Kat Husar Mikey Reilly Hannah Johnson Eric Fawcett nvm presentation Outline Please ignore spelling errors OPkaauidf.IR ctaria9Yafionsm Stanley old of acyclic Xg C D C 1 orientations Zaslaushy definitions examples of cycles in signed directed graphs generalization of Stanley to signed graphs nests mm Stanley's theorem steps to proving this From Dr Chomutov's Lecture notes 1.1 every proper coloring has an associated orientation 1.3 µ g C 1 PH H 1.2 ICH Ba 1.3 goal E E O E e e and Definitions variables er G is a finite graph without loops or multredges es G is the set of vertices of graph G G vcG A B GD E3 is the set of X X G of G XCG se e es edges eyes SB A e unordered pair e EX of vertices where Utr is u v p Ivan 5 an 9 1 611 5 the number of vertices in p 1VCG l G e AO e not acyclic l X G 1 the number of G edges in g cycle ABCA r Les assignment of a direction to each edge orientation an denoted by 9 v T v u u or acyclic an orientation of G has no directed acyclic cycles es
evaluated at 7 colors X G X X X is the of G chromatic polynomial of proper colorings in 7 colors X C IN for A 6,71 73 k is any map K K 1,2 is ie a coloring O will be a certain orientation ICA with is XCX a different condition has KlukkCv is where an edge U v improper coloring www.moo orientation 0 Proposition 1 I Every proper coloring has associated an X t is equal to thenumber of pairs 1K 01 13 K is any map K recall to 1,2 or coloring of G with 0 orientation the following is conditions an orientation 0 is acyclic 1 the is redundant but it helps to transition to It yes this then K in the orientation 0 121 If y Ktv Utv K 01 Lets do to understand an example 6 be µ Ext and let Let 1 3 kills thus 1,333 I am going to pick K 3 2 nodgetinitthfegriennotationow improper colorings go 1 Lets orient this graph based on the coloring IF the the 0 j µ Z of CK 01 Then XC3
This should give you intuition about the proof example so I will breifly touch upon the proof but after this should go presentation you back and if you truly care about math convince yourself proof 1 I note that condition 121 forces the graph to be acyclic Assume to get fake contradiction that there is since 1 3 a cycle a too ffff v wz u w but then condition 2 based on Kev Kiwi KI V is Wrong This But since Nukkas Thus condition 2 forces acyclicity rt If orientation 0 then KcalKH ur in the coloring K is proper note that condition implies that the 2 since Kla NIV EMB k u 4 If KcalKH ur in the orientation 0 then µ v o area Z 3 2 2 Thus if you a pair have X K Ol K must be proper K proper coloring then it coresponds to a unique 1K 01 pair 3 if Conversley is a Lets orient this graph based on the coloring of K Thus of pairs K Ol of proper XD BqEhE colorings
than u Now are going to ICT define we X is equal to thenumber of pairs 1K 01 73 K is any map K 44081,2 recall or coloring 0 of G with orientation the following is conditions an orientation 0 is acyclic 1 the not redundant anymore then K ul in the orientation 0 121 If v Utv for X E II where 720 and Theorem 1.2 C 1 PATH III pike.ffie this will be proved using induction but first properties of 7TH some recall that 7171 hasthe following properties or threaten LIX lil X aka with 7 colors Iii Xo we 71 76,1710 217 of I Ciii to to eat touch L 11.4 contraction deletion re I and so has similar properties 7 lil I 171 7 Iii 7 171.7 2171 we iii 76 76 e 7 X Gye 7 contraction deletion follow from the i and definition ii so just need to prove liii
let e Gu v3 y Hojo K 73 V G e I z with orientation G O is the orientation of G e e must be acyclic that corresponds with K based on the definition of 7cm lol µ I is the orientation of G with O v f G with orientation orientation of G with Oz is the v U y µ qV K on VCG VCG e note defined is claim 6 with orientation z either one of 10 K and Oz N for every pair CO K for ICT or both fits with the work definition only one of 0 and Oz t 2 ICGle ICG 7 so G ICGle worn e you need to t IC Gk G e count these twice both 0 and Oz since proof work Kcal Kcr case 1 but KID Kcal Oz not compatible u V contradiction the following cycle exists assume to get 0 acyclic a d w u v s a K u which is false but then Kcu Case 2 Ktv Hu similar to case 1 but Oz and 0 is not compatible is acyclic
Klut Kcv case and Oz are compatible with condition 12 0 both at least 1 of 0 and Oz is acyclic a contradiction that both have cycles assume to get 0 d k i u u v n d Q e v n c v a u d k d O i u c e v v n n a T 9 THIS contradicts that 0 is acyclic BAD IS last step 0 and Oz are in Both acyclic proof for ICGle pairs K O KYO't Make a bijection g R both 0 and Q K Gte 73 1,2 are acyclic is acyclic i let z be the vertex that X G et XCGle results from um contracting 0 define the bijection we VCGlet su v3 ftp.r.ly O K'Cw Kew for q4z7 KlW KCv1 G e in O iff 0 in w wz w we Cole 7 GA 2 ICGle 7 G e ICGle CG e 1 ICGle to Hta i i
Induction time I 17 for X E II where 720 and p C HP XI 2 Huerffie complete I and so has similar properties 7 lil I 111 7 Induct on sum vertices and of edges Iii Xo we 71 76,1710 217 iii 76 76 e 7 X Gye 7 contraction deletion note if 7 D 1 111 XC 1 7 le T base case you have a bunch of verticies that ptg Ek unconnected just for assume property 2 apply f IIP C y T 7 G is a graphs t p't g htt th case htt 1 ICG le 7 ICG 71 G e has 9 ledges 7 16 e 1 Bed note f 1 P NG edges _k so verticiest f 1 P zygge y 7 e BiH Gre has g ledges pl note and p 1 vertices so verticiest X Gk f 1 7 7 X G e edges k t f 1 P'XCG 71 DATED an A on It.EE tEEE What was the point of this entire proof P C l 7Gt 1 Stanley's Theorem acyclic orientations then all the K are if 7 1 1.2 c 1 Px f W Ig EMB the same and compatible with0 III number of pairs 1K is equal to the K isany map K b't 81,2 13 recall fk 0 or coloring O f 1 so pairs orientation of G with the following conditions O is an in the orientation 0 is acyclic orientations acyclic not redundant anymore 21 If u in the orientation 0 then KUHN C 1 P C l g 1
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