Solutions to Homwork 5 Pro. Silvia Frnánz Disrt Mathmatis Math 53A, Fall 2008. [3.4 #] (a) Thr ar x olor hois or vrtx an x or ah o th othr thr vrtis. So th hromati polynomial is P (G, x) =x (x ) 3. () With x hois or an x or 2 an 3 w hav P (G, x) =x (x ) 2. () Thr ar x hois or, x hois or 2, x 2 or 3, an x or ah o th othr thr vrtis. Thus P (G, x) =x (x ) 4 (x 2). () Thr ar x hois or, x hois or 2, x 2 or 3, x 3 hois or 4, an x 2 or 5 an 6. Thus P (G, x) =x (x ) (x 2) 3 (x 3). 6 2 3 2 3 2 4 3 4 (a) () 4 2 3 5 () 5 () 2. [3.4 #2] (a) P (G, 3) = 3 (3 ) 3 =24. () P (G, 3) = 3 (3 ) 2 =2. () P (G, 3) = 3 (3 ) 4 (3 2) = 48. () P (G, 3) = 3 (3 ) (3 2) 3 (3 3) = 0 3. [3.4 #3] Th map M low an rprsnt y th graph G on its right a a Looking at ithr M or G thr ar x olor hois or an x or any o th othr 4 rgions/vrtis. Thn P (M,x) =P (G, x) =x (x ) 4. 4. [3.4 #4] Thr ar x hois or vrtx an x or all othrs atr. Thn P (L n,x)= x (x ) n. 2 3 4 5 n-3 n-2 n- n... 5. [3.4 #5] W us P (,x)=p (K 3,x)=x (x ) (x 2) an P (,x)=p (I,x)=x. Part (a)isusinallothrparts. (a) P (,x) = P (,x) P (,x)=p (,x) P (,x) P (,x) = P (,x)p (,x) 2P (,x) = P (,x)(p (,x) 2) = x (x ) (x 2) 2.
() P (,x) = P (,x) P (,x)=p (,x) P (,x) P (,x) = P (,x)p (,x) P (,x) P (,x) = P ( µ,x)(x ) P (,x) = P (,x) P (,x) (x ) P (,x) µ = P (,x)p(,x) P (,x) (x ) P (,x) = P (,x)(x ) 2 P (,x) = x (x ) 3 (x 2) x (x ) (x 2) 2 = x (x 2) (x ) x 2 3x +3 () P (,x) = P (,x) P (,x) = P (,x)p(,x) P (,x) µ = P (,x)(x ) = P (,x)p(,x) P (,x) (x ) = P (,x)(x ) 2 = x (x ) 3 (x 2) 2. () Th graph in part () is quivalnt to x (x ) (x 2) 2. in part (a). Thn its hromati polynomial is 6. [3.4 #6] From th prvious xris w hav P (,x)=p(k 3,x)(P (I,x) 2). Using rution thorms w hav P (K 3,x)=P (L 3,x) P (K 2,x)=P (K 2,x) P (I,x) P (K 2,x) P (K 2,x) = P (K 2,x)(P (I,x) 2) = (P (I 2,x) P (I,x)) (P (I,x) 2). Thn P (,x)=(p (I 2,x) P (I,x)) (P (I,x) 2) 2 (not that sin P (I n,x)=x n thn this agrs with th answr in 3.4 #5 (a)). 7. [3.4 #7] Using th rution thorm or Z n w hav P (Z n,x)=p (L n,x) P (Z n,x). Thn P (L n,x)=p (Z n,x)+p (Z n,x). 8. [3.4 #8] Th graph G assoiat to th map is a Thn P (M,x) =P (G, x) =P (L 4,x) P (I,x) P (L 4,x)=P (L 4,x)(x ) = x (x ) 4 using Exris 3.4 #4. 2
9. [3.4 #3] (a) an () ail Thorm 3.3 (not onsutiv powrs), () ails Thorm 3.2., () an (g) ail Thorm 3.3 (thy o not altrnat signs), () an () ail Thorm 3.2.. So non o ths polynomials oul a hromati polynomial. 0. [3.4 #2] Not. Hr W n nots th whl on n +vrtis. (W prviously us W n to not th whl on n vrtis.) Solution. Lt F n th graph on n + vrtis otain y joining on vrtx to all vrtis o L n. n+... n-3 n 2 3 4 5 n-2 n- Thn, using th Funamntal Rution Thorm, P (W n,x)=p (F n,x) P (W n,x). W irst gt th hromati polynomial o F n. I x olors ar availal thn thr ar x olor hois or vrtx n +, x or vrtx, anx 2 or vrtis 2, 4,...,n. Thn P (F n,x)=x(x ) (x 2) n. Thus w an writ or all k 4 an P (W k,x)=x (x ) (x 2) k P (W k,x) P (W 3,x) = x (x ) (x 2) 2 P (K 3,x) = x (x ) (x 2) 2 x (x ) (x 2). Now onsir th ollowing sum ( ) k P (W k,x) = = = k=3 ( ) k x (x ) (x 2) k ( ) k P (W k,x) k=4 ³ k=4 +( ) 3 x (x ) (x 2) 2 x (x ) (x 2) ( ) k x (x ) (x 2) k + ( ) k P (W k,x) k=4 k=4 +( ) 3 x (x ) (x 2) 2 +( ) 2 x (x ) (x 2) n X ( ) k x (x ) (x 2) k + ( ) j P (W j,x) k=2 Not that th two sums o hromati polynomials anl, xpt or th trm ( ) n P (W n,x) on th lt. Hn ( ) n P (W n,x) = ( ) k x (x ) (x 2) k = x (x ) (x 2) (( ) (x 2)) k 2 k=2 j=3 n 2 X = x (x ) (x 2) (( ) (x 2)) j j=0 3 k=2
P n 2 j=0 (( ) (x 2))j is a gomtri sum. It is known that i a 6= thn a m + a m +... + a 2 + a += am+.usinga =( ) (x 2) w hav a n 2 X (( ) (x 2)) j = ( )n (x 2) n ( ) (x 2) j=0 Thror = ( )n (x 2) n ( ) (x ) = ( )n (x 2) n +. x P (W n,x) = ( ) n x (x ) (x 2) ( )n (x 2) n + ³ x = ( ) n x (x 2) ( ) n (x 2) n + = x (x 2) n +( ) n x (x 2). Solution 2. 3 2 n+ n n- W hav x ways to olor vrtx n +an, sin this vrtx is ajant to all othr vrtis, w hav x olors lt to olor vrtis, 2,..., n. Sin th graph inu y vrtis, 2,...,n is a Z n, w hav P (Z n,x ) ways to olor thm. 4 n-2 So P (W n,x)=xp (Z n,x ). 6 5 Not. Comining th two prvious solutions w gt th hromati polynomial o Z n,namly P (Z n,x ) = (x 2) n +( ) n (x 2), or quivalntly, P (Z n,x)=(x ) n +( ) n (x ).. [3.5 #7] (a) Thr is no spanning tr aus th graph is isonnt. (-h) Answrs may vary. () or () () () () (g) (h) 4
2. [3.5 #9] Lt F is a orst on n vrtis an with k omponnts. W prov that F has n k vrtis. By inition, th k omponnts T,T 2,...T k ar trs. Th numr o gs i o ah tr T i is on lss than th orrsponing numr o vrtis n i,thatis, i = n i. Also n + n 2 +... + n k = V (F ) = n. Thn E (F ) = + 2 +... + k = n +n 2 +... + n k = n k. 3. [3.5 #] Any tr with 5 gs maximizs th numr o vrtis. Thn th maximum numr o vrtis is 5 + = 6. (Any onnt graph on mor than 6 vrtis ontains a spanning tr with mor than 5 gs. So th graph itsl annot hav only 5 gs.) 4. [3.5 #2] Th minimum numr o gs is ahiv y a tr. (Not that lting an g o a iruit in a onnt graph rus th numr o gs an kps th graph onnt.) Th th numr o gs is 25 =24. 5. [3.5 #3] As in 3.5 #, ah onnt omponnt ontains a spanning tr. Th union o ths spanning trs is a orst an thus th graph ontains at last th numr o gs in this orst. So, y prolm 3.5 #9, a graph on n vrtis an with 3 onnt omponnts has at last n 3 gs. That is, n 3 5. Son is at most 8. Aorstwith3omponnts an 5 gs has xatly 8 vrtis, maximizing th numr o vrtis. Thn th maximum numr o vrtis is 8. 6. [3.5 #8] Lt T a tr on n vrtis. W prov P (T,x)=x (x ) n () y inution on n. I n =thn T = I an so P (T,x) =P (I,x)=x = x (x ) vriying (). Suppos Thorm 3.7 guarants that T has a vrtx v o gr. Lt α thginintinv. NotthatT 00 α is a tr on n vrtis an T 0 α is orm y T 00 α an an isolat vrtx. Thn P (T,x) = P Tα,x 0 P Tα,x 00 = P Tα,x 00 P (I,x) P Tα,x 00 = P Tα,x 00 (x ) By th inution hypothsis P (Tα,x)=x 00 (x ) (n ) = x (x ) n 2 an thus P (T,x)= x (x ) n. 7. [3.5 #9] Lt T any tr. Using P (T,x) =x (x ) n w hav P (T,) = 0 an P (T,2) = 2. Thn th hromati numr o T is 2 (th smallst positiv intgr valu o x or whih P (T,x) 6= 0). 8. [3.5 #20] Rall that a graph is ipartit i its st o vrtis an partition into two isjoint sts V an V 2 suh that all gs o th graph join on vrtx in V to on in V 2. Lt T a tr an u any o its vrtis. To show that /T is ipartit w partition th vrtis o T in two lasss V o an V vn as ollows. Thorm 3.8 guarants that or any othr vrtx v o T thr is a uniqu simpl hain rom u to v. Thnv V o i th lngth o this hain is o an v V vn ithlngthisvn(thnu V vn ). W laim that all gs o T must join vrtis in irnt lasss... Not that i w olor th vrtis o V o lu an th vrtis in V vn r, thn thr ar no monohromati gs an thus i T has at last two vrtis thn χ (T )=2. 5
9. [3.7 #a] In ah as th rows an olumns rprsnt th vrtis o th orrsponing igraph in alphatial orr rom top to ottom an lt to right. A (D )=A(D 2 )= A (D 3)= A (D 7 )= A (D 4 )= 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A (D 5 )= 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A (D 6 )= 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A (D 8 )= 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 20. [3.7 #2] (a) {a, } {, } {, } {, a} {, } {,h} {h, g} {g,} a 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 g 0 0 0 0 0 0 h 0 0 0 0 0 0 () {a, } {, } {, } {, } {,g} {g, h} {a, } {, } {, g} {, h} a 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 g 0 0 0 0 0 0 0 h 0 0 0 0 0 0 0 0 6
() {a, } {, } {, } {,} {g, h} {a, } {, } {, g} {, h} a 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 g 0 0 0 0 0 0 0 h 0 0 0 0 0 0 0 () {a, } {, } {, } {, } {,g} {g, h} {a, } {, } {, g} {, h} {, } {, g} a 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 g 0 0 0 0 0 0 0 0 h 0 0 0 0 0 0 0 0 0 0 () {a, } {, } {, } {, } {,g} {g,h} {a, } {, g} {, h} {, } a 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 g 0 0 0 0 0 0 0 h 0 0 0 0 0 0 0 0 () {a, } {a, } {a, } {, } {, } {, } {, } {, } {, g} {,g} a 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 g 0 0 0 0 0 0 0 0 7
(g) a g h i j k l {a, } {, } {, } {, } {g, h} {i, j} {j, k} {k, l} {a, g} {, h} {, i} {, j} {, k} {,l} 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 (h) {a, } {, } {a, } {, } {, } a 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2. [3.7 #3] x u v w y v y w 22. [3.7 #7] A (G )= A (G 3 )= u 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A (G 2 )= A (G 4 )= z 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x 8