Exercise sheet 6: Solutions

Transcription

1 Eerise sheet 6: Solutions Cvet emptor: These re merel etended hints, rther thn omplete solutions. 1. If grph G hs hromti numer k > 1, prove tht its verte set n e prtitioned into two nonempt sets V 1 nd V 2, suh tht χ(g[v 1 ]) + χ(g[v 2 ]) = k. Solution. We know tht G n e olored properl k olors. Let the olor lsses (of verties) in one suh oloring e C 1, C 2,..., C k. Then it suffies to set V 1 = C 1 nd V 2 = C 2 C k. Indeed, it s es to see tht χ(g[v 1 ]) = 1 nd χ(g[v 2 ]) = k Let G 1 nd G 2 e two grphs on the sme verte set. Prove tht the hromti numer of their union is t most χ(g 1 )χ(g 2 ). Solution. If i : V (G i ) {1,..., χ(g i )} re proper olorings of G i for i = 1, 2, then : V (G) {1,..., χ(g 1 )} {1,..., χ(g 2 )} given (v) = ( 1 (v), 2 (v)) is proper oloring of G = G 1 G 2 with χ(g 1 ) χ(g 2 ) olors. 3. Prove tht χ(g) n α(g) + 1, where α(g) is the independene numer of given grph G. Solution. B oloring α(g) independent verties in olor 0 nd the remining n α(g) verties ll in different olors 1, 2,..., n α(g), we get proper oloring. 4. Given grph G, let P G (k) denote the numer of proper k-olorings of G (we ssume tht the set of olors is {1, 2,..., k}). Prove tht P G (k) = P G e (k) P G/e (k) for ever edge e E(G). Prove tht P G (k) is polnomil in k of degree V (G). (This polnomil is lled the hromti polnomil of G).

2 Solution. Notie tht the set of proper olorings of G is equl preisel to the set of proper olorings of G e eept for the olorings tht give the sme olor to the endpoints of e. However, the numer of the ltter olorings is equl to the numer of proper olorings of G/e. B indution on the numer of edges of G. If there re no edges, then the numer of proper k olorings is k V (G), whih is polnomil of degree V (G). If there is t lest one edge e, then indution P G e (k) is polnomil of degree V (G) nd P G/e (k) is polnomil of degree V (G) 1, hene from we onlude tht P G (k) is lso polnomil of degree V (G). 5. Let G e grph with hromti numer lrger thn k, whose verte set is prtitioned into sets X nd Y. Suppose tht G[X] nd G[Y ] hve hromti numer k. Prove tht there re t lest k edges etween X nd Y. Solution. Suppose there were t most k 1 edges etween X nd Y. Fi olorings of X nd Y with the sme set of olors {1,..., k}. If no two verties of X nd Y with the sme olor re djent we hve proper k-oloring of G, whih would e ontrdition. We show indution on k tht we n lws renme the olors in Y (i.e. hoose permuttion of {1,..., k}) so s to void hving monohromti edges. Sine there re t most k 1 edges there is olor, w.l.o.g. let it e 1, suh tht no X Y edge hs two verties of tht olor. Now if there is n X Y edge with one verte of olor 1, we n ppl indution on the set of verties olored 2, 3,..., k whih hve k 2 edges etween them. Thus we n suppose tht the olor-1 verties re not inident to n X Y edge. Then we n hoose olor lss in Y, let it e 2, tht hs verte inident to n X Y edge nd swp olors 1 nd 2 in Y. Now we re in the previous se nd therefore done indution. Alterntive solution: Suppose there were t most k 1 edges etween X nd Y. Fi proper k-olorings of X nd Y. Consider iprtite grph B whose verties in one prt re the k olor lsses from X nd in the other prt the olor lsses from Y, with n edge etween two verties iff there is NO edge etween the orresponding olor lsses in G. Note tht B hs t lest k 2 k + 1 edges. We wnt to show tht there is perfet mthing in B. This would esil provide proper k-oloring of G nd there the desired ontrdition. To prove the lim we use the mrrige theorem. Tke n i verties in one

3 prt of B nd suppose the hve in totl j < i neighors in B. Then E(B) ij +(k i)k i(i 1)+(k i)k < k 2 k +1. Contrdition. Therefore, we must hve j i, i.e. the mrrige ondition is stisfied nd we hve perfet mthing. 6. In plne finite numer of ongruent irles re given. These irles re mutull noninterseting (the m e eternll tngent). Prove (without using the Four olor theorem or its reltives!) tht one n use t most four olors to olor these irles so tht two irles tngent to eh other re of different olors. Wht is the smllest numer of irles tht requires four olors? Solution. The proof is indution on the numer of irles. Consider the leftmost irles (there might e more thn one of them) nd mong them hoose the topmost one. The hosen irle touhes t most 3 other irles. All the irles eept for the hosen one n e properl olored 4 olors 1,2,3,4 indution. Then we n sfel olor the hosen irle one of the olors tht is free. Sine no one solved this prt ompletel, we ll give full solution (lso we d like to pologie to the students for this eerise tht turned out to e too tehnil nd thus inpproprite ). Consider the smllest set S of irles tht requires 4 olors. Eh irle in S touhes t lest 3 other irles, sine otherwise we ould remove irle tht touhes t most 2 other irles nd get smller set tht lso requires 4 olors. Clim: there must e t lest 11 irles in S. Suppose it s not the se, i.e. we hve t most 10 irles. Let P denote the set of the enters of the irles nd w.l.o.g. suppose tht the rdius of eh irle is 1/2. If we onnet ever two points from P tht orrespond to two touhing irles, we get plne grph G whose edges re stright line segments of length 1. Also note tht n two points from P re t distne t lest 1. Before we ontinue with the proof we need to introdue the following notion. A sugrph of G tht looks like the grph from Figure 1 is lled dimond, its verties nd d re lled its middle verties, while nd re its top verties. The ke step in the proof is the following geometri lemm. Lemm. Ever verte of the onve hull of P is middle point of dimond.

4 q p l 2 v d n l 1 m Figure 1: Dimond; Proof of Lemm Proof of Lemm: Let v e verte of the onve hull. It hs degree 3 in G nd let,, e its neighors. We ssume tht v is horiontl, is ove v nd s elow. The gol is to show tht v = v = 60. We ll do this in two steps, first showing tht t lest one of these ngles is 60, nd then tht oth hve to e equl 60. Suppose tht oth v nd v re greter thn 60. Consider the horiontl lines l 1 nd l 2 through nd. Point hs two neighors m nd n elow or on l 1 nd point hs two neighors p nd q ove or on l 2 (we ssume the re leled s in Figure 1). Point m is t distne 1 from nd possil n, ut definitel from no other point tht we hve so fr. Therefore, there is new verte t distne 1 from m. Similrl, there is new point P suh tht q = 1 nd sine mq > 2, whih is es to see. We hve lred 10 points. Suppose = 1. Then point n lies inside the le mv in G nd sine it hs degree t lest 3, it hs to e t distne 1 from two more verties prt from. Clerl vn > 1. Moreover, mn = 1 immeditel gives ontrdition, so the onl option is n = n = 1, ut then n lies on l 1 nd would e ove l 1 ontrditing m = 1. Therefore, the initil ssumption = 1 nnot hold. So we hve > 1 nd lso > 1. Sine is of degree t lest 3, we must hve

5 n = p = 1, whih implies tht n l 1 nd p l 2. Then mn nd pq re dimonds, ut nd would e of degree t most 2 (see Figure 2). Contrdition. We hve shown tht t lest one of the ngles hs to e 60. W.l.o.g. q q r s p p v v n n m m Figure 2: Proof of Lemm suppose v = 60 nd we ssume for ontrdition tht v > 60. Agin point hs neighor m elow or on l 1 nd point hs two neighors p nd q ove or on l 2. Furthermore, m hs nother neighor n nd q lso hs new neighor r tht is ove or on the line through q prllel to p. Finll, r must hve one more neighor s nd we don t hve n more points in P (see Figure 2). Also rp = 1 is the onl hne for r to e of degree t lest 3. Similrl, m = 1. The onl potentil neighors of n in G, eept for m, re nd s. Thus, n = ns = 1. Then we would lso hve pq = sp = 1, ut then our grph G is neessril isomorphi to the grph in Figure 3, whih is 3-olorle. Hving the lemm t our disposl, it s not diffiult to finish the proof. Ever interior ngle of onv(p ) is t lest 120, thus the onve hull hs t lest 6 verties. Let k e the verties of onv(p ) listed ounterlokwise. Due to its minimlit, we see tht G hs to e 2- onneted, nd hene ll its fes re ounded les. The outer

6 v 3 v 2 v 4 v 1 v 5 Figure 3: Three olorle; The smllest emple tht needs 4 olors fe ertinl ontins ll the verties 1,..., k nd, moreover, for eh side whose length is > 1 (i.e. tht is not n edge of G) there is one more verte on the oundr of the outer fe (etween nd ). Sine we hve t most 10 points, we onlude tht t lest 2 sides of onv(p ) re of length one. Consider two ses. 1 Two onseutive sides nd +2 re of length i 1 t Figure 4: Cse 1 B Lemm we know tht,, P (see Figure 4). If one of the points or is verte of the onve hull, then t P nd no mtter how we dd t most 3 points to the wheel G[,, +2,,,, t] we

7 hve 3-olorle grph. Thus we n ssume tht nd re not verties of the onve hull. Then we hve tht 1 +3 > +2, so the dimonds t 1 nd +3 (tht eist Lemm, see Figure 4) n shre t most one verte (nmel, middle verte). Also eh of them n shre t most one (top) verte with {,, +2,,, }. This would give t lest 11 verties. Contrdition. 2 No two onseutive sides of the onve hull re of length Figure 5: Cse 2 Let = 1. As efore,, P (see Figure 5). Now 1 +2 > 2 nd the dimonds orresponding to 1 nd +2 ( Lemm) hve no verte in ommon. In totl we would hve gin t lest 11 verties (see Figure 5). Contrdition. The proof of the lim is finished nd we know now tht S 11. Finll, we ehiit set of 11 irles tht indeed requires 4 olors. The onstrution is given Figure 3. Suppose there is proper 3-oloring of the irles. In proper 3-oloring two top verties of dimond lws hve the sme olor, hene (v 1 ) = (v 2 ) = (v 3 ) = (v 4 ) = (v 5 ), ut it implies tht (v 1 ) = (v 5 ), whih is ontrdition. If ou spot n mistkes on this sheet, plese drop n emil to filip.mori@epfl.h.