Planar graphs. Definition. A graph is planar if it can be drawn on the plane in such a way that no two edges cross each other.

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1 Plaar graphs Defiitio. A graph is plaar if it ca be draw o the plae i such a way that o two edges cross each other. Example: Face 1 Face 2 Exercise: Which of the followig graphs are plaar? K, P, C, K,m, B Defiitio. A face of a plaar graph is a maximal sectio of the plae i which ay two poits ca be joit by a curve that does ot itersect ay part of G. Defiitio. Degree of a face is the umber of edges i the boudary surroudig the face The degree of Face 1 is 3 ad the degree of Face 2 is 7

2 Euler s Formula Theorem. Let G be a coected plaar graph with vertices, m edges ad f faces, the -m+f=2 Proof. By iductio o m. For m=0, G=K 1, a graph with 1 vertex ad 1 face. Assume the formula is true for ay coected plaar graph with fewer the m edges ad let G have m edges. if G is a tree, the m=-1 ad f=1 ad the formula is true. if G is ot a tree, cosider a cycle C ad a edge e i C. The graph G-e is coected, it has the same umber of vertices, oe edge fewer, ad oe face fewer. By the iductio hypothesis, i G-e we have -(m-1)+(f-1)=2. Therefore, i G we have -m+f=2.

3 Euler s Formula Theorem. Let G be a coected plaar graph with vertices, m edges ad f faces, the -m+f=2 Corollary. If G is a coected plaar graph with 3 vertices ad m edges, the m3-6. If additioally G has o triagles, the m2-4. Proof. If we trace aroud all faces, we ecouter each edge exactly twice. Deotig the umber of faces of degree k by f k, we coclude that k kf k =2m. Sice the degree of ay face i a simple plaar graph is at least 3, we have 3f=3 k3 f k k3 kf k =2m. Together with the Euler's formula, this proves that m 3-6. If additioally, G has o triagles, the 4f=4 k4 f k k4 kf k =2m, ad therefore m2-4.

4 Euler s Formula Theorem. Let G be a coected plaar graph with vertices, m edges ad f faces, the -m+f=2 Corollary. If G is a coected plaar graph with 3 vertices ad m edges, the m3-6. If additioally G has o triagles, the m2-4. Corollary. K 5 ad K 3,3 are ot plaar. Proof. For K 5 we have m>3-6, sice =5 ad m=10. Therefore, K 5 is ot plaar. K 3,3 is ot plaar, because it has o triagles ad m>2-4 (=6, m=9).

5 Euler s Formula Theorem. Let G be a coected plaar graph with vertices, m edges ad f faces, the -m+f=2 Corollary. If G is a coected plaar graph with 3 vertices ad m edges, the m3-6. If additioally G has o triagles, the m2-4. Corollary. K 5 ad K 3,3 are ot plaar. Corollary. Every plaar graph has a vertex of degree at most 5. Proof. Let G be a coected plaar graph with vertices ad m edges. The vv ( G) deg( v) 2m 2(3 6) Therefore, G has a vertex of degree at most

6 Euler s Formula Theorem. Let G be a coected plaar graph with vertices, m edges ad f faces, the -m+f=2 Corollary. If G is a coected plaar graph with 3 vertices ad m edges, the m3-6. If additioally G has o triagles, the m2-4. Corollary. K 5 ad K 3,3 are ot plaar. Corollary. Every plaar graph has a vertex of degree at most 5. Corollary. The umber of plaar labeled graphs with vertices is at most 3 6.

7 Euler s Formula Fact 1. The umber of labeled graphs with vertices ad m edges is m 2 1) / ( Fact 2. k p k p k p p p p k k p p p k p k p k p 1!...! 1) 1)...( ( )!!(! Corollary. The umber of plaar labeled graphs with vertices is at most 3 6. Proof. Sice for a plaar graph m3-6, the umber of plaar labeled graphs with vertices does ot exceed ) ( ) ( 2) 1) / ( ( 2 1) / ( i i i i i i i

8 Subgraphs, iduced subgraphs, spaig subgraphs, miors Defiitio. A graph H is a subgraph of a graph G if V(H) V(G) ad E(H) E(G). I other words, H is a subgraph of G if H ca be obtaied from G by deletig some vertices ad/or edges. Which of the followig graphs,,, are subgraphs of P 5 Defiitio. H is a iduced subgraph of G if H ca be obtaied from G by deletig some vertices (but ot edges). Defiitio. H is a spaig subgraph of G if H ca be obtaied from G by deletig some edges (but ot vertices).

9 Subgraphs, iduced subgraphs, spaig subgraphs, miors a b ab cotractio of edge ab Defiitio. A graph H is a mior of a graph G if H ca be obtaied from G by vertex deletios, edge deletios ad edges cotractios.

10 Kuratowski Theorem Theorem. A graph G is plaar if ad oly if G does ot cotai K 5 ad K 3,3 as miors. Peterse graph

11 Kuratowski Theorem Theorem. A graph G is plaar if ad oly if G does ot cotai K 5 ad K 3,3 as miors. Peterse graph

12 Kuratowski Theorem Theorem. A graph G is plaar if ad oly if G does ot cotai K 5 ad K 3,3 as miors. Peterse graph

13 Kuratowski Theorem Theorem. A graph G is plaar if ad oly if G does ot cotai K 5 ad K 3,3 as miors. Peterse graph

14 Kuratowski Theorem Theorem. A graph G is plaar if ad oly if G does ot cotai K 5 ad K 3,3 as miors. Exercise: apply Kuratowski Theorem to show that this graph is ot plaar

15 Colorig graphs A vertex colorig is a assigmet of colors to the vertices such that o edge coects vertices of the same color. The miimum umber of colors eeded to color a graph is the chromatic umber of the graph. Exercise. What is the chromatic umber of K, P, C, B? The k-colorig problem is the problem of determiig if a graph admits a colorig with at most k colors.

16 Four color problem Is it possible to color a plaar graph with at most 4 colors? The cojecture was first proposed i 1852 whe Fracis Guthrie, while tryig to color the map of couties of Eglad, oticed that oly four differet colors were eeded. There were several early failed attempts at provig the theorem. Oe proof of the theorem was give by Alfred Kempe i 1879, which was widely acclaimed; aother proof was give by Peter Guthrie Tait i It was ot util 1890 that Kempe's proof was show icorrect by Percy Heawood, ad 1891 that Tait's proof was show icorrect by Julius Peterse each false proof stood uchalleged for 11 years. I 1890, i additio to exposig the flaw i Kempe's proof, Heawood proved that all plaar graphs are five-colorable; see five color theorem. It was ot util 1976 that the four-color cojecture was fially prove by Keeth Appel ad Wolfgag Hake at the Uiversity of Illiois. Their proof reduced the ifiitude of possible graphs to 1,936 reducible cofiguratios (later reduced to 1,476) which had to be checked oe by oe by computer. The computer program ra for hudreds of hours. This reducibility part of the work was idepedetly double checked with differet programs ad computers. However, the uavoidability part of the proof was over 500 pages of had writte couter-couter-examples, much of which was Hake's teeage so Lippold verifyig graph colorigs.

17 Six color theorem Theorem. Every plaar graph ca be colored with at most 6 colors. Proof. By iductio o the umber of vertices. Obviously, every graph with at most 6 vertices ca be colored with at most 6 colors. If a plaar graph G has more tha 6 vertices, delete from G ay vertex v of degree at most 5. By iductio hypothesis G-v is 6-colorable. The eighbors of v use at most 5 differet colors. Therefore, the uused color ca be used to color v.

18 Five color theorem Theorem. Every plaar graph ca be colored with at most 5 colors. x Proof. Assume without loss of geerality that y 1 is ot adjacet to y 2 y 1 y 2 y 3 y 4 y 5

19 Five color theorem Theorem. Every plaar graph ca be colored with at most 5 colors. x Proof. Assume without loss of geerality that y 1 is ot adjacet to y 2 y 1 y 2 y 3 y 4 y 5

20 Five color theorem Theorem. Every plaar graph ca be colored with at most 5 colors. Proof. x Assume without loss of geerality that y 1 is ot adjacet to y 2 y 1 y 2 y 3 y 4 y 5

21 Five color theorem Theorem. Every plaar graph ca be colored with at most 5 colors. Proof. x y 1 y 2 y 3 y 4 y Assume without loss of geerality that y 1 is ot adjacet to y 2

22 Five color theorem Theorem. Every plaar graph ca be colored with at most 5 colors. x Proof. Assume without loss of geerality that y 1 is ot adjacet to y 2 y 1 y 2 y 3 y 4 y

23 Five color theorem Theorem. Every plaar graph ca be colored with at most 5 colors. x 5 Proof. Assume without loss of geerality that y 1 is ot adjacet to y 2 y 1 y 2 y 3 y 4 y