Cse 20 discrete mathematics
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CSE 20 – Discrete Mathematics

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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,…


San diego road graph

San diego road graph


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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