Today. Graph Coloring. Planar graphs and maps. Given G = ( V , E ) - PowerPoint PPT Presentation

Today. Graph Coloring. Planar graphs and maps. Given G = ( V , E ) , a coloring of a G assigns colors to vertices V Planar graph coloring ≡ map coloring. where for each edge the endpoints have different colors. Planar Five Color theorem. Types of graphs. Complete Graphs. Trees. Hypercubes. Notice that the last one, has one three colors. Fewer colors than number of vertices. Fewer colors than max degree node. Four color theorem is about planar graphs! Interesting things to do. Algorithm! 1 / 24 2 / 24 3 / 24 Six color theorem. Five color theorem: prelimnary. Five color theorem Theorem: Every planar graph can be colored with five colors. Preliminary Observation: Connected components of vertices with two Theorem: Every planar graph can be colored with six colors. colors in a legal coloring can switch colors. Preliminary Observation: Connected components of vertices with two Proof: Proof: Again with the degree 5 vertex. Again recurse. colors in a legal coloring can switch colors. Recall: e ≤ 3 v − 6 for any planar graph where v > 2. Assume neighbors are colored all differently. From Euler’s Formula. Otherwise done. Total degree: 2 e Switch green to blue in component. v ≤ 2 ( 3 v − 6 ) Average degree: ≤ 2 e ≤ 6 − 12 v . Done. Unless blue-green path to blue. v Switch orange to red in its component. There exists a vertex with degree < 6 or at most 5. Look at only green and blue. Done. Unless red-orange path to red. Connected components. Remove vertex v of degree at most 5. Can switch in one component. Planar. = ⇒ paths intersect at a vertex! Inductively color remaining graph. Or the other. Color is available for v since only five neighbors... What color is it? . ······ . and only five colors are used. . Must be blue or green to be on that path. Must be red or orange to be on that path. Contradiction. Can recolor one of the neighbors. And recolor “center” vertex. 4 / 24 5 / 24 6 / 24

Four Color Theorem Complete Graph. K 4 and K 5 Theorem: Any planar graph can be colored with four colors. K n complete graph on n vertices. All edges are present. Proof: Not Today! Everyone is my neighbor. Each vertex is adjacent to every other vertex. K 5 is not planar. How many edges? Cannot be drawn in the plane without an edge crossing! Prove it! We did! Each vertex is incident to n − 1 edges. Sum of degrees is n ( n − 1 ) . = ⇒ Number of edges is n ( n − 1 ) / 2. Remember sum of degree is 2 | E | . 7 / 24 8 / 24 9 / 24 A Tree, a tree. Trees. Equivalence of Definitions. Definitions: Theorem: A connected graph without a cycle. “G connected and has | V |− 1 edges” ≡ A connected graph with | V |− 1 edges. “G is connected and has no cycles.” A connected graph where any edge removal disconnects it. Graph G = ( V , E ) . A connected graph where any edge addition creates a cycle. Lemma: If v is a degree 1 in connected graph G , G − v is connected. Binary Tree! Proof: Some trees. For x � = v , y � = v ∈ V , there is path between x and y in G since connected. and does not use v (degree 1) = ⇒ G − v is connected. no cycle and connected? Yes. y v | V |− 1 edges and connected? Yes. More generally. removing any edge disconnects it. Harder to check. but yes. Adding any edge creates cycle. Harder to check. but yes. To tree or not to tree! x 10 / 24 11 / 24 12 / 24

Proof of only if. Proof of if Tree’s fall apart. v Thm: Thm: Thm: There is one vertex whose removal disconnects | V | / 2 nodes “G connected and has | V |− 1 edges” ≡ from each other. “G is connected and has no cycles” “G is connected and has no cycles.” = ⇒ “G connected and has | V |− 1 edges” Proof: Proof of = ⇒ : By induction on | V | . Walk from a vertex using untraversed edges. Base Case: | V | = 1. 0 = | V |− 1 edges and has no cycles. Until get stuck. Induction Step: Claim: Degree 1 vertex. Claim: There is a degree 1 node. Proof of Claim: Idea of proof. Proof: First, connected = ⇒ every vertex degree ≥ 1. Can’t visit more than once since no cycle. Sum of degrees is 2 | V |− 2 Point edge toward bigger side. Entered. Didn’t leave. Only one incident edge. Remove center node. Average degree 2 − 2 / | V | Removing node doesn’t create cycle. Not everyone is bigger than average! New graph is connected. By degree 1 removal lemma, G − v is connected. Removing degree 1 node doesn’t disconnect from Degree 1 lemma. G − v has | V |− 1 vertices and | V |− 2 edges so by induction By induction G − v has | V |− 2 edges. = ⇒ no cycle in G − v . G has one more or | V |− 1 edges. And no cycle in G since degree 1 cannot participate in cycle. 13 / 24 14 / 24 15 / 24 Hypercubes. Recursive Definition. Hypercube: Can’t cut me! Complete graphs, really connected! But lots of edges. | V | ( | V |− 1 ) / 2 Trees, few edges. ( | V |− 1 ) but just falls apart! A 0-dimensional hypercube is a node labelled with the empty string of bits. Hypercubes. Really connected. | V | log | V | edges! Thm: Any subset S of the hypercube where | S | ≤ | V | / 2 has ≥ | S | An n -dimensional hypercube consists of a 0-subcube (1-subcube) Also represents bit-strings nicely. 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 ) G = ( V , E ) with the additional edges ( 0 x , 1 x ) . Terminology: | V | = { 0 , 1 } n , ( S , V − S ) is cut. | E | = { ( x , y ) | x and y differ in one bit position. } ( E ∩ S × ( V − S )) - cut edges. 101 111 01 11 Restatement: for any cut in the hypercube, the number of cut edges 001 011 0 1 is at least the size of the small side. 110 00 10 000 100 010 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 / 24 17 / 24 18 / 24

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 | . 19 / 24 20 / 24 21 / 24 Induction Step. Case 2. Hypercubes and Boolean Functions. 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. The cuts in the hypercubes are exactly the transitions from 0 sets to 1 Recall Case 1: | S 0 | , | S 1 | ≤ | V | / 2 set on boolean functions on { 0 , 1 } n . | S 1 | ≤ | V 1 | / 2 since | S | ≤ | V | / 2. Have a nice weekend! = ⇒ ≥ | S 1 | edges cut in E 1 . Central area of study in computer science! | S 0 | ≥ | V 0 | / 2 = ⇒ | V 0 − S | ≤ | V 0 | / 2 Yes/No Computer Programs ≡ Boolean function on { 0 , 1 } n = ⇒ ≥ | V 0 |−| S 0 | edges cut in E 0 . 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. 22 / 24 23 / 24 24 / 24