking for N(a 1 a 2 a 3 a 4 ). N = x 4

98 Solution Let a i be the property that edge e i has both incident vertices colored with the same color. So, what we will want is none of the a i to be true. That is, we are looking for N(a 1 a 2 a 3 a 4 ). N = x 4 N(a 1 ) = x 3

99 Solution Let a i be the property that edge e i has both incident vertices colored with the same color. So, what we will want is none of the a i to be true. That is, we are looking for N(a 1 a 2 a 3 a 4 ). N = x 4 N(a 1 ) = x 3 N(a i ) = 4x 3

100 Solution Let a i be the property that edge e i has both incident vertices colored with the same color. So, what we will want is none of the a i to be true. That is, we are looking for N(a 1 a 2 a 3 a 4 ). N = x 4 N(a 1 ) = x 3 N(a i ) = 4x 3 N(a 1 a 2 ) = x 2

101 Solution Let a i be the property that edge e i has both incident vertices colored with the same color. So, what we will want is none of the a i to be true. That is, we are looking for N(a 1 a 2 a 3 a 4 ). N = x 4 N(a 1 ) = x 3 N(a i ) = 4x 3 N(a 1 a 2 ) = x 2 N(a i a j ) = 6x 2

102 Solution Let a i be the property that edge e i has both incident vertices colored with the same color. So, what we will want is none of the a i to be true. That is, we are looking for N(a 1 a 2 a 3 a 4 ). N = x 4 N(a 1 ) = x 3 N(a i ) = 4x 3 N(a 1 a 2 ) = x 2 N(a i a j ) = 6x 2 N(a 1 a 2 a 3 ) = x

103 Solution Let a i be the property that edge e i has both incident vertices colored with the same color. So, what we will want is none of the a i to be true. That is, we are looking for N(a 1 a 2 a 3 a 4 ). N = x 4 N(a 1 ) = x 3 N(a i ) = 4x 3 N(a 1 a 2 ) = x 2 N(a i a j ) = 6x 2 N(a 1 a 2 a 3 ) = x N(a i a j a k ) = 4x

104 Solution Let a i be the property that edge e i has both incident vertices colored with the same color. So, what we will want is none of the a i to be true. That is, we are looking for N(a 1 a 2 a 3 a 4 ). N = x 4 N(a 1 ) = x 3 N(a i ) = 4x 3 N(a 1 a 2 ) = x 2 N(a i a j ) = 6x 2 N(a 1 a 2 a 3 ) = x N(a i a j a k ) = 4x N(a 1 a 2 a 3 a 4 ) = x

105 Solution Let a i be the property that edge e i has both incident vertices colored with the same color. So, what we will want is none of the a i to be true. That is, we are looking for N(a 1 a 2 a 3 a 4 ). N = x 4 Therefore, we have N(a 1 ) = x 3 N(a i ) = 4x 3 N(a 1 a 2 ) = x 2 N(a i a j ) = 6x 2 N(a 1 a 2 a 3 ) = x N(a i a j a k ) = 4x N(a 1 a 2 a 3 a 4 ) = x P C4 (x) = N(a 1a 2a 3a 4) = x 4 4x 3 +6x 2 4x+x = x 4 4x 3 +6x 2 3x

106 Example Example Find P G (x) for the following graph by using the Principle of Inclusion-Exclusion. e 1 e 4 e 3 e 2 e 5

107 Solution N =

108 Solution N = x 4 N(a 1 ) =

109 Solution N = x 4 N(a 1 ) = x 3 i N(a i ) =

110 Solution N = x 4 N(a 1 ) = x 3 i N(a i ) = 5x 3 N(a 1 a 2 ) =

111 Solution N = x 4 N(a 1 ) = x 3 i N(a i ) = 5x 3 N(a 1 a 2 ) = x 2 i j N(a i a j ) =

112 Solution N = x 4 N(a 1 ) = x 3 i N(a i ) = 5x 3 N(a 1 a 2 ) = x 2 i j N(a i a j ) = 10x 2 N(a 1 a 2 a 3 ) =

113 Solution N = x 4 N(a 1 ) = x 3 i N(a i ) = 5x 3 N(a 1 a 2 ) = x 2 i j N(a i a j ) = 10x 2 N(a 1 a 2 a 3 ) = x 2 N(a i a j a k ) = i j k

114 Solution N = x 4 N(a 1 ) = x 3 i N(a i ) = 5x 3 N(a 1 a 2 ) = x 2 i j N(a i a j ) = 10x 2 N(a 1 a 2 a 3 ) = x 2 N(a i a j a k ) = 2x 2 + 8x i j k

115 Solution N(a 1 a 2 a 3 a 4 ) =

116 Solution N(a 1 a 2 a 3 a 4 ) = x N(a i a j a k a l ) = i j k l

117 Solution N(a 1 a 2 a 3 a 4 ) = x N(a i a j a k a l ) = 5x i j k l N(a 1 a 2 a 3 a 4 a 5 ) =

118 Solution N(a 1 a 2 a 3 a 4 ) = x N(a i a j a k a l ) = 5x i j k l N(a 1 a 2 a 3 a 4 a 5 ) = x

119 Solution N(a 1 a 2 a 3 a 4 ) = x N(a i a j a k a l ) = 5x i j k l N(a 1 a 2 a 3 a 4 a 5 ) = x Therefore, N(a 1a 2a 3a 4a 5) = x 4 5x x 2 (2x 2 + 8x) + 5x x So, P G (x) = x 4 5x 3 + 8x 2 4x