Lecture 6. Planar graphs. Euler’s Formula. A graph that can be drawn in the plane without edge crossings. Euler’s Formula. Faces: connected regions of the plane. Planar Six and then Five Color theorem. How many faces for Planar? Yes for Triangle. Types of graphs. triangle? 2 Four node complete? Yes. Complete Graphs. complete on four vertices or K 4 ? 4 (complete ≡ every edge present. K n is n -vertex complete graph. ) Trees (a little more.) bipartite, complete two/three or K 2 , 3 ? 3 Five node complete or K 5 ? No! Why? Later. Hypercubes. v is number of vertices, e is number of edges, f is number of faces. Euler’s Formula: Connected planar graph has v + f = e + 2. Triangle: 3 + 2 = 3 + 2! K 4 : 4 + 4 = 6 + 2! Two to three nodes, bipartite? Yes. K 2 , 3 : 5 + 3 = 6 + 2! Three to three nodes, complete/bipartite or K 3 , 3 . No. Why? Later. Examples = 3! Proven! Not!!!! 1 / 21 2 / 21 3 / 21 Euler and Polyhedron. Euler and non-planarity of K 5 and K 3 , 3 Proving non-planarity for K 3 , 3 Greeks knew formula for polyhedron. . Euler: v + f = e + 2 for connected planar graph. We consider simple graphs where v ≥ 3. Consider Face edge Adjacencies. K 3 , 3 ? Edges? 9. Vertices. 6. Faces? 6. Edges? 12. Vertices? 8. e ≤ 3 ( v ) − 6 for planar graphs. Euler: Connected planar graph: v + f = e + 2. 9 ≤ 3 ( 6 ) − 6? Sure! 8 + 6 = 12 + 2. Each face is adjacent to at least three edges. Step in proof of K 5 : faces are adjacent to ≥ 3 edges. Greeks couldn’t prove it. Induction? Remove vertice for polyhedron? ≥ 3 f face-edge adjacencies. Polyhedron without holes ≡ Planar graphs. For K 3 , 3 every cycle is of even length or ≥ 4. Each edge is adjacent to (at most) two faces. For Convex Polyhedron: ≤ 2 e face-edge adjacencies. Finish in homework! Or f ≤ 2 Surround by sphere. = ⇒ 3 f ≤ 2 e for any planar graph with v > 2. 3 e . Project from internal point polytope to sphere: drawing on sphere. Plug into Euler: v + 2 3 e ≥ e + 2 = ⇒ e ≤ 3 v − 6 Project Sphere-N onto Plane: drawing on plane. K 5 Edges? e = 4 + 3 + 2 + 1 = 10. Vertices? v = 5. Euler proved formula thousands of years later! 10 �≤ 3 ( 5 ) − 6 = 9. = ⇒ K 5 is not planar. 4 / 21 5 / 21 6 / 21
Planarity and Euler Euler’s formula. Graph Coloring. Euler: Connected planar graph has v + f = e + 2. Proof: Induction on e . Given G = ( V , E ) , a coloring of G assigns colors to vertices V where Base: e = 0, v = f = 1. for each edge the endpoints have different colors. Induction Step: If it is a tree. e = v − 1, f = 1, v + 1 = ( v − 1 )+ 2. Yes. These graphs cannot be drawn in the plane without edge crossings. If not a tree. Euler’s Formula: v + f = e + 2 for any planar drawing. Find a cycle. Remove edge. = ⇒ for simple planar graphs: e ≤ 3 v − 6. . . Idea: Face is a cycle in graph of length 3. . f 1 Outer face. Count face-edge incidences. Notice that the last one, has one three colors. = ⇒ for bipartite simple planar graphs: e ≤ 2 v − 4. Joins two faces. Fewer colors than number of vertices. Idea: face is a cycle in graph of length 4. New graph: v -vertices. e − 1 edges. f − 1 faces. Planar. Fewer colors than max degree node. Count face-edge incidences. v +( f − 1 ) = ( e − 1 )+ 2 by induction hypothesis. Therefore v + f = e + 2. Interesting things to do. Algorithm! Proved absolutely no drawing can work for these graphs. Quick: So......so ...Cool! v + 1 = ( v − 1 )+ 2, add edge: f → f + 1, e → e + 1. 7 / 21 8 / 21 9 / 21 Planar graphs and maps. Six color theorem. Five color theorem: prelimnary. Planar graph coloring ≡ map coloring. Theorem: Every planar graph can be colored with six colors. Preliminary Observation: Connected components of vertices with two Proof: colors in a legal coloring can switch colors. Recall: e ≤ 3 v − 6 for any planar graph where v > 2. From Euler’s Formula. Total degree: 2 e v ≤ 2 ( 3 v − 6 ) Average degree: = 2 e ≤ 6 − 12 v . v There exists a vertex with degree < 6 or at most 5. Look at only green and blue. Connected components. Remove vertex v of degree at most 5. Can switch in one component. Inductively color remaining graph. Or the other. Color is available for v since only five neighbors... and only five colors are used. Four color theorem is about planar graphs! 10 / 21 11 / 21 12 / 21
Five color theorem Four Color Theorem Complete Graph. Theorem: Every planar graph can be colored with five colors. Preliminary Observation: Connected components of vertices with two colors in a legal coloring can switch colors. Proof: Again with the degree 5 vertex. Again recurse. Assume neighbors are colored all differently. Otherwise one of 5 colors is available. = ⇒ Done! Theorem: Any planar graph can be colored with four colors. K n complete graph on n vertices. Switch green and blue in green’s component. Proof: Not Today! All edges are present. Done. Unless blue-green path to blue. Everyone is my neighbor. Switch orange and red in oranges component. Each vertex is adjacent to every other vertex. Done. Unless red-orange path to red. How many edges? Planar. = ⇒ paths intersect at a vertex! Each vertex is incident to n − 1 edges. What color is it? Sum of degrees is n ( n − 1 ) = 2 | E | . ······ . . Must be blue or green to be on that path. = ⇒ Number of edges is n ( n − 1 ) / 2. Must be red or orange to be on that path. Contradiction. Can recolor one of the neighbors. Gives an available color for center vertex! 13 / 21 14 / 21 15 / 21 K 4 and K 5 Tree’s fall apart. Hypercubes. Complete graphs, really connected! But lots of edges. | V | ( | V |− 1 ) / 2 Thm: There is one vertex whose removal disconnects | V | / 2 nodes Trees, few edges. ( | V |− 1 ) from each other. but just falls apart! Hypercubes. Really connected. | V | log | V | edges! Also represents bit-strings nicely. G = ( V , E ) | V | = { 0 , 1 } n , Idea of proof. | E | = { ( x , y ) | x and y differ in one bit position. } Point edge toward bigger side. 101 111 K 5 is not planar. Remove center node: node with no outgoing arc. (Hotel 01 11 Cannot be drawn in the plane without an edge crossing! 001 011 California.) 0 1 Prove it! We did! All the neighbors in components that are smaller than | V | / 2. 110 100 00 10 000 010 2 n vertices. number of n -bit strings! n 2 n − 1 edges. 2 n vertices each of degree n total degree is n 2 n and half as many edges! 16 / 21 17 / 21 18 / 21
Recursive Definition. Hypercube: Can’t cut me! Cuts in graphs. 8 4 11 A 0-dimensional hypercube is a node labelled with the empty string of bits. 7 3 Thm: Any subset S of the hypercube where | S | ≤ | V | / 2 has ≥ | S | 5 An n -dimensional hypercube consists of a 0-subcube (1-subcube) edges connecting it to V − S ; | E ∩ S × ( V − S ) | ≥ | S | which is a n − 1-dimensional hypercube with nodes labelled 0 x (1 x ) 10 9 Terminology: with the additional edges ( 0 x , 1 x ) . ( S , V − S ) is cut. 6 1 2 ( E ∩ S × ( V − S )) - cut edges. S is red, V − S is blue. Restatement: for any cut in the hypercube, the number of cut edges What is size of cut? is at least the size of the small side. Number of edges between red and blue. 4. Hypercube: any cut that cuts off x nodes has ≥ x edges. 19 / 21 20 / 21 21 / 21 Proof of Large Cuts. Induction Step Idea Induction Step Thm: For any cut ( S , V − S ) in the hypercube, the number of cut edges is at least the size of the small side. Thm: For any cut ( S , V − S ) in the hypercube, the number of cut Use recursive definition into two subcubes. edges is at least the size of the small side, | S | . Two cubes connected by edges. Proof: Induction Step. Thm: For any cut ( S , V − S ) in the hypercube, the number of cut Recursive definition: Case 1: Count edges inside edges is at least the size of the small side. Case 2: Count inside and across. H 0 = ( V 0 , E 0 ) , H 1 = ( V 1 , E 1 ) , edges E x that connect them. subcube inductively. Proof: H = ( V 0 ∪ V 1 , E 0 ∪ E 1 ∪ E x ) Base Case: n = 1 V= { 0,1 } . S = S 0 ∪ S 1 where S 0 in first, and S 1 in other. S = { 0 } has one edge leaving. | S | = φ has 0. Case 1: | S 0 | ≤ | V 0 | / 2 , | S 1 | ≤ | V 1 | / 2 Both S 0 and S 1 are small sides. So by induction. Edges cut in H 0 ≥ | S 0 | . Edges cut in H 1 ≥ | S 1 | . Total cut edges ≥ | S 0 | + | S 1 | = | S | . 22 / 21 23 / 21 24 / 21
Induction Step. Case 2. Hypercubes and Boolean Functions. Summary. Thm: For any cut ( S , V − S ) in the hypercube, the number of cut edges is at least the size of the small side, | S | . Proof: Induction Step. Case 2. | S 0 | ≥ | V 0 | / 2. Recall Case 1: | S 0 | , | S 1 | ≤ | V | / 2 The cuts in the hypercubes are exactly the transitions from 0 sets to 1 We did lots today! set on boolean functions on { 0 , 1 } n . | S 1 | ≤ | V 1 | / 2 since | S | ≤ | V | / 2. Euler, coloring, types of graphs. = ⇒ ≥ | S 1 | edges cut in E 1 . S 1 Central area of study in computer science! S 0 | S 0 | ≥ | V 0 | / 2 = ⇒ | V 0 − S | ≤ | V 0 | / 2 And Isoperimetric inequality for Hypercubes. Yes/No Computer Programs ≡ Boolean function on { 0 , 1 } n = ⇒ ≥ | V 0 |−| S 0 | edges cut in E 0 . Have a nice weekend! Central object of study. Edges in E x connect corresponding nodes. = ⇒ = | S 0 |−| S 1 | edges cut in E x . Total edges cut: ≥ | S 1 | + | V 0 |−| S 0 | + | S 0 |−| S 1 | = | V 0 | | V 0 | = | V | / 2 ≥ | S | . Also, case 3 where | S 1 | ≥ | V | / 2 is symmetric. 25 / 21 26 / 21 27 / 21
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