# Fractional Colorings and Zykov Products of graphs

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1 Fractional Colorings and Zykov Products of graphs Who? Nichole Schimanski When? July 27, 2011

2 Graphs A graph, G, consists of a vertex set, V (G), and an edge set, E(G). V (G) is any finite set E(G) is a set of unordered pairs of vertices

3 Graphs A graph, G, consists of a vertex set, V (G), and an edge set, E(G). V (G) is any finite set E(G) is a set of unordered pairs of vertices Example Figure: Peterson graph

4 Subgraphs A subgraph H of a graph G is a graph such that V (H) V (G) and E(H) E(G).

5 Subgraphs Example A subgraph H of a graph G is a graph such that V (H) V (G) and E(H) E(G). Figure: Subgraph of the Peterson graph

6 Subgraphs An induced subgraph, H, of G is a subgraph with property that any two vertices are adjacent in H if and only if they are adjacent in G.

7 Subgraphs Example An induced subgraph, H, of G is a subgraph with property that any two vertices are adjacent in H if and only if they are adjacent in G. Figure: Induced Subgraph of Peterson graph

8 Independent Sets A set of vertices, S, is said to be independent if those vertices induce a graph with no edges.

9 Example Independent Sets A set of vertices, S, is said to be independent if those vertices induce a graph with no edges. Figure: Independent set

10 Example Independent Sets A set of vertices, S, is said to be independent if those vertices induce a graph with no edges. Figure: Independent set The set of all independent sets of a graph G is denoted I (G).

11 Weighting I (S) A weighting of I (G) is a function w : I (G) R 0.

12 Example Weighting I (S) A weighting of I (G) is a function w : I (G) R 0. e d a c b S w(s) {a} 1/3 {b} 1/3 {c} 1/3 {d} 1/3 {e} 1/3 {a,c} 1/3 {a,d} 1/3 {b,d} 1/3 {b,e} 1/3 {e,c} 1/3 Figure: C 5 and a corresponding weighting

13 Fractional k-coloring A fractional k-coloring of a graph, G, is a weighting of I (G) such that S I (G) w(s) = k; and

14 Fractional k-coloring A fractional k-coloring of a graph, G, is a weighting of I (G) such that S I (G) w(s) = k; and For every v V (G), S I (G) v S w(s) = w[v] 1

15 Example Fractional k-coloring e d a c b S w(s) {a} 1/3 {b} 1/3 {c} 1/3 {d} 1/3 {e} 1/3 {a,c} 1/3 {a,d} 1/3 {b,d} 1/3 {b,e} 1/3 {e,c} 1/3 Figure: A fractional coloring of C 5 with weight 10/3

16 Example Fractional k-coloring e d a c b S w(s) {a} 1/3 {b} 1/3 {c} 1/3 {d} 1/3 {e} 1/3 {a,c} 1/3 {a,d} 1/3 {b,d} 1/3 {b,e} 1/3 {e,c} 1/3 Figure: A fractional coloring of C 5 with weight 10/3 S I (G) w(s) = 10/3

17 Example Fractional k-coloring e d a c b S w(s) {a} 1/3 {b} 1/3 {c} 1/3 {d} 1/3 {e} 1/3 {a,c} 1/3 {a,d} 1/3 {b,d} 1/3 {b,e} 1/3 {e,c} 1/3 Figure: A fractional coloring of C 5 with weight 10/3 S I (G) w(s) = 10/3 w[v] = 1 for every v V (G)

18 Fractional Chromatic Number The fractional chromatic number, χ f (G), is the minimum possible weight of a fractional coloring.

19 Example Fractional Chromatic Number The fractional chromatic number, χ f (G), is the minimum possible weight of a fractional coloring. e d a c b S w(s) {a} 0 {b} 0 {c} 0 {d} 0 {e} 0 {a,c} 1/2 {a,d} 1/2 {b,d} 1/2 {b,e} 1/2 {e,c} 1/2 Figure: A weighting C 5

20 Example Fractional Chromatic Number e d a c b S w(s) {a} 0 {b} 0 {c} 0 {d} 0 {e} 0 {a,c} 1/2 {a,d} 1/2 {b,d} 1/2 {b,e} 1/2 {e,c} 1/2 Figure: A fractional 5/2-coloring of C 5

21 Example Fractional Chromatic Number e d a c b S w(s) {a} 0 {b} 0 {c} 0 {d} 0 {e} 0 {a,c} 1/2 {a,d} 1/2 {b,d} 1/2 {b,e} 1/2 {e,c} 1/2 Figure: A fractional 5/2-coloring of C 5 S I (G) w(s) = 5/2

22 Example Fractional Chromatic Number e d a c b S w(s) {a} 0 {b} 0 {c} 0 {d} 0 {e} 0 {a,c} 1/2 {a,d} 1/2 {b,d} 1/2 {b,e} 1/2 {e,c} 1/2 Figure: A fractional 5/2-coloring of C 5 S I (G) w(s) = 5/2 w[v] = 1 for every v V (G)

23 Fractional Chromatic Number How do we know what the minimum is?

24 Fractional Chromatic Number How do we know what the minimum is? Linear Programming

25 Fractional Chromatic Number How do we know what the minimum is? Linear Programming Formulas

26 Zykov Product of Graphs

27 Zykov Product of Graphs The Zykov product Z(G 1, G 2,..., G n ) of simple graphs G 1, G 2,..., G n is formed as follows.

28 Zykov Product of Graphs The Zykov product Z(G 1, G 2,..., G n ) of simple graphs G 1, G 2,..., G n is formed as follows. Take the disjoint union of G i

29 Zykov Product of Graphs The Zykov product Z(G 1, G 2,..., G n ) of simple graphs G 1, G 2,..., G n is formed as follows. Take the disjoint union of G i Example Figure: Drawings of P 2 and P 3

30 Zykov Product of Graphs The Zykov product Z(G 1, G 2,..., G n ) of simple graphs G 1, G 2,..., G n is formed as follows. Take the disjoint union of G i For each (x 1,..., x n ) V (G 1 ) V (G 2 )... V (G n ) add a new vertex adjacent to the vertices {x 1,..., x n }

31 Zykov Product of Graphs The Zykov product Z(G 1, G 2,..., G n ) of simple graphs G 1, G 2,..., G n is formed as follows. Take the disjoint union of G i For each (x 1,..., x n ) V (G 1 ) V (G 2 )... V (G n ) add a new vertex adjacent to the vertices {x 1,..., x n } Example Figure: Constructing Z(P 2, P 3 )

32 Zykov Product of Graphs The Zykov product Z(G 1, G 2,..., G n ) of simple graphs G 1, G 2,..., G n is formed as follows. Take the disjoint union of G i For each (x 1,..., x n ) V (G 1 ) V (G 2 )... V (G n ) add a new vertex adjacent to the vertices {x 1,..., x n } Example Figure: Z(P 2, P 3 )

33 Zykov Graphs

34 Zykov Graphs The Zykov graphs, Z n, are formed as follows: Set Z 1 as a single vertex

35 Zykov Graphs The Zykov graphs, Z n, are formed as follows: Set Z 1 as a single vertex Define Z n := Z(Z 1,..., Z n 1 ) for all n 2 Figure: Drawing of Z 1

36 Zykov Graphs The Zykov graphs, Z n, are formed as follows: Set Z 1 as a single vertex Define Z n := Z(Z 1,..., Z n 1 ) for all n 2 Figure: Drawings of Z 1 and Z 2

37 Zykov Graphs The Zykov graphs, Z n, are formed as follows: Set Z 1 as a single vertex Define Z n := Z(Z 1,..., Z n 1 ) for all n 2 Figure: Drawings of Z 1, Z 2, and Z 3

38 Zykov Graphs The Zykov graphs, Z n, are formed as follows: Set Z 1 as a single vertex Define Z n := Z(Z 1,..., Z n 1 ) for all n 2 Figure: Drawing of Z 4

39 Jacobs Conjecture

40 Jacobs Conjecture Corollary For n 1, χ f (Z n+1 ) = χ f (Z n ) + 1 χ f (Z n )

41 Jacobs Conjecture Corollary For n 1, χ f (Z n+1 ) = χ f (Z n ) + 1 χ f (Z n ) Example χ f (Z 1 ) = 1

42 Jacobs Conjecture Corollary For n 1, χ f (Z n+1 ) = χ f (Z n ) + 1 χ f (Z n ) Example χ f (Z 1 ) = 1 χ f (Z 2 ) = = 2

43 Jacobs Conjecture Corollary For n 1, χ f (Z n+1 ) = χ f (Z n ) + 1 χ f (Z n ) Example χ f (Z 1 ) = 1 χ f (Z 2 ) = = 2 χ f (Z 3 ) = = 5 2

44 Verifying χ f (C 5 )

45 Verifying χ f (C 5 ) Notice that Figure: Z 3 and C 5

46 Verifying χ f (C 5 ) Notice that Figure: Z 3 and C 5 So, χ f (Z 3 ) = χ f (C 5 ) = 5/2

47 The Main Result: Theorem 1

48 Theorem The Main Result: Theorem 1 For n 2, let G 1,..., G n be graphs. Set G := Z(G 1,..., G n ) and χ i = χ f (G i ). Suppose also that the graphs G i are numbered such that χ i χ i+1. Then ( n n ) χ f (G) = max χ n, 2 + (1 ) 1χk i=2 k=i

49 Theorem Example The Main Result: Theorem 1 For n 2, let G 1,..., G n be graphs. Set G := Z(G 1,..., G n ) and χ i = χ f (G i ). Suppose also that the graphs G i are numbered such that χ i χ i+1. Then ( n n ) χ f (G) = max χ n, 2 + (1 ) 1χk i=2 k=i ( ( χ f (Z(P 2, P 3 )) = max 2, )) 2 = max(2, 5 2 ) = 5 2.

50 Lower Bound: χ f (G) max (χ n, f (n))

51 Lower Bound: χ f (G) max (χ n, f (n)) Lemma The fractional chromatic number of a subgraph, H, is at most equal to the fractional chromatic number of a graph, G.

52 Lower Bound: χ f (G) max (χ n, f (n)) Lemma Conclusion The fractional chromatic number of a subgraph, H, is at most equal to the fractional chromatic number of a graph, G. χ f (G) χ n

53 Lower Bound: χ f (G) max (χ n, f (n)) Lemma Let G be a graph and w a weighting of X I (G). Then, for every induced subgraph H of G, there exists x V (H) such that w[x] 1 χ f (H) w(s). S X

54 Lower Bound: χ f (G) max (χ n, f (n)) Start with w, a χ f -coloring of G and x 1 V (G 1 ).

55 Lower Bound: χ f (G) max (χ n, f (n)) Start with w, a χ f -coloring of G and x 1 V (G 1 ). Construct F 1 = {S I (G) : x 1 S} with the property S F 1 w(s) = w[x 1 ] 1.

56 Lower Bound: χ f (G) max (χ n, f (n)) Start with w, a χ f -coloring of G and x 1 V (G 1 ). Construct F 1 = {S I (G) : x 1 S} with the property S F 1 w(s) = w[x 1 ] 1. Construct F 2 = {S I (G) : S {x 1, x 2 } } with the property ( ) S F 2 w(s) χ 2 S F 1 w(s).

57 Lower Bound: χ f (G) max (χ n, f (n)) Continue this process so that for all k {1,..., n},

58 Lower Bound: χ f (G) max (χ n, f (n)) Continue this process so that for all k {1,..., n}, F k = {S I (G) : S {x 1,..., x k } } with the property ( ) S F k w(s) χ k S F k 1 w(s)

59 Lower Bound: χ f (G) max (χ n, f (n)) Continue this process so that for all k {1,..., n}, F k = {S I (G) : S {x 1,..., x k } } with the property ( ) S F k w(s) χ k S F k 1 w(s) It follows that w(s) 1 + S F n n n ) (1 1χk = f (n) 1. i=2 k=i

60 Lower Bound: χ f (G) max (χ n, f (n)) Conclusion χ f (G) f (n)

61 Upper Bound: χ f (G) max (χ n, f (n)) Special Sets and Cool Weightings

62 Upper Bound: χ f (G) max (χ n, f (n)) Special Sets and Cool Weightings Special Sets Let M (G) I (G) be the set of all maximal independent sets of G

63 Upper Bound: χ f (G) max (χ n, f (n)) Special Sets and Cool Weightings Special Sets Let M (G) I (G) be the set of all maximal independent sets of G and for each i {1,..., n}, F i := {S M (G) S V (G j ) = if and only if j < i}

64 Upper Bound: χ f (G) max (χ n, f (n)) Special Sets and Cool Weightings Special Sets Let M (G) I (G) be the set of all maximal independent sets of G and for each i {1,..., n}, F i := {S M (G) S V (G j ) = if and only if j < i} Weightings w i : I (G i ) R 0, a χ f (G i )-coloring of each G i

65 Upper Bound: χ f (G) max (χ n, f (n)) Special Sets and Cool Weightings Special Sets Let M (G) I (G) be the set of all maximal independent sets of G and for each i {1,..., n}, F i := {S M (G) S V (G j ) = if and only if j < i} Weightings w i : I (G i ) R 0, a χ f (G i )-coloring of each G i p i : I (G i ) R 0 where p i := w i (S)/χ i

66 Upper Bound: χ f (G) max (χ n, f (n)) Special Sets and Cool Weightings Special Sets Let M (G) I (G) be the set of all maximal independent sets of G and for each i {1,..., n}, F i := {S M (G) S V (G j ) = if and only if j < i} Weightings w i : I (G i ) R 0, a χ f (G i )-coloring of each G i p i : I (G i ) R 0 where p i := w i (S)/χ i p : n i=1 F i R 0 where p(s) := n i=1 p i(s V (G i ))

67 Upper Bound: χ f (G) max (χ n, f (n)) The Final Weighting Final Weighting We construct a fractional max(χ n, f (n))-coloring of G defined by the weighting (χ i χ i 1 )p(s), S F i w(s) = max(0, f (n) χ n ), S = V 0 0, otherwise

68 Upper Bound: χ f (G) max (χ n, f (n)) The Final Weighting works! We can show, S I (G) w(s) = max(χ n, f (n)) w[x] 1 for all x V (G) So, w is a fractional max(χ n, f (n))-coloring of G. Conclusion χ f (G) max (χ n, f (n))

69 Results Theorem For n 2, let G 1,..., G n be graph. Suppose also that the graphs G i are numbered such that χ i χ i+1. Then ( n n ) χ f (Z(G 1,..., G n )) = max χ n, 2 + (1 ) 1χk i=2 k=i Corollary For every n 2, χ f (Z n+1 ) = χ f (Z n ) + 1 χ f (Z n )

70 Jacobs Conjecture - Proved! Corollary For every n 2, χ f (Z n+1 ) = χ f (Z n ) + 1 χ f (Z n )

71 Jacobs Conjecture - Proved! Corollary For every n 2, χ f (Z n+1 ) = χ f (Z n ) + 1 χ f (Z n ) Proof. By induction on n 2, we prove χ n+1 = f (n) = χ n + χ 1 n.

72 Jacobs Conjecture - Proved! Corollary For every n 2, χ f (Z n+1 ) = χ f (Z n ) + 1 χ f (Z n ) Proof. By induction on n 2, we prove χ n+1 = f (n) = χ n + χ 1 n. Base Case: χ f (Z 1 ) = 1 and f (1) = 2 = χ 2

73 Jacobs Conjecture - Proved! Corollary For every n 2, χ f (Z n+1 ) = χ f (Z n ) + 1 χ f (Z n ) Proof. By induction on n 2, we prove χ n+1 = f (n) = χ n + χ 1 n. Base Case: χ f (Z 1 ) = 1 and f (1) = 2 = χ 2 Inductive Hypothesis: Suppose χ n = f (n 1).

74 Jacobs Conjecture - Proved! Proof. Inductive Hypothesis: Suppose χ n = f (n 1).

75 Jacobs Conjecture - Proved! Proof. Inductive Hypothesis: Suppose χ n = f (n 1). Then f (n) = 2 + = 2 + n ) (1 1χk i=2 k i (1 1χn ) (f (n 1) 1) = χ n + 1 χ n

76 Jacobs Conjecture - Proved! Proof. Inductive Hypothesis: Suppose χ n = f (n 1). Then f (n) = 2 + = 2 + n ) (1 1χk i=2 k i (1 1χn ) (f (n 1) 1) = χ n + 1 χ n Since χ n+1 = max(χ n, f (n)), we have χ n+1 = χ n + 1 χ n.

77 Questions?

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