The University of Sydney MATH2969/2069. Graph Theory Tutorial 5 (Week 12) Solutions 2008

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1 Th Univrsity o Syny MATH2969/2069 Grph Thory Tutoril 5 (Wk 12) Solutions (i) Lt G th isonnt plnr grph shown. Drw its ul G, n th ul o th ul (G ). (ii) Show tht i G is isonnt plnr grph, thn G is onnt. Du tht (G ) is not isomorphi to G. G (i) G (G ) (ii) A onnt grph is on in whih thr xists pth twn ny pir o vrtis. Lt G isonnt plnr grph, n onsir its ul G. In G, thr is lrly pth twn ny two vrtis whih orrspon to s within on o th omponnts o G. Lt v th vrtx in G orrsponing to th ininit o G. Thn v is jnt to t lst on vrtx (orrsponing to ) in h omponnt o G. So i u n w r two vrtis o G whih orrspon to s in irnt omponnts o G, thn thr is pth rom u to w, vi v. Hn G is onnt. Sin G is onnt, its ul (G ) is lso onnt. ut G is isonnt, n so (G ) is not isomorphi to G. 2. A rtin polyhron hs s whih r tringls n pntgons, with h tringl surroun y pntgons n h pntgon surroun y tringls. I vry vrtx hs th sm gr, p sy, show tht 1 = 1 p Du tht p = 4, n tht thr r 20 tringls n 12 pntgons. Cn you onstrut suh polyhron? Stt th ul rsult. Lt T th numr o tringls, n P th numr o pntgons. Thn th numr o s, = T + P. Sin h tringl is oun y 3 gs, n h pntgon y 5, w hv 3T + 5P = 2. ut h g joins xtly on tringl n xtly on pntgon,

2 2 so 3T = 5P. Hn, T = 2 6 = 3, P = 2 10 = 5, = P + T = = Also, ssuming vry vrtx hs gr p, w hv pv = 2, or v = 2/p. Now sustitut into Eulr s ormul: Divi y 2 n simpliy: 2 p = 2. 1 = 1 p 7 30 Clrly, 1 must positiv numr, so w must hv 1 p 7 0, or p 30/7. 30 ut p must n intgr 3. So th only possil vlus or p r 3 or 4. I p = 3, = 10, ut thn v = 2/p = 20/3, whih is not n intgr. (Atully, it is sy nough to s tht it is not possil to put togthr tringls n pntgons in th wy spii, suh tht th gr o h vrtx is thr. Try rwing it n s!) I p = 4, = 60 n T = 20 n P = 12, s rquir. Suh polyhron my onstrut in two hlvs: W thn stith togthr long th gs whih orrspon in th lling. g It is rtinly possil to rw th ull grph, with littl mor ort: h i j j i h g Not tht th outsi rgion is on o th pntgons. Th ul rsult is otin y rpling vrtis y s n s y vrtis: A rtin polyhron hs vrtis whih hv grs 3 n 5, with h vrtx o gr 3 jnt to 3 vrtis o gr 5, n h vrtx o gr

3 3 5 jnt to 5 vrtis o gr 3. I vry is oun y th sm numr o gs, thn tht numr is 4, n thr r 20 vrtis o gr 3 n 12 vrtis o gr 5. (Eh vrtx in th ul grph orrspons to rgion in th originl grph, n n g inint to two jnt rgions in th originl is n g in th ul joining th orrsponing vrtis o th ul.) 3. Dtrmin th hromti numr o h o th ollowing grphs: g k g h j i k l g h j i g (i) (ii) (iii) (iv) A grph G is k-olourl i its vrtis n olour using k olours in suh wy tht no two jnt vrtis hv th sm olour. Th hromti numr o simpl grph G, writtn χ(g), is in to th smllst intgr k or whih G is k-olourl. Usul ts: I G ontins ( sugrph isomorphi to) K n, thn χ(g) n. I G ontins n o iruit, thn χ(g) 3. I G hs t lst two vrtis n no o iruits, thn χ(g) = 2. (i) This grph ontins svrl tringls, so χ(g) 3. On th othr hn w n sily in 3-olouring (g.,,, r;,, whit; g lu), so χ(g) = 3. (ii) Th grph hs no o yls, so χ(g) = 2. (A 2-olouring is sily oun (g.,,,, g, i whit,,,, h, j, k lk). (iii) Sin thr r tringls, χ(g) 3. W n in 3-olouring (g.,,, h, k r;,, i, l whit;,, g, j lu), so χ(g) = 3. (iv) Sin thr r tringls, (g., {,, }), χ(g) 3, ut is G 3-olourl? I so, without loss o gnrlity lt,, r, whit, lu rsptivly. Thn, jnt to oth n, must whit. Also, jnt to oth n, must lu. Also, jnt to oth n, must r. Thn, howvr, ourth olour is n or g, whih is jnt to, n. Hn χ(g) = For h o th ollowing grphs, wht os rooks Thorm tll you out th hromti numr o th grph? Fin th hromti numr o h grph. (i) Th omplt grph K 20. (ii) Th iprtit grph K 10,20. (iii) A yl with 20 gs. (iv) A yl with 29 gs.

4 (v) Th u grph Q 3. (vi) Th ul o Q 3. 4 Lt (G) th mximum o th grs o th vrtis o grph G. rooks Thorm stts tht, or ll grphs othr thn omplt grphs or o yls, th hromti numr o grph G, χ(g) (G). For omplt grphs, n or o yls, χ(g) = 1 + (G). (i) (K 20 ) = 19, χ(k 20 ) = = 20. (Clrly, h o th 20 vrtis must olour irntly or propr olouring.) (ii) (K 10,20 ) = 20, so rooks Thorm sys tht χ(k 10,20 ) 20. O ours, th hromti numr o ny iprtit grph is 2, so χ(k 10,20 ) = 2. (iii) In yl, ll vrtis hv gr 2. y rooks Thorm, th hromti numr o yl with n vn numr o gs is t most 2, n lrly 2 olours r rquir, so th hromti numr is qul to 2. (iv) y rooks Thorm, th hromti numr o yl with n o numr o gs is 3. (v) Q 3 is rgulr o gr 3, n so y rooks Thorm χ(q 3 ) 3. In t, Q 3 is iprtit grph with hromti numr 2. (vi) Th ul o Q 3 is th othron, in whih vry vrtx hs gr 4, so y rooks Thorm its hromti numr is t most 4. Not tht th grph ontins tringls, so its hromti numr is t lst 3. It is sy to in 3-olouring, n so χ(th othron) = (i) Dtrmin th minimum numr o olours rquir to olour th s o Q 3 in suh wy tht joining s hv irnt olour. (ii) pt prt (i) or th ul o Q 3. Th grph Q 3 is th polyhrl grph orrsponing to u. It is lr tht t lst 3 olours r n to olour th s o u so tht no two joining s hv th sm olour. Th igrm low inits on wy in whih to olour th s o th u using thr olours (whr =r, =lu, G=grn). Th othron is th ul o th u. For n othron only two olours r nssry. Th igrm shows 2-olouring o th s o n othron. G G Not tht th minimum numr o olours rquir to olour th s o th u is qul to th hromti numr o th othron, n vi vrs.

5 5 6. Show tht simpl onnt plnr grph with 17 gs n 10 vrtis nnot proprly olour with two olours. (Hint: Show tht suh grph must ontin tringl.) Suppos suh grph h no tringls. Thn, sin it is not tr, h must oun y t lst 4 gs, n so 4 2, or 2. Howvr, y Eulr s ormul, = 2 v + = = 9, so 2 = 2 9 > 17 = ontrition. Hn th grph hs t lst on tringl n so is not 2-olourl. 7. Lt T tr with t lst 2 vrtis. Prov tht χ(t)=2. Not tht w n t lst two olours to proprly olour T, so i w prov tht T is 2-olourl thn χ(t)=2. Us inution on th numr o vrtis. I thr r 2 vrtis, thn th tr is lrly 2-olourl. Suppos tht tr with k vrtis is 2-olourl. Lt T tr with k + 1 vrtis, n rmov rom T on o its vrtis with gr 1. (Evry tr hs t lst two vrtis with gr 1.) This lvs tr with k vrtis, whih is 2-olourl y th inution hypothsis. Colour th smllr tr with 2 olours, rstor th rmov vrtx n olour it with th olour not us on th (on) vrtx to whih it is jnt. Hn T is 2-olourl, n th rsult ollows. 8. Hurt kps iv vritis (A,, C, D, E) o snks in oxs in his prtmnt. Som vritis ttk othr vritis, n n t kpt togthr. In th tl, n strisk inits tht vritis n t kpt togthr. Wht is th minimum numr o oxs n? Construt grph with vritis o snks s vrtis, n gs joining vritis whih nnot kpt togthr. Thn th minimum numr o oxs rquir is th hromti numr o th grph. Sin th grph ontins tringls, t lst 3 olours r n. A 3-olouring n oun, s shown. Hn th minimum numr o oxs rquir is thr. E LUE D WHITE A C D E A C D E A ED WHITE C LUE 9. Dtrmin th numr o wys in whih h o th ollowing grphs n proprly olour, givn λ irnt olours. (i) Th omplt grph K 6. (ii) Th str grph K 1,5. (iii) Th linr grph L 6.

6 6 ll tht olouring o grph is n ssignmnt o olour (rom som st o vill olours) to h vrtx, with th onition tht no two jnt vrtis my riv th sm olour. ll lso tht th hromti polynomil o grph G, writtn P G (λ), is th untion in th on intgr vril λ, suh tht i λ is ny intgr 0, P G (λ) is th numr o wys o olouring G i λ olours r vill. Th untion P G (λ) is polynomil with intgr oiints. (i) G = K 6 = With λ olours vill, n strting with no vrtis olour, olour ny vrtx (λ hois), thn nothr vrtx (only λ 1 hois, th son vrtx ing jnt to th irst), thn nothr (only λ 2 hois, th thir vrtx ing jnt to oth th prvious), t., thr ing λ 5 wys o olouring th inl vrtx. (Not tht t h stp th numr o hois os not pn on th prviously m hois.) y th prout prinipl, th totl numr o olourings is thror P G (λ) = λ (6) = λ(λ 1)(λ 2)(λ 3)(λ 4)(λ 5). (ii) G = K 1,5 = = First olour th vrtx o gr 5 (λ hois), thn h o th othr 5 vrtis (λ 1 hois h, s h is jnt to just on lry olour vrtx). Hn P G (λ) = λ(λ 1) 5. (O ours, K 1,5 is tr with 6 vrtis, n i T is tr with n vrtis thn P T (λ) = λ(λ 1) n 1.) (iii) G = L 6 = This is nothr tr on 6 vrtis, n so th polynomil is s in prt (ii). Or on n pro s ollows: Colour th vrtis in orr rom th lt, strting with gr 1 vrtx (λ hois), n ontinuing with n jnt vrtx, t. (λ 1 hois h). Hn P G (λ) = λ(λ 1) 5. Not tht tking th vrtis in rully hosn orr ws importnt hr. Strting with th two gr 1 vrtis, or xmpl (λ hois h), woul vntully ln us in iiultis, s th lst vrtx to olour woul jnt to 2 lry olour vrtis, n th numr

7 7 o hois woul pn on whthr thy h n olour th sm or irntly. 10. (i) Fin th hromti polynomils o h o th six onnt simpl grphs on our vrtis. (ii) Vriy tht h o th polynomils in (i) hs th orm λ 4 λ 3 +λ 2 λ whr is th numr o gs n n r positiv onstnts. G P G (λ) = L 4 λ(λ 1) 3 = λ 4 3λ 3 + 3λ 2 λ = K 1,3 λ(λ 1) 3 = λ 4 3λ 3 + 3λ 2 λ = C 4 λ(λ 1)(λ 2 3λ + 3) = λ 4 4λ 3 + 6λ 2 3λ λ(λ 1) 2 (λ 2) = λ 4 4λ 3 + 5λ 2 2λ λ(λ 1)(λ 2) 2 = λ 4 5λ + 8λ 2 4λ = K 4 λ(λ 1)(λ 2)(λ 3) = λ 4 6λ λ 2 6λ All th polynomils r sily oun, xpt tht or C 4. As w sw in lturs, th polynomil or C 4 n oun y ing togthr th numr o wys o olouring C 4 in whih pir o non-jnt vrtis r olour th sm, n th numr o wys in whih tht pir o vrtis is olour irntly. This mtho givs P C4 (λ) = λ(λ 1) 2 + λ(λ 1)(λ 2) 2. Th vriitions rquir in (ii) r y insption. 11. Fin th hromti polynomils o K 1,n, K 2,n n K 3,n. x G = K 1,n =... y 1 y 2 y n Th grph is tr on (n + 1) vrtis, so P G (λ) = λ(λ 1) n. x 1 x 2 G = K 2,n =... y 1 y 2 y n Colour x 1 n x 2 irst, ut thn, to olour y 1, y 2,..., y n, w n to onsir two mutully xlusiv ss: Cs 1: x 1 n x 2 hv n olour irntly: λ hois or x 1 thn λ 1 hois or x 2. In this s h o y 1, y 2,..., y n my thn olour in λ 2 wys: λ(λ 1)(λ 2) n wys ll togthr.

8 8 Cs 2: x 1 n x 2 hv n givn th sm olour: λ hois or tht olour. In this s h o y 1, y 2,..., y n my olour in λ 1 wys: λ(λ 1) n wys ll togthr. Using th ition prinipl or omining mutully xlusiv ss, w gt: P G (λ) = λ(λ 1)(λ 2) n + λ(λ 1) n. x 1 x 2 x 3 G = K 3,n =... y 1 y 2 y n Hr, i olouring x 1, x 2, x 3 irst, thr r thr mutully xlusiv ss to onsir: Cs 1: All o x 1, x 2, x 3 hv n givn irnt olours: λ hois or x 1, λ 1 or x 2 n λ 2 or x 3. In this s thr r λ 3 hois or h o y 1, y 2,..., y n : λ(λ 1)(λ 2)(λ 3) n wys ll togthr. Cs 2: Two o x 1, x 2, x 3 hv n givn th sm olour, th othr irnt olour. Not tht thr r ( 3 2) = 3 hois or whih two hv th sm olour, λ or wht olour tht is, thn λ 1 or th irnt olour. In this s thr r λ 2 hois or h o y 1, y 2,..., y n : 3λ(λ 1)(λ 2) n wys ll togthr. Cs 3: All o x 1, x 2, x 3 hv n givn th sm olour: λ hois or tht olour. In this s thr r λ 1 hois or h o y 1, y 2,..., y n : λ(λ 1) n wys ll togthr. Comining ths y th ition prinipl: P G (λ) = λ(λ 1)(λ 2)(λ 3) n + 3λ(λ 1)(λ 2) n + λ(λ 1) n. 12. Stt two rution ormuls or hromti polynomils. Us whihvr sms pproprit to lult th hromti polynomil or h o th two givn grphs. Also trmin th hromti numr o h grph. For simpl grph G, th two orms r P G (λ) = P G+ (λ) + P G (λ) (1) P G (λ) = P G (λ) P G (λ) (2) whr, in (1), G + is G with nw g (onnting two vrtis whih r non-jnt in G), n, in (2), G is G with n xisting g rmov. In h s G is G ontrt long, i.., is th rsult o intiying th two vrtis join y, n sy, s singl vrtx. Any vrtx jnt to ithr or or oth in G is jnt to in G, n ll othr jnis r prsrv.

9 9 W know th hromti polynomils o th omplt grph K n, n o ny n-vrtx tr T n : λ(λ 1) (λ n + 1) n λ(λ 1) n 1 rsptivly. So it is pproprit to us (1) on nrly omplt grphs, n (2) on grphs with w iruits, s in h s only w pplitions r n to ru to sums/irns o known polynomils. Not tht w r using h grph igrm to mn th hromti polynomil o th grph in th ollowing. Using (1): = + = ( + ) + Using (2): = = K 6 + K 5 + K 5 = λ(λ 1)(λ 2)(λ 3)(λ 4)(λ 5) + 2λ(λ 1)(λ 2)(λ 3)(λ 4) = λ(λ 1)(λ 2)(λ 3) 2 (λ 4). = ( ) ( ) ( ) ( ) = T 6 T 5 + = T 6 T 5 + ( ) T 5 + T 4 T 3 = T 6 2T 5 + 2T 4 T 3 K 3 = λ(λ 1) 5 2λ(λ 1) 4 + 2λ(λ 1) 3 λ(λ 1) 2 λ(λ 1)(λ 2) = (vntully) λ(λ 1)(λ 2) 2 (λ 2 2λ + 2). ll tht th hromti numr o grph is th minimum numr o olours with whih th vrtis o th grph my proprly olour (i.., with no pir o jnt vrtis hving th sm olour). Th irst grph ontins K 5 s sugrph, so ns 5 olours. Insption shows sixth olour is not n. Hn th hromti numr is 5. (Altrntivly, osrv tht 5 is th irst positiv intgr whih is not zro o th hromti polynomil.) Th son grph ontins o lngth iruits, so ns 3 olours. Insption shows 4th olour is not n. Hn th hromti numr is 3. (Altrntivly, osrv tht 3 is th irst positiv intgr whih is not zro o th hromti polynomil.)

10 13. Fin th hromti polynomil o C 5, th yl with 5 vrtis. Using th ormul P G (λ) = P G (λ) P G (λ) on C 5, w otin 10 P C5 (λ) = P T5 (λ) P C4 (λ) = P T5 (λ) [P T4 (λ) P C3 (λ)] (T n nots tr on n vrtis.) = λ(λ 1) 4 λ(λ 1) 3 + λ(λ 1)(λ 2) = λ(λ 1)(λ 2)(λ 2 2λ + 2) 14. Explin why th hromti polynomil P G (λ) o plnr grph G nnot ontin trm (λ k) or ny k 4. I P G (λ) ontins th trm (λ k), thn P G (k) = 0, n thr r zro wys to olour th grph with k olours. In othr wors, th grph is not k-olourl. Sin ny plnr grph is 4-olourl, n thror k-olourl or ll k 4, its hromti polynomil nnot ontin trm (λ k) or ny k Fin th hromti inx (or g-hromti numr) o th grph G, whr G is: () () ll tht grph G is proprly g-olour whn its gs r ssign olours n no two gs inint t th sm vrtx hv th sm olour. ll lso tht th hromti inx o G (lso ll th g-hromti numr o G), χ (G), is th minimum numr o olours rquir to g-olour G proprly. A thorm o Vizing stts tht or simpl grph G, χ (G) is ithr or + 1, whr is th mximum vrtx gr o G. To trmin whih o ths two is orrt, w might in wy o g-olouring G proprly with just olours, or prov it is impossil with just olours. () Hr G hs mximum gr 3 (ll vrtis hving gr 3 xpt, whih hs gr 2). So oviously t lst 3 olours r n to g olour G proprly (g., th thr gs inint to rquir thr irnt olours). y Vizing s thorm, χ (G) is ithr 3 or 4. Try to olour G proprly with just 3 olours:, G n, sy. Without loss o gnrlty, [, ] =, [, ] = G n [, ] =. Hn [, ] =, so [, ] = n [, ] = G. Now ourth olour is n to olour [, ] proprly. Hn χ (G) = 4.

11 11 () Vizing s thorm hr shows χ (G) is ithr 4 (th mximum vrtx gr) or 5. Try using just 4 olours:,, G n Y, sy. Without loss o gnrlty, [, ] =, [, ] = G, [, ] = n [, ] = Y. Thn [, ] nnot G or, so is ithr or Y y symmtry w n ssum without loss o gnrlity [, ] =. Hn [, ] = G, n so [, ] = Y n thn [, ] =, n now ith olour is n or [, ]. Hn χ (G) = Fin th hromti inx o th u, n o th othron. Th grphs r s shown: y Vizing s thorm, th u grph hs hromti inx 3 or 4 (3 ing th mximum vrtx gr). Similrly th othrl grph hs hromti inx 4 or 5. In t th lowr vlu is orrt in h s, s th ollowing g olourings (thik, thin, ott, sh) show: