CCS Disrete II Professor: Pri Brtlett Leture 4: Grph Theory n the Four-Color Theorem Week 4 UCSB 2015 Through the rest of this lss, we re going to refer frequently to things lle grphs! If you hen t seen grphs efore, we efine them here: Definition. A grph G with n erties n m eges onsists of the following two ojets: 1. set V = { 1,... n }, the memers of whih we ll G s erties, n 2. set E = {e 1,... e m }, the memers of whih we ll G s eges, where eh ege e i is n unorere pir of istint elements in V, n no unorere pir is repete. For gien ege e = {, w}, we will often refer to the two erties, w ontine y e s its enpoints. Exmple. The following pir (V, E) efines grph G on fie erties n fie eges: V = {1, 2, 3, 4, 5}, E = {{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 1}}. Something mthemtiins like to o to quikly represent grphs is rw them, whih we n o y tking eh ertex n ssigning it point in the plne, n tking eh ege n rwing ure etween the two erties represente y tht ege. For exmple, one wy to rw our grph G is the following: 1 5 2 4 3 We oul lso rw our grph like this: 1 4 3 2 5 In generl, ll we re out for our grphs is their erties n their eges; we on t usully re out how they re rwn, so long s they onsist of the sme erties onnete i the sme eges. Also, we usully will not re out how we lel the erties of grph: i.e. we will usully skip the lelings on our grphs, n just rw them s erties onnete y eges. Some grphs get speil nmes: 1
Definition. The yle grph on n erties, C n, is the grph on the ertex set { 1, 2,... n } with ege set E(C n ) = {{ 1, 2 }, { 2, 3 },... { n 1, n }, { n, 1 }}. The yle grphs C n n e rwn s n-gons, s epite elow:... Definition. The pth grph on n erties, P n, is the grph on the ertex set { 1, 2,... n } with ege set E(C n ) = {{ 1, 2 }, { 2, 3 },... { n 1, n }}. The pth grphs P n n e rwn s pths of length n, s epite elow:... Definition. The omplete grph K n. The omplete grph on n erties, K n, is the simple grph on the ertex set { 1, 2,... n } tht hs eery possile ege: in other wors, E(K n ) = {{ i, j } : i j}. We rw seerl of these grphs elow: Eery ertex in K n hs egree n 1, s it hs n ege onneting it to eh of the other n 1 erties; s well, K n hs n(n 1)/2 eges in totl in it, y the egree-sum formul. (Expliitly: eery ertex hs egree n 1 n there re n erties, therefore the sum of the egrees of K n s erties is n(n 1). We e shown tht this quntity is twie the numer of eges in the grph; iiing y 2 then tells us tht the numer of eges in K n is n(n 1)/2, s lime.) Definition. The omplete iprtite grph K n,m. The omplete iprtite grph on n + m erties with prt sizes n n m, K n,m, is the following grph: V (K n,m ) = { 1, 2,... n, w 1, w 2,... w m }. E(K n,m ) onsists of ll of the eges etween the n-prt n the m-prt; in other wors, E(K n,m ) = {( i, w j ) : 1 i n, 1 j m}. The erties i ll he egree m, s they he preisely m eges leing them (one to eery ertex w j ); similrly, the erties w j ll he egree n. By either the egree-sum formul or just ounting, we n see tht there re nm eges in K n,m. 2
Definition. Gien grph G n nother grph H, we sy tht H is sugrph of G if n only if V (H) V (G) n E(H) E(G). Definition. Gien grph G, we ll G onnete if for ny two erties x, y V (G), there is pth tht strts t x n ens t y in our grph G. Definition. If grph G hs no sugrphs tht re yle grphs, we ll G yli. A tree T is grph tht s oth onnete n yli. In tree, lef is ertex whose egree is 1. Exmple. The following grph is tree: 1 The Four-Color Theorem Grph theory got its strt in 1736, when Euler stuie the Seen Briges of Königserg prolem. Howeer, I lim tht it first lossome in ernest in 1852 when Guthrie me up with the Four-Color Prolem. Theorem. Tke ny mp, whih for our purposes is wy to prtition the plne R 2 into olletion of onnete regions R 1,... R n with ontinuous ounries. There is some wy to ssign eh region R i to olor in the set {R, G, B, Y }, suh tht if two regions R i, R j re touhing (i.e. they shre some nonzero length of ounry etween them,) then those two regions must reeie ifferent olors. We re going to proe this theorem in this lss! First, some kgroun: Definition. Tke ny mp M. We n turn this into grph s follows: Assign to eh region R i ertex i. Connet i to j with n ege if the regions R i, R j re touhing. We ll this grph the ul grph to M. We gie n exmple here: Exmple. Consier the following mp: 3
This mp onsists of 14 regions. If you ount, you n see tht the figure rwn onsists thirteen tringles; s well, we he the outer region onsisting of eerything else left oer, whih forms ery strnge 15-gon. Now, tke eh region, n ssign to it ertex. As well, onnet two regions shring orer with n ege: this will gie you the following grph, with eges gien y the she tel lines n erties gien y the yellow ots: Note how we he rwn the eges so tht they onnet two jent ountries y treling through the orer tht they shre! This osertion is useful to rell when thinking out our seon efinition: Definition. We sy tht grph G is plnr if we n rw it in the plne so tht none of its eges interset. Sometimes, it will help to think of plnrity in the following wy: 4
Definition. We ll onnete grph G plnr if we n rw it on the sphere S 2 in the following fshion: Eh ertex of G is represente y point on the sphere. Eh ege in G is represente y ontinuous pth rwn on the sphere onneting the points orresponing to its erties. These pths o not interset eh other, exept for the triil sitution where two pths shre ommon enpoint. We ll suh rwing plnr emeing of G on the sphere. It is not hr to see tht this efinition is equilent to our erlier efinition of plnrity. Simply use the stereogrphi projetion mp (rwn elow) to trnslte ny grph on the plne to grph on the sphere: By rwing lines from the north pole (0,0,1) through points either in the xy-plne or on the surfe of the sphere, we n trnslte grphs rwn on the sphere (in re) to grphs rwn in the plne (in yellow.) Definition. For ny onnete plnr grph G, we n efine fe of G to e onnete region of R whose ounry is gien y the eges of G. For exmple, the following grph hs four fes, s lele: f 1 f 3 f 2 f 4 Notie tht we lwys he the outsie fe in these rwings, whih n e esy to forget out when rwing our grphs on the plne. This is one reson why I like to think out these grphs s rwn on the sphere; in this setting, there is no outsie fe, s ll of the fes re eqully nturl to work with. 5
f 4 f 4 f 1 f 2 f 3 f 1 f 3 f 2 This osertion hs nie ompnying lemm: Lemm. Tke ny onnete plnr grph G, n ny fe F of G. Then G n e rwn on the plne in suh wy tht F is the outsie fe of G. Proof. Tke plnr emeing of G on the unit sphere. Rotte this rwn-upon sphere so tht the fe F ontins the north pole (0, 0, 1) of the sphere. Now, perform stereogrphi projetion to rete plnr emeing of G in R 2. By onstrution, the fe F is now the outsie fe, whih proes our lim. It ers noting tht not ll grphs re plnr: Proposition. The grph K 5 is not plnr. Proof. Drw 5-yle on the sphere. If the eges of this 5-yle o not interset eh other, then the resulting pentgon prtitions the sphere into two prts, eh prt of whih is oune y this pentgon. Tke either one of these prts; notie tht within tht prt, we n rw t most two noninterseting eges onneting nonjent erties in tht prt. Consequently, it is impossile to rw the itionl 5 eges require to rete K 5 without using oerlpping eges. Therefore it is impossile to fin plnr emeing of K 5 on the sphere, s lime! For mps, something you my he notie is the following: Osertion. The ul grph to ny mp M is onnete n plnr. Proof. On the homework! The reson we re out this is tht it gies us the following more grph-theoreti wy to esrie the four-olor theorem: Theorem. Tke ny onnete plnr grph on finitely mny erties. There is wy to ssign eh of its erties one of the four olors {R, G, B, Y } suh tht no ege in this grph hs oth enpoints olore the sme olor. In generl, this onept of oloring omes up ll the time in grph theory! We gie it nme here: 6
Definition. A grph G is lle k-olorle if there is olletion of k istint olors tht we n mp the erties of G to, so tht no ege in G hs oth enpoints olore the sme olor. Gien grph G, we efine the hromti numer of G, χ(g), s the smllest numer k suh tht G is k-olorle. This gies us one lst rephrsing of the four-olor theorem: Theorem. If G is onnete plnr grph on finitely mny erties, then χ(g) 4.... So. Before we n strt Kempe s proof, we nee one lst it of kgroun, whih is the onept of Euler hrteristi: Theorem. (Euler hrteristi.) Tke ny onnete grph tht hs een rwn in R 2 s plnr grph. Then, if V is the numer of erties, E is the numer of eges, n F is the numer of fes in this grph, we he the following reltion: V E + F = 2. Proof. We will tully proe stronger lim: we will show tht ny plnr multigrph ( grph, ut where we llow multiple eges etween erties, n lso eges tht strt n en t the sme ertex) stisfies the V E + F = 2 formul. For the rest of this proof, we will ssume tht grph n multigrph re synonymous; one we re one with this proof, though, we will stop ssuming this. f 5 1 f 6 6 f 4 f 1 f 2 f 3 2 3 5 4 We proee y inution on the numer of erties. Suppose tht V = 1. Then our grph looks like something of the following form:......... 7
I lim tht V E + F = 2 for ny of these grphs, n proe it y seon inution on the numer of eges. For zero-ege grph, this is esy; we he one ertex, no eges n one fe, we he V E + F = 1 0 + 1 = 2. Now, ssume i inution tht eery one-ertex multigrph on n eges hs V E + F = 2. Tke ny grph on one ertex with n + 1 eges. Pik one of these eges, n look t it. I lim tht this ege orers extly two fes. To see why, tke ny ege, n ssign n orienttion to it (i.e. if our ege is {x, y}, then orient the ege so tht we trel from x to y.) If you o this, then our ege hs two sies, the left- n right-hn sies, if we trel long it i this orienttion. y y (left) (right) x (left) (right) x There re two possiilities, s rwn oe: either the left- n right-hn sies re ifferent, or they re the sme. This tells us tht our ege either orers one or two fes! To see tht we he extly two, we now rell tht our ege (euse our grph hs extly one ertex) must strt n en t the sme ertex. In other wors, it is lose loop: i.e. its outsie is ifferent from its insie! In other wors, our left- n right-hn sies re ifferent, n our ege seprtes two istint fes. Therefore, eleting this ege oes the following things to the grph: it ereses our ege ount y 1, n lso ereses our fe ount y 1 (s we merge two fes when we elete this ege.) In other wors, eleting this ege oes not hnge V E + F! But y inution we know tht V E + F = 2 for ll 1-ertex grphs on n eges, whih is wht we get if we elete this ege from n + 1-ege grph. So we re one! This settles our se se for our lrger inution on V, the numer of erties. We now go to the seon phse of n inutie proof: we show how to reue lrger ses to smller ses! To o this, onsier the following opertion, lle ege ontrtion. Tke ny ege with two istint enpoints. Delete this ege, n omine its two enpoints together: this gies us new grph! We rw exmples of this proess elow: we strt with grph on six erties, n ontrt one y one the eges lele in re t eh step. 1 1 1 6 2 6 2 6 2 3 5 4 5 4 5 1 6 2 6 2 2 8
Contrting n ege ereses the numer of erties y 1 t eh step, s it squishes together two jent erties into one ertex. It lso ereses the numer of eges y 1 t eh step, s we re ontrting n ege to point! Finlly, it neer hnges the numer of fes; if two fes were istint efore this proess hppens, they sty istint, s we re not mking ny uts in ny of our ounries (n inste re just shrinking them prtilly it!) But this mens tht V E+F is still onstnt! Therefore, y inution, if V E+F hols for eery n-ertex multigrph, it hols for ny n + 1-ertex multigrph y just ontrting n ege! This finishes our inution, n thus our proof. Using this theorem, we n proe the following useful lemm, whih is the only prt of the Euler hrteristi property tht we nee for our grph: Lemm. Tke ny onnete plnr grph G. Then there is some ertex in our grph with egree t most 5. Proof. We proee y ontrition. Assume tht eery ertex hs t lest egree 6; we will rete ontrition to the lim tht V E + F = 2. First, onsier the sum G eg(). On one hn, this is twie the numer of eges in G: this is euse eh ege shows up twie in this sum (one for eh enpoint when we re lulting eg().) On the other hn, if eh ertex hs egree t lest 6, we he G eg() G 6 = 6V. Consequently, we he 2E 6V, n therefore E/3 V. Similrly: notie tht eery fe F of our plnr grph must he t lest three eges ouning it, euse our fes re me out of eges in our grph. Also, if we sum oer ll fes the numer of eges in eh fe, we get gin t most twie the numer of eges; this is euse eh ege is in t most two fes (s isusse erlier!) Therefore, we he 2E f G feeg(f) f G 3 = 3F, n therefore tht 2E/3 F. Therefore, we he 2 = V E + F E/3 E + 2E/3 = 0, whih is lerly impossile. Therefore, we he ontrition, n n onlue tht our initil ssumption tht ll erties he egree t lest 6 is flse! 2 Kempe s Proof With this nottion set up, Kempe s proof of the four-olor theorem is tully firly strightforwr! We gie it here. 9
Proof. We proee y ontrition. Assume not: tht there re onnete plnr grphs on finitely mny erties tht nee t lest 5 olors to e olore properly. Consequently, there must e some smllest onnete plnr grph G, in terms of the numer of its erties, tht nees t lest fie olors to olor its erties! Pik suh grph G. Notie tht if we remoe ny ertex from G, we he grph on smller numer of erties thn G. Consequently, the grph G \ {} n e olore with four olors! Let e the ertex in G with egree t most 5. Delete from G: this lees us grph tht we n four-olor. Do so. Our gol is now the following: to k in n (y possily hnging the oloring of G \ {}) gie one of our four olors, so tht we he four-oloring of G! This will proe tht our initil ssumption tht G n exist tht nees fie olors is flse, n therefore proe our theorem. We proee y ses, onsiering s possile egrees. 1. hs egree 1, 2 or 3. In these ses, note tht when we k in, it is jent to t most three olors! So there is some fourth olor left oer. Gie tht olor. G-{} 2. hs egree 4. In this se, there re two possiilities: In the four neighors,,, of, some olor is not use. In this se, we re in the sme kin of sitution s oe: just olor with the olor tht oesn t show up in its neighors? In the four neighors,,, of, eh olor is use extly one. So, up to the nmes of the olors, we re in the following sitution: First, notie tht without losing ny generlity we my ssume tht there re eges. To see why, notie the following: () Aing these eges oes not hnge the ssume property tht χ(g) 5, s extr eges only mkes it hrer for us to olor grph. 10
() We n lwys rw in these eges if they o not exist: for exmple, if the ege {, } i not exist, we oul it in without reking plnrity y simply rwing pth tht is ery lose to the two eges {, } n {, }. This pth will not ross other eges, s there re no other eges leing. () Therefore, euse we he presere χ(g) 5 n G s plnrity, we n put these eges into G without hnging ny of our rguments thus fr! By the rgument oe, we n now ssume our grph looks like the following: Now, o the following: for ny two olors C 1, C 2, let G C1,C 2 enote the sugrph of G gien y tking ll of the erties in G tht re olore either C 1 or C 2, long with ll of the eges tht onnet C 1 erties to C 2 erties. Look t the re-yellow sugrph G RY. In this grph, there re two possiilities: () There is no pth from to in this grph. In other wors, efine A RY s the sugrph of G RY gien y tking ll of the G RY erties tht he pths to, long with ll of the eges in our grph etween suh erties. A RY 11
Suppose tht we swith the olors re n yellow in the sugrph A RY. Does it rete ny issues with our oloring? Let s hek. No ege etween two erties in A RY is roken (i.e. hs oth enpoints me the sme olor) y this proess; efore it h one re n one yellow enpoint, n now it hs one yellow n one re enpoint. As well, no ege tht inoles no erties in A RY is roken y this proess, s we i not hnge the olors of either of their enpoints! A RY A YR Finlly, onsier ny ege with one enpoint in A RY n nother enpoint not in A RY. In orer for this ege to he one enpoint in A RY n nother not in A RY, one enpoint must e re or yellow (the enpoint in our set) n the other must e green or lue (the enpoint not in our set!) So if we swith re n yellow in A RY, this ege is lso not roken! No eges re roken y this swp; therefore we still he li oloring. Furthermore, in this oloring, hs no neighors tht re re; so we n olor re n he four-oloring of our entire grph G! () Alterntely, () oes not hppen. In this se, there is pth from to me entirely of re-yellow erties linke y eges. In this se: look t the grph G GB. D GB 12
In prtiulr, notie tht there nnot e pth from to long green-lue eges, euse our grph is plnr n ny suh pth woul he to ross our re-yellow eges! Therefore, we n efine D GB to e the olletion of ll of the G GB erties tht he pths to, long with the eges in our grph etween suh erties. As note oe, / B GB. Now, swith the olors G n B in D GB! This uses no onflits, y extly the sme rgument s oe, n yiels grph where hs no green neighor; therefore, we n gie the olor green, n he proper four-oloring s esire. D GB D BG 3. hs egree 5. Agin, s efore, we n ssume tht ll four of the olors in our grph our on G s neighors, euse if they o not we n simply gie whiheer olor is missing. Agin, s efore, we n ssume tht the neighors of re onnete y the following pentgonl struture.: e e This is euse of the following: Aing eges to our grph will neer mke it esier to olor grph: ll they o is gie us more onitions on wht erties he to he ifferent olors, whih only mkes oloring hrer. Furthermore we n these eges without reking plnrity y simply rwing them ritrrily lose to the -eges. Up to symmetry n olorings, then, we re in the following sitution: 13
e This is euse we he to repet one olor (so it might s well e re,) we he to use ll of the other olors (so we he green, lue n yellow in some orer,) re nnot our on two jent erties (euse there re eges etween jent erties,) n therefore up to rottion n flipping we he the oe. Do the following: () First, look t the G GB sugrph. Either the ertex is not onnete to in this sugrph, in whih se we n o the swithing-trik tht we isusse erlier. Otherwise, is onnete to in G GB, n we he green-lue hin from to. () Now, look t the G GY sugrph. Similrly, either the ertex is not onnete to e in this sugrph, in whih se we n o the swithing-trik tht we isusse erlier, or it is, n we he green-yellow hin from to e. If we were le to swith in either of the two ses oe, then hs only three olors mongst its neighors, n we n olor it with whteer olor remins. Otherwise, we re in the following se: e Do the following: () First, look t the G RB sugrph. Beuse of the green-yellow hin, the erties n re not onnete to eh other. Therefore, we n swith re n lue in the -onnete prt of this sugrph! 14
() Now, look t the G RY sugrph. Beuse of the green-lue hin, the erties n e re not onnete to eh other. Therefore, we n swith re n yellow in the -onnete prt of this sugrph! e e This yiels grph where hs no re neighor: onsequently, we n olor re, whih gies us proper four-oloring! This proes our lim. 3 Plot Twist This proof... is in ft Kempe s fmous flwe proof of the four-olor theorem, whih stoo for 11+ yers efore eing isproen! In prtiulr, it ws isproen: i.e. the proof you e re oe is flse! Guess wht your HW is for toy? 4 Grphs: Terminology We trnsition from the four-olor theorem to more reful isussion of the efinitions n terms we nee to isuss grphs. We strt y isussing wht it mens for two grphs to e the sme, in the sme sense tht we tlke out groups or fiels or etor spes or sets eing the sme: isomorphisms! Whih is to sy the following: In our first exmples of grphs, we lele ll of our erties euse this is prt of the efinition of wht grph *is* olletion of lele erties n eges etween them. Howeer, when we h to tully o triky mthemtis the four-olor theorem we stoppe leling our grphs! An in generl, I lim tht we on t relly re out the lelings of most of our grphs. For exmple, onsier the following pir of grphs: 15
These grphs re, in one sense, ifferent; the first grph hs n ege onneting 1 to 2, where the seon grph oes not. Howeer, in nother sense, these grphs re representing the sme sitution: they re oth epiting the grph skethe out y pentgon! For grphs like the ones in our mengerie, we on t re so muh out the leling of the erties; rther, the interesting fetures of these grphs re the intersetions of their eges n erties. In other wors, we wnt to sy tht oth of the grphs elow re the Petersen grph 1 : een though they initilly look rther ifferent, there is wy of releling the erties on the seon grph so tht (i, j) is n ege in the first grph iff it s n ege in the relele seon grph. How n we o this? Wht notion n we introue tht will llow us to regr suh grphs s eing the sme, in well-efine sense? Well, onsier the following: Definition. We sy tht two grphs G 1, G 2 re isomorphi if n only if there is mp σ : V (G 1 ) V (G 2 ) suh tht σ mthes eh element of V (G 1 ) to unique element of V (G 2 ), n ie-ers: in other wors, σ is wy of releling G 1 s erties with G 2 s lels, n ie-ers. { i, j } is n ege in G 1 if n only if {σ( 1 ), σ( 2 )} is n ege in G 2. We will often regr two isomorphi grphs s eing the sme, n therefore refer to grphs like K n or the Petersen grph without speifying or worrying out wht the erties re lele. A onept tht s muh more interesting (gien the ie of isomorphism) is the onept of sugrph, whih we efine elow: 1 The Petersen grph P The Petersen grph P is grph on ten erties, rwn elow: The erties in P ll he egree three; y ounting or the egree-sum formul, P hs 15 eges. notorious s eing the ounterexmple or exmple to numer of onjetures in grph theory! It is 16
Definition. Gien grph G n nother grph H, we sy tht H is sugrph of G if n only if V (H) V (G) n E(H) E(G). Exmple. The Petersen grph hs the isjoint union of two pentgons C 5 C 5 s sugrph, whih we she in re elow: In generl, when we sk if grph H is sugrph of grph G, we won t mention leling of H s erties; in this sitution, we re tully sking whether there is *ny* sugrph of G tht is isomorphi to H. For exmple, one question we oul sk is the following: wht kins of grphs ontin n tringle (i.e. C 3 ) s sugrph? Or, more generlly, wht kins of grphs ontin n o yle (i.e. C 2k+1 ) s sugrph? We nswer this in the next setion: 5 Clssifying Biprtite Grphs Definition. We ll grph G iprtite if n only if we n rek the set V (G) up into two prts V 1 (G) n V 2 (G), suh tht eery ege e E(G) hs one enpoint in V 1 (G) n one enpoint in V 2 (G). Alterntely, we sy tht grph is iprtite iff there is some wy to olor G s erties re n lue i.e. to tke eery ertex in G n ssign it either the olor lue or olor re, ut not oth or neither so tht eery ege hs one lue enpoint n one re enpoint. Exmple. The following grph is iprtite, with inite prtition (V 1, V 2 ): 17
Howeer, there re grphs tht re not iprtite; for exmple, C 3, the tringle, is not iprtite! This is not ery hr to see: in ny prtition of C 3 s erties into two sets V 1 n V 2, one of the two sets V 1 or V 2 hs to ontin two erties of our tringle. Therefore, there is n ege in C 3 with oth enpoints in one of our prtitions; so this prtition oes not mke C 3 iprtite. Beuse this hols for eery possile prtition, we n onlue tht no suh prtition exists i.e. C 3 is not iprtite! In generl, we n sy muh more: Proposition 1. C 2k+1 is not iprtite. Proof. We will proe this proposition with proof y ontrition. In other wors, we will ssume tht C n is iprtite, n from there we ll eue something we know to e flse; from there, we n onlue tht our ssumption must not he een true in the first ple (s it le us to something flse,) n therefore tht C n is not iprtite. To o this: s stte, we ll suppose for ontrition tht C 2k+1 is iprtite. Then, there must e some wy of oloring the erties { 1,... 2k+1 } of C n re n lue, so tht no ege is monohrome (i.e. hs two re enpoints or two lue enpoints.) How o we o this? Well: look t k+1. k+1 hs to e either re or lue: without ny loss of generlity 2, we n ssume tht it s re. Then, euse no ege in C 1 is monohrome, we speifilly know tht none of k+1 s neighors n e re: in other wors, they oth he to e lue! So oth k n k+2 re lue. Similrly, we know tht neither of k or k+2 s neighors n e lue: so oth k 1 n k+3 he to e re! Repeting this proess, we n see tht k+1 eing re fores k, k+2 to e lue, whih fores k 1, k+3 to e re, whih fores k 2, k+4 to e lue, whih fores... whih fores 1, 2k+1 to oth e the sme olor. But there is n ege etween 1 n n in C 2k+1! This ontrits the efinition of iprtite: therefore, we e rehe ontrition. Consequently, C n nnot e iprtite. This llows us to tully lssify lrge numer of grphs s not eing iprtite: Proposition 2. If grph G hs sugrph isomorphi to C 2k+1, then G is not iprtite. 2 The phrse without loss of generlity is something mthemtiins re oerly fon of. In generl, it s use in situtions where there is some sort of symmetry to the sitution tht llows you to ssume tht ertin sitution hols: for exmple, in this use, we re ssuming tht 1 is re euse it hs to e either re or lue, n if it ws lue we oul just swith the olors re n lue through the entire proof. 18
Proof. Suppose tht G ontins sugrph H tht s not iprtite. Then, for ny oloring of H s erties, there is some ege in H tht s monohrome. Therefore, euse ny oloring of G s erties into two prts will lso olor H s erties, we know tht ny oloring of G s erties with the olors re n lue will rete monohrome ege; therefore, G nnot e iprtite. Is this it? Or re there other wys in whih grph n fil to e iprtite? Surprisingly, s it turns out, there isn t: Proposition 3. A grph G on n erties is iprtite if n only if none of its sugrphs re isomorphi to n o yle. Proof. Our erlier proposition proe the if iretion of this lim: i.e. if grph is iprtite, it oesn t he ny o yles s sugrphs. We fous now on the only if iretion: i.e. gien grph tht oesn t ontin ny o yles, we seek to show tht it is iprtite. First, note the following efinitions: Definition. A grph G is lle onnete iff for ny two erties, w V (G), there is pth onneting n w. Definition. Gien grph G, iie it into sugrphs H 1,... H k suh tht eh of the sugrphs H i re onnete, n for ny two H i, H j s there ren t ny eges with one enpoint in H i n one enpoint in H j. These prts H i re lle the onnete omponents of G; grph G is onnete if it hs only one onnete omponent. Definition. For grph G n two erties, w we efine the istne (, w) etween n w s the numer of eges of the smllest pth onneting n w. For onnete grph, this quntity is lwys efine, (, ) = 0, n (, w) > 0 for ny w. Tke our grph G, n iie it into its onnete omponents H 1,... H k. If we n fin re-lue oloring of eh onnete omponent H i tht shows it s iprtite, we n omine ll of these olorings to get oloring of ll of G; euse there re no eges etween the onnete omponents, this omine oloring woul show tht G itself is iprtite! Therefore, it suffies to just show tht ny onnete grph H on n erties without ny o yles in it is iprtite. To o this, tke ny ertex y V (H), n onstrut the following sets: N 0 = {w : (, y) = 0} N 1 = {w : (, y) = 1} N 2 = {w : (, y) = 2}... N n = {w : (, y) = n} 19
First, notie tht eery ertex shows up in t lest one of these sets, s H is onnete n hs n erties (n thus, ny pth in H hs length n.) Furthermore, no ertex shows up in more thn one of these sets, euse istne is well-efine. Finlly, notie tht for ny x N k n ny pth P gien y y = 0 e 01 1 e 12... e k 1,k k = x, eh of the erties j lies in N j. This is euse eh of these hs pth of length j from y to j (just tke our pth n ut it off t j ), n hs no shorter pth (euse if there ws shorter pth, we oul use it to get from y to x in less thn k steps, n therefore (y, x) woul not e k.) Now, olor ll of the erties in the een N-sets re, n ll of the erties in the o N-sets lue. We lim tht there re no monohromti eges. To see this, tke ny ege { 1, 2 } in our grph H. Let (y, 1 ) = k n (y, 2 ) = l, P 1 e pth of length k onneting 1 with y, n P 2 e pth of length l onneting 2 with y. These pths my interset repetely, so tke x to e the furthest-wy ertex from y tht s in oth of these pths. Let P 1 e the pth tht we get y strting P 1 t x n proeeing to 1, n P 2 e the pth tht we get y strting P 2 t x n proeeing to 2. There re two possilities. Either x is one of 1 or 2, in whih se (euse there s n ege from 1 to 2 ) the istne from y to 1 is either one greter or one less thn the istne from y to 2. In either se, 1 n 2 he ifferent olors (euse our olors lternte etween re n lue s our istne inrese,) so this ege is not monohrome. Otherwise, x is neither 1 or 2. In this se, look t the yle forme y oing the following: Strt t x, n proee long P 1. One we get to 1, trel long the ege { 1, 2 }. Now, go kwrs long P 2 k to x. This is yle, euse P 1 n P 2 on t shre ny erties in ommon prt from x. Wht is its length? Well, the length of P 1 is just (y, 1) (y, x), the length of P 2 is just (y, 2 ) (y, x), n the length of single ege is just 1; so, the totl length of this pth is (y, 1 ) (y, x) + (y, 2 ) (y, x) + 1 = (y, 1 ) + (y, 2 ) (2(y, x) + 1). We know tht this nnot e o, euse our grph hs no o yles; so the numer oe is een! Beuse (2(y, x) + 1) is o, this mens tht (y, 1 ) + (y, 2 ) must lso 20
e o; in other wors, extly one of (y, 1 ), (y, 2 ) n e o,n extly one n e een. But this mens speifilly tht extly one must e lue n one must e re (uner our oloring sheme,) so our ege must not e monohromti. Therefore, our grph hs no monohromti eges; so it s iprtite! 21