CSE 20 – Discrete Mathematics

1. 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 http://peerinstruction4cs.org.Permissions beyond the scope of this license may be available at http://peerinstruction4cs.org. CSE 20 – Discrete Mathematics Dr. Cynthia Bailey Lee Dr. Shachar Lovett

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

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

4. San diego road graph

5. Graph terminology • The fruits are the • Graphs • Vertices • Edges • Loops • None/other/more than one

6. Graph terminology • The arrows are the • Graphs • Vertices • Edges • Loops • None/other/more than one

7. Graph terminology • Is this graph • Undirected • Directed • Both • Neither • None/other/more than one

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

9. 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)

10. 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

11. 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 • Each vertex v has degree(v) edges adjacent to it, so this list has sum of degrees many elements • 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.

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 • Is this graph Eulerian? • Yes • No

13. 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 • Yes • No

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

15. Eulerian graphs • Degree of a vertex: number of edges adjacent to it • Euler’s theorem: a graph is Eulerianiff 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 ?

16. 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

17. 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

18. Proving Euler’s theorem: necessary part • Proof of Euler’s theorem (necessary part): Let G be a graph with an Euler path: v1,v2,v3,….,vk where (vi,vi+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 v1 or vk), we count 2 edges in the path (one going in and one going out). So, any vertex which is not v1 or vk 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.

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

20. 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 x2 is irrational then x is irrational.

21. Example 2 • Theorem: If x2 is irrational then x is irrational. • Proof: by contradiction. Assume that • There exists x where both x,x2 are rational • There exists x where both x,x2 are irrational • There exists x where x is rational and x2 irrational • There exists x where x is irrational and x2 rational • None/other/more than one

22. Example 2 • Theorem: If x2 is irrational then x is irrational. • Proof: by contradiction. Assume that there exists x where x is rational and x2 irrational. Try by yourself first

23. Example 2 • Theorem: If x2 is irrational then x is irrational. • Proof: by contradiction. Assume that there exists x where x is rational and x2 irrational. Since x is rational x=a/b where a,b are integers. But then x2=a2/b2. Both a2,b2 are also integers and hence x2 is rational. A contracition.

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

25. 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. ]

26. 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 c2/d2=6. So c2=6d2 must be divisible by 2. Which means c is divisible by 2. Which means c2 is divisible by 4. But 6 is not divisible by 4, so d2 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.