10.1. Definitions
10. Extremal graph theory
Problem definition
Paths and cycles
Complete subgraphs
Let us examine the following “forbidden subgraph“ problems: At most
how many edges are in a graph of order n which does not contain
a path of length n ?
a cycle of length at least (at most) l ?
a complete graph Kr ?
k+1 independent edges ?
a Kuratowsky graph ?
etc…
(*)
Why are these problems extremal?
extremal?
An acyclic graph of order n has at most n – 1 edges, and the extremextremal graphs are trees of order n.
If we are looking for the extremal graph which does contain k+1
independent edges, then in ex(n;
ex(n;F )we
)we have F=(k+1)K2.
Let C denote an arbitrary cycle of a graph G with order n. Then
ex(n;C)
ex(n;C) = n – 1, since a graph order n contains at most n – 1 edges
to remain acyclic.
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10.2. Paths and Cycles
For a given graph G denotes the minimal degree of vertices and g
is the girth and denotes the length of a shortest cycle.
cycle.
Theorem 10.1.: For g
n 0 ( g ,δ ) =
1+
3 and
Usually,
Usually, let ex(n;F)
ex(n;F) denote the extremal graph of order n which does
not contain a given subgraph F, then the graph has ex(n;F)
ex(n;F) edges.
Examples:
Examples:
2
If for a given class of graphs a certain graph parameter,
parameter, say the
number of edges or the minimum degree, is at most some number f,
then the graphs for which equality holds are extremal graphs of the
inequality.
inequality.
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Graph Theory 8
3 put
δ
{ ( δ − 1 )( g − 1 ) 2 − 1} if g is odd
δ −2
{ ( δ − 1 ) g 2 − 1}
δ −2
Then a graph G with minimum degree
n0(g, ) vertices.
2
if g is even.
and girth g has at least
Proof.
We will examine separately the two cases if g is even or odd.
odd.
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1
Suppose that g = 2d+1, d
1.
Let x
G:
V(G) and y V(G).
In a connected graph G the distance d(u,v)
u,v) between two vertices u
and v is the minimum of the lengths of the u−v paths of G.
We can classify the vertices into different classes accordind to their
distance from the vertex x:
y
x
Since deg(x)
, so there are at least vertices at distance 1
deg(x)
from x.
We can reach the vertices are distance 2 from x through vertices
which are at distance 1 from x. So we have at least ( -1)
vertices at distance 2.
The number of vertices at distance 3 is at least ( -1)2.
The number of vertices at distance d is at least ( -1)d-1.
Suppose the opposite.
opposite. Then the length of both paths (l
(l1 and l2)
d.
So, l1 + l2 2d and so we have a cycle in G with length at most 2d,
2d,
which contradicts that g = 2d+1.
5
Suppose that g = 2d, d
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1.
So n ≥ 1 + δ + δ ( δ − 1 ) + ... + δ ( δ − 1 )
δ
= 1+
(( − 1 )(g −1 ) 2 − 1)
δ −2
d- 1
There is no two x–y paths of length at most d.
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Let us looking for an extremal graph G0 with parameters and g .
Pick two adjacent vertices, say x and y.
The above theorem shows that G0 is regular of degree .
Then there are at least 2( -1) vertices at distance 1 from {x,y},
x,y},
Furthermore,
Furthermore,
and at least 2( -1)2 vertices at distance 2, and so on, …,
if g = 2d+1 then G0 has diameter d and
if g = 2d then every vertex is within distance d – 1 of each pair of
adjacent vertices.
at least 2( -1)d-1 vertices at distance d-1 from {x,y}.
x,y}.
d- 1
So n = 2 + 2( δ − 1 ) + ... + 2( δ − 1 )
2
(( − 1 )g 2 − 1)
=
δ −2
We call G0 a Moore graph of degree and girth g, or,
or, if g = 2d+1,
2d+1, a
Moore graph of degree and diameter d.
This completes the proof.
proof.
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2
A Moore graph with =3 and g=7
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A Moore graph with =4 and g=6
9
Let us see now what we can say about long cycles and paths in a
graph.
The length of a longest cycle in a graph G is called as circumference.
circumference.
Example:
Example: the circumference of a Hamiltonian graph is n, while the
longest path is n-1.
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Theorem 10.3.: Every nonnon-Hamiltonian connected graph contains at
least as long paths as the circumference of the graph.
Proof.
Let C = x1x2…xl be a longest cycle. l < n since the graph is not
Hamiltonian.
Hamiltonian.
G
x2
x3
x4
x1
Is there any connection between the length of a circumference and
the length of a longest path?
y
xl
x5
x6
x l -2
x l -1
Since l < n then there exists a vertex y V(G) which is not on C and
is adjacent with a vertex of C, say x1, and then yx1x2…xl is a path of
length l.
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3
Lemma 10.4.: Let G be a connected graph of order n 3 such that for
any two nonnon-adjacent vertices x and y we have d(x)+
d(x)+ d(y)
d(y) k.
If k = n then G is Hamiltonian.
Hamiltonian.
If k < n then G contains a path of length k, and a cycle of length
at least (k+2)/2.
Pl=x1x2…xixlxl-1…xi+1
x1
x2
x3
xi
xi+1
We define the following two sets:
sets:
(x1)={
)={xi: x1xi E(G)}
E(G)}
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The first two assertions of the theorem follow from this inequality:
inequality:
if k = n this is impossible,
impossible, so G is Hamiltonian,
Hamiltonian,
if k < n then P has length l – 1 k.
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E(G)}
E(G)}. Then
Theorem 10.5.: Let G be a graph of order n without a path of
length k 1. Then
k −1
e( G ) ≤
n.
2
A graph is an extremal graph (that is equality holds for it) iff all its
components are complete graphs of order k.
k/2 , where z denotes the
Put t = max {i: x1xi
These two sets are disjoint subsets of {x2,x3,…,xl} and so
k d(x1) + d(xl) l – 1 n – 1.
The Lemma 10.4. has the following consequence:
consequence:
It remains to prove the statement about cycles.
cycles.
Assume that d(x1) d(xl) so d(x1)
least integer not less than z.
(xl)={
)={xi+1: xixl E(G)}
E(G)}
Suppose that G is not Hamiltonian.
Hamiltonian. Let P = x1x2…xl be a longest
path in G.
Since P is maximal, so all the neigbours of x1 and xl are in P.
The path P cannot contain vertices xi and xi+1 such that x1 is adjacent
to xi+1 and xl is adjacent to xi.
Since, otherwise x1x2…xixlxl-1…xi+1 is a cycle of length l.
xl
Proof.
xl-1
Proof.
x1
x2
x3
x4
t ≥ d ( x1 ) + 1 ≥
xt
xl-1
xl
We fix k and apply induction on n.
k
k+2
+1 ≥
2
2
If n k then the statement is trivial.
Let n > k and suppose that the statetemnt is true for all n' < n.
If G is disconnected,
disconnected, the induction hypothesis implies the result.
and so G contains a cycle of length t: C = x1x2x3…xt .
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4
10.3. Complete Subgraphs
So we can suppose that G is connected.
If G contains Kk (a complete k order graph) then it has a path of
length k.
So we can suppose that G does not contain Kk. So by the Lemma
10.4.
10.4. there exist two nonnon-adjacent vertices x and y, for which d(x)
d(x) +
d(y)
d(y) < k.
So, at least one of them,
them, say x
k −1
.
d( x ) ≤
2
Since G – x is not an extremal graph. So
k −1
k −1 k −1
n
+
( n −1) =
e( G ) ≤ d ( x ) + e( G − x ) <
2
2
2
which completes the proof.
proof.
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Let we realize that Tr-1(n) is the complete (r – 1)-partite graph of
order n whose classes are as equal as possible:
possible: there are nk vertices in
the kth class and n1 n2 … nr-1 n1+1.
+1.
Theorem 10.6.: If G is an (r
(r – 1)-partite graph of order n and maximmaximal size then G = Tr-1(n).
What is the maximal number of edges in a graph of order n not
containing a Kr (a complete graph of order n)? ex(n;K
ex(n;Kr)?
If G is (r
(r – 1)-partite then it does not contain a Kr since every vertex
class of G contains at most one vertex of a complete subgraph.
Consequence:
Consequence: ex(n;K
ex(n;Kr)
of order n.
the maximal size of an (r
(r – 1)-partite graph
Is there a unique (r – 1)-partite graph of order n that has maximal
size? How looks like this graph?
Let Tr-1(n) be the (r
(r – 1)-partite complete graph which has in k. class
n + k −1
r −1
vertices.
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ith Class
n-(mi+mj) further
vertices
Proof.
jth Class
The number of ignored edges: n – (mi+mj) + mi = n – mj
The number of new edges:
(mj–1) + (n – (mi + mj),
So the difference is:
mj–1+n–
1+n–mi–mj–n+mj = mj – 1– mi
mi +2 – 1 – mi
= 1.
Let G be a graph with maximal size and suppose the classes of G are
not as equal as possible.
possible.
Say,
Say, there are mi vertices in the ith class and mj in the jth class and mj
mi + 2.
Transfer one vertex from the jth class to the ith class.
class.
So we would increase the number of edges.
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5
n=7 r=4
5
2,3
1,2,2
1,1,1,2
1,1,1,1,1
6
3,3
2,2,2
1,1,2,2
1,1,1,1,2
7
3,4
2,2,3
1,2,2,2
1,1,1,2,2
If k = 1 then n1 = 2
If k = 2 then n2 = 2
If k = 3 then n3 = 3
21
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Consequence:
Consequence: ex(n;K
ex(n;K r)
n
4
t2 ( n ) =
22
Theorem 10.7.: Let G be a graph with vertex set V that does not
contain Kr. Then there is an (r
(r – 1)-partite graph H with vertex set V
such that for every vertex z V we have dG(z)
(z) dH(z).
(z).
If G is not a complete (r – 1)1)-partite graph then there is at least one
vertex z for which the inequality above is strict.
strict.
Denote the number of edges in Tr-1(n) by tr-1(n).
Example:
Example:
The Turan graph T3(7).
tr-1(n).
The fundamental theorem of Turán states that this trivial inequality,
inequality,
in fact,
fact, an equality for every n and r.
r
First, we shall show that the degree sequence of a graph without a K
is dominated by the degree sequence of an (r
(r-1)-partite graph. In
view of the remarks above this will imply Turán'
Turán's theorem.
theorem.
We shall apply induction on r.
For r = 2 there is nothing to prove since G is the empty graph ,
which is 1-partite.
Assume now that r
values of r.
3 and the assertion holds for smaller
We say that a sequence a1,a2,… an dominates the sequence
b1,b2,…,bn if ai bi, 1 i n.
Proof.
Tr-1(n) vertices
Graph Theory 8
2
3
4
5
6
7
8
2 3
4
1,1 1,2 2,2
1,1,1 1,1,2
1,1,1,1
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6
= W, and
G0 does not contain a K r-1 otherwise with x it would form a K r. By
the induction hypothesis we can replace G0 by an (r
(r – 2)-partite
graph H0 with vertex set W in such a way that dG0(y) dH0(y) for
every y W.
Add H0 the vertices in V–W and join each vertex in V–W to each
vertex in W. Denote this graph by H.
Let us examine the degrees:
degrees:
If z V–W then dH(z)
(z) = dH(x)
(x) = dG(x)
(x)
If z
W then dH (z) = dH(z)
(z) + n - |W |
0
(x)
G(x)
Pick a vertex x V for which dG(x)
(x) is maximal. Let
let G0 = G(W).
dG(z).
(z).
dG(z)
(z)
What can we say about the extremal case e(G)
e(G) = e(H)?
e(H)?
Since e(G)
e(G) = e(H)
e(H) then e(G0) = e(H0), so by the induction hypothesis
G0 is a complete (r – 2)-partite graph, which implies that G is a
complete (r–1)-partite graph.
We will prove that H has the required properties.
properties.
H does not contain K r, since a complete subgraph of H has at most
one vertex in V–W and at most r – 2 vertices in W.
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Now,
Now, as a consequence,
consequence, we can state the Turán'
Turán's theorem:
theorem:
Theorem 10.8. (Turán
): ex(n;K
(Turán):
ex(n;Kr) = tr-1(n) and Tr-1(n) is the unique
graph of order n and size tr-1(n) that does not contain a complete
graph of order r.
Proof.
Since Tr-1(n) is the unique (r-1)-partite graph of order n and
maximum size,
size, so both assertions follow from the Theorem 10.7.
Graph Theory 8
Problem of Zarankiewicz:
Zarankiewicz: If G2(m,n) is a bipartite graph with m
vertices in the first class and n vertices in the second one. What is
the maximum size of a graph G2(m,n) if it does not contain a
complete bipartite graph G2(s,t)? The maximum usually denoted by
z(m,
z(m, n ; s, t).
There are not known exact values for z(m,
z(m, n ; s, t) but upper bounds
can be given. The following lemma seems ato imply a very good
upper bound for the function z.
Lemma 10.9.: Let m, n, s, t, r, k be integers,
integers, 2 s m,
m, 2 t n, 0
k, 0 r < m,
m, and let G = G2(m,n)
(m,n) be a graph of size z = my = km + r
without a K(s,t)
K(s,t) subgraph having s vertices in the first class and t in
the second. Then
m
Graph Theory 8
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26
Graph Theory 8
y
t
≤(m−r)
k
t
+r
k +1
t
≤ ( s −1)
n
t
.
28
7
x∈V1
u for u
d(x) = z = my = km + r , and f(u) = ut
As
Proof.
Denote by V1 and V2 the the vertex classes of G.
is a convex function of
t, the inequality (*) implies the statement of the theorem.
theorem.
We say that a subset T of t elements of V2 belongs to a vertex x V1 if
x is joined to every vertex in T. We call this set as t-set.
set.
The number of t-sets belonging to a vertex x is
d( x )
t
Theorem 10.10.:
z ( m , n ; s , t ) ≤ ( s − 1 )1 t ( n − t + 1 )m 1− 1 t + ( t − 1 )m .
.
Proof.
Since the assumption on G is exactly that each t-set in V2 belongs to
at most s–1 vertices of V1, we find that
d( x )
x∈V1
Graph Theory 8
t
≤ ( s −1)
n
(*)
Let G=G2(m,n) be an extremal graph for the function z(m,n;s,t),
z(m,n;s,t), that
is let G be a bipartite graph of size z(m,n;s,t)
z(m,n;s,t) = my without a K(s,t)
K(s,t)
subgraph. As y n, the inequality of the Lemma 10.9. implies
( y − ( t − 1 )) t ≤ ( s − 1 )( n − ( t − 1 )t m − 1 .
t
29
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8