Summing degree sequences work out degree sequence, and sum.

Summing degree sequenceswork out degree sequence, and sum

The handshaking lemma

weak version:

The sum of degrees of a graph G is even

strong version:

The sum of degrees of a graph G is double the number of edges

why? proof? GEv

GVv

2)deg(

Proof by induction on the number of edges

...

If G has 3 edges, then




GEv

GVv

2)deg(

GEv

GVv

2)deg(

GEv

GVv

2)deg(

GEv

GVv

2)deg(

Proof by inductionbegin the induction:


Proof:

If G has no edges, then all the degrees of its vertices must be zero.

The sum of the degrees is therefore = 2 E(G)

GEv

GVv

2)deg(

Proof by inductionrough thinking for next step:

If for graphs with no edges

then for graphs with one edge.

Step:

If G has one edge, then let G’ be G with the edge removed. G’ has no edges, so

Replace the edge - two vertices increase degree by 1,

so sum increases by 2.

GEv

GVv

2)deg(

GEv

GVv

2)deg(

'2)deg(

'

GEvGVv

2)deg()deg(

'

GVvGVv

vv

GEGEGEvv

GVvGVv

21'22'22)deg()deg('

Proof by inductionrough thinking for next step:

If for graphs with one edges

then for graphs with two edges.

Step:

If G has two edges, then let G’ be G with an edge removed. G’ has one edge, so



GEv

GVv

2)deg(

GEv

GVv

2)deg(

'2)deg(

'

GEvGVv

2)deg()deg(

'

GVvGVv

vv

GEGEGEvv

GVvGVv

21'22'22)deg()deg('

Proof by inductiongeneral step:

If for graphs with k edges

then for graphs with k+1 edges.

Inductive proof:

If G has k+1 edges, then let G’ be G with an edge removed. G’ has k edges, so



GEv

GVv

2)deg(

GEv

GVv

2)deg(

'2)deg(

'

GEvGVv

2)deg()deg(

'

GVvGVv

vv

GEGEGEvv

GVvGVv

21'22'22)deg()deg('

Degree sequences

when is there a graph with a given degree sequence?

<1, 1, 2, 2>

<1, 1, 2, 3>

<1, 1, 2, 14>

Degree sequences

Given a degree sequence

- check that the sum is even

(if not, quote the handshaking lemma)

- apply a rule, iteratively:

<1, 1, 2, 2, 4> -> <0, 0, 1, 1> -> <1, 1>

<1, 1> -> <0> -> stop

if this process ends up with <0>, then a graph is drawable with the given degree sequence.

why?

Graphs

Properties which remain the same under isomorphism:

number of nodes

number of edges

connectedness

degree sequence

if all of these are equal, the graphs may or may not be isomorphic

Graphs

More properties :

Eulerian

Hamiltonian

plane

planar

NB: “Euler” is pronounced “oiler”

Eulerian?

try to visit all edges exactly once using a cycle

Eulerian definition

A graph G is Eulerian

if and only if

there is a cycle which includes all edges once

(called an “Euler cycle”)

Eulerian theorem

A graph G is Eulerian

if and only if

all vertices have even degree.

Hamiltonian?

try to visit all vertices exactly once using a cycle

Hamiltonian definition

A graph G is Hamiltonian

if and only if

there is a cycle which visits all vertices once

(called a “Hamiltonian cycle”)

no known theorem!

Plane?

A graph is plane if there are no edge-crossings in the drawing

Plane vs. Planar

plane not plane not plane

planar planar not planar

In general - a graph is planar if it’s isomorphic to a plane graph.

So all plane graphs are planar.

And all non-planar graphs are not plane.

Complete graphs

Complete graphs

Kn have n vertices and all possible edges

(how many edges?)

Kr,s have r + s vertices, split into two sets,

and all possible edges between the r-set and the s-set.

(how many edges?)

Complete graphs

Which of Kn and/or Kr,s are Eulerian/Hamiltonian/planar

Answer for K1 K2 K3 K4 K5

and for K1,1 K1,2 K2,2 K2,3 K3,3

Start by drawing them.

Summing degree sequences work out degree sequence, and sum.

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