Office Hours. COS 341 Discrete Math. Office Hours. Homework 8. Currently, my office hours are on Friday, from 2:30 to 3:30.

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1 Oce Hours Curretly, my oce hours are o Frday, rom :30 to 3:30. COS 31 Dscrete Math 1 Oce Hours Curretly, my oce hours are o Frday, rom :30 to 3:30. Nobody seems to care. Chage oce hours? Tuesday, 8 PM to 9 PM. Homework 8 Due o Wedesday at the begg o class. No collaborato! Questo 3: Never crosses tsel s the key. Questo : Assume > the theorem s ot true or. For some values o >, the boud may ot be a teger. It does t matter the umber o crossgs wll be strctly greater tha that. 3

2 Jorda curve theorem: From last class Ay Jorda curve dvdes the plae to two parts, the teror ad the exteror. K 5 s ot plaar. K 3,3 s ot plaar. -Coected Graphs Recall that a graph s -coected t has at least 3 vertces, ad by deletg ay sgle vertex we obta a coected graph. We also kow the ollowg: A graph G s -coected ad oly t ca be created rom a tragle K 3 by a sequece o edge subdvsos ad edge sertos. 5 6 Faces ad Cycles Theorem: Let G be a -vertex-coected plaar graph. The every ace ay plaar drawg o G s a rego o some cycle o G. Faces ad Cycles Theorem: Let G be a -vertex-coected plaar graph. The every ace ay plaar drawg o G s a rego o some cycle o G. We do eed t to be -vertex-coected. 7 8

3 Faces ad Cycles Proo: by ducto o umber o vertces Base case: 3 oly -coected graph s the tragle oe cycle, two regos: OK. Hypothess: assume true or o 1, wth 0 > 3. Let s prove t s true or o. -coected graph G wth at least vertces. Faces ad Cycles Take a plaar -coected graph G wth > 3 vertces. Ca be bult rom a tragle by a sequece o edge sertos ad subdvsos. Oe o these must be true: a There s a edge e such that G G e s -coected. b There s a graph G V, E ad there s a edge e E such that the subdvso o e creates G. I ether case, G s a smaller -coected graph. By the ductve hypothess, every ace ay plaar drawg o G s a rego o some cycle o G Faces ad Cycles Case a: there s a edge e such that G G e s -coected. Let e {v,w}. There s a ace F G correspodg to a cycle that cotas both v ad w. v α 1 w α v α 1 ad α are arcs the cycle The arc correspodg to e dvdes F to two aces, each correspodg to a deret cycle. v α 1 w αe v v e w α v α 1 F v αe F α Faces ad Cycles Case b: there s a graph G V, E wth a edge e E such that the subdvso o e creates G. Each ace o G s a rego o some cycle G. Subdvdg e amouts to drawg a vertex sde the edge. Ths exteds the legth o the cycles e partcpates, but does t chage the property. w 11 1

4 Combatoral Characterzato Combatoral Characterzato Every subgraph o a plaar graph must be plaar: caot cota K 5 caot cota K 3, Combatoral Characterzato Every subgraph o a plaar graph must be plaar: caot cota K 5 caot cota K 3,3 More geerally: o subgraph o a plaar graph ca be a subdvso o a o-plaar graph. caot cota a subdvso o K 5 caot cota a subdvso o K 3,3 Combatoral Characterzato Every subgraph o a plaar graph must be plaar: caot cota K 5 caot cota K 3,3 More geerally: o subgraph o a plaar graph ca be a subdvso o a o-plaar graph. caot cota a subdvso o K 5 caot cota a subdvso o K 3,3 Is that eough? 15 16

5 Combatoral Characterzato Kuratowsk s theorem: A graph G s plaar ad oly t has o subgraph somorphc to a subdvso o K 3,3 or to a subdvso o K 5. We ca test a graph s plaar wthout actually drawg t: we ust have to very there are volatg subgraphs. There are aster ways o testg plaarty, though. Euler s Formula Theorem: Let G V, E be a coected plaar graph, ad let be the umber o aces o ay plaar drawg o G. The V E. The umber o aces does ot deped o the plaar drawg, ust o the graph tsel Euler s Formula Proo by ducto o E. Base case: E 0 sgle vertex, sgle ace: V E E > 0 ad G does ot cota a cycle t s a tree: V E V V 1 1. E > 0 ad G V, E cotas a cycle: Some edge e belogs to a cycle; remove t. The resultg graph G obeys the ormula: V E Clearly, V V ad E E 1. e was adacet to two aces by Jorda that become oe: 1 V E V E 1 1 V E dmesoal covex bodes; te umber o aces; aces are cogruet copes o the same regular polygo; same umber o aces meet at each vertex; also kow as Platoc Solds. 0

6 Tetrahedro: aces Hexahedro a.k.a. cube: 6 aces Octahedro: 8 aces Dodecahedro: 1 aces Icosahedro: 0 aces Are there more? [mages rom mathworld.wolram.com] Every covex polytope ca be coverted to a plaar graph: Fd a sphere such that: ceter o sphere sde polytope; sphere cotas the whole polytope. Proect the polytope oto the sphere: we get a graph o the surace o a sphere; that graph ca be coverted to a plaar graph wth a stereographc proecto. Vertces, aces, ad edges o the polytope become vertces, aces, ad edges o a plaar graph. 1 Tetrahedro Cube 3

7 Octahedro Dodecahedro 5 6 Icosahedro Parameters o a regular covex polytope: k: umber o sdes each polygo ace d: umber o aces that meet at each vertex : vertces m: edges : aces Lookg at the vertces: Every edge appears exactly two vertces: d m m/d 7 8

8 Parameters o a regular covex polytopes: k: umber o sdes each ace d: umber o aces that meet at each vertex : vertces m: edges : aces Lookg at the aces: Every edge appears exactly two aces: k m m/k Parameters o a regular covex polytopes: k: umber o sdes each ace d: umber o aces that meet at each vertex : vertces m: edges : aces Lookg at the whole graph: It s plaar, so we ca apply Euler s ormula: m 9 30 So we have: m/k m/d m Substtutg ad the thrd equato: m m/d m m/k dvdg by m ad rearragg d k m So every regular polytope must obey I partcular, 1 1 > d k I both d ad k, we would have: d k 1 Se ether d3 or k3 or both. 1 1 d k 1 1 m 31 3

9 Assume d3: d k m k m k 6 m The rght-had sde s postve, so k < 6. k {3,, 5} Assume k3: d k m d 3 m d 6 m The rght-had sde s postve, so d < 6. d {3,, 5} 33 3 So the oly possbltes are: d k m Polytope tetrahedro cube dodecahedro octahedro cosahedro Theorem: Number o Edges Let G V,E be a plaar graph wth at least 3 vertces. The E 3 V 6. I the graph s maxmal o edge ca be added wthout volatg plaarty, the equalty holds: E 3 V 6. It suces to prove the secod statemet; the graph s ot maxmal, we ca always add edges utl t becomes oe

10 Lemma: Number o Edges Every maxmal plaar graph G s a tragulato every ace s a tragle. Proo: we show that G s ot a tragulato, t s always possble to add a edge wthout volatg plaarty. Three cases to cosder: G s dscoected. I G s coected but ot -coected. G s -coected. Number o Edges Case 1: G s ot coected: A edge ca be added betwee two compoets. V 1 V V 3 V Number o Edges Case 1: G s ot coected: A edge ca be added betwee two compoets. V 1 V Number o Edges Case : G s coected, but ot -coected: There s a vertex v whose removal dscoects G. Let V 1, V,..., V k be the resultg compoets k >. A edge ca be added betwee compoets assocated wth edges draw ext to each other aroud v. V 3 v V V V 1 V 3 V 39 0

11 Number o Edges Case : G s coected, but ot -coected: There s a vertex v whose removal dscoects G. Let V 1, V,..., V k be the resultg compoets k >. A edge ca be added betwee compoets assocated wth edges draw ext to each other aroud v. Number o Edges Case 3: G s -coected. Every ace s bouded by a cycle. Take ay ace wth or more edges: v 1 v v V v 3 V 1 v V 3 V 1 Number o Edges Case 3: G s -coected. Every ace s bouded by a cycle. Take ay ace wth or more edges: Number o Edges Case 3: G s -coected. Every ace s bouded by a cycle. Take ay ace wth or more edges: v 1 v v 1 v v 3 v 3 v v I v 1 ad v 3 are ot coected, you ca add a edge betwee them. 3 I v 1 ad v 3 are coected, v ad v ca t be. So you ca add a edge betwee v ad v.

12 Number o Edges So every maxmal plaar graph s a tragulato. Because every ace s a tragle ad every edge s cdet to exactly two aces, we have: 3 E E /3. Usg ths value Euler s ormula: V E V E E /3 V E /3 E 3 V 6. Tragle-Free Plaar Graphs Theorem: Let GV,E be a plaar graph wth o tragles.e., wthout K 3 as a subgraph ad at least 3 vertces. The E V. Proo smlar to the prevous oe Cosder a maxmal tragle-ree plaar graph G; we ca always add edges utl t becomes oe. G s clearly coected. Corollary: there exsts a vertex o degree at most Tragle-Free Plaar Graphs Assume G s coected, but ot -coected. There s a vertex v whose removal dscoects G. Let V 1, V,..., V k be the resultg compoets k >. Edges ca be added betwee these compoets wthout volatg plaarty. But we could create a tragle we oed vertces that are adacet to v. I every V s a sgle vertex, the G s a tree: E V 1 E V 3 E V V because G has at least three vertces E V the equalty holds 7 Tragle-Free Plaar Graphs Now cosder the case whch compoet V 1 has at least two vertces. Cosder a ace F havg both a vertex o V 1 ad a vertex o some other V o ts boudary. V 1 must have at least oe edge {v 1,v } o the boudary o F. We ca t have both v 1 ad v coected to v or these vertces would costtute a tragle. So a edge ca be added betwee oe o these vertces ad a vertex V. G s ot maxmal a cotradcto. Maxmal tragle-ree plaar graphs must be -coected. 8

13 Tragle-Free Plaar Graphs G s a -coected, maxmal tragle-ree plaar graph. -coected: every ace s a rego o a cycle. Tragle-ree: every cycle has at least edges. Coutg edges rom aces: E E / From Euler s ormula: V E V E E / E V. Corollary: there exsts a vertex o degree at most 3. 9 Scores o Plaar Graphs Theorem: Let GV,E be a -coected plaar graph wth at least 3 vertces. Dee: : umber o vertces o degree ; : umber o aces some xed drawg o G bouded by cycles o legth. The we have Scores o Plaar Graphs Why s ths relevat? We ca rewrte as The rst... cotas oly egatve terms. The secod... cotas oly postve terms. So Amog other thgs, ths meas that there are at least 3 vertces o degree at most 5 every plaar graph. 51 Proo o the theorem: Scores o Plaar Graphs Obvous acts: ad V From Euler s ormula: E V E E 5

14 53 Scores o Plaar Graphs Proo o the theorem: From prevous slde: Coutg edges rom the aces: Coutg edges rom the vertces: E E E 5 Scores o Plaar Graphs Proo o the theorem: From the prevous slde: Addg ad, we get the al expresso: