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
If G has 2 edges, then
If G has 1 edges, then
If G has 0 edges, then
GEv
GVv
2)deg(
GEv
GVv
2)deg(
GEv
GVv
2)deg(
GEv
GVv
2)deg(
Proof by induction on the number of edges
...
If G has 3 edges, then
If G has 2 edges, then
If G has 1 edges, then
If G has 0 edges, then
GEv
GVv
2)deg(
GEv
GVv
2)deg(
GEv
GVv
2)deg(
GEv
GVv
2)deg(
Proof by inductionbegin the induction:
If G has 0 edges, then
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
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 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
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('
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.