· Quick Proof. The sum of the vertex degrees is equal to (A) the total number of vertices, jVj....

Lecture 5: Graphs.

Graphs!

EulerDefinitions: model.Fact!Euler Again!!

Planar graphs.Euler Again!!!!

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Lecture 5: Graphs.

Graphs!Euler

Definitions: model.Fact!Euler Again!!


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Lecture 5: Graphs.

Graphs!EulerDefinitions: model.

Fact!Euler Again!!


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Lecture 5: Graphs.

Graphs!EulerDefinitions: model.Fact!

Euler Again!!Planar graphs.

Euler Again!!!!

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Lecture 5: Graphs.

Graphs!EulerDefinitions: model.Fact!Euler Again!!


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Lecture 5: Graphs.


Planar graphs.

Euler Again!!!!

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Lecture 5: Graphs.



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Konigsberg bridges problem.

Can you make a tour visiting each bridge exactly once?

“Konigsberg bridges” by Bogdan Giusca - License.

A

B

C

D

Can you draw a tour in the graph where you visit each edge once?

Yes? No?We will see!

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https://commons.wikimedia.org/wiki/File:Konigsberg_bridges.png#/media/File:Konigsberg_bridges.png




A

B

C

D

Can you draw a tour in the graph where you visit each edge once?Yes?

No?We will see!

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A

B

C

D

Can you draw a tour in the graph where you visit each edge once?Yes? No?

We will see!

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A

B

C

D

Can you draw a tour in the graph where you visit each edge once?Yes? No?We will see!

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Graphs: formally.

AB C

D

AB C

D

Graph:

G = (V ,E).V - set of vertices.{A,B,C,D}

E ⊆ V ×V - set of edges.{{A,B},{A,B},{A,C},{B,C},{B,D},{B,D},{C,D}}.

For CS 70, usually simple graphs.No parallel edges.

Multigraph above.

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Graphs: formally.

AB C

D

AB C

D

Graph: G = (V ,E).

V - set of vertices.{A,B,C,D}



Multigraph above.

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Graphs: formally.

AB C

D

AB C

D

Graph: G = (V ,E).V - set of vertices.

{A,B,C,D}E ⊆ V ×V - set of edges.{{A,B},{A,B},{A,C},{B,C},{B,D},{B,D},{C,D}}.


Multigraph above.

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Graphs: formally.

AB C

D

AB C

D

Graph: G = (V ,E).V - set of vertices.{A,B,C,D}



Multigraph above.

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Graphs: formally.

AB C

D

AB C

D


E ⊆ V ×V -

set of edges.{{A,B},{A,B},{A,C},{B,C},{B,D},{B,D},{C,D}}.


Multigraph above.

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Graphs: formally.

AB C

D

AB C

D


E ⊆ V ×V - set of edges.

{{A,B},{A,B},{A,C},{B,C},{B,D},{B,D},{C,D}}.For CS 70, usually simple graphs.

No parallel edges.Multigraph above.

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Graphs: formally.

AB C

D

AB C

D


E ⊆ V ×V - set of edges.{{A,B}

,{A,B},{A,C},{B,C},{B,D},{B,D},{C,D}}.For CS 70, usually simple graphs.


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Graphs: formally.

AB C

D

AB C

D


E ⊆ V ×V - set of edges.{{A,B},{A,B}

,{A,C},{B,C},{B,D},{B,D},{C,D}}.For CS 70, usually simple graphs.


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Graphs: formally.

AB C

D

AB C

D


E ⊆ V ×V - set of edges.{{A,B},{A,B},{A,C},

{B,C},{B,D},{B,D},{C,D}}.For CS 70, usually simple graphs.


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Graphs: formally.

AB C

D

AB C

D




Multigraph above.

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Graphs: formally.

AB C

D

AB C

D



For CS 70, usually simple graphs.


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Graphs: formally.

AB C

D

AB C

D




Multigraph above.

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Directed Graphs

12 3

4

G = (V ,E).

V - set of vertices.{1,2,3,4}

E ordered pairs of vertices.{(1,2),(1,3),(1,4),(2,4),(3,4)}

One way streets.Tournament: 1 beats 2, ...Precedence: 1 is before 2, ..

Social Network: Directed? Undirected?Friends. Undirected.Likes. Directed.

Sometimes both ways!

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Directed Graphs

12 3

4

G = (V ,E).V - set of vertices.

{1,2,3,4}E ordered pairs of vertices.{(1,2),(1,3),(1,4),(2,4),(3,4)}




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Directed Graphs

12 3

4

G = (V ,E).V - set of vertices.{1,2,3,4}





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Directed Graphs

12 3

4


E ordered pairs of vertices.

{(1,2),(1,3),(1,4),(2,4),(3,4)}




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Directed Graphs

12 3

4


E ordered pairs of vertices.{(1,2),

(1,3),(1,4),(2,4),(3,4)}




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Directed Graphs

12 3

4


E ordered pairs of vertices.{(1,2),(1,3),

(1,4),(2,4),(3,4)}




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Directed Graphs

12 3

4


E ordered pairs of vertices.{(1,2),(1,3),(1,4),

(2,4),(3,4)}




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Directed Graphs

12 3

4






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Directed Graphs

12 3

4



One way streets.

Tournament: 1 beats 2, ...Precedence: 1 is before 2, ..



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Directed Graphs

12 3

4



One way streets.Tournament:

1 beats 2, ...Precedence: 1 is before 2, ..



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Directed Graphs

12 3

4



One way streets.Tournament: 1 beats 2,

...Precedence: 1 is before 2, ..



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Directed Graphs

12 3

4



One way streets.Tournament: 1 beats 2, ...Precedence:

1 is before 2, ..



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Directed Graphs

12 3

4



One way streets.Tournament: 1 beats 2, ...Precedence: 1 is before 2,

..



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Directed Graphs

12 3

4






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Directed Graphs

12 3

4




Social Network:

Directed? Undirected?Friends. Undirected.Likes. Directed.


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Directed Graphs

12 3

4




Social Network: Directed?

Undirected?Friends. Undirected.Likes. Directed.


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Directed Graphs

12 3

4




Social Network: Directed? Undirected?

Friends. Undirected.Likes. Directed.


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Directed Graphs

12 3

4




Social Network: Directed? Undirected?Friends.

Undirected.Likes. Directed.


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Directed Graphs

12 3

4




Social Network: Directed? Undirected?Friends. Undirected.

Likes. Directed.Sometimes both ways!

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Directed Graphs

12 3

4




Social Network: Directed? Undirected?Friends. Undirected.Likes.

Directed.Sometimes both ways!

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Directed Graphs

12 3

4






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Graph Concepts and Definitions.Graph: G = (V ,E)

neighbors, adjacent, degree, incident, in-degree, out-degree

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Neighbors of 10? 1, 5, 7, 8.u is neighbor of v if {u,v} ∈ E .

Edge {10,5} is incident to vertex 10 and vertex 5.Edge {u,v} is incident to u and v .

Degree of vertex 1? 2Degree of vertex u is number of incident edges.Equals number of neighbors in simple graph.

Directed graph?In-degree of 10? 1 Out-degree of 10? 3

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Neighbors of 10?

1, 5, 7, 8.u is neighbor of v if {u,v} ∈ E .




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910

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910

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Neighbors of 10? 1,

5, 7, 8.u is neighbor of v if {u,v} ∈ E .




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910

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910

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Neighbors of 10? 1, 5,

7, 8.u is neighbor of v if {u,v} ∈ E .




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910

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910

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Neighbors of 10? 1, 5, 7,

8.u is neighbor of v if {u,v} ∈ E .




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910

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910

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Neighbors of 10? 1, 5, 7, 8.

u is neighbor of v if {u,v} ∈ E .Edge {10,5} is incident to vertex 10 and vertex 5.

Edge {u,v} is incident to u and v .Degree of vertex 1? 2

Degree of vertex u is number of incident edges.Equals number of neighbors in simple graph.


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910

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Edge {10,5} is incident to

vertex 10 and vertex 5.Edge {u,v} is incident to u and v .



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1 2

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910

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1

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910

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Degree of vertex 1?

2Degree of vertex u is number of incident edges.Equals number of neighbors in simple graph.


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Degree of vertex 1? 2

Degree of vertex u is number of incident edges.Equals number of neighbors in simple graph.


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910

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910

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Degree of vertex 1? 2Degree of vertex u is number of incident edges.

Equals number of neighbors in simple graph.


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Directed graph?

In-degree of 10? 1 Out-degree of 10? 3

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Directed graph?In-degree of 10?

1 Out-degree of 10? 3

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Directed graph?In-degree of 10? 1

Out-degree of 10? 3

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Directed graph?In-degree of 10? 1 Out-degree of 10?

3

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Quick Proof.The sum of the vertex degrees is equal to

(A) the total number of vertices, |V |.(B) the total number of edges, |E |.(C) What?

Not (A)! Triangle.

Not (B)! Triangle.

What? For triangle number of edges is 3, the sum of degrees is 6.

Could it always be...2|E |? ..or 2|V |?How many incidences does each edge contribute? 2.

2|E | incidences are contributed in total!What is degree v? incidences contributed to v !sum of degrees is total incidences ... or 2|E |.

Thm: Sum of vertex degress is 2|E |.

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(A) the total number of vertices, |V |.

(B) the total number of edges, |E |.(C) What?

Not (A)! Triangle.

Not (B)! Triangle.





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(A) the total number of vertices, |V |.(B) the total number of edges, |E |.

(C) What?

Not (A)! Triangle.

Not (B)! Triangle.





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Not (A)! Triangle.

Not (B)! Triangle.





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Not (A)!

Triangle.

Not (B)! Triangle.





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Not (A)! Triangle.

Not (B)! Triangle.





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Not (A)! Triangle.

Not (B)!

Triangle.





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Not (A)! Triangle.

Not (B)! Triangle.





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Not (A)! Triangle.

Not (B)! Triangle.

What?

For triangle number of edges is 3, the sum of degrees is 6.




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Not (A)! Triangle.

Not (B)! Triangle.





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Not (A)! Triangle.

Not (B)! Triangle.


Could it always be...

2|E |? ..or 2|V |?How many incidences does each edge contribute? 2.



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Not (A)! Triangle.

Not (B)! Triangle.


Could it always be...2|E |? ..

or 2|V |?How many incidences does each edge contribute? 2.



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Not (A)! Triangle.

Not (B)! Triangle.


Could it always be...2|E |? ..or 2|V |?

How many incidences does each edge contribute? 2.2|E | incidences are contributed in total!

What is degree v? incidences contributed to v !sum of degrees is total incidences ... or 2|E |.


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Not (A)! Triangle.

Not (B)! Triangle.


Could it always be...2|E |? ..or 2|V |?How many incidences does each edge contribute?

2.2|E | incidences are contributed in total!



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Not (A)! Triangle.

Not (B)! Triangle.





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Not (A)! Triangle.

Not (B)! Triangle.



2|E | incidences are contributed in total!



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Not (A)! Triangle.

Not (B)! Triangle.



2|E | incidences are contributed in total!What is degree v?

incidences contributed to v !sum of degrees is total incidences ... or 2|E |.


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Not (A)! Triangle.

Not (B)! Triangle.



2|E | incidences are contributed in total!What is degree v? incidences contributed to v !

sum of degrees is total incidences ... or 2|E |.Thm: Sum of vertex degress is 2|E |.

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Not (A)! Triangle.

Not (B)! Triangle.



2|E | incidences are contributed in total!What is degree v? incidences contributed to v !sum of degrees is total incidences

... or 2|E |.Thm: Sum of vertex degress is 2|E |.

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Not (A)! Triangle.

Not (B)! Triangle.





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Paths, walks, cycles, tour.

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A path in a graph is a sequence of edges.

Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!

Path: (v1,v2),(v2,v3), . . .(vk−1,vk ).Quick Check! Length of path? k vertices or k −1 edges.

Cycle: Path with v1 = vk . Length of cycle? k −1 vertices and edges!Path is usually simple. No repeated vertex!

Walk is sequence of edges with possible repeated vertex or edge.Tour is walk that starts and ends at the same node.

Quick Check!Path is to Walk as Cycle is to ?? Tour!

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Path?

{1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!





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Path? {1,10}, {8,5}, {4,5} ?

No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!





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Path? {1,10}, {8,5}, {4,5} ? No!

Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!Path: (v1,v2),(v2,v3), . . .(vk−1,vk ).

Quick Check! Length of path? k vertices or k −1 edges.Cycle: Path with v1 = vk . Length of cycle? k −1 vertices and edges!Path is usually simple. No repeated vertex!



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Path? {1,10}, {8,5}, {4,5} ? No!Path?

{1,10}, {10,5}, {5,4}, {4,11}? Yes!Path: (v1,v2),(v2,v3), . . .(vk−1,vk ).




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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}?

Yes!Path: (v1,v2),(v2,v3), . . .(vk−1,vk ).




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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!





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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!

Path: (v1,v2),(v2,v3), . . .(vk−1,vk ).




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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!

Path: (v1,v2),(v2,v3), . . .(vk−1,vk ).Quick Check!

Length of path? k vertices or k −1 edges.Cycle: Path with v1 = vk . Length of cycle? k −1 vertices and edges!Path is usually simple. No repeated vertex!



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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!

Path: (v1,v2),(v2,v3), . . .(vk−1,vk ).Quick Check! Length of path?

k vertices or k −1 edges.Cycle: Path with v1 = vk . Length of cycle? k −1 vertices and edges!Path is usually simple. No repeated vertex!



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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!

Path: (v1,v2),(v2,v3), . . .(vk−1,vk ).Quick Check! Length of path? k vertices

or k −1 edges.Cycle: Path with v1 = vk . Length of cycle? k −1 vertices and edges!Path is usually simple. No repeated vertex!



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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!





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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!


Cycle: Path with v1 = vk .

Length of cycle? k −1 vertices and edges!Path is usually simple. No repeated vertex!



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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!


Cycle: Path with v1 = vk . Length of cycle?

k −1 vertices and edges!Path is usually simple. No repeated vertex!



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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!


Cycle: Path with v1 = vk . Length of cycle? k −1 vertices and edges!

Path is usually simple. No repeated vertex!Walk is sequence of edges with possible repeated vertex or edge.Tour is walk that starts and ends at the same node.


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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!


Cycle: Path with v1 = vk . Length of cycle? k −1 vertices and edges!Path is usually simple.

No repeated vertex!Walk is sequence of edges with possible repeated vertex or edge.Tour is walk that starts and ends at the same node.


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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!





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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!



Walk is sequence of edges with possible repeated vertex or edge.

Tour is walk that starts and ends at the same node.


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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!





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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!




Quick Check!

Path is to Walk as Cycle is to ?? Tour!

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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!




Quick Check!Path is to Walk as Cycle is to ??

Tour!

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Path? {1,10}, {8,5}, {4,5} ? No!Path? {1,10}, {10,5}, {5,4}, {4,11}? Yes!





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Directed Paths.

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Path: (v1,v2),(v2,v3), . . .(vk−1,vk ).Paths, walks, cycles, tours ... are analagous to undirected now.

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Directed Paths.

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Path: (v1,v2),(v2,v3), . . .(vk−1,vk ).

Paths, walks, cycles, tours ... are analagous to undirected now.

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Directed Paths.

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Path: (v1,v2),(v2,v3), . . .(vk−1,vk ).Paths,

walks, cycles, tours ... are analagous to undirected now.

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Directed Paths.

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Path: (v1,v2),(v2,v3), . . .(vk−1,vk ).Paths, walks,

cycles, tours ... are analagous to undirected now.

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Directed Paths.

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Path: (v1,v2),(v2,v3), . . .(vk−1,vk ).Paths, walks, cycles,

tours ... are analagous to undirected now.

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Directed Paths.

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Path: (v1,v2),(v2,v3), . . .(vk−1,vk ).Paths, walks, cycles, tours

... are analagous to undirected now.

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Directed Paths.

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Path: (v1,v2),(v2,v3), . . .(vk−1,vk ).Paths, walks, cycles, tours ... are analagous to undirected now.

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Connectivity: undirected graph.

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u and v are connected if there is a path (or walk) between u and v .

A connected graph is a graph where all pairs of vertices areconnected.

If one vertex x is connected to every other vertex.Is graph connected? Yes? No?

Proof:Any u,v : path from u to x and then from x to v is u−v walk.

May not be simple!Either modify definition to walk.Or cut out cycles. .

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If one vertex x is connected to every other vertex.

Is graph connected? Yes? No?



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If one vertex x is connected to every other vertex.Is graph connected?

Yes? No?



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If one vertex x is connected to every other vertex.Is graph connected? Yes?

No?



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Proof:

Any u,v : path from u to x and then from x to v is u−v walk.


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Proof:Any u,v :

path from u to x and then from x to v is u−v walk.


9 / 20


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May not be simple!

Either modify definition to walk.Or cut out cycles. .

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May not be simple!Either modify definition to walk.

Or cut out cycles. .

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May not be simple!Either modify definition to walk.Or cut out cycles.

.

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Connected Components

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Is graph above connected?

Yes!

How about now? No!

Connected Components? {1},{10,7,5,8,4,3,11},{2,9,6}.Connected component - maximal set of connected vertices.

Quick Check: Is {10,7,5} a connected component? No.

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Is graph above connected? Yes!

How about now? No!



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How about now?

No!



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How about now? No!



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How about now? No!

Connected Components?

{1},{10,7,5,8,4,3,11},{2,9,6}.Connected component - maximal set of connected vertices.


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How about now? No!

Connected Components? {1},{10,7,5,8,4,3,11},{2,9,6}.

Connected component - maximal set of connected vertices.Quick Check: Is {10,7,5} a connected component? No.

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How about now? No!



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How about now? No!


Quick Check: Is {10,7,5} a connected component?

No.

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How about now? No!



10 / 20

Finally..back to Euler!

An Eulerian Tour is a tour that visits each edge exactly once.

Theorem: Any undirected graph has an Eulerian tour if and only if allvertices have even degree and is connected.Proof of only if: Eulerian =⇒ connected and all even degree.

Eulerian Tour is connected so graph is connected.Tour enters and leaves vertex v on each visit.Uses two incident edges per visit. Tour uses all incident edges.Therefore v has even degree.

When you enter, you can leave.For starting node, tour leaves first ....then enters at end.

Not The Hotel California.

11 / 20

https://www.youtube.com/watch?v=Sm3b4_XLCOU



Theorem: Any undirected graph has an Eulerian tour if and only if allvertices have even degree and is connected.

Proof of only if: Eulerian =⇒ connected and all even degree.




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Eulerian Tour is connected so graph is connected.

Tour enters and leaves vertex v on each visit.Uses two incident edges per visit. Tour uses all incident edges.Therefore v has even degree.



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Eulerian Tour is connected so graph is connected.Tour enters and leaves vertex v on each visit.

Uses two incident edges per visit. Tour uses all incident edges.Therefore v has even degree.



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Eulerian Tour is connected so graph is connected.Tour enters and leaves vertex v on each visit.Uses two incident edges per visit.

Tour uses all incident edges.Therefore v has even degree.



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Eulerian Tour is connected so graph is connected.Tour enters and leaves vertex v on each visit.Uses two incident edges per visit. Tour uses all incident edges.

Therefore v has even degree.



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When you enter,

you can leave.For starting node, tour leaves first ....then enters at end.


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When you enter, you can leave.

For starting node, tour leaves first ....then enters at end.Not The Hotel California.

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When you enter, you can leave.For starting node,

tour leaves first ....then enters at end.Not The Hotel California.

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When you enter, you can leave.For starting node, tour leaves first

....then enters at end.Not The Hotel California.

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Finding a tour!

Proof of if: Even + connected =⇒ Eulerian Tour.We will give an algorithm.

First by picture.

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1. Start walk from v (1) on “unused” edges... till you get back to v .2. Remove tour, C.3. Let G1, . . . ,Gk be connected components.

Each is touched by C: Vi ∩C 6= φ .Why? G was connected.

Let vi be (first) node in Gi touched by C.Example: v1 = 1, v2 = 10, v3 = 4, v4 = 2.

4. Recurse on G1, . . . ,Gk starting from vi5. Splice together.

1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!

Proof of if: Even + connected =⇒ Eulerian Tour.We will give an algorithm. First by picture.

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1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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1. Start walk from v (1) on “unused” edges

... till you get back to v .2. Remove tour, C.3. Let G1, . . . ,Gk be connected components.




1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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1. Start walk from v (1) on “unused” edges... till you get back to v .

2. Remove tour, C.3. Let G1, . . . ,Gk be connected components.




1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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1. Start walk from v (1) on “unused” edges... till you get back to v .2. Remove tour, C.

3. Let G1, . . . ,Gk be connected components.Each is touched by C: Vi ∩C 6= φ .

Why? G was connected.Let vi be (first) node in Gi touched by C.

Example: v1 = 1, v2 = 10, v3 = 4, v4 = 2.4. Recurse on G1, . . . ,Gk starting from vi5. Splice together.

1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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Each is touched by C: Vi ∩C 6= φ .

Why? G was connected.Let vi be (first) node in Gi touched by C.


1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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Each is touched by C: Vi ∩C 6= φ .Why?

G was connected.Let vi be (first) node in Gi touched by C.


1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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Let vi be (first) node in Gi touched by C.


1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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Let vi be (first) node in Gi touched by C.Example: v1 = 1,

v2 = 10, v3 = 4, v4 = 2.4. Recurse on G1, . . . ,Gk starting from vi5. Splice together.

1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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Let vi be (first) node in Gi touched by C.Example: v1 = 1, v2 = 10,

v3 = 4, v4 = 2.4. Recurse on G1, . . . ,Gk starting from vi5. Splice together.

1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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Let vi be (first) node in Gi touched by C.Example: v1 = 1, v2 = 10, v3 = 4,

v4 = 2.4. Recurse on G1, . . . ,Gk starting from vi5. Splice together.

1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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4. Recurse on G1, . . . ,Gk starting from vi

5. Splice together.1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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1,10

,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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1,10,7,8,5,10

,8,4,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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1,10,7,8,5,10 ,8,4

,3,11,4 5,2,6,9,2 and to 1!

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Finding a tour!


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5

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1,10,7,8,5,10 ,8,4,3,11,4

5,2,6,9,2 and to 1!

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Finding a tour!


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3

4

5

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1,10,7,8,5,10 ,8,4,3,11,4 5,2

,6,9,2 and to 1!

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Finding a tour!


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1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2

and to 1!

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Finding a tour!


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1,10,7,8,5,10 ,8,4,3,11,4 5,2,6,9,2 and to 1!

12 / 20

Recursive/Inductive Algorithm.1. Take a walk from arbitrary node v , until you get back to v .

Claim: Do get back to v !Proof of Claim: Even degree. If enter, can leave except for v .

2. Remove cycle, C, from G.Resulting graph may be disconnected. (Removed edges!)Let components be G1, . . . ,Gk .Let vi be first vertex of C that is in Gi .

Why is there a vi in C?G was connected =⇒

a vertex in Gi must be incident to a removed edge in C.

Claim: Each vertex in each Gi has even degree and is connected.Prf: Tour C has even incidences to any vertex v .

3. Find tour Ti of Gi starting/ending at vi . Induction.4. Splice Ti into C where vi first appears in C.

Visits every edge once:Visits edges in C exactly once.By induction for all edges in each Gi .

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Claim: Do get back to v !

Proof of Claim: Even degree. If enter, can leave except for v .







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Claim: Do get back to v !Proof of Claim: Even degree.

If enter, can leave except for v .







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Claim: Do get back to v !Proof of Claim: Even degree. If enter, can leave

except for v .







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2. Remove cycle, C, from G.

Resulting graph may be disconnected. (Removed edges!)Let components be G1, . . . ,Gk .Let vi be first vertex of C that is in Gi .






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2. Remove cycle, C, from G.Resulting graph may be disconnected. (Removed edges!)

Let components be G1, . . . ,Gk .Let vi be first vertex of C that is in Gi .






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2. Remove cycle, C, from G.Resulting graph may be disconnected. (Removed edges!)Let components be G1, . . . ,Gk .

Let vi be first vertex of C that is in Gi .Why is there a vi in C?

G was connected =⇒a vertex in Gi must be incident to a removed edge in C.




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Why is there a vi in C?

G was connected =⇒a vertex in Gi must be incident to a removed edge in C.




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Claim: Each vertex in each Gi has even degree

and is connected.Prf: Tour C has even incidences to any vertex v .



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Claim: Each vertex in each Gi has even degree and is connected.

Prf: Tour C has even incidences to any vertex v .



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3. Find tour Ti of Gi

starting/ending at vi . Induction.4. Splice Ti into C where vi first appears in C.


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3. Find tour Ti of Gi starting/ending at vi .

Induction.4. Splice Ti into C where vi first appears in C.


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3. Find tour Ti of Gi starting/ending at vi . Induction.

4. Splice Ti into C where vi first appears in C.


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Visits every edge once:Visits edges in C

exactly once.By induction for all edges in each Gi .

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Visits every edge once:Visits edges in C exactly once.

By induction for all edges in each Gi .

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Break time!

Well admin time!

Must choose homework option or test only: soon after recieving hw 1scores.

Test Option: don’t have to do homework. Yes!!Should do homework. No need to write up.

Homework Option: have to do homework. Bummer!

The truth: mostly test, both options!Variance mostly in exams for A/B range.most homework students get near perfect scores on homework.

How will I do?

Mostly up to you.

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https://piazza.com/class/jqq2huwjsw25sy?cid=173

Break time!

Well admin time!


Test Option: don’t have to do homework.

Yes!!Should do homework. No need to write up.



How will I do?

Mostly up to you.

14 / 20


Break time!

Well admin time!


Test Option: don’t have to do homework. Yes!!

Should do homework. No need to write up.Homework Option: have to do homework. Bummer!


How will I do?

Mostly up to you.

14 / 20


Break time!

Well admin time!


Test Option: don’t have to do homework. Yes!!Should do homework.

No need to write up.Homework Option: have to do homework. Bummer!


How will I do?

Mostly up to you.

14 / 20


Break time!

Well admin time!





How will I do?

Mostly up to you.

14 / 20


Break time!

Well admin time!



Homework Option: have to do homework.

Bummer!


How will I do?

Mostly up to you.

14 / 20


Break time!

Well admin time!





How will I do?

Mostly up to you.

14 / 20


Break time!

Well admin time!




The truth:

mostly test, both options!Variance mostly in exams for A/B range.most homework students get near perfect scores on homework.

How will I do?

Mostly up to you.

14 / 20


Break time!

Well admin time!




The truth: mostly test,

both options!Variance mostly in exams for A/B range.most homework students get near perfect scores on homework.

How will I do?

Mostly up to you.

14 / 20


Break time!

Well admin time!




The truth: mostly test, both options!

Variance mostly in exams for A/B range.most homework students get near perfect scores on homework.

How will I do?

Mostly up to you.

14 / 20


Break time!

Well admin time!




The truth: mostly test, both options!Variance mostly in exams for A/B range.

most homework students get near perfect scores on homework.

How will I do?

Mostly up to you.

14 / 20


Planar graphs.A graph that can be drawn in the plane without edge crossings.

Planar? Yes for Triangle.Four node complete? Yes.

Note: Complete means every possible edge is present, also clique.Five node complete or K5? No! Why? Later.

Two to three nodes, bipartite? Yes.Three to three nodes, complete/bipartite or K3,3. No. Why? Later.

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Planar?

Yes for Triangle.Four node complete? Yes.



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Planar? Yes for Triangle.

Four node complete? Yes.Note: Complete means every possible edge is present, also clique.

Five node complete or K5? No! Why? Later.


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Planar? Yes for Triangle.Four node complete?

Yes.Note: Complete means every possible edge is present, also clique.



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Note: Complete means every possible edge is present, also clique.



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Note: Complete means every possible edge is present, also clique.Five node complete or K5?

No! Why? Later.


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Note: Complete means every possible edge is present, also clique.Five node complete or K5? No!

Why? Later.


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Note: Complete means every possible edge is present, also clique.Five node complete or K5? No! Why?

Later.


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Two to three nodes, bipartite?

Yes.Three to three nodes, complete/bipartite or K3,3. No. Why? Later.

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Two to three nodes, bipartite? Yes.

Three to three nodes, complete/bipartite or K3,3. No. Why? Later.

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Two to three nodes, bipartite? Yes.Three to three nodes, complete/bipartite or K3,3.

No. Why? Later.

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Two to three nodes, bipartite? Yes.Three to three nodes, complete/bipartite or K3,3. No.

Why? Later.

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Two to three nodes, bipartite? Yes.Three to three nodes, complete/bipartite or K3,3. No. Why?

Later.

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Euler’s Formula.

Faces: connected regions of the plane.

How many faces fortriangle? 2complete on four vertices or K4? 4bipartite, complete two/three or K2,3? 3

v is number of vertices, e is number of edges, f is number of faces.

Euler’s Formula: Connected planar graph has v + f = e+2.

Triangle: 3+2 = 3+2!K4: 4+4 = 6+2!K2,3: 5+3 = 6+2!

Examples = 3! Proven! Not!!!!

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Euler’s Formula.


How many faces for

triangle? 2complete on four vertices or K4? 4bipartite, complete two/three or K2,3? 3



Triangle: 3+2 = 3+2!K4: 4+4 = 6+2!K2,3: 5+3 = 6+2!


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Euler’s Formula.


How many faces fortriangle?

2complete on four vertices or K4? 4bipartite, complete two/three or K2,3? 3



Triangle: 3+2 = 3+2!K4: 4+4 = 6+2!K2,3: 5+3 = 6+2!


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Euler’s Formula.


How many faces fortriangle? 2

complete on four vertices or K4? 4bipartite, complete two/three or K2,3? 3



Triangle: 3+2 = 3+2!K4: 4+4 = 6+2!K2,3: 5+3 = 6+2!


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Euler’s Formula.


How many faces fortriangle? 2complete on four vertices or K4?

4bipartite, complete two/three or K2,3? 3



Triangle: 3+2 = 3+2!K4: 4+4 = 6+2!K2,3: 5+3 = 6+2!


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Euler’s Formula.


How many faces fortriangle? 2complete on four vertices or K4? 4

bipartite, complete two/three or K2,3? 3



Triangle: 3+2 = 3+2!K4: 4+4 = 6+2!K2,3: 5+3 = 6+2!


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Euler’s Formula.


How many faces fortriangle? 2complete on four vertices or K4? 4bipartite, complete two/three or K2,3?

3



Triangle: 3+2 = 3+2!K4: 4+4 = 6+2!K2,3: 5+3 = 6+2!


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Euler’s Formula.





Triangle: 3+2 = 3+2!K4: 4+4 = 6+2!K2,3: 5+3 = 6+2!


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Euler’s Formula.





Triangle:

3+2 = 3+2!K4: 4+4 = 6+2!K2,3: 5+3 = 6+2!


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Euler’s Formula.





Triangle: 3+2 = 3+2!

K4: 4+4 = 6+2!K2,3: 5+3 = 6+2!


16 / 20

Euler’s Formula.





Triangle: 3+2 = 3+2!K4:

4+4 = 6+2!K2,3: 5+3 = 6+2!


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Euler’s Formula.





Triangle: 3+2 = 3+2!K4: 4+4 = 6+2!

K2,3: 5+3 = 6+2!


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Euler’s Formula.





Triangle: 3+2 = 3+2!K4: 4+4 = 6+2!K2,3:

5+3 = 6+2!


16 / 20

Euler’s Formula.





Triangle: 3+2 = 3+2!K4: 4+4 = 6+2!K2,3: 5+3 = 6+2!


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Euler’s Formula.





Triangle: 3+2 = 3+2!K4: 4+4 = 6+2!K2,3: 5+3 = 6+2!

Examples = 3!

Proven! Not!!!!

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Euler’s Formula.





Triangle: 3+2 = 3+2!K4: 4+4 = 6+2!K2,3: 5+3 = 6+2!

Examples = 3! Proven!

Not!!!!

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Euler’s Formula.





Triangle: 3+2 = 3+2!K4: 4+4 = 6+2!K2,3: 5+3 = 6+2!


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Euler and Polyhedron.Greeks knew formula for convex polyhedron.

.

Faces? 6. Edges? 12. Vertices? 8.

Euler: Connected planar graph: v + f = e+2.8+6 = 12+2.

Greeks couldn’t prove it. Induction? Remove vertice for polyhedron?

Convex Polyhedron without holes ≡ Planar graphs.

Surround by sphere.Project from point inside polytope onto sphere.Sphere ≡ Plane! Topologically.

Euler proved formula thousands of years later!

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.

Faces?

6. Edges? 12. Vertices? 8.






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.

Faces? 6. Edges?

12. Vertices? 8.






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.

Faces? 6. Edges? 12.

Vertices? 8.






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.

Faces? 6. Edges? 12. Vertices?

8.






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Euler: Connected planar graph: v + f = e+2.

8+6 = 12+2.





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Greeks couldn’t prove it.

Induction? Remove vertice for polyhedron?




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Greeks couldn’t prove it. Induction?

Remove vertice for polyhedron?




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.







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Convex Polyhedron without holes

≡ Planar graphs.



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Convex Polyhedron without holes ≡

Planar graphs.



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Surround by sphere.

Project from point inside polytope onto sphere.Sphere ≡ Plane! Topologically.


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Surround by sphere.Project from point inside polytope onto sphere.

Sphere ≡ Plane! Topologically.


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Surround by sphere.Project from point inside polytope onto sphere.Sphere

≡ Plane! Topologically.


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Surround by sphere.Project from point inside polytope onto sphere.Sphere ≡ Plane!

Topologically.


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Euler proved formula thousands of years later!17 / 20

Euler and non-planarity of K5 and K3,3

Euler: v + f = e+2 for connected planar graph.We consider simple graphs where v ≥ 3.Consider Face edge Adjacencies.

Each face is adjacent to at least three edges.≥ 3f face-edge adjacencies.

Each edge is adjacent to (at most) two faces.≤ 2e face-edge adjacencies.

=⇒ 3f ≤ 2e for any planar graph. Or f ≤ 23e.

Plug into Euler: v + 23e ≥ e+2 =⇒ e ≤ 3v −6

K5 Edges? 4+3+2+1 = 10. Vertices? 5.10 6≤ 3(5)−6 = 9. =⇒ K5 is not planar.

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Euler: v + f = e+2 for connected planar graph.

We consider simple graphs where v ≥ 3.Consider Face edge Adjacencies.






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Euler: v + f = e+2 for connected planar graph.We consider simple graphs where v ≥ 3.

Consider Face edge Adjacencies.






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Each face is adjacent to at least three edges.

≥ 3f face-edge adjacencies.Each edge is adjacent to (at most) two faces.

≤ 2e face-edge adjacencies.=⇒ 3f ≤ 2e for any planar graph. Or f ≤ 2

3e.



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Each edge is adjacent to (at most) two faces.

≤ 2e face-edge adjacencies.=⇒ 3f ≤ 2e for any planar graph. Or f ≤ 2

3e.



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=⇒ 3f ≤ 2e

for any planar graph. Or f ≤ 23e.



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=⇒ 3f ≤ 2e for

any planar graph. Or f ≤ 23e.



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=⇒ 3f ≤ 2e for any

planar graph. Or f ≤ 23e.



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=⇒ 3f ≤ 2e for any planar

graph. Or f ≤ 23e.



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=⇒ 3f ≤ 2e for any planar graph.

Or f ≤ 23e.



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Plug into Euler:

v + 23e ≥ e+2 =⇒ e ≤ 3v −6


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Plug into Euler: v + 23e ≥ e+2

=⇒ e ≤ 3v −6


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K5

Edges? 4+3+2+1 = 10. Vertices? 5.10 6≤ 3(5)−6 = 9. =⇒ K5 is not planar.

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K5 Edges?

4+3+2+1 = 10. Vertices? 5.10 6≤ 3(5)−6 = 9. =⇒ K5 is not planar.

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K5 Edges? 4+3+2+1

= 10. Vertices? 5.10 6≤ 3(5)−6 = 9. =⇒ K5 is not planar.

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K5 Edges? 4+3+2+1 = 10.

Vertices? 5.10 6≤ 3(5)−6 = 9. =⇒ K5 is not planar.

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K5 Edges? 4+3+2+1 = 10. Vertices?

5.10 6≤ 3(5)−6 = 9. =⇒ K5 is not planar.

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K5 Edges? 4+3+2+1 = 10. Vertices? 5.

10 6≤ 3(5)−6 = 9. =⇒ K5 is not planar.

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K5 Edges? 4+3+2+1 = 10. Vertices? 5.10 6≤ 3(5)−6 = 9.

=⇒ K5 is not planar.

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Proving non-planarity for K3,3

Euler’s formula =⇒ 3f ≤ 2e

for any planar graph.

K3,3? Edges? 9. Vertices. 6.

9≤ 3(6)−6? Sure!

Proof doesn’t work. Let’s fix this.

But no cycles that are triangles. Face is of length ≥ 4.

Because all cycles are even length; bipartite or edges only gobetween two groups.

.... 4f ≤ 2e for any bipartite planar graph.Euler: v + 1

2e ≥ e+2 =⇒ e ≤ 2v −4 for bipartite planar graph

9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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Euler’s formula =⇒ 3f ≤ 2e for any planar graph.


9≤ 3(6)−6? Sure!






9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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K3,3?

Edges? 9. Vertices. 6.

9≤ 3(6)−6? Sure!






9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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K3,3? Edges?

9. Vertices. 6.

9≤ 3(6)−6? Sure!






9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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K3,3? Edges? 9.

Vertices. 6.

9≤ 3(6)−6? Sure!






9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!






9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6?

Sure!






9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!






9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!

Proof doesn’t work.

Let’s fix this.





9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!






9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!


But no cycles that are triangles.

Face is of length ≥ 4.




9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!






9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!



Because all cycles are even length;

bipartite or edges only gobetween two groups.



9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!






9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!




.... 4f ≤ 2e

for any bipartite planar graph.Euler: v + 1


9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!




.... 4f ≤ 2e for

any bipartite planar graph.Euler: v + 1


9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!




.... 4f ≤ 2e for any

bipartite planar graph.Euler: v + 1


9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!




.... 4f ≤ 2e for any bipartite

planar graph.Euler: v + 1


9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!




.... 4f ≤ 2e for any bipartite planar graph.

Euler: v + 12e ≥ e+2 =⇒ e ≤ 2v −4 for bipartite planar graph

9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!





2e ≥ e+2

=⇒ e ≤ 2v −4 for bipartite planar graph

9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!






9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!






9 6≤ 2(6)−4.

=⇒ K3,3 is not planar!

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9≤ 3(6)−6? Sure!






9 6≤ 2(6)−4. =⇒ K3,3 is not planar!

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Oh my goodness..what have we done!

Graphs.

Basics.Connectivity.Algorithm for Eulerian Tour.

Planar Graphs.

Euler’s formula.

Non-planarity of K5 and K3,3.

Next Time: prove Euler’s formula.

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Graphs.Basics.

Connectivity.Algorithm for Eulerian Tour.

Planar Graphs.

Euler’s formula.



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Graphs.Basics.

Connectivity.

Algorithm for Eulerian Tour.

Planar Graphs.

Euler’s formula.



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Graphs.Basics.

Connectivity.Algorithm for Eulerian Tour.

Planar Graphs.

Euler’s formula.



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· Quick Proof. The sum of the vertex degrees is equal to (A) the total number of vertices, jVj....

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Transcript of · Quick Proof. The sum of the vertex degrees is equal to (A) the total number of vertices, jVj....