This chapter covers special properties of planar graphs.

Transcription

1 Chptr 21 Plnr Grphs This hptr ovrs spil proprtis of plnr grphs Plnr grphs A plnr grph is grph whih n b rwn in th pln without ny gs rossing. Som piturs of plnr grph might hv rossing gs, but it s possibl to rrw th pitur to limint th rossings. For xmpl, lthough th usul piturs of K 4 n Q 3 hv rossing gs, it s sy to rrw thm so tht no gs ross. For xmpl, plnr pitur of Q 3 is shown blow. Howvr, if you fil roun with rwings of K 3,3 or K 5, thr osn t sm to b ny wy to limint th rossings. W ll s how to prov tht ths two grphs rn t plnr. 230

2 CHAPTER 21. PLANAR GRAPHS 231 C b A B Why shoul w r? Plnr grphs hv som intrsting mthmtil proprtis,.g. thy n b olor with only 4 olors. Also, s w ll s ltr, w n us fts bout plnr grphs to show tht thr r only 5 Pltoni solis. Thr r lso mny prtil pplitions with grph strutur in whih rossing gs r nuisn, inluing sign problms for iruits, subwys, utility lins. Two rossing onntions normlly mns tht th gs must b run t iffrnt hights. This isn t big issu for ltril wirs, but it rts xtr xpns for som typs of lins.g. burying on subwy tunnl unr nothr (n thrfor pr thn you woul orinrily n). Ciruits, in prtiulr, r sir to mnuftur if thir onntions liv in fwr lyrs Fs Whn plnr grph is rwn with no rossing gs, it ivis th pln into st of rgions, ll fs. By onvntion, w lso ount th unboun r outsi th whol grph s on f. Th bounry of f is th subgrph ontining ll th gs jnt to tht f n bounry wlk is los wlk ontining ll of thos gs. Th gr of th f is th minimum lngth of bounry wlk. For xmpl, in th figur blow, th lfthn grph hs thr fs. Th bounry of f 2 hs gs f, f,,, so this f hs gr 4. Th bounry of f 3 (th unboun f) hs gs b, f, f,,, b, so f 3 hs gr 6.

3 CHAPTER 21. PLANAR GRAPHS 232 b f b f Th righthn grph bov hs spik g stiking into th mil of f 1. Th bounry of f 1 hs gs bf, f,,,, b. Howvr, ny bounry wlk must trvrs th spik twi,.g. on possibl bounry wlk is bf, f,,,,, b, in whih is us twi. So th gr of f 1 in th righthn grph is 7. In suh ss, it my hlp to think of wlking long insi th f just nxt to th bounry, rthr wlking long th bounris thmslvs. Noti tht th bounry wlk for suh f is not yl. Suppos tht w hv grph with gs, v nos, n f fs. W know tht th Hnshking thorm hols, i.. th sum of no grs is 2. For plnr grphs, w lso hv Hnshking thorm for fs: th sum of th f grs is 2. To s this, noti tht typil g forms prt of th bounry of two fs, on to h si of it. Th xptions r gs, suh s thos involv in spik, tht ppr twi on th bounry of singl f. Finlly, for onnt plnr grphs, w hv Eulr s formul: v +f = 2. W ll prov tht this formul works Trs Bfor w try to prov Eulr s formul, lt s look t on spil typ of plnr grph: fr trs. In grph thory, fr tr is ny onnt grph with no yls. Fr trs r somwht lik norml trs, but thy on t hv signt root no n, thrfor, thy on t hv lr nstor-snnt orring to thir nots. A fr tr osn t ivi th pln into multipl fs, bus it osn t 1 You n sily gnrliz Eulr s formul to hnl grphs with mor thn on onnt omponnts.

4 CHAPTER 21. PLANAR GRAPHS 233 ontin ny yls. A fr tr hs only on f: th ntir pln surrouning it. So Eulr s thorm rus to v = 1, i.. = v 1. Lt s prov tht this is tru, by inution. Proof by inution on th numbr of nos in th grph. Bs: If th grph ontins no gs n only singl no, th formul is lrly tru. Inution: Suppos th formul works for ll fr trs with up to n nos. Lt T b fr tr with n + 1 nos. W n to show tht T hs n gs. Now, w fin no with gr 1 (only on g going into it). To o this strt t ny no r n follow wlk in ny irtion, without rpting gs. Bus T hs no yls, this wlk n t rturn to ny no it hs lry visit. So it must vntully hit n: th no t th n must hv gr 1. Cll it p. Rmov p n th g oming into it, mking nw fr tr T with n nos. By th inutiv hypothsis, T hs n 1 gs. Sin T hs on mor g thn T, T hs n gs. Thrfor our formul hols for T Proof of Eulr s formul W n now prov Eulr s formul (v + f = 2) works in gnrl, for ny onnt plnr grph. Proof: by inution on th numbr of gs in th grph. Bs: If = 0, th grph onsists of singl no with singl f surrouning it. So w hv = 2 whih is lrly right. Inution: Suppos th formul works for ll grphs with no mor thn n gs. Lt G b grph with n + 1 gs. Cs 1: G osn t ontin yl. So G is fr tr n w lry know th formul works for fr trs.

5 CHAPTER 21. PLANAR GRAPHS 234 Cs 2: G ontins t lst on yl. Pik n g p tht s on yl. Rmov p to rt nw grph G. Sin th yl sprts th pln into two fs, th fs to ithr si of p must b istint. Whn w rmov th g p, w mrg ths two fs. So G hs on fwr fs thn G. Sin G hs n gs, th formul works for G by th inution hypothsis. Tht is v + f = 2. But v = v, = 1, n f = f 1. Substituting, w fin tht So v ( 1) + (f 1) = 2 v + f = Som orollris of Eulr s formul Corollry 1 Suppos G is onnt plnr grph, with v nos, gs, n f fs, whr v 3. Thn 3v 6. Proof: Th sum of th grs of th fs is qul to twi th numbr of gs. But h f must hv gr 3. So w hv 3f 2. Eulr s formul sys tht v + f = 2, so f = v + 2 n thus 3f = 3 3v + 6. Combining this with 3f 2, w gt 3 3v So 3v 6. W n lso us this formul to show tht th grph K 5 isn t plnr. K 5 hs fiv nos n 10 gs. This isn t onsistnt with th formul 3v 6. Unfortuntly, this mtho won t hlp us with K 3,3, whih isn t plnr but os stisfy this qution. W n lso us this Corollry 1 to riv usful ft bout plnr grphs:

6 CHAPTER 21. PLANAR GRAPHS 235 Corollry 2 If G is onnt plnr grph, G hs no of gr lss thn six. Proof: This is lrly tru if G hs on or two nos. If G hs t lst thr nos, thn suppos tht th gr of h no ws t lst 6. By th hnshking thorm, 2 quls th sum of th grs of th nos, so w woul hv 2 6v. But orollry 1 sys tht 3v 6, so 2 6v 12. W n t hv both 2 6v n 2 6v 12. So thr must hv bn no with gr lss thn six. If our grph G isn t onnt, th rsult still hols, bus w n pply our proof to h onnt omponnt iniviully. So w hv: Corollry 3 If G is plnr grph, G hs no of gr lss thn six K 3,3 is not plnr Whn ll th yls in our grph hv t lst four nos, w n gt tightr rltionship btwn th numbrs of nos n gs. Corollry 4 Suppos G is onnt plnr grph, with v nos, gs, n f fs, whr v 3. n if ll yls in G hv lngth 4, thn 2v 4. Proof: Th sum of th grs of th fs is qul to twi th numbr of gs. But h f must hv gr 4 bus ll yls hv lngth 4. So w hv 4f 2, so 2f. Eulr s formul sys tht v + f = 2, so v + 2 = f, so 2 2v + 4 = 2f. Combining this with 2f, w gt tht 2 2v + 4. So 2v 4. This rsult lts us show tht K 3,3 isn t plnr. All th yls in K 3,3 hv t lst four nos. But K 3,3 hs 9 gs n 6 nos, whih isn t onsistnt with this formul. So K 3,3 n t b plnr.

7 CHAPTER 21. PLANAR GRAPHS Kurtowski s Thorm Th two xmpl non-plnr grphs K 3,3 n K 5 wrn t pik rnomly. It turns out tht ny non-plnr grph must ontin subgrph losly rlt to on of ths two grphs. Spifilly, w ll sy tht grph G is subivision of nothr grph F if th two grphs r isomorphi or if th only iffrn btwn th grphs is tht G ivis up som of F s gs by ing xtr gr 2 nos in th mil of th gs. For xmpl, in th following pitur, th righthn grph is subivision of th lfthn grph. A B A B F G E C C W n now stt our thorm prisly. Clim 57 Kurtowski s Thorm: A grph is nonplnr if n only if it ontins subgrph tht is subivision of K 3,3 or K 5. This ws prov in 1930 by Kzimirz Kurtowski, n th proof is pprntly somwht iffiult. So w ll just s how to pply it. For xmpl, hr s grph known s th Ptrsn grph (ftr nish mthmtiin nm Julius Ptrsn).

8 CHAPTER 21. PLANAR GRAPHS 237 A B b E C This isn t plnr. Th offning subgrph is th whol grph, xpt for th no B (n th gs tht onnt to B): A b E C This subgrph is subivision of K 3,3. To s why, first noti tht th no b is just subiviing th g from to, so w n lt it. Or,

9 CHAPTER 21. PLANAR GRAPHS 238 formlly, th prvious grph is subivision of this grph: A E C In th sm wy, w n rmov th nos A n C, to limint unnssry subivisions: E Now form th pitur bit n w s tht w hv K 3,3.

10 CHAPTER 21. PLANAR GRAPHS 239 E 21.8 Coloring plnr grphs On pplition of plnr grphs involvs oloring mps of ountris. Two ountris shring borr 2 must b givn iffrnt olors. W n turn this into grph oloring problm by ssigning grph no to h ountry. W thn onnt two nos with n g xtly whn thir rgions shr borr. This grph is ll th ul of our originl mp. Bus th mps r plnr, ths ul grphs r lwys plnr. Plnr grphs n b olor muh mor sily thn othr grphs. For xmpl, w n prov tht thy nvr rquir mor thn 6 olors: Proof: by inution on th numbr of nos in G. Bs: Th plnr grph with just on no hs mximum gr 0 n n b olor with on olor. Inution: Suppos tht ny plnr grph with < k nos n b olor with 6 olors. Lt G b plnr grph with k nos. 2 Two rgions touhing t point r not onsir to shr borr.

11 CHAPTER 21. PLANAR GRAPHS 240 By Corollry 3, G hs no of gr lss thn 6. Lt s pik suh no n ll it v. Rmov som no v (n its gs) from G to rt smllr grph G. G is plnr grph with k 1 nos. So, by th inutiv hypothsis, G n b olor with 5 olors. Bus v hs lss thn 6 nighbors, its nighbors r only using 5 of th vilbl olors. So thr is spr olor to ssign to v, finishing th oloring of G. It s not hr, but bit mssy, to upgr this proof to show tht plnr grphs rquir only fiv olors. Four olors is muh hrr. Wy bk in 1852, Frnis Guthri hypothsiz tht ny plnr grph oul b olor with only four olors, but it took 124 yrs to prov tht h ws right. Alfr Kmp thought h h prov it in 1879 n it took 11 yrs for nothr mthmtiin to fin n rror in his proof. Th Four Color Thorm ws finlly prov by Knnth Appl n Wolfgng Hkn t UIUC in Thy ru th problm mthmtilly, but wr lft with 1936 spifi grphs tht n to b hk xhustivly, using omputr progrm. Not vryon ws hppy to hv omputr involv in mthmtil proof, but th proof hs om to b pt s lgitimt Applition: Pltoni solis A ft ting bk to th Grks is tht thr r only fiv Pltoni solis: ub, ohron, ttrhron, ioshron, othron. Ths r onvx polyhr whos fs ll hv th sm numbr of sis (k) n whos nos ll hv th sm numbr of gs going into thm (). To turn Pltoni soli into grph, imgin tht it s m of strthy mtril. Mk smll hol in on f. Put your fingrs into tht f n pull siwys, strthing tht f rlly big n mking th whol thing flt. For xmpl, n othron (8 tringulr sis) turns into th following grph. Noti tht it still hs ight fs, on for h f of th originl soli, h with thr sis.

12 CHAPTER 21. PLANAR GRAPHS 241 Grphs of polyhr r slightly spil plnr grphs. Polyhr rn t llow to hv xtr nos prtwy long gs, so h no in th grph must hv gr t lst thr. Also, sin th fs must b flt n th gs stright, h f ns to b boun by t lst thr gs. So, if G is th grph of Pltoni soli, ll th nos of G must hv th sm gr 3 n ll fs must hv th sm gr k 3. Now, lt s o som lgbr to s why thr r so fw possibilitis for th strutur of suh grph. By th hnshking thorm, th sum of th no grs is twi th numbr of gs. So, sin th grs r qul to, w hv v = 2 n thrfor v = 2 By th hnshking thorm for fs, th sum of th f grs is lso twi th numbr of gs. Tht is kf = 2. So f = 2 k Eulr s formul sys tht v +f = 2, so v+f = 2+ >. Substituting th bov qutions into this on, w gt: k >

13 CHAPTER 21. PLANAR GRAPHS 242 iviing both sis by 2: k > 1 2 If w nlyz this qution, w isovr tht n k n t both b lrgr thn 3. If thy wr both 4 or bov, th lft si of th qution woul b 1. Sin w know tht n k r both 3, this implis tht on of th 2 two is tully qul to thr n th othr is som intgr tht is t lst 3. Suppos w st to b 3. Thn th qution boms > 1. So 3 k 2 1 > 1, whih mns tht k n t b ny lrgr thn 5. Similrly, if k is 3, k 6 thn n t b ny lrgr thn 5. This lvs us only fiv possibilitis for th grs n k: (3, 3), (3, 4), (3, 5), (4, 3), n (5, 3). Eh of ths orrspons to on of th Pltoni solis.