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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 G = (V, E) be a loop free connected undirected graph
Let {a, b} be an edge in G
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Consider a problem
Let G = (V, E) be a loop free connected undirected graph
Let {a, b} be an edge in G
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
Let G = (V, E) be a loop free connected undirected graph
Let {a, b} be an edge in G
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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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}
Usually the easiest way to prove such a problem is to show
that both:
{Something} implies {something else}
And also
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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}
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?
Recall that ( ) is the same as (( ) ( ))
If you dont recall this, how could you check?
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Review Why Does This Template Work?
Recall that ( ) is the same as (( ) ( ))
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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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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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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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
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
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
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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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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Proving one of the implications
, ,
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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Proving one of the implications
, ,
For now, simply consider {a,b} as a way to get from a to b
Note that if {a,b} is part of a cycle then
P i f th i li ti
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Proving one of the implications
, ,
For now, simply consider {a,b} as a way to get from a to b
Note that if {a,b} is part of a cycle then there must also be another way toget from a to b!
Proving one of the implications
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Proving one of the implications
, ,
For now, simply consider {a,b} as a way to get from a to b
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
Proving one of the implications
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Proving one of the implications
, ,
How does this notion relate to our implication above?
Proving one of the implications
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Proving one of the implications
, ,
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
Proving one of the implications
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Proving one of the implications
, ,
If we remove {a,b}, this path from I to C no longer exists
Proving one of the implications
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Proving one of the implications
, ,
If we remove {a,b}, this path from I to C no longer exists
But since it is part of a cycle, we can always get from a to b a different way
Proving one of the implications
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Proving one of the implications
, ,
If we remove {a,b}, this path from I to C no longer exists
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
Proving one of the implications
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Proving one of the implications
, ,
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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Now for the other implication
, ,
If removing {a,b} does not disconnect graph, what do we know about
the relationship(s) between a and b?
Now for the other implication
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Now for the other implication
, ,
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}
Now for the other implication
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Now for the other implication
, ,
So consider:
1. A path from a to b that does not include {a, b}
2. {a, b}
What can we do with these?
Now for the other implication
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Now o t e ot e p cat o
, ,
So consider:
1. A path from a to b that does not include {a, b}
2. {a, b}
What can we do with these?
Walk from a to b using 1
Then walk from b to a using 2
What is this called?
Now for the other implication
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p
, ,
So consider:
1. A path from a to b that does not include {a, b}
2. {a, b}
What can we do with these?
Walk from a to b using 1
Then walk from b to a using 2
What is this called?
A closed walk!
And what can we form if we have a closed walk?
Now for the other implication
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p
, ,
So consider:
1. A path from a to b that does not include {a, b}
2. {a, b}
What can we do with these?
Walk from a to b using 1
Then walk from b to a using 2
What is this called?
A closed walk!
A cycle (using the methodology shown earlier)
Now for the other implication
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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)