Today. Complete Graph. K 4 and K 5 Types of graphs. K n complete - PowerPoint PPT Presentation

Today. Complete Graph. K 4 and K 5 Types of graphs. K n complete graph on n vertices. Complete Graphs. All edges are present. Trees. Everyone is my neighbor. Hypercubes. 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! Each vertex is incident to n − 1 edges. Prove it! Alas. Different course. Sum of degrees is n ( n − 1 ) . = ⇒ Number of edges is n ( n − 1 ) / 2. Remember sum of degree is 2 | E | . 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 ) . Lemma: If v is a degree 1 in connected graph G , G − v is connected. A connected graph where any edge addition creates a cycle. Proof: Binary Tree! 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

Proof of only if. Proof of if Tree’s fall apart. v Thm: Thm: Thm: Removing a single disconnects | V | / 2 nodes from each “G connected and has | V |− 1 edges” ≡ “G is connected and has no cycles” = ⇒ “G connected and has other. “G is connected and has no cycles.” | 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: Proof: First, connected = ⇒ every vertex degree ≥ 1. Idea of proof. 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. Average degree 2 − 2 / | V | Remove center node. 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. Hypercubes. Recursive Definition. Hypercube: Can’t cut me! Complete graphs, really connected! But lots of edges. | V | ( | V |− 1 ) / 2 Trees, But few edges. ( | V |− 1 ) A 0-dimensional hypercube is a node labelled with the empty string of just falls apart! bits. Hypercubes. Really connected. | V | log | V | edges! Thm: Any subset S of the hypercube where | S | ≤ | V | / 2 has An n -dimensional hypercube consists of a 0-subcube (1-subcube) Also represents bit-strings nicely. ≥ | S | 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 ) Terminology: with the additional edges ( 0 x , 1 x ) . | 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 001 011 0 1 edges is at least the size of the small side. 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!

Proof of Large Cuts. Induction Step Idea Induction Step Thm: For any cut ( S , V − S ) in the hypercube, the number of cut Thm: For any cut ( S , V − S ) in the hypercube, the number of edges is at least the size of the small side. cut edges is at least the size of the small side, | S | . Use recursive definition into two subcubes. Proof: Induction Step. Two cubes connected by edges. Thm: For any cut ( S , V − S ) in the hypercube, the number of Recursive definition: Case 1: Count edges inside cut 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 | . 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. Recall Case 1: | S 0 | , | S 1 | ≤ | V | / 2 The cuts in the hypercubes are exactly the transitions from 0 | S 1 | ≤ | V 1 | / 2 since | S | ≤ | V | / 2. sets to 1 set on boolean functions on { 0 , 1 } n . 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 = ⇒ ≥ | V 0 |−| S 0 | edges cut in E 0 . Yes/No Computer Programs ≡ Boolean function on { 0 , 1 } n Edges in E x connect corresponding nodes. Central object of study. = ⇒ = | 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.