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CSE 20 – Discrete Mathematics

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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

- Graphs
- Some theorems on graphs

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

- The fruits are the
- Graphs
- Vertices
- Edges
- Loops
- None/other/more than one

- The arrows are the
- Graphs
- Vertices
- Edges
- Loops
- None/other/more than one

- Is this graph
- Undirected
- Directed
- Both
- Neither
- None/other/more than one

- Which of the following is not a correct graph

A.

B.

C.

D. None/other/more than one

- 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)

- 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

- 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.

- 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

- 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

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

- 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 ?

- 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

- 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

- 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 v1vk then both have odd degrees. If v1=vk is the same vertex this is also has even degree. QED.

Another example (student self-study)

- 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

- E.g (proved in textbook)

- 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

- 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

- 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.

- Theorem: is irrational
- Proof (by contradiction).
THIS ONE IS MORE TRICKY.

TRY BY YOURSELF FIRST IN GROUPS.

- 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. ]

- 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.