List Coloring Graphs

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1 Lit Coloring Graph February 6, 004 LIST COLORINGS AND CHOICE NUMBER Thomaen Long Grotzch girth 5 verion Thomaen Long Let G be a connected planar graph of girth at leat 5. Let A be a et of vertice in G uch that each vertex of A i on the outercircuit. Aume that either (i) G(A) ha no edge or (ii) G(A) ha preciely one edge xy and G ha no -path from x to a vertex of A. Aume that L i a color aignment uch that L(v) for each vertex in G and L(v) 3 for each vertex in V (G)\A. Let u, w be any adjacent vertice in G both on the outer face boundary and let c(u), c(w) be ditinct color in L(u) and L(w) repectively. Then c can be extended to a lit coloring of G. Grötzch girth 5 verion Every planar graph G of girth at leat 5 i 3-colorable. Moreover, if G ha an outer cycle of length 5 then any 3-coloring of the outer cycle can be extended to a 3-coloring of G. Thomaen Long Grotzch girth 5 verion Aume G i a graph with girth at leat 5. If G doe not have an outercycle of length 5 then it doe not have any precolored vertice. In thi cae, we give every vertex the ame lit of ize 3 and ue Thomaen (Long). 1

2 If G ha an outercycle C : v 1, v, v 3, v 4, v 5, v 1, we can aume it i precolored. Let v 1 and v play the role of u and w. A = {v 3, v 5 } L(v 3 ) = {c(v 3 ), c(v )} L(v 5 ) = {c(v 5 ), c(v 1 )} L(v 4 ) = {c(v 3 ), c(v 4 ), c(v 5 )} All other lit will contain the ame three color. Thomaen Short Let G be a plane graph of girth at leat 5. Let c be a 3-coloring of a path or cycle P : v 1, v,..., v q, 1 q 6 uch that all vertice of P are on the outer face boundary. For all v V (G), let L(v) be it lit of color. If v P then L(v) = {c(v)}. Otherwie L(v). If v i not on the outer face boundary then L(v) = 3. There are no edge joining vertice whoe lit have at mot color, except the edge of P. Then c can be extended to an L-coloring of G. Show ome picture. Thomaen Short (B) Thi i eaier. Aume G i a graph with girth at leat 5. If G doe not have an outercycle of length 5 then it doe not have any precolored vertice. In thi cae, we give every vertex the ame lit of ize 3 and ue Thomaen (Short). If G ha an outercycle C : v 1, v, v 3, v 4, v 5, v 1, we can aume it i precolored. Let P = C. Then give the ame three color to the lit of all other vertice. Apply Thomaen (Short). A theorem of E, and Hull Defn 1 A defective coloring with defect d i a coloring of the vertice uch that for each color cla C, the maximum degree of the induced graph on C i d. If there exit a k-coloring of a graph with defect d, we ay the graph i (k, d)-colorable. A d-defective L-lit coloring i an L-lit coloring

3 which i defective with defect d. We ay a graph G i (k, d)-chooable if for every k-lit aignment L, G i d-defective L-lit colorable. NOTICE: A defective coloring with defect 0 i a proper coloring. Theorem 1 All planar graph are (3, )-chooable. In the pirit of Thomaen Theorem Given a connected plane graph G with outercircuit C = (v 1, v,..., v k ), and lit aignment L uch that for v V (C), L(v) and otherwie L(v) 3, any precoloring c of the vertice v 1 and v, can be extended to a -defective L-lit coloring of G in uch a way that If c(v 1 ) = c(v ) then def(v 1 ) = def(v ) = 1. If c(v 1 ) c(v ) then def(v 1 ) 1 and def(v ) = 0. Current reearch quetion: Are all planar graph (4,1)-chooable? Notice: Theorem implie Theorem 1. Proof of Theorem : We may aume G i inner triangulated. Cut vertex: Chooe an endblock B with cut vertex v. Let G 1 =< (V (G)\V (B)) {v} > and G = B. Suppoe wlog, G 1 contain both v 1 and v. We can color G 1 by the induction aumption. Next, we color G. We let the one precolored vertex be called v and precolor a neighbor of it, call it v 1 on the outer circuit with a color in it lit not equal to c(v ). Now color G by induction. We are aured that v get no new defect, o that if it already ha a defect() in G 1, we didn t created any new one. Aume no cut vertex, o C i a cycle and G i -connected. Let a L(v 3 ) c(v ). Let A = {v V (C) : a L(v)}. (Note: A include v 1 if c(v 1 ) = a. Cae 1 The ubgraph < V (C) > ha no chord among vertice in A. 3

4 Let I = Int(C) and for all v A, let I(v) = N(v) I. Set I(A) = v A I(v) Let G =< V (G)\A > and create a new lit aignment L for G a follow: For v I(A), let L (v) = L(v)\{a}, otherwie, let L (v) = L(v). Apply the induction hypothei to G. The remaining vertice and be colored with color a. Cae The ubgraph < V (C) > ha a chord v i v j uch that v i, v j A. Let C be the larget cycle uch that V (C ) V (C), v 1, v V (C ), and it ha no chord with both vertice in A. Let G be the ubgraph with outer circuit C. We apply Cae 1 to G. Each component K i,j of G\G i attached to G by a chord v i v j of < V (C) >. For each component K i,j, et G i,j =< V (K i,j {v i, v j } >. In the coloring of G, v i and v j have the ame color, a. Then when we apply induction to each G i,j with precolored vertice v i and v j, they have the ame color and o will get no new defect Erdö, Rubin and Taylor General Graph Characterized all -chooable graph There exit a c uch that, for every n, ch(k n,n ) > c ln n. If G i connected, not K n, not an odd cycle then ch(g) i at mot it maximum degree. Conjectured that G planar implie ch(g) 5. Stirling Formula ( ) n n ( ) n n ( πn < n! < πn ) e e n 1 Thu we have: 4

5 Lower bound: n i k i n k ( ) ( n k ( n e n < < k k) k for all i [k 1]. ) k Upper bound on ( ) n k ( ) ( ) n e n k < k k Notice ( ) n k n n k = (1 + k n k )n k < e k and n! k!(n k)! < ( k e ( ) n n e πn ) k ( ) n k πk n k π(n k) e n n k n n k ( e n = πk(n k) k k (n k) n k k To get the lat inequality we notice that if k n, n = 1 + k, o n k n k that n 1 and if k > n, we exchange the role of n k and k in the πk(n k) above argument. Theorem 3 There exit a c uch that for every n, ch(k n,n ) > c ln n ) k For k, we chooe the larget integer uch that n ( ) k 1 k. If we let = 1 ln n. Then 1 ln n, ln n, ln n 1. We ee that e 1 > (1 + 1 ) and o, ( n e 1 = e e 1 > e ) ( 1 = e By Stirling formula, ( ) ( ) e( 1) 1 >. 5 )

6 Thu, k 1 ln n. Let N be the et of all ( ) k 1 k k-ubet of [k 1]. We ee that n ( ) k 1 k = N. For V1 and V, the partite et of K n,n, aign the lit N to both V 1 and V. If there are any vertice left over, jut repeat ome lit. Conider any choice of color for the vertice in V 1, call the et of choen color T 1. We call uch a et of color a tranveral of N becaue for each N N, T 1 N. The ize of T 1 mut be at leat k. If T 1 k 1, then [k 1]\T 1 k. But then there exit a k-ubet of [k 1] in [k 1]\T 1 that i not covered by the tranveral. Contradiction. Similarly, the ize of T, the tranveral of color ued for V mut be at leat k. But then T 1 T ince there are only k 1 color. Hence another contradiction. Thi i a lit aignment of ize at leat 1 ln n. That can t be properly colored, hence ch(k n,n ) 1 ln n. Theorem 4 If G i connected, not K n, not an odd cycle then ch(g) i at mot it maximum degree. Let G be a connected graph, not K n and not an odd cycle. Let d = (G). Aume v S(v) i an arbitrary d-lit aignment of G. Suppoe G ha a pair of adjacent vertice x and y uch that S(x) S(y) and one of x, y i not a cut vertex. Wlog, aume y i not a cut vertex. Chooe c(y) S(y)\S(x). Since G y i connected, we can define a panning tree of G with root x. So that we can form a lit of the vertice, v 1 = x, v, v 3,..., v n 1, v n = y uch that for each i > 1, v i i adjacent to at leat one vertex proceeding it in the lit. We color the vertice in the order: v n 1, v n,..., v, realizing that lit of ize d color will uffice. When we color x, we realize that even though it ha poibly d neighbor that have already been colored, one of them, namely y, will not be uing a color from L(x). So there i a color available for x. Now uppoe that for every pair of adjacent vertice x and y uch that S(x) S(y) both x and y are cut vertice. If there i at leat one uch pair, we conider the block tructure of G. Conider the auxiliary bipartite graph B(G) that ha the cut vertice in one 6

7 partite et and the block in the other. Since G i connected, thi i a tree. Chooe an end-block B with aociated cut vertex v. We know that B v i nonempty and none of the vertice of B v are cut vertice. So all the vertice in B v mut have the ame lit aigned to them a v, namely L = L(v). We know that G =< (V (G)\V (B)) {v} > i connected. Form a lit of the vertice in G, v 1 = v, v, v 3,..., v p, a before, o that each vertex ha a neighbor to it left, except of coure for v. Now we can color the vertice in the lit v 1 = v, v, v 3,..., v p, a before, going from right to left. In the end, we know that we can color v with a color c from L ince it ha at leat one neighbor in B, and thoe vertice are not in the lit. By Brook Theorem we can color B uing the d color in L. Then if v end up with color c c in thi coloring, we exchange the color cla c with the color cla c of B. Now we can aume that for every pair of adjacent vertice, x and y, S(x) = S(y). By tranitivity, all lit are the ame, and we can ue Brook theorem. 7

LINEAR ALGEBRA METHOD IN COMBINATORICS. Theorem 1.1 (Oddtown theorem). In a town of n citizens, no more than n clubs can be formed under the rules

LINEAR ALGEBRA METHOD IN COMBINATORICS 1 Warming-up example Theorem 11 (Oddtown theorem) In a town of n citizen, no more tha club can be formed under the rule each club have an odd number of member each