Polynomials in graph theory

Transcription

1 Polynomials in graph theory Alexey Bogatov Department of Software Engineering Faculty of Mathematics and Mechanics Saint Petersburg State University JASS 2007 Saint Petersburg Course 1: Polynomials: Their Power and How to Use Them

2 2 / 67 Outline 1 Introduction 2 Algebraic methods of counting graph colorings 3 Counting graph colorings in terms of orientations 4 Probabilistic restatement of Four Color Conjecture 5 Arithmetical restatement of Four Color Conjecture 6 The Tutte polynomial

3 3 / 67 Outline 1 Introduction 2 Algebraic methods of counting graph colorings 3 Counting graph colorings in terms of orientations 4 Probabilistic restatement of Four Color Conjecture 5 Arithmetical restatement of Four Color Conjecture 6 The Tutte polynomial

4 / 67 Introduction to Coloring Problem We are going to color maps on an island (or on a sphere). Countries are planar regions. In case of proper coloring 2 neighboring countries must have different colors.

5 4 / 67 Introduction to Coloring Problem We are going to color maps on an island (or on a sphere). Countries are planar regions. In case of proper coloring 2 neighboring countries must have different colors.

6 Introduction to Coloring Problem We are going to color maps on an island (or on a sphere). Countries are planar regions. In case of proper coloring 2 neighboring countries must have different colors. Terminological remark: 4 / 67 proper coloring vs coloring coloring vs assignment of colors

7 5 / 67 Introduction to Coloring Problem Boundaries are Jordan curves (Jordan curve is a continuous image of a segment [a, b]).

8 5 / 67 Introduction to Coloring Problem Boundaries are Jordan curves (Jordan curve is a continuous image of a segment [a, b]). It can be proved that countries may be colored in 6 colors.

9 / 67 Introduction to Coloring Problem If we consider countries as vertices of graph and connect neighboring countries by an edge, then we can reformulate the problem in terms of coloring the graph.

10 / 67 Introduction to Coloring Problem If we consider countries as vertices of graph and connect neighboring countries by an edge, then we can reformulate the problem in terms of coloring the graph. The problem can easily be reduced to the case of 3-valent graph.

11 / 67 Introduction to Coloring Problem If we consider countries as vertices of graph and connect neighboring countries by an edge, then we can reformulate the problem in terms of coloring the graph. The problem can easily be reduced to the case of 3-valent graph. Line graph: its vertices correspond to edges of initial graph, 2 vertices are connected by edge iff 2 corresponding edges of initial graph are incidental

12 7 / 67 Introduction to Coloring Problem 1852, Guthrie: The Four Color Conjecture (4CC). 1976, Appel, Haken: a computer-to-computer proof (cannot be checked by human).

13 7 / 67 Introduction to Coloring Problem 1852, Guthrie: The Four Color Conjecture (4CC). 1976, Appel, Haken: a computer-to-computer proof (cannot be checked by human). Searching for simplified reformulations where proofs can be checked by a human being. Polynomials are the main instrument of counting colorings.

14 8 / 67 Outline 1 Introduction 2 Algebraic methods of counting graph colorings 3 Counting graph colorings in terms of orientations 4 Probabilistic restatement of Four Color Conjecture 5 Arithmetical restatement of Four Color Conjecture 6 The Tutte polynomial

15 / 67 Algebraic methods of counting graph colorings χ G (p) is the number of (proper) colorings of (vertices of) graph G in p colors.

16 10 / 67 Algebraic methods: the theorem M G (p) = N p (x k, x l ) (v k,v l ) E N p (y, z) = p 1 y p 1 z y p 2 z 2... yz p 1

17 10 / 67 Algebraic methods: the theorem M G (p) = N p (x k, x l ) (v k,v l ) E N p (y, z) = p 1 y p 1 z y p 2 z 2... yz p 1 Operator R p replaces the exponent of each variable x p by its value modulo p: x i x 3p+1 i

18 10 / 67 Algebraic methods: the theorem M G (p) = N p (x k, x l ) (v k,v l ) E N p (y, z) = p 1 y p 1 z y p 2 z 2... yz p 1 Operator R p replaces the exponent of each variable x p by its value modulo p: x i x 3p+1 i Theorem For any graph G = (V, E), V = m, E = n p N χ G (p) = p m n (R p M p (G))(0,..., 0).

19 11 / 67 Algebraic methods: Proof of the theorem Polynomial R p M p (G) can be uniquely determined by choosing appropriate p m values of set of variables.

20 11 / 67 Algebraic methods: Proof of the theorem Polynomial R p M p (G) can be uniquely determined by choosing appropriate p m values of set of variables. Let c 0 = 1, c 1 = ω,..., c p 1 = ω p 1 be colors (where ω is the primitive root of 1 of degree p).

21 11 / 67 Algebraic methods: Proof of the theorem Polynomial R p M p (G) can be uniquely determined by choosing appropriate p m values of set of variables. Let c 0 = 1, c 1 = ω,..., c p 1 = ω p 1 be colors (where ω is the primitive root of 1 of degree p). µ : V C = {c 0,..., c p 1 } is a coloring of the graph G.

22 12 / 67 Algebraic methods: Proof of the theorem (cont.) Using interpolation theorem we get: R p M p (G) = µ (R p M p (G))(µ(v 1 ),..., µ(v m ))P µ (the sum is taken through all p m colorings), m P µ = S p (x k, µ(v k )), S p (x, c q ) = k=1 0 l p 1, l q x c l c q c l.

23 12 / 67 Algebraic methods: Proof of the theorem (cont.) Using interpolation theorem we get: R p M p (G) = µ (R p M p (G))(µ(v 1 ),..., µ(v m ))P µ (the sum is taken through all p m colorings), m P µ = S p (x k, µ(v k )), S p (x, c q ) = k=1 0 l p 1, l q x c l c q c l. For arguments from C values of M p (G) and R p M p (G) are the same, so we have R p M p (G) = µ M p (G)(µ(v 1 ),..., µ(v m ))P µ

24 13 / 67 Algebraic methods: Proof of the theorem (cont.) N p (y, y) = p 1 y p... y p = 0,

25 13 / 67 Algebraic methods: Proof of the theorem (cont.) N p (y, y) = p 1 y p... y p = 0, N p (y, z) = p y p z p y = p if y z, y z

26 13 / 67 Algebraic methods: Proof of the theorem (cont.) N p (y, y) = p 1 y p... y p = 0, N p (y, z) = p y p z p y = p if y z, y z M p (G)(µ(v 1 ),..., µ(v m )) = p n for a proper coloring, else M p (G)(µ(v 1 ),..., µ(v m )) = 0. We came to R p M p (G) = p n µ P µ where sum is taken through χ G (p) proper colorings.

27 14 / 67 Algebraic methods: Proof of the theorem (cont.) Substituting x 1 =... = x m = 0: m P µ (0,..., 0) = S p (0, µ(v k )), k=1

28 14 / 67 Algebraic methods: Proof of the theorem (cont.) Substituting x 1 =... = x m = 0: m P µ (0,..., 0) = S p (0, µ(v k )), k=1 S p (0, c q ) = 0 l p 1, l q c l c q c l = 0 l p 1, l q 1 1 cq c l = 1 l p ω l

29 14 / 67 Algebraic methods: Proof of the theorem (cont.) Substituting x 1 =... = x m = 0: m P µ (0,..., 0) = S p (0, µ(v k )), k=1 S p (0, c q ) = 0 l p 1, l q c l c q c l = 0 l p 1, l q 1 1 cq c l = 1 l p ω l So S p (0, c q ) = S p (0) is independent of c q.

30 15 / 67 Algebraic methods: Proof of the theorem (cont.) (R p M p (G))(0,..., 0) = p n S p (0) m χ G (p) If we substitute any specific graph (e.g. K 1 that has 1 vertex and 0 edges) we get S p (0): m = 1, n = 0, (R p M p (K 1 ))(0,..., 0) = 1, χ K1 (p) = p.

31 15 / 67 Algebraic methods: Proof of the theorem (cont.) (R p M p (G))(0,..., 0) = p n S p (0) m χ G (p) If we substitute any specific graph (e.g. K 1 that has 1 vertex and 0 edges) we get S p (0): m = 1, n = 0, (R p M p (K 1 ))(0,..., 0) = 1, χ K1 (p) = p. Therefore S p (0) = p 1 and χ G (p) = p m n (R p M p (G))(0,..., 0).

32 6 / 67 Algebraic methods: Colorings of 3-valent graphs Let G be a planar 3-valent graph (each vertex has 3 adjacent vertices).

33 6 / 67 Algebraic methods: Colorings of 3-valent graphs Let G be a planar 3-valent graph (each vertex has 3 adjacent vertices). 3-valent graph T can be represented as T = { e i1, e j1, e k1,..., e i2n, e j2n, e k2n } (2n vertices, 3n edges).

34 Algebraic methods: Colorings of 3-valent graphs Let G be a planar 3-valent graph (each vertex has 3 adjacent vertices). 3-valent graph T can be represented as T = { e i1, e j1, e k1,..., e i2n, e j2n, e k2n } (2n vertices, 3n edges). 6 / 67 λ G (p) is the number of (proper) colorings of edges of G in p colors.

35 17 / 67 Algebraic methods: Colorings of 3-valent graphs L(x p, x q, x r ) = (x p x q )(x q x r )(x r x p )

36 17 / 67 Algebraic methods: Colorings of 3-valent graphs L(x p, x q, x r ) = (x p x q )(x q x r )(x r x p ) M(G) = 2n l=1 L(x il, x jl, x kl )

37 17 / 67 Algebraic methods: Colorings of 3-valent graphs L(x p, x q, x r ) = (x p x q )(x q x r )(x r x p ) M(G) = R 3 M(G)(x 1,..., x 3n ) = 2n l=1 L(x il, x jl, x kl ) d 1,...,d 3n {0,1,2} c d1,...,d 3n x d x d 3n 3n

38 17 / 67 Algebraic methods: Colorings of 3-valent graphs L(x p, x q, x r ) = (x p x q )(x q x r )(x r x p ) M(G) = R 3 M(G)(x 1,..., x 3n ) = 2n l=1 L(x il, x jl, x kl ) d 1,...,d 3n {0,1,2} c d1,...,d 3n x d x d 3n 3n Theorem For any planar 3-valent graph G λ G (3) = (R 3 M(G))(0,..., 0) = c 0,...,0.

39 8 / 67 Algebraic methods: Proof of the theorem Here coloring is defined as ν : E {1, ω, ω 2 } where ω = 1+i 3 2, the primitive cubic root of 1.

40 8 / 67 Algebraic methods: Proof of the theorem Here coloring is defined as ν : E {1, ω, ω 2 } where ω = 1+i 3 2, the primitive cubic root of 1. By interpolation theorem: R 3 M(G) = ν M(G)(ν(e 1 ),..., ν(e 3n ))P ν, P ν = 3n k=1 S 3 (x k, ν(e k )),

41 8 / 67 Algebraic methods: Proof of the theorem Here coloring is defined as ν : E {1, ω, ω 2 } where ω = 1+i 3 2, the primitive cubic root of 1. By interpolation theorem: R 3 M(G) = ν M(G)(ν(e 1 ),..., ν(e 3n ))P ν, P ν = 3n k=1 S 3 (x k, ν(e k )), summation is taken through all 3 3n colorings, but really through proper ones because

42 Algebraic methods: Proof of the theorem Here coloring is defined as ν : E {1, ω, ω 2 } where ω = 1+i 3 2, the primitive cubic root of 1. By interpolation theorem: R 3 M(G) = ν M(G)(ν(e 1 ),..., ν(e 3n ))P ν, 8 / 67 P ν = 3n k=1 S 3 (x k, ν(e k )), summation is taken through all 3 3n colorings, but really through proper ones because = 2n l=1 M(G)(ν(e 1 ),..., ν(e 3n )) = (ν(e il ) ν(e jl ))(ν(e jl ) ν(e kl ))(ν(e kl ) ν(e il )) equals 0 if ν is not a proper coloring.

43 19 / 67 Algebraic methods: Proof of the theorem cont. If the coloring ν is proper then there are 1, ω, ω 2 between ν(e p ), ν(e q ), ν(e r ), so L(ν(e p ), ν(e q ), ν(e r )) = ±i3 3, and M(G)(ν(e 1 ),..., ν(e 3n )) = ±3 3n.

44 9 / 67 Algebraic methods: Proof of the theorem cont. If the coloring ν is proper then there are 1, ω, ω 2 between ν(e p ), ν(e q ), ν(e r ), so L(ν(e p ), ν(e q ), ν(e r )) = ±i3 3, and M(G)(ν(e 1 ),..., ν(e 3n )) = ±3 3n. The proper sign is +, as can be proven by induction on n (Exercise).

45 9 / 67 Algebraic methods: Proof of the theorem cont. If the coloring ν is proper then there are 1, ω, ω 2 between ν(e p ), ν(e q ), ν(e r ), so L(ν(e p ), ν(e q ), ν(e r )) = ±i3 3, and M(G)(ν(e 1 ),..., ν(e 3n )) = ±3 3n. The proper sign is +, as can be proven by induction on n (Exercise). Then R 3 M(G) = 3 3n ν P ν, here are λ G (3) summands. (R 3 M(G))(0,..., 0) = 3 3n (S 3 (0)) 3n λ G (3) = λ G (3).

46 20 / 67 Outline 1 Introduction 2 Algebraic methods of counting graph colorings 3 Counting graph colorings in terms of orientations 4 Probabilistic restatement of Four Color Conjecture 5 Arithmetical restatement of Four Color Conjecture 6 The Tutte polynomial

47 21 / 67 Orientations: Definitions 1 G = (V, E), f : V Z. G is f -choosable if S : V 2 Z, S(v) = f (v) v there exists a proper coloring c : V Z such that v c(v) S(v).

48 21 / 67 Orientations: Definitions 1 G = (V, E), f : V Z. G is f -choosable if S : V 2 Z, S(v) = f (v) v there exists a proper coloring c : V Z such that v c(v) S(v). G is k-choosable (k Z) if f k. Minimal k for which G is k-choosable is referred to as a choice number of G.

49 22 / 67 Orientations: Definitions 2 Let χ(g), χ (G) be chromatic numbers of G and line graph of G resp., and ch(g), ch (G) choice numbers.

50 22 / 67 Orientations: Definitions 2 Let χ(g), χ (G) be chromatic numbers of G and line graph of G resp., and ch(g), ch (G) choice numbers. Obviously for any graph G holds ch(g) χ(g).

51 22 / 67 Orientations: Definitions 2 Let χ(g), χ (G) be chromatic numbers of G and line graph of G resp., and ch(g), ch (G) choice numbers. Obviously for any graph G holds ch(g) χ(g). There exist graphs with ch(g) > χ(g).

52 Orientations: Definitions 2 Let χ(g), χ (G) be chromatic numbers of G and line graph of G resp., and ch(g), ch (G) choice numbers. Obviously for any graph G holds ch(g) χ(g). There exist graphs with ch(g) > χ(g). Figure: S(u i ) = S(v i ) = {1, 2, 3} \ {i} 2 / 67 But there is a conjecture claiming that G ch (G) = χ (G).

53 23 / 67 Orientations: Definitions 3. Main Theorem Consider oriented graphs (digraphs). Eulerian graph: for each vertex its indegree equals its outdegree.

54 23 / 67 Orientations: Definitions 3. Main Theorem Consider oriented graphs (digraphs). Eulerian graph: for each vertex its indegree equals its outdegree. Even (odd) graph: a graph with even (odd) number of edges.

55 23 / 67 Orientations: Definitions 3. Main Theorem Consider oriented graphs (digraphs). Eulerian graph: for each vertex its indegree equals its outdegree. Even (odd) graph: a graph with even (odd) number of edges. EE(D): a number of even Eulerian subgraphs of graph D. EO(D): a number of odd Eulerian subgraphs of graph D.

56 23 / 67 Orientations: Definitions 3. Main Theorem Consider oriented graphs (digraphs). Eulerian graph: for each vertex its indegree equals its outdegree. Even (odd) graph: a graph with even (odd) number of edges. EE(D): a number of even Eulerian subgraphs of graph D. EO(D): a number of odd Eulerian subgraphs of graph D. d + D (v): outdegree of vertex v in D.

57 23 / 67 Orientations: Definitions 3. Main Theorem Consider oriented graphs (digraphs). Eulerian graph: for each vertex its indegree equals its outdegree. Even (odd) graph: a graph with even (odd) number of edges. EE(D): a number of even Eulerian subgraphs of graph D. EO(D): a number of odd Eulerian subgraphs of graph D. d + D (v): outdegree of vertex v in D. Theorem D = (V, E) being a digraph, V = n, d i = d + D (v i), f (i) = d i + 1 i {1,..., n}, EE(D) EO(D) D is f -choosable.

58 24 / 67 Orientations: Proof of the theorem Lemma Let P(x 1,..., x n ) be polynomial in n variables over Z, for 1 i n the degree of P in x i doesn t exceed d i, S i Z : S i = d i + 1. If (x 1,..., x n ) S 1... S n P(x 1,..., x n ) = 0 then P 0. (Exercise proof by induction.)

59 25 / 67 Orientations: Proof of the theorem (cont.) Graph polynomial of undirected graph G: f G (x 1,..., x n ) = (x i x j ) i<j,v i v j E

60 25 / 67 Orientations: Proof of the theorem (cont.) Graph polynomial of undirected graph G: f G (x 1,..., x n ) = (x i x j ) i<j,v i v j E Monomials of that polynomial are in natural correspondence with orientations of G.

61 25 / 67 Orientations: Proof of the theorem (cont.) Graph polynomial of undirected graph G: f G (x 1,..., x n ) = (x i x j ) i<j,v i v j E Monomials of that polynomial are in natural correspondence with orientations of G. We call edge v i v j decreasing if i > j.

62 25 / 67 Orientations: Proof of the theorem (cont.) Graph polynomial of undirected graph G: f G (x 1,..., x n ) = (x i x j ) i<j,v i v j E Monomials of that polynomial are in natural correspondence with orientations of G. We call edge v i v j decreasing if i > j. Orientation is even if it has even number of decreasing edges, else it s odd.

63 26 / 67 Orientations: Proof of the theorem (cont.) DE(d 1,..., d n ) and DO(d 1,..., d n ) are sets of even and odd orientations; here non-negative numbers d i correspond to outdegrees of vertices.

64 26 / 67 Orientations: Proof of the theorem (cont.) DE(d 1,..., d n ) and DO(d 1,..., d n ) are sets of even and odd orientations; here non-negative numbers d i correspond to outdegrees of vertices. Then evidently holds Lemma f G (x 1,..., x n ) = d 1,...,d n 0 ( DE(d 1,..., d n ) DO(d 1,..., d n ) ) n i=1 x d i i

65 27 / 67 Orientations: Proof of the theorem (cont.) Let us further take D 1, D 2 DE(d 1,..., d n ) DO(d 1,..., d n ). D 1 D 2 denotes set of edges in D 1 that have the opposite direction in D 2.

66 27 / 67 Orientations: Proof of the theorem (cont.) Let us further take D 1, D 2 DE(d 1,..., d n ) DO(d 1,..., d n ). D 1 D 2 denotes set of edges in D 1 that have the opposite direction in D 2. Mapping D 2 D 1 D 2 is a bijection between DE(d 1,..., d n ) DO(d 1,..., d n ) and set of Eulerian subgraphs of D 1.

67 28 / 67 Orientations: Proof of the theorem (cont.) If D 1 is even then it maps even orientations to even subgraphs and odd ones to odd ones.

68 28 / 67 Orientations: Proof of the theorem (cont.) If D 1 is even then it maps even orientations to even subgraphs and odd ones to odd ones. If D 1 is odd then it maps even to odd and odd to even.

69 28 / 67 Orientations: Proof of the theorem (cont.) If D 1 is even then it maps even orientations to even subgraphs and odd ones to odd ones. If D 1 is odd then it maps even to odd and odd to even. Thus we get DE(d 1,..., d n ) DO(d 1,..., d n ) = EE(D 1 ) EO(D 1 ) (it s the coefficient of the monomial x d i i in f G ).

70 29 / 67 Orientations: Proof of the theorem (cont.) Recall the statement of the theorem. Suppose there is no such coloring.

71 29 / 67 Orientations: Proof of the theorem (cont.) Recall the statement of the theorem. Suppose there is no such coloring. Then (x 1,..., x n ) S 1... S n f G (x 1,..., x n ) = 0.

72 29 / 67 Orientations: Proof of the theorem (cont.) Recall the statement of the theorem. Suppose there is no such coloring. Then (x 1,..., x n ) S 1... S n f G (x 1,..., x n ) = 0. Let Q i (x i ) be Q i (x i ) = s S i (x i s) = x d i +1 i d i q ij x j i. j=0

73 0 / 67 Orientations: Proof of the theorem (cont.) If x i S i then x d i +1 i = d i j=0 q ijx j i.

74 0 / 67 Orientations: Proof of the theorem (cont.) If x i S i then x d i +1 i = d i j=0 q ijx j i. We are going to replace in f G each occurrence of x f i i, f i > d i, by a linear combination of smaller powers (using the above equality). So we get polynomial f G.

75 0 / 67 Orientations: Proof of the theorem (cont.) If x i S i then x d i +1 i = d i j=0 q ijx j i. We are going to replace in f G each occurrence of x f i i, f i > d i, by a linear combination of smaller powers (using the above equality). So we get polynomial f G. (x 1,..., x n ) S 1... S n f G (x 1,..., x n ) = 0 and by first Lemma f G 0.

76 0 / 67 Orientations: Proof of the theorem (cont.) If x i S i then x d i +1 i = d i j=0 q ijx j i. We are going to replace in f G each occurrence of x f i i, f i > d i, by a linear combination of smaller powers (using the above equality). So we get polynomial f G. (x 1,..., x n ) S 1... S n f G (x 1,..., x n ) = 0 and by first Lemma f G 0. But coefficient of n i=1 x d i i in f G is nonzero, and it remains the same in f G due to homogeneity of f G. We come to a contradiction.

77 31 / 67 Orientations: Corollaries 1 Corollary If undirected graph G has orientation D satisfying EE(D) EO(D) in which maximal outdegree is d then G is (d + 1)-colorable. (Evident)

78 32 / 67 Orientations: Corollaries Definition Set of vertices S V is called independent if vertices in S can be colored in the same color.

79 32 / 67 Orientations: Corollaries Definition Set of vertices S V is called independent if vertices in S can be colored in the same color. Corollary If undirected graph G with vertices V = {v 1,..., v n } has orientation D satisfying EE(D) EO(D), d 1... d n is ordered sequence of outdegrees of its vertices then k : 0 k < n G has an independent set of size at least n k d k (Exercise)

80 33 / 67 Outline 1 Introduction 2 Algebraic methods of counting graph colorings 3 Counting graph colorings in terms of orientations 4 Probabilistic restatement of Four Color Conjecture 5 Arithmetical restatement of Four Color Conjecture 6 The Tutte polynomial

81 34 / 67 Probabilistic restatement: introduction G is a 3-valent biconnected undirected graph with 2n vertices, 3n edges. Its undirected line graph F G has 3n vertices (each of degree 4) and 6n edges.

82 34 / 67 Probabilistic restatement: introduction G is a 3-valent biconnected undirected graph with 2n vertices, 3n edges. Its undirected line graph F G has 3n vertices (each of degree 4) and 6n edges.

83 Probabilistic restatement: introduction G is a 3-valent biconnected undirected graph with 2n vertices, 3n edges. Its undirected line graph F G has 3n vertices (each of degree 4) and 6n edges. 34 / 67 We assign the same probability to each of 2 6n orientations that can be attached to F G.

84 35 / 67 Probabilistic restatement: introduction Consider 2 orientations F G and F G.

85 35 / 67 Probabilistic restatement: introduction Consider 2 orientations F G and F G. Orientations have the same parity if they differ in direction of even number of edges.

86 35 / 67 Probabilistic restatement: introduction Consider 2 orientations F G and F G. Orientations have the same parity if they differ in direction of even number of edges. Orientations F G, F G are equivalent modulo 3 if for every vertex its outdegree in F G equals modulo 3 its outdegree in F G.

87 35 / 67 Probabilistic restatement: introduction Consider 2 orientations F G and F G. Orientations have the same parity if they differ in direction of even number of edges. Orientations F G, F G are equivalent modulo 3 if for every vertex its outdegree in F G equals modulo 3 its outdegree in F G. Event A G : 2 randomly chosen orientations have the same parity. Event B G : 2 randomly chosen orientations are equivalent modulo 3.

88 36 / 67 Probabilistic restatement: the theorem Theorem For any biconnected planar 3-valent graph G having 2n vertices ( ) 27 n P (B G A G ) P(B G ) = χg (4)

89 36 / 67 Probabilistic restatement: the theorem Theorem For any biconnected planar 3-valent graph G having 2n vertices ( ) 27 n P (B G A G ) P(B G ) = χg (4) So 4CC is equivalent to the statement that there is a positive correlation between A G and B G.

90 37 / 67 Probabilistic restatement: proof of the theorem Let us examine 2 graph polynomials: M = e i e j L G (x i x j ) (here L G is a set of edges of F G ), M = e i e j L G (x 2 i x 2 j ).

91 37 / 67 Probabilistic restatement: proof of the theorem Let us examine 2 graph polynomials: M = e i e j L G (x i x j ) (here L G is a set of edges of F G ), M = e i e j L G (x 2 i x 2 j ). Orientations that are equivalent modulo 3 correspond to equal (up to sign) monomials in R 3 M. (Operator R 3, as usually, reduces all powers modulo 3.)

92 37 / 67 Probabilistic restatement: proof of the theorem Let us examine 2 graph polynomials: M = e i e j L G (x i x j ) (here L G is a set of edges of F G ), M = e i e j L G (x 2 i x 2 j ). Orientations that are equivalent modulo 3 correspond to equal (up to sign) monomials in R 3 M. (Operator R 3, as usually, reduces all powers modulo 3.) Let us count m 0, that is, free term of R 3 (M M ) in two different ways.

93 38 / 67 Probabilistic restatement: proof of the theorem The first way. P(A G ) = 1 2 P (B G A G ) P(B G ) = 2P(A G B G ) P(B G )

94 38 / 67 Probabilistic restatement: proof of the theorem The first way. P(A G ) = 1 2 P (B G A G ) P(B G ) = 2P(A G B G ) P(B G ) If 2 orientations are not equivalent modulo 3 then they contribute 0 into probabilities and 0 into m 0.

95 38 / 67 Probabilistic restatement: proof of the theorem The first way. P(A G ) = 1 2 P (B G A G ) P(B G ) = 2P(A G B G ) P(B G ) If 2 orientations are not equivalent modulo 3 then they contribute 0 into probabilities and 0 into m 0. If they are equivalent modulo 3 and have the same parity then they contribute 2 12n into probabilities and 1 into m 0.

96 38 / 67 Probabilistic restatement: proof of the theorem The first way. P(A G ) = 1 2 P (B G A G ) P(B G ) = 2P(A G B G ) P(B G ) If 2 orientations are not equivalent modulo 3 then they contribute 0 into probabilities and 0 into m 0. If they are equivalent modulo 3 and have the same parity then they contribute 2 12n into probabilities and 1 into m 0. If they are equivalent but have the different parity they contribute 2 12n and 1 respectively.

97 38 / 67 Probabilistic restatement: proof of the theorem The first way. P(A G ) = 1 2 P (B G A G ) P(B G ) = 2P(A G B G ) P(B G ) If 2 orientations are not equivalent modulo 3 then they contribute 0 into probabilities and 0 into m 0. If they are equivalent modulo 3 and have the same parity then they contribute 2 12n into probabilities and 1 into m 0. If they are equivalent but have the different parity they contribute 2 12n and 1 respectively. Finally we have m 0 = 2 12n (P (B G A G ) P(B G )).

98 39 / 67 Probabilistic restatement: proof cont. The second way deals with Tait colorings. We have a map colored in 4 colors (say, α, β, γ, δ) and construct a coloring of edges by the following rule.

99 39 / 67 Probabilistic restatement: proof cont. The second way deals with Tait colorings. We have a map colored in 4 colors (say, α, β, γ, δ) and construct a coloring of edges by the following rule. An edge which separates α from β or γ from δ gets color 1. An edge which separates α from γ or β from δ gets color 2. An edge which separates α from δ or β from γ gets color 3.

100 Probabilistic restatement: proof cont. 40 / 67

101 40 / 67 Probabilistic restatement: proof cont. For any vertex three edges incidental to it have 3 different colors.

102 40 / 67 Probabilistic restatement: proof cont. For any vertex three edges incidental to it have 3 different colors. And vice versa: for given Tait coloring of edges (in 3 colors) we may reconstruct coloring of vertices in 4 colors.

103 Probabilistic restatement: proof cont. For any vertex three edges incidental to it have 3 different colors. And vice versa: for given Tait coloring of edges (in 3 colors) we may reconstruct coloring of vertices in 4 colors. 40 / 67 Thus number of Tait colorings is χ G (4) 4.

104 1 / 67 Probabilistic restatement: proof cont. We are going to determine R 3 (M M ) = R 3 M by choosing appropriate 3 3n values of variables.

105 1 / 67 Probabilistic restatement: proof cont. We are going to determine R 3 (M M ) = R 3 M by choosing appropriate 3 3n values of variables. Let x j (j {1,..., 3n}) take values 1, ω, ω 2 where ω = 1+i 3 2 (cubic root of 1).

106 1 / 67 Probabilistic restatement: proof cont. We are going to determine R 3 (M M ) = R 3 M by choosing appropriate 3 3n values of variables. Let x j (j {1,..., 3n}) take values 1, ω, ω 2 where ω = 1+i 3 2 (cubic root of 1). Applying interpolation formula we see R 3 M = µ (R 3 M)(ω µ(v 1),..., ω µ(v 3n) )P µ = µ M(ω µ(v 1),..., ω µ(v 3n) )P µ (R eliminated due to the choice of colors, summation through Tait colorings enough),

107 41 / 67 Probabilistic restatement: proof cont. We are going to determine R 3 (M M ) = R 3 M by choosing appropriate 3 3n values of variables. Let x j (j {1,..., 3n}) take values 1, ω, ω 2 where ω = 1+i 3 2 (cubic root of 1). Applying interpolation formula we see R 3 M = µ (R 3 M)(ω µ(v 1),..., ω µ(v 3n) )P µ = µ M(ω µ(v 1),..., ω µ(v 3n) )P µ (R eliminated due to the choice of colors, summation through Tait colorings enough), P µ = 3n k=1 (x k ω µ(v k)+1 )(x k ω µ(v k)+2 ) (ω µ(v k) ω µ(v k)+1 )(ω µ(v k) ω µ(v k)+2 )

108 42 / 67 Probabilistic restatement: proof cont. We break 12n factors in M into 2n groups, each group looking like (x i x j )(x j x k )(x i x k )(x 2 i x 2 j )(x 2 j x 2 k )(x 2 i x 2 k ) (i < j < k, edges e i, e j, e k are incidental to the same vertex).

109 42 / 67 Probabilistic restatement: proof cont. We break 12n factors in M into 2n groups, each group looking like (x i x j )(x j x k )(x i x k )(x 2 i x 2 j )(x 2 j x 2 k )(x 2 i x 2 k ) (i < j < k, edges e i, e j, e k are incidental to the same vertex). Such a product for Tait coloring equals 27. Therefore m 0 = 3 6n µ P µ(0,..., 0).

110 42 / 67 Probabilistic restatement: proof cont. We break 12n factors in M into 2n groups, each group looking like (x i x j )(x j x k )(x i x k )(x 2 i x 2 j )(x 2 j x 2 k )(x 2 i x 2 k ) (i < j < k, edges e i, e j, e k are incidental to the same vertex). Such a product for Tait coloring equals 27. Therefore m 0 = 3 6n µ P µ(0,..., 0). P µ (0,..., 0) can be easily computed and equals 3 3n.

111 42 / 67 Probabilistic restatement: proof cont. We break 12n factors in M into 2n groups, each group looking like (x i x j )(x j x k )(x i x k )(x 2 i x 2 j )(x 2 j x 2 k )(x 2 i x 2 k ) (i < j < k, edges e i, e j, e k are incidental to the same vertex). Such a product for Tait coloring equals 27. Therefore m 0 = 3 6n µ P µ(0,..., 0). P µ (0,..., 0) can be easily computed and equals 3 3n. 2 12n (P (B G A G ) P(B G )) = 3 3n χg (4) 4 We have proved the theorem.

112 43 / 67 Outline 1 Introduction 2 Algebraic methods of counting graph colorings 3 Counting graph colorings in terms of orientations 4 Probabilistic restatement of Four Color Conjecture 5 Arithmetical restatement of Four Color Conjecture 6 The Tutte polynomial

113 44 / 67 Arithmetical restatement: The Main Theorem Formulation Theorem p, q N, 4q linear functions A k (m, c 1,..., c p ), B k (m, c 1,..., c p ), C k (m, c 1,..., c p ), D k (m, c 1,..., c p ), k {1,..., q} such that 4CC is equivalent to the following statement: m, n c 1,..., c p E(n, m, c 1,..., c p ) 0 mod 7, E(n, m, c 1,..., c p ) = ( Ak (m, c 1,..., c p ) + 7 n B k (m, c 1,..., c p ) C k (m, c 1,..., c p ) + 7 n D k (m, c 1,..., c p ) ).

114 45 / 67 Arithmetical restatement: Value of The Main Theorem Having E(n, m, c 1,..., c p ) we can take G(m, n) whose values are never divisible by 7, arbitrary integer-valued functions F (n, m, c 1,..., c p ),

115 45 / 67 Arithmetical restatement: Value of The Main Theorem Having E(n, m, c 1,..., c p ) we can take G(m, n) whose values are never divisible by 7, arbitrary integer-valued functions F (n, m, c 1,..., c p ), and c 1... c p E(n, m, c 1,..., c p )F (n, m, c 1,..., c p ) = G(m, n) implies the 4CC.

116 6 / 67 Arithmetical restatement: Reformulations We are to come to main theorem from original 4CC step by step using reformulations.

117 6 / 67 Arithmetical restatement: Reformulations We are to come to main theorem from original 4CC step by step using reformulations. Firstly, we will color not countries but their capitals (2 capitals are connected by road iff countries are neighbors).

118 6 / 67 Arithmetical restatement: Reformulations We are to come to main theorem from original 4CC step by step using reformulations. Firstly, we will color not countries but their capitals (2 capitals are connected by road iff countries are neighbors). Then we introduce internal and external (as a whole ranked) edges (G = V, E I, E X ). Ends of internal edge should have the same color, ends of external edge should be colored differently. Now the 4CC sounds as follows:

119 46 / 67 Arithmetical restatement: Reformulations We are to come to main theorem from original 4CC step by step using reformulations. Firstly, we will color not countries but their capitals (2 capitals are connected by road iff countries are neighbors). Then we introduce internal and external (as a whole ranked) edges (G = V, E I, E X ). Ends of internal edge should have the same color, ends of external edge should be colored differently. Now the 4CC sounds as follows: If a planar graph with ranked edges can be colored in some number of colors (more precisely in 6 colors it s always possible) then it can be colored in 4 colors.

120 7 / 67 Arithmetical restatement: Reformulations Then we say: we have a graph with vertices (V ) colored somehow and edges (E). 2 colorings are equivalent if they induce the same division of E in 2 groups (internal and external). For every coloring we are searching for the equivalent one consisting of 4 colors.

121 8 / 67 Arithmetical restatement: Reformulations The next term to introduce is spiral graph. It has infinitely many vertices k. 2 vertices i, j are connected by edge iff i j = 1 or i j = n. We color this construct in colors from {0,..., 6} so that finitely many vertices have color greater than 0.

122 8 / 67 Arithmetical restatement: Reformulations The next term to introduce is spiral graph. It has infinitely many vertices k. 2 vertices i, j are connected by edge iff i j = 1 or i j = n. We color this construct in colors from {0,..., 6} so that finitely many vertices have color greater than 0. 4-coloring λ for a given coloring µ is made paying attention to 3 properties:

123 8 / 67 Arithmetical restatement: Reformulations The next term to introduce is spiral graph. It has infinitely many vertices k. 2 vertices i, j are connected by edge iff i j = 1 or i j = n. We color this construct in colors from {0,..., 6} so that finitely many vertices have color greater than 0. 4-coloring λ for a given coloring µ is made paying attention to 3 properties: 1 λ(k) = 0 µ(k) = 0 2 λ(k) = λ(k + 1) µ(k) = µ(k + 1) 3 λ(k) = λ(k + n) µ(k) = µ(k + n)

124 9 / 67 Arithmetical restatement: Reformulations We can represent our colorings as a natural number in base-7 notation: m = k=0 µ(k)7k, l = k=0 λ(k)7k. Our requirements to λ imply that:

125 49 / 67 Arithmetical restatement: Reformulations We can represent our colorings as a natural number in base-7 notation: m = k=0 µ(k)7k, l = k=0 λ(k)7k. Our requirements to λ imply that: 1 there are no 7-digits 5 and 6 in l 2 the k-th digit of l equals 0 the k-th digit of m equals 0 3 and 2 more (for (k + 1) and (k + n))

126 0 / 67 Arithmetical restatement: Reformulations We can view m as 6 i=1 im i so that m i = µ(k)=i 7k. Now we need 2 more definitions.

127 0 / 67 Arithmetical restatement: Reformulations We can view m as 6 i=1 im i so that m i = µ(k)=i 7k. Now we need 2 more definitions. b Z + is Bool if its 7-digits are either 0 or 1.

128 0 / 67 Arithmetical restatement: Reformulations We can view m as 6 i=1 im i so that m i = µ(k)=i 7k. Now we need 2 more definitions. b Z + is Bool if its 7-digits are either 0 or 1. Boolean numbers a and b are said to be orthogonal (a b) if they never have 1 in the same position.

129 0 / 67 Arithmetical restatement: Reformulations We can view m as 6 i=1 im i so that m i = µ(k)=i 7k. Now we need 2 more definitions. b Z + is Bool if its 7-digits are either 0 or 1. Boolean numbers a and b are said to be orthogonal (a b) if they never have 1 in the same position. We introduce c ij = µ(k)=i,λ(k)=j 7k and know that Bool(c ij ), c ij c i j if i, j i, j. m i = 4 j=1 c ij, l j = 6 i=1 c ij

130 0 / 67 Arithmetical restatement: Reformulations We can view m as 6 i=1 im i so that m i = µ(k)=i 7k. Now we need 2 more definitions. b Z + is Bool if its 7-digits are either 0 or 1. Boolean numbers a and b are said to be orthogonal (a b) if they never have 1 in the same position. We introduce c ij = µ(k)=i,λ(k)=j 7k and know that Bool(c ij ), c ij c i j if i, j i, j. m i = 4 j=1 c ij, l j = 6 i=1 c ij Now conditions on 4-coloring are following (not counting those we ve already seen):

131 50 / 67 Arithmetical restatement: Reformulations We can view m as 6 i=1 im i so that m i = µ(k)=i 7k. Now we need 2 more definitions. b Z + is Bool if its 7-digits are either 0 or 1. Boolean numbers a and b are said to be orthogonal (a b) if they never have 1 in the same position. We introduce c ij = µ(k)=i,λ(k)=j 7k and know that Bool(c ij ), c ij c i j if i, j i, j. m i = 4 j=1 c ij, l j = 6 i=1 c ij Now conditions on 4-coloring are following (not counting those we ve already seen): 1 7c ij c ij, j j 2 7c ij c i j, i i 3 7 n c ij c ij, j j 4 7 n c ij c i j, i i

132 51 / 67 Arithmetical restatement: Reformulations Theorem (E. E. Kummer) A prime number p comes into the factorization of the binomial coefficient ( ) a + b a with the exponent equal to the number of carries performed while computing sum a + b in base-p positional notation. (Exercise)

133 52 / 67 Arithmetical restatement: Reformulations The first corollary: All 7-digits of a are less or equal to 3 iff ( ) 2a 0 mod 7. a

134 52 / 67 Arithmetical restatement: Reformulations The first corollary: All 7-digits of a are less or equal to 3 iff ( ) 2a 0 mod 7. a The second corollary: Bool(a) ( 2a a ) ( 4a 2a ) 0 mod 7.

135 Arithmetical restatement: Reformulations The first corollary: All 7-digits of a are less or equal to 3 iff ( ) 2a 0 mod 7. a The second corollary: Bool(a) The third corollary: ( 2a a ) ( 4a 2a ) 0 mod / 67 [ a b Bool(a)&Bool(b) ) ( ) 4(a + b) 2(a + b) ( 2(a + b) a + b ] 0 mod 7.

136 53 / 67 Arithmetical restatement: Reformulations The last corollary: Bool(a)&Bool(b) [ a b ( 4(a + b) 2(a + b) ) ] 0 mod 7.

137 53 / 67 Arithmetical restatement: Reformulations The last corollary: Bool(a)&Bool(b) [ a b ( 4(a + b) 2(a + b) ) ] 0 mod 7. The last reformulation: n, m Z + c ij Z +, i {1,..., 6}, j {1,..., 4}: none of 986 binomial coefficients is divisible by 7: ( 2cij c ij ), ( 4cij 2c ij ( 4(7cij + c ij ) 2(7c ij + c ij ) ) ( 4(ci, j + c i j ) 2(c i j + c i j ) ) (, j j 4(7cij + c, i j) 2(7c ij + c i j) ), i, j i, j, ), i i,

138 4 / 67 Arithmetical restatement: Reformulations The last reformulation cont.: n, m Z + c ij Z +, i {1,..., 6}, j {1,..., 4}: none of 986 binomial coefficients is divisible by 7: ( 4(7 n c ij + c ij ) 2(7 n c ij + c ij ) ) (, j j 4(7, n c ij + c i j) 2(7 n c ij + c i j) ( m C where C = ) ( C, m 6 i=1 ), 4 j=1 c ij. ), i i,

139 55 / 67 Outline 1 Introduction 2 Algebraic methods of counting graph colorings 3 Counting graph colorings in terms of orientations 4 Probabilistic restatement of Four Color Conjecture 5 Arithmetical restatement of Four Color Conjecture 6 The Tutte polynomial

140 56 / 67 The Tutte polynomial: definitions Let G = (V, E) be a (multi)graph (may have loops and multiple edges). We define 2 operations:

141 6 / 67 The Tutte polynomial: definitions Let G = (V, E) be a (multi)graph (may have loops and multiple edges). We define 2 operations: 1 cut : G e, where e E (delete the edge e) 2 fuse : G \ e, where e E (delete the edge e and join vertices incident to e)

142 57 / 67 The Tutte polynomial: definitions k(g) is a number of components in G.

143 57 / 67 The Tutte polynomial: definitions k(g) is a number of components in G. Then we introduce 2 terms: 1 rank of graph (V, E) is r(g) = V k(g) 2 nullity of graph (V, E) is n(g) = E V + k(g)

144 57 / 67 The Tutte polynomial: definitions k(g) is a number of components in G. Then we introduce 2 terms: 1 rank of graph (V, E) is r(g) = V k(g) 2 nullity of graph (V, E) is n(g) = E V + k(g) For any spanning subgraph F we write k F, r F, n F. Then S(G; x, y) = x r E r F y n F = x k F k E y n F F E(G) F E(G) is called rank-generating polynomial.

145 58 / 67 The Tutte polynomial: theorem 1 Theorem (x + 1)S(G e; x, y), e is a bridge, S(G; x, y) = (y + 1)S(G e; x, y) e is a loop, S(G e; x, y) + S(G \ e; x, y), otherwise. Furthermore, S(E n ; x, y) = 1 for an empty graph E n with n vertices.

146 The Tutte polynomial: theorem 1 Theorem (x + 1)S(G e; x, y), e is a bridge, S(G; x, y) = (y + 1)S(G e; x, y) e is a loop, S(G e; x, y) + S(G \ e; x, y), otherwise. Furthermore, S(E n ; x, y) = 1 for an empty graph E n with n vertices. 58 / 67 This can be easily proved if we form two groups of F s (subsets of E(G)): those which include e (the edge to be eliminated) and those which do not and investigate simple properties of rank and nullity. (Exercise)

147 59 / 67 The Tutte polynomial: main definition The Tutte polynomial is defined as follows: T G (x, y) = S(G; x 1, y 1) = (x 1) r E r F (y 1) n F F E(G)

148 59 / 67 The Tutte polynomial: main definition The Tutte polynomial is defined as follows: T G (x, y) = S(G; x 1, y 1) = (x 1) r E r F (y 1) n F F E(G) Of course the appropriate statement holds: xt G e (x, y), T G (x, y) = yt G e (x, y) T G e (x, y) + T G\e (x, y), e is a bridge, e is a loop, otherwise.

149 60 / 67 The Tutte polynomial: theorem 2 Theorem There is a unique map U from the set of multigraphs to the ring of polynomials over Z of variables x, y, α, σ, τ such that U(E n ) = U(E n ; x, y, α, σ, τ) = α n and xu G e (x, y), e is a bridge, U(G) = yu G e (x, y) e is a loop, σu G e (x, y) + τu G\e (x, y), otherwise. Furthermore, U(G) = α k(g) σ n(g) τ r(g) T G (αx/τ, y/σ).

150 61 / 67 The Tutte polynomial: theorem 2, proof sketch U(G) is a polynomial of σ and τ because deg x T G (x, y) = r(g), deg y T G (x, y) = n(g).

151 61 / 67 The Tutte polynomial: theorem 2, proof sketch U(G) is a polynomial of σ and τ because deg x T G (x, y) = r(g), deg y T G (x, y) = n(g). The uniqueness of U is implied by constructive definition.

152 61 / 67 The Tutte polynomial: theorem 2, proof sketch U(G) is a polynomial of σ and τ because deg x T G (x, y) = r(g), deg y T G (x, y) = n(g). The uniqueness of U is implied by constructive definition. It can be checked simply that U(E n ) = α n and recurrent equalities hold for U(G) = α k(g) σ n(g) τ r(g) T G (αx/τ, y/σ).

153 62 / 67 The Tutte polynomial: corollary Definition If p G (x) is the number of proper colorings of vertices of graph G in x colors then p G (x) is called the chromatic function of G.

154 62 / 67 The Tutte polynomial: corollary Definition If p G (x) is the number of proper colorings of vertices of graph G in x colors then p G (x) is called the chromatic function of G. Corollary p G (x) = ( 1) r(g) x k(g) T G (1 x, 0) So the chromatic function is actually the chromatic polynomial.

155 3 / 67 The Tutte polynomial: corollary, proof sketch and e E(G) p G (x) = p En (x) = x n x 1 x p G e(x), e is a bridge, 0 e is a loop, p G e (x) p G\e (x), otherwise. p G (x) = U(G; x 1, 0, x, 1, 1) = x k(g) ( 1) r(g) T G (1 x, 0) x

156 64 / 67 List of exercises 1 Colorings of 3-valent graph, proper sign is + M(G)(ν(e 1 ),..., ν(e 3n )) = ±3 3n,

157 64 / 67 List of exercises 1 Colorings of 3-valent graph, proper sign is + M(G)(ν(e 1 ),..., ν(e 3n )) = ±3 3n, 2 Let P(x 1,..., x n ) be polynomial in n variables over Z, for 1 i n the degree of P in x i doesn t exceed d i, S i Z : S i = d i + 1. If (x 1,..., x n ) S 1... S n P(x 1,..., x n ) = 0 then P 0.

158 4 / 67 List of exercises 1 Colorings of 3-valent graph, proper sign is + M(G)(ν(e 1 ),..., ν(e 3n )) = ±3 3n, 2 Let P(x 1,..., x n ) be polynomial in n variables over Z, for 1 i n the degree of P in x i doesn t exceed d i, S i Z : S i = d i + 1. If (x 1,..., x n ) S 1... S n P(x 1,..., x n ) = 0 then P 0. 3 If undirected graph G with vertices V = {v 1,..., v n } has orientation D satisfying EE(D) EO(D), d 1... d n is ordered sequence of outdegrees of its vertices then k : 0 k < n G has an independent set of size at least n k d k+1 +1.

159 65 / 67 List of exercises 4 (Kummer theorem) A prime number p comes ( into) the factorization of the a + b binomial coefficient with the exponent equal to a the number of carries performed while computing sum a + b in base-p positional notation.

160 65 / 67 List of exercises 4 (Kummer theorem) A prime number p comes ( into) the factorization of the a + b binomial coefficient with the exponent equal to a the number of carries performed while computing sum a + b in base-p positional notation. 5 (x + 1)S(G e; x, y), e is a bridge, S(G; x, y) = (y + 1)S(G e; x, y) e is a loop, S(G e; x, y) + S(G \ e; x, y), otherwise.

161 List of exercises 4 (Kummer theorem) A prime number p comes ( into) the factorization of the a + b binomial coefficient with the exponent equal to a the number of carries performed while computing sum a + b in base-p positional notation. 5 (x + 1)S(G e; x, y), e is a bridge, S(G; x, y) = (y + 1)S(G e; x, y) e is a loop, S(G e; x, y) + S(G \ e; x, y), otherwise. 65 / 67 6 p G (x) = x 1 x p G e(x), e is a bridge, 0 e is a loop, p G e (x) p G\e (x), otherwise.

162 Literature 6 / 67 1 Yu. Matiyasevich. Some Algebraic methods of Counting Graph Colorings (In Russian) 2 Yu. Matiyasevich. One Criterion of Colorability of Graph Vertices in Terms of Orientations (In Russian) 3 Yu. Matiyasevich. One Probabilistic Restatement of the Four Color Conjecture (In Russian) 4 Yu. Matiyasevich. Some Probabilistic Restatements of the Four Color Conjecture Yu. Matiyasevich. Some Arithmetical Restatements of the Four Color Conjecture. 6 N. Alon. Combinatorial Nullstellensatz. 7 N. Alon, M. Tarsi. Colorings and orientations of graphs B. Bollobás. Modern Graph Theory. Chapter X. The Tutte Polynomial. 9 B. Bollobás, O. Riordan. A Tutte Polynomial for Coloured Graphs

163 Thank you for your attention! Danke für Ihre Aufmerksamkeit! 7 / 67 Any questions?