CSE 20 – Discrete Mathematics Dr. Cynthia Bailey Lee Dr. Shachar Lovett Peer Instruction in Discrete Mathematics by Cynthia Leeis licensed under a Creative Commons Attribution- NonCommercial-ShareAlike 4.0 International License. Based on a work at Permissions beyond the scope of this license may be available at LeeCreative Commons Attribution- NonCommercial-ShareAlike 4.0 International Licensehttp://peerinstruction4cs.org

Today’s Topics: 1. Graphs 2. Some theorems on graphs 2

Graphs  Model relations between pairs of objects  Basic ingredient in many algorithms: Network routing, GPS guidance, Simulation of chemical reactions,… 3

San diego road graph 4

Graph terminology  The fruits are the A. Graphs B. Vertices C. Edges D. Loops E. None/other/more than one 5

Graph terminology  The arrows are the A. Graphs B. Vertices C. Edges D. Loops E. None/other/more than one 6

Graph terminology  Is this graph A. Undirected B. Directed C. Both D. Neither E. None/other/more than one 7

Graph terminology  Which of the following is not a correct graph 8 A.B.C. D. None/other/more than one

Our first graph theorem  We already saw a theorem about graphs!  Recall: in any group of 6 people, either there are 3 that form a club, or 3 that are strangers  Any graph with 6 vertices contains either a triangle (3 vertices all connected) or an empty triangle (3 vertices not connected) 9

Our second graph theorem  Let G be an undirected graph  Degree of a vertex – number of edges adjacent to it (e.g. touch it)  Denote it by degree(v)  Theorem: in any undirected graph, the sum of all the degrees is even  Try and prove yourself first 10

Our second graph theorem  Theorem: in any undirected graph, the sum of all the degrees is even  Proof: Consider pairs (v,e) with v a vertex and e an edge adjacent to it. Create a list of all such pairs. How many elements this list has? We calculate it in two ways 1. Each vertex v has degree(v) edges adjacent to it, so this list has sum of degrees many elements 2. Each edge has 2 vertices adjacent to it, so this list has twice the number of edges many elements So, sum of degrees = twice the number of edges, hence it must be even. QED. 11

Eulerian graphs  Let G be an undirected graph  A graph is Eulerian if it can drawn without lifting the pen and without repeating edges  Is this graph Eulerian? A. Yes B. No 12

Eulerian graphs  Let G be an undirected graph  A graph is Eulerian if it can drawn without lifting the pen and without repeating edges  What about this graph A. Yes B. No 13

Eulerian graphs  How can we check if a graph is Eulerian? A. Check all possible paths B. Stare and guess C. Be brave and do some math 14

Eulerian graphs  Degree of a vertex: number of edges adjacent to it  Euler’s theorem: a graph is Eulerian iff the number of vertices with odd degrees is either 0 or 2 (eg all vertices or all but two have even degrees)  Does it work for and ? 15

Proving Euler’s theorem  Euler’s theorem gives a necessary and sufficient condition for a graph to be Eulerian  All degrees are even  Two degrees odd, rest are even  Will prove in class that this is necessary  Take-home challenge: prove that this is also sufficient 16

Proving Euler’s theorem: necessary part  Euler’s theorem (necessary part): If a graph G is Eulerian then all degrees are even; or two degrees are odd and rest are even Try to prove it first yourself 17

Proving Euler’s theorem: necessary part  Proof of Euler’s theorem (necessary part): Let G be a graph with an Euler path: v 1,v 2,v 3,….,v k where (v i,v i+1 ) are edges in G; vertices may appear more than once; and each edge of G is accounted for exactly once. The degree of a vertex of G is the number of edges it has. For any internal vertex in the path (eg not v 1 or v k ), we count 2 edges in the path (one going in and one going out). So, any vertex which is not v 1 or v k must have an even degree. If v1  vk then both have odd degrees. If v1=vk is the same vertex this is also has even degree. QED. 18

Proof by contradiction Another example (student self-study) 19

Example 2  A number x is rational if x=a/b for integers a,b.  E.g. 3=3/1, 1/2, -3/4, 0=0/1  A number is irrational if it is not rational  E.g (proved in textbook)  Theorem: If x 2 is irrational then x is irrational. 20

Example 2  Theorem: If x 2 is irrational then x is irrational.  Proof: by contradiction. Assume that A. There exists x where both x,x 2 are rational B. There exists x where both x,x 2 are irrational C. There exists x where x is rational and x 2 irrational D. There exists x where x is irrational and x 2 rational E. None/other/more than one 21

Example 2  Theorem: If x 2 is irrational then x is irrational.  Proof: by contradiction. Assume that there exists x where x is rational and x 2 irrational. 22 Try by yourself first

Example 2  Theorem: If x 2 is irrational then x is irrational.  Proof: by contradiction. Assume that there exists x where x is rational and x 2 irrational. 23 Since x is rational x=a/b where a,b are integers. But then x 2 =a 2 /b 2. Both a 2,b 2 are also integers and hence x 2 is rational. A contracition.

Example 3  Theorem: is irrational  Proof (by contradiction). THIS ONE IS MORE TRICKY. TRY BY YOURSELF FIRST IN GROUPS. 24

Example 3  Theorem: is irrational  Proof (by contradiction).  Assume not. Then there exist integers a,b such that  Squaring gives So also is rational since [So, to finish the proof it is sufficient to show that is irrational. ] 25

Example 3  Theorem: is irrational  Proof (by contradiction).  is rational … is rational.  =c/d for positive integers c,d. Assume that d is minimal such that c/d= Squaring gives c 2 /d 2 =6. So c 2 =6d 2 must be divisible by 2. Which means c is divisible by 2. Which means c 2 is divisible by 4. But 6 is not divisible by 4, so d 2 must be divisible by 2. Which means d is divisible by 2. So both c,d are divisible by 2. Which means that (c/2) and (d/2) are both integers, and (c/2) / (d/2) = Contradiction to the minimality of d. 26