Counterexamples to the List Square Coloring Conjecture

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1 Counterexamples to the List Square Coloring Conjecture Boram Park National Institute for Mathematical Sciences (NIMS), Korea Joint work with Seog-Jin Kim Konkuk University, Korea

2 Introduction Definition. The square of a graph G is the graph G 2 obtained from G by adding edges joining the vertices at distance 2 in G. The square of a Peterson graph is K 10.

3 Introduction Definition. If χ (H)=χ(H), then H is called chromatic-choosable. Problem. Which graphs are chromatic-choosable?

4 Introduction Definition. If χ (H)=χ(H), then H is called chromatic-choosable. Problem. Which graphs are chromatic-choosable? Example of chromatic-choosable graphs Complete graphs K n, and cycle C n Petersen graph - Brooks Theorem χ (G) Δ(G) if G =K n or C 2k+1. Obha sconjecture: graph G with n(g) 2χ(G)+1. Line graph of K m,m (Galvin, 1995) Complete bipartite K m,n is not chromatic-choosable.

5 Introduction Definition. If χ (H)=χ(H), then H is called chromatic-choosable. Problem. Which graphs are chromatic-choosable? Example of chromatic-choosable graphs Complete graphs K n, and cycle C n Petersen graph - Brooks Theorem χ (G) Δ(G) if G =K n or C 2k+1. Obha sconjecture: graph G with n(g) 2χ(G)+1. Line graph of K m,m (Galvin, 1995) Complete bipartite K m,n is not chromatic-choosable. List Square Coloring Conjecture (Kosothcka and Woodall, 2001) χ (G 2 )=χ(g 2 ) for every graph G.

6 Introduction Conjectures of chromatic-choosable graphs List Coloring Conjecture χ (L(G))=χ(L(G)) for any graph G. - proposed independently by Vizing, by Gupa, by Albertson and Collins, and by Bollobás and Harris.

7 Introduction Conjectures of chromatic-choosable graphs List Coloring Conjecture χ (L(G))=χ(L(G)) for any graph G. - proposed independently by Vizing, by Gupa, by Albertson and Collins, and by Bollobás and Harris. The Dinitz Problem (1980): Is it true that χ (L(K m,m ))=m for all m? Theorem [Galvin, 1995] For every bipartite G, χ (L(G))=χ(L(G)). Theorem [Haggkvist and Janssen, 1997] For any integer n, χ (L(K n )) n. - This implis that LCC holds for K n when n is odd. Theorem [Borodin, Kostochka, Woodall, 2001] If G planar with Δ 12, χ (L(G))=χ(L(G)).

8 Introduction Conjectures of chromatic-choosable graphs List Total Coloring Conjecture For any graph G, χ (T(G))=χ(T(G)).

9 Introduction Conjectures of chromatic-choosable graphs List Total Coloring Conjecture For any graph G, χ (T(G))=χ(T(G)). List Square Coloring Conjecture χ (G 2 )=χ(g 2 ) for any graph G. LSCC implies LTCC H = thegraph obtainedby subdivision ofeach edge of G H 2 =T(G), where T(G) is the totalgraph of G. χ (H 2 )=χ(h 2 ) implies that χ (T(G))=χ(T(G)). If LSCC holds a special class of bipartite,then LTCC holds. e 2 y e 2 e 1 e 1 G = y H 01 01

10 Introduction Graphs satisfying the LSCC Only a few graph classes are known to be satisfied the LSCC G=inflation of C n with n 11 (Kostochka Woodall, 2001) Definition. H is an inflation of G if (1) V(H)=V 1 V n, (2) if V and y V j, then y E(H) if and only if =j or j E(G). K 2,3 -minorfree graphs with Δ 6 (Hetherington Woodall, 2011) In general, it is hard to check if χ (G 2 )=χ(g 2 ) or not Computing χ(g 2 ) is difficult, exact value of χ (H) is known only for a few classes of graphs

11 Introduction Graphs satisfying the LSCC List Square Coloring Conjecture χ (G 2 )=χ(g 2 ) for anygraph G. LSCC provides many conjectures from coloring version to list version Conjecture. (Wegner) χ(g 2 ) 3Δ 2 graph. if G is a planar Question. Is χ (G 2 ) 3Δ 2 for every planar G? (Thomassen) χ(g 2 ) 7 if G is a subcubicplanar. Question. Is χ (G 2 ) 7 if G is a subcubic planar?

12 Main Result Theorem For each prime n 3, there is a graph G such that G 2 =K n (2n 1). - K n (2n 1) is the completemultipartitegraph with (2n 1) partite sets in which each partiteset has size n.

13 Main Result Theorem For each prime n 3, there is a graph G such that G 2 =K n (2n 1). - K n (2n 1) is the completemultipartitegraph with (2n 1) partite sets in which each partiteset has size n. 2n 1=χ(K n (2n 1) ) for any n 3

14 Main Result Theorem For each prime n 3, there is a graph G such that G 2 =K n (2n 1). - K n (2n 1) is the completemultipartitegraph with (2n 1) partite sets in which each partiteset has size n. 2n 1=χ(K n (2n 1) ) for any n 3 χ (K n (2n 1) ) 3n 2 for any n 3.

15 Main Result Theorem For each prime n 3, there is a graph G such that G 2 =K n (2n 1). - K n (2n 1) is the completemultipartitegraph with (2n 1) partite sets in which each partiteset has size n. 2n 1=χ(K n (2n 1) ) for any n 3 χ (K n (2n 1) ) 3n 2 for any n 3. χ (K n (2n 1) ) χ (K n (2n 1) ) n 1 for n 3.

16 Main Result Theorem For each prime n 3, there is a graph G such that G 2 =K n (2n 1). - K n (2n 1) is the completemultipartitegraph with (2n 1) partite sets in which each partiteset has size n. 2n 1=χ(K n (2n 1) ) for any n 3 χ (K n (2n 1) ) 3n 2 for any n 3. χ (K n (2n 1) ) χ (K n (2n 1) ) n 1 for n 3. Observation. The gap χ (G 2 ) χ(g 2 ) as n.

17 Construction of counterexample Orthogonal Latin Squares Definition. Latin square is an n n array filled with n different symbols, each occurring exactly once in each row and exactly oncein each column. Definition. Two Latin squares L 1 =[ j ] and L 2 =[b j ] are orthogonalif the ordered pairs( j,b j ) are distint for all and j = (1,1) (2,3) (3,2) (2,2) (3,1) (1,3) (3,3) (1,2) (2,1)

18 Construction of counterexample Orthogonal Latin Squares Definition. Latin square is an n n array filled with n different symbols, each occurring exactly once in each row and exactly oncein each column. Definition. Two Latin squares L 1 =[ j ] and L 2 =[b j ] are orthogonalif the ordered pairs( j,b j ) are distint for all and j = (1,1) (2,3) (3,2) (2,2) (3,1) (1,3) (3,3) (1,2) (2,1) From now on, let n be an odd prime. For each 1 n 1, let L be a Latin square of order n defined by L (j,k)=j+ (k 1)(mod n), for(j,k) [n] [n].

19 Construction of counterexample Orthogonal Latin Squares Then {L 1,L 2,...,L n 1 } is a family of mutually orthogonal Latin squares. L (j,k) = j+ (k 1)(mod n), for(j,k) [n] [n] L 1 = L 2 = L 3 = L 4 =

20 Construction of counterexample Orthogonal Latin Squares Then {L 1,L 2,...,L n 1 } is a family of mutually orthogonal Latin squares. L (j,k) = j+ (k 1)(mod n), for(j,k) [n] [n] L 1 = L 2 = L 3 = L 4 = There are several families of mutually orthogonal Latin squares. We use {L 1,L 2,...,L n 1 } to construct G. Not every family of mutually orthogonal Latin squares work well to construct of counterexample G.

21 Construction of counterexample Construction of G {L 1,L 2,...,L n 1 }: L (j,k)=j+ (k 1)(mod n) We define a graph G as follows: V(G)=P 1 P n Q 1 Q n 1, where P = {,1,,2,...,,n } {1,...n} Q j = { j,1, j,2,..., j,n } j {1,...n 1} E(G)=E 1 E 2 such that E 1 = {,j k,l (j,k) :1 k n}, E 2 = where [n 1] j [n] { y:,y T j }, j [n] T j ={ 1,j, 2,j,..., n,j } for 1 j n.

22 Construction of counterexample Construction of G P 1 P 2 P 3 P n 2,1 3,1 n,1 (n 1),1 Q 1 Q 2 Q n 1 T j ={ 1,j, 2,j,..., n,j } forms a clique in G N G ( j ) is defined from the family of mutually orthogonallatin squares {L 1,L 2,...,L n 1 }

23 Construction of counterexample Construction of G P 1 P 2 P 3 P n 2,1 3,1 n,1 (n 1),1 Q 1 Q 2 Q n 1 T j ={ 1,j, 2,j,..., n,j } forms a clique in G N G ( j ) is defined from the family of mutually orthogonallatin squares {L 1,L 2,...,L n 1 }

24 Construction of counterexample Construction of G P 1 P 2 P 3 2,1 3,1 L 1 = L 2 = ,2 N G ( )={,, 3,3 } N G ( )={,, 3,1 } N G ( )={, 2,1, 3,2 } 3,3 N G ( 2,1 )={,, 3,2 } N G ( )={, 2,1, 3,3 } N G ( )={,, 3,1 } 2,1 E(G)=E 1 E 2 such that E 1 = {,j k,l (j,k):k [n]}, (,j) [n 1] [n] E 2 = j [n] { y:,y T j}, where T j ={ 1,j, 2,j,..., n,j } for 1 j n. Q 1 Q 2 N G ( )={,, 3,3 } from the1st row of L 1

25 Construction of counterexample Construction of G Why are L 1,L 2,...,L n 1? P 1 P 2 P 3 2,1 3,1 {L L (j,k)=j+ (k 1)(mod n)} satisfies the following property. 3,2 3,3 2,1 Q 1 Q

26 Construction of counterexample Construction of G Why are L 1,L 2,...,L n 1? P 1 P 2 P 3 2,1 3,1 {L L (j,k)=j+ (k 1)(mod n)} satisfies the following property. 3,2 3,3 N G (,s ) N G ( j,t ) =1 when =j -For any sth row of L and tth row L j, they must have exactly one common entries. 2,1 Q 1 Q

27 Construction of counterexample Construction of G Why are L 1,L 2,...,L n 1? P 1 P 2 P 3 2,1 3,1 {L L (j,k)=j+ (k 1)(mod n)} satisfies the following property. 3,2 3,3 N G (,s ) N G ( j,t ) =1 when =j -For any sth row of L and tth row L j, they must have exactly one common entries. 2,1 N G (,s ) N G ( j,t ) =1 when =j and s =t -There is a Latin square L k such thatfor some th row, th column and jth column entries are s and t. Q 1 L 1 = Q L 2 =

28 Proof of G 2 =K n (2n 1) P 1 P 2 P 3 2,1 3,1 L 1 = L 2 = ,2 3,3 and y are adjacentin G 2 for all P and y Q j. 2,1 P 1 P 2 P 3 induces K 3,3,3 in G 2. Q 1 Q 2 inducesk 3,3 in G 2. Q 1 Q 2

29 Proof of G 2 =K n (2n 1) P 1 P 2 P 3 2,1 3,1 L 1 = L 2 = ,2 3,3 and y are adjacentin G 2 for all P and y Q j. 2,1 P 1 P 2 P 3 induces K 3,3,3 in G 2. Q 1 Q 2 inducesk 3,3 in G 2. Q 1 Q 2 and y are adjacent in G 2 for all P and y Q j. Enough to show that,j is adjacent in G 2 to all P j.

30 Proof of G 2 =K n (2n 1) P 1 P 2 P 3 2,1 3,1 L 1 = L 2 = ,2 3,3 and y are adjacentin G 2 for all P and y Q j. 2,1 P 1 P 2 P 3 induces K 3,3,3 in G 2. Q 1 Q 2 inducesk 3,3 in G 2. Q 1 Q 2 and y are adjacent in G 2 for all P and y Q j. Enough to show that,j is adjacent in G 2 to all P j. reaches to P j in distancetwo in G. is adjacent to all P j in G 2.

31 Proof of G 2 =K n (2n 1) P 1 P 2 P 3 2,1 3,1 L 1 = L 2 = ,2 3,3 and y are adjacentin G 2 for all P and y Q j. 2,1 P 1 P 2 P 3 induces K 3,3,3 in G 2. Q 1 Q 2 inducesk 3,3 in G 2. Q 1 Q 2 and y are adjacent in G 2 for all P and y Q j. Enough to show that,j is adjacent in G 2 to all P j. reaches to P j in distancetwo in G. is adjacent to all P j in G 2.

32 Proof of G 2 =K n (2n 1) P 1 P 2 P 3 2,1 3,1 3,2 and y are adjacentin G 2 for all P and y Q j. 3,3 P 1 P 2 P 3 induces K 3,3,3 in G 2. 2,1 Q 1 Q 2 inducesk 3,3 in G 2. Q 1 Q 2 P 1 P 2 P 3 induces K 3,3,3 in G 2. Each P is an independentset in G 2-1,j and 1,k do not haveacommonneighborby the way of defining the edges

33 Proof of G 2 =K n (2n 1) P 1 P 2 P 3 2,1 3,1 3,2 and y are adjacentin G 2 for all P and y Q j. 3,3 P 1 P 2 P 3 induces K 3,3,3 in G 2. 2,1 Q 1 Q 2 inducesk 3,3 in G 2. Q 1 Q 2 P 1 P 2 P 3 induces K 3,3,3 in G 2. Each P is an independentset in G 2-1,j and 1,k do not haveacommonneighborby the way of defining the edges From the property of our orthogonal Latin squares, N G (,j ) N G ( s,t ) =1 ( =s, j =t).

34 Proof of G 2 =K n (2n 1) P 1 P 2 P 3 2,1 3,1 3,2 and y are adjacentin G 2 for all P and y Q j. 3,3 P 1 P 2 P 3 induces K 3,3,3 in G 2. 2,1 Q 1 Q 2 inducesk 3,3 in G 2. Q 1 Q 2 Q 1 Q 2 induces K 3,3 in G 2 Each Q is an independentset in G 2 -,j and,k haveno commonneighborsinceeach column of a Latin square has distinct elements.

35 Proof of G 2 =K n (2n 1) P 1 P 2 P 3 2,1 3,1 3,2 and y are adjacentin G 2 for all P and y Q j. 3,3 P 1 P 2 P 3 induces K 3,3,3 in G 2. 2,1 Q 1 Q 2 inducesk 3,3 in G 2. Q 1 Q 2 Q 1 Q 2 induces K 3,3 in G 2 Each Q is an independentset in G 2 -,j and,k haveno commonneighborsinceeach column of a Latin square has distinct elements. N G (,j ) N G ( s,t ) =1 when =s from the property of our special Latin squares.

36 Further Research Questions Summary Theorem For each prime n 3, there is a graph G such that G 2 =K n (2n 1). - K n (2n 1) is the completemultipartitegraph with (2n 1) partitesetsin whicheachpartiteset hassize n. Theorem [Vetrik, 2012] For a completemultipartitegraph K n r with n,r 2, χ l (K n r )>(n 1) 2r 1. n χ(k n (2n 1) )=2n 1<3n 2 χ (K n (2n 1) ) for any n 3

37 Further Research Questions Summary Theorem For each prime n 3, there is a graph G such that G 2 =K n (2n 1). - K n (2n 1) is the completemultipartitegraph with (2n 1) partitesetsin whicheachpartiteset hassize n. Theorem [Vetrik, 2012] For a completemultipartitegraph K n r with n,r 2, χ l (K n r )>(n 1) 2r 1. n χ(k n (2n 1) )=2n 1<3n 2 χ (K n (2n 1) ) for any n 3 Not only χ (G 2 ) =χ(g 2 ), Observation. The gap χ (G 2 ) χ(g 2 ) as n.

38 Further Research Questions Noel suggested the following questions: Question. Does there exist a function ƒ(k)=o(k 2 ) such that for every graph G, χ G 2 ƒ χ G 2? Question. Does there exist a constant c such that every graph G satisfies χ G 2 cχ G 2 logχ G 2?

39 Further Research Questions Noel suggested the following questions: Question. Does there exist a function ƒ(k)=o(k 2 ) such that for every graph G, χ G 2 ƒ χ G 2? Question. Does there exist a constant c such that every graph G satisfies χ G 2 cχ G 2 logχ G 2? of graph powers

40 Further Research Questions Theorem [Kim, Kwon, Park, 2013+] For any positive integer m 2 and for any positive integer s, there exists a graph G such that χ (G m ) sm sm 1 =χ(g m ). Question. Does there exist a function ƒ(k)=o(k m ) such that for every graph G, χ G m ƒ χ G m?

41 Further Research Questions Thank you very much!

Coloring. Basics. A k-coloring of a loopless graph G is a function f : V (G) S where S = k (often S = [k]).

Coloring. Basics. A k-coloring of a loopless graph G is a function f : V (G) S where S = k (often S = [k]). Coloring Basics A k-coloring of a loopless graph G is a function f : V (G) S where S = k (often S = [k]). For an i S, the set f 1 (i) is called a color class. A k-coloring is called proper if adjacent