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cse 20 discrete mathematics

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

today s topics
Today’s Topics:
  • Graphs
  • Some theorems on graphs
graphs
Graphs
  • Model relations between pairs of objects
  • Basic ingredient in many algorithms: Network routing, GPS guidance, Simulation of chemical reactions,…
graph terminology
Graph terminology
  • The fruits are the
  • Graphs
  • Vertices
  • Edges
  • Loops
  • None/other/more than one
graph terminology1
Graph terminology
  • The arrows are the
  • Graphs
  • Vertices
  • Edges
  • Loops
  • None/other/more than one
graph terminology2
Graph terminology
  • Is this graph
  • Undirected
  • Directed
  • Both
  • Neither
  • None/other/more than one
graph terminology3
Graph terminology
  • Which of the following is not a correct graph

A.

B.

C.

D. None/other/more than one

our first graph theorem
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)
our second graph theorem
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
our second graph theorem1
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.

eulerian graphs
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
eulerian graphs1
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
eulerian graphs2
Eulerian graphs
  • How can we check if a graph is Eulerian?
  • Check all possible paths
  • Stare and guess
  • Be brave and do some math
eulerian graphs3
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 ?
proving euler s theorem
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
proving euler s theorem necessary part
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

proving euler s theorem necessary part1
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.

proof by contradiction
Proof by contradiction

Another example (student self-study)

example 2
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.
example 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
example 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

example 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.

example 3
Example 3
  • Theorem: is irrational
  • Proof (by contradiction).

THIS ONE IS MORE TRICKY.

TRY BY YOURSELF FIRST IN GROUPS.

example 31
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. ]

example 32
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.

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