a proof involving graphs

7/29/2019 A Proof Involving Graphs

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A Proof Involving GraphsCCNY CSc 104 Dave Chisholm [email protected]

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Consider a problem

Let G = (V, E) be a loop free connected undirected graph


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Consider a problem


Let {a, b} be an edge in G


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Consider a problem



Prove

{a,b} is part of a cycle if and only if its removal does not

disconnect G


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Consider a problem



Prove

{a,b} is part of a cycle if and only if its removal does not

disconnect G

Note we are not removing a and b from the set of vertices

Just removing {a,b} from the set of edges

(This is problem 11.1.9 from Pg. 519)


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First glance at the problem

The first thing to notice is that this problem fits a template:

PROVE {Something} IF AND ONLY IF {something else}


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Usually the easiest way to prove such a problem is to show

that both:

{Something} implies {something else}

And also


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Usually the easiest way to prove such a problem is to show

that both:

{Something} implies {something else}

And also

{Something else} implies {something}

Recall that ( ) is the same as (( ) ( ))

This approach works for all such proofs, not just graphs!


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Review Why Does This Template Work?


If you dont recall this, how could you check?


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Review Why Does This Template Work?


A truth table, of course!

Notice the truth values are the same for both statements

A B AB BA (AB) (BA) A B

T T T TT T

T F F T F F

F T T F F F

F F T T T T


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So what do we need to show here?

Lets just use our template and plug in

Who can finish this?

1.

2.


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1. , ,

2.


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1. , ,

2. , ,


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Great!

We can now just prove each of these implications one-at-a-time

Two smaller problems are usually easier to solve than one large one!

But each of these two statements are still tricky

Lets review a few graph concepts


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Graph Concepts

A graph is a collection of ?


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Graph Concepts

A graph is a collection ofvertices and ?


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Graph Concepts

A graph is a collection of vertices and edges


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Graph Concepts

In a connectedgraph, ?


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Graph Concepts

In a connected graph, every vertex can be reached from

every other one


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Graph Concepts

A loop is ?


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Graph Concepts

A loop is an edge from one vertex to itself

Often we ignore loops for simplicity


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Walks

Consider the types of walks we can have

Name

May have

repeated

vertices

May have

repeated

edges

Must start and

end on same

vertex

Open Walk Yes Yes No

Closed Walk Yes Yes Yes

Trail (Open) Yes No No

Circuit (Closed) Yes No Yes

Path (Open) No No No

Cycle (Closed) No No Yes

This is table 11.1 from page 516


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Walks

Consider the types of walks we can have

Note that any cycle is also a circuit which is also a closedwalk, but not visa versa

Name

May have

repeated

vertices

May have

repeated

edges

Must start and

end on same

vertex









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Walks

Likewise for open walks...

Any path is also a trail which is also an open walk But an open walk is not necessarily also a trail

Name

May have

repeated

vertices

May have

repeated

edges

Must start and

end on same

vertex









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More on Walks

More interestingly, note that we can turn any less specific

walk into a shorter, more specific walk


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More on Walks

More interestingly, note that we can turn any less specific

walk into a shorter, more specific walk

Consider the following closed walk (starting from vertexa)


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Walks

How can we turn this closed walk into a circuit?


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Walks

How can we turn this closed walk into a circuit?

Lets just remove this part!


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Walks

How can we turn this circuit into a cycle?


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Walks

How can we turn this circuit into a cycle?

Lets just remove this part!

All of this also applies to undirected graphs the pointed arrows were simply shown for

clarity


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Walks

We are left with a cycle

More importantly, now we have enough understanding to tackle our proof


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Um, what were we proving again?

Recall we need to show the two statements below are true

.

,

,

. , ,

U h i i ?


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Um, what were we proving again?

Recall we need to show the two statements below are true

.

,

,

. , ,

Personally, I find the second statement more intuitive

So lets start by showing that one

P i f h i li i


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Proving one of the implications

, ,

Consider the graph below

P i f th i li ti


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, ,

For now, simply consider {a,b} as a way to get from a to b

P i f th i li ti


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, ,


Note that if {a,b} is part of a cycle then

P i f th i li ti


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, ,


Note that if {a,b} is part of a cycle then there must also be another way toget from a to b!



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, ,


Note that if {a,b} is part of a cycle then there must also be another way toget from a to b!

All of this also applies to a way to get from b to a



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, ,

How does this notion relate to our implication above?



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, ,

A connected graph must have at least one path between all vertices

Some of these paths might include {a,b}

Such as the one below from I to C



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, ,

If we remove {a,b}, this path from I to C no longer exists



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, ,


But since it is part of a cycle, we can always get from a to b a different way



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, ,


But since it is part of a cycle, we can always get from a to b a different way

Meaning there will still be a path from I to C!

We just replace {a,b} in the original path with the remainder of the cycle



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, ,

Thus the implication above is true

Since any two vertices connected via a path that uses {a,b}

Are also connected via another path that uses the remainder of the cycle

Now for the other implication


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, ,

If removing {a,b} does not disconnect graph, what do we know about

the relationship(s) between a and b?



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, ,

If removing {a,b} does not disconnect graph, what do we know about

the relationship(s) between a and b?

There must be a path from a to b that does not include {a, b}

Otherwise, our graph would be disconnected after removing {a, b}



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, ,

So consider:

1. A path from a to b that does not include {a, b}

2. {a, b}

What can we do with these?



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Now o t e ot e p cat o

, ,

So consider:


2. {a, b}


Walk from a to b using 1

Then walk from b to a using 2

What is this called?



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p

, ,

So consider:


2. {a, b}





A closed walk!

And what can we form if we have a closed walk?



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p

, ,

So consider:


2. {a, b}





A closed walk!

A cycle (using the methodology shown earlier)



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p

, ,

Thus if removing {a,b} does not disconnect the graph, then there must

be some cycle that includes {a,b}

Weve now shown both implications are true, and thus our original

equivalence statement (aka if and only if) is true.

That was really long!


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y g

That was a very long proof

But only because I broke it out into very gradual slides :)

What would I expect in a homework?

Consider the following slide, which contains an informal write-up:

A Shorter Write-up


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p

To prove this equivalence, we must first show that

[{a,b} is part of a cycle] -> [removal of {a,b} does not disconnect graph]

Note that a cycle on a through {a,b} can be divided into two parts:

- {a,b}

- and a path from a to b that does not contain {a,b}

So any path connecting two vertices x and y that includes {a,b} can be replaced by another path that instead

uses the other path from a to b that does not contain {a,b}

Thus removing {a,b} does not disconnect the graph

We must also show that

[removal of {a,b} does not disconnect graph] -> [{a,b} is part of a cycle]

Note that if the graph is still connected after removing {a,b} then we know that

- There must be a path from b to a that does not contain {a,b}

If we take this path starting at b and ending at a, and then travel along {a,b} to return to b, we have formed a

closed path. We can always form a cycle from a closed path by eliminating redundant edges.

QED

Another approach


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The textbook answer key showed ! ! instead of

This is equivalent, but a bit terse and confusing

Nevertheless Id accept it Shown below for reference

(Yuck)

a proof involving graphs

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