Graph Theory
Judith Stumpp
7. November 2013
Bei dem folgenden Skript handelt es sich um einen Mitschrieb der Vorlesung Graph Theory vom Wintersemester 2011/2012. Sie wurde gehalten von Prof. Maria Axenovich Ph.D. . Der Mitschrieb erhebt weder
Anspruch auf Vollständigkeit, noch auf Richtigkeit!
1
Kapitel 1
Definitions
V
2
The graph is a pair V, E. V is a finite set and E ⊆
and E the set of edges.
a pair of elements in V . V is called the set of vertices
1
Visualize: G = (V, E), V = {1, 2, 3, 4, 5}, E = {{1, 2}, {1, 3}, {2, 4}}
5
3
4
History: word: Sylvester (1814-1897) and Cayley (1821-1895)
Euler - developed graph theory
Königsberg bridges (today Kaliningrad in Russia):
A
A
D
D
C
B
C
B
Problem: Travel through each bridge once, come back to the original point.
Impossible!
Notations:
• Kn = (V,
V
2
) - complete graph on n vertices |V | = n
v1
v2
v6
v3
v5
K5
K3
v4
C 6 = v1 v2 v3 v4 v5 v6
• Cn - cycle on n vertices
V = {v1 , v2 , . . . , vn }, E = {{v1 , v2 }, {v2 , v3 }, . . . , {vn−1 , vn }, {vn , v1 }}
• Pn - path on n vertices (Note: P n . path on n edges (Diestel))
V = {v1 , v2 , . . . , vn }, E = {{v1 , v2 }, {v2 , v3 }, . . . , {vn−1 , vn }}
2
2
• Let P be a path from v1 to vn . The subpath of P from vi to vj is vi P vj and the subpath from vi+1 to
◦
vj is v i P vj .
• En = (V, ∅), |V | = n isolated vertices
m
n
Km,n
E10
• Kn,m = (A ∪ B, A × B), A ∩ B = ∅ complete bipartite graph
, E = {{{i, j}, {k, l} : {i, j} ∩ {k, l} = ∅}
• Peterson graph: V = {1,2,3,4,5}
2
{1, 2}
{3, 5}
{4, 5}
{1, 3}
{1, 4}
{3, 4}
{2, 5}
{2, 4}
{2, 3}
{1, 5}
• Kneser Graph K(n, k)= ( Vk , E)
|V | = n, E = {{A, B} : A, B ∈ Vk and A ∩ B = ∅}.
|V |
V
V
k is the set of k-element subsets of V, | k | = k
• Qn - hypercube of dimension n.
Qn = {2{1,2,...,n} , E}, E = {{A, B} : |A4B| = 1}
(A4B := (A ∪ B) − (A ∩ B))
V - set of binary n-tuples E - pairs of binary tuples different in 1 position
Q1
V = {∅, {1}} = {(0), (1)}
Q2
V = {(0, 0), (0, 1), (1, 0), (1, 1)}
Q3
V = {(0, 0, 0), . . . , (1, 1, 1)}
(1,1,1)
(1,1,0)
(1,1)
1
0
Qn :
n
2
(0,0,0)
weight n-1
1001
0001
n
3
n
2
n
(0,0)
weight 2
weight 1 (# 1’s in a binary tuple)
(0, . . . , 0)
3
Qn−1
(0,1,1)
(0,0,1)
(0,1,0)
(1, . . . , 1)
n
n−1
n
(1,0,0)
(1,0)
(0,1)
(1,0,1)
Qn−1
Parameter: Let G = (V, E) be a graph. The order of G ist the number of vertices (|V |) and the size of G
is the number of edges (|E|).
If the order of G is n, then 0 ≤ size(G) ≤ n2 .
If e = {x, y} ∈ E, x is adjacent to y and x is incident to e.
There is a n × n matrix A of
. . , vn }, E) which is called the adjacent matrix.
 G = ({v1 , . 
0 1 1 0
v1
1 0 1 1
v2

v4 A = 
For v3
1 1 0 0.
0 1 0 0
Subgraph: H ⊆ G, H = (V 0 , E 0 ), G = (V, E), V 0 ⊆ V, E 0 ⊆ E
v1
v1
v2
v3
v2
v3
v4
H ⊆ H is an induced subgraph of G if H ⊆ G and for v1 , v2 ∈ V (H): {v1 , v2 } ∈ E(H) ⇔ {v1 , v2 } ∈ E(G).
ind
In the upper example it is no induced subgraph.
An induced subgraph is obtained from G by deleting vertices. E.g.:
v1
v1
v2
v3
v4
v1
v2
v3
v1
v2
v4
v3
v2
v4
v3
v4
Let G = (V, E) and G0 = (V 0 , E 0 ) be graphs. Then we define G ∪ G0 := (V ∪ C 0 , E ∪ E 0 ) and G ∩ G0 :=
(V ∩ C 0 , E ∩ E 0 ).
G[X] := (X, {{x, y} : x, y ∈ X, {x, y} ∈ E(G)}) is called the subgraph of G induced by a vertex set
X ⊆ V (G). E.g.:
G[{1, 2, 3, 4}] 2
G
2
4
1
3
5
4
1
3
A degree d(v) = deg v of a vertex is the number of edges incident to that vertex.
v1
v2
v3
v4
deg v1 = 2, deg v2 = 3, deg v3 = 2, deg v4 = 1
In this example the degree sequence is (2, 3, 2, 1), the minimum degree δ(G) is 1 and the maximum degree
∆(G) is 3.
n
P
Apparently |E(G)| = 12
deg vi is true.
i=1
Thus
n
P
deg vi is even and therefore the number of vertices with odd degree is even.
i=1
d(G) :=
1
n
n
P
deg vi is called the average degree of G.
i=1
4
Extremal graph theorem: We’ll prove that if G has n vertices and >
triangle.
l 2m
n
4
edges ⇒ G has a
Let A, B ⊆ V, A ∩ B = ∅. P is an A-B-path if P = v1 . . . vk , V (P ) ∩ A = {v1 } and V (P ) ∩ B = {vk }.
A graph is connected if any two vertices are linked by a path. A maximal connected subgraph of a
graph is a connected component.
A connected graph without cycles is called a tree. A graph without cycles (acyclic graph) is called a
forest.
Other „special named“ graphs:
star
caterpillar
spider
broom
Proposition: If a graph G has a minimum degree δ(G) ≥ 2 then G has a path of length δ(G) and
a cycle with at least δ(G) + 1 vertices.
proof: Let P = (x0 , . . . , xk ) be a longest path in G. Then all neighbors of xk are in V (P ) (y is a
neighbor of x if {x, y} ∈ E). In particular k ≥ δ(G).
Let i = min{j ∈ {0, . . . , k} : {xk , xj } ∈ E}. Then xi xk xk−1 . . . xi is a cycle of length at least δ + 1.
≥δ
xk
xi
x0
The girth of a graph G is the length of a smallest cycle in G.
The distance dG (v, w) of v, w ∈ G is the length of the smallest path between them (min ∅ = ∞).
The diameter of G is max{dG (v, w) : v, w ∈ G}.
Proposition: Every nontrivial tree T has a leaf.
proof: Assume T has no leaves. T has no isolated vertices ⇒ δ(T ) ≥ 2 ⇒ Cn ⊆ T
- A tree T of order n ≥ 1 has n − 1 edges.
proof: T = K1 X
Assume it holds for all trees of order < n.
Let v be a leaf of T , T 0 := T − v.
⇒ |T 0 | = n − 1 < n.
T 0 is acyclic.
Let v 0 , w ∈ T 0 . ∃P v 0 = v0 , v1 , . . . , vn = w ⊆ T .
To show: vi 6= v for all i = 0, . . . , n
v0 , vn 6= v because v0 , vn ∈ T 0 , v 6∈ T 0
vi 6= v (i = 1, . . . , n − 1) because dT (vi ) ≥ 2, vi is not a leaf.
5
⇒ P ⊆ T 0 connecting v0 and w ⇒ T 0 connected ⇒ T is a tree.
With induction hypothesis T 0 has (n − 1) − 1 edges. Thus T has (n − 1) − 1 + 2 = n − 1 edges. A walk is an alternating sequence v0 e0 v1 e1 . . . vn of vertices and edges so that ei = vi vi+1 for all
n = 0, . . . , n − 1. Compared to a path it is allowed to pass edges and vertices more than once. If
v0 = vn , then the walk is a closed walk.
- If G has a u-v-walk (between vertices u, v) ⇒ G has a u-v-path.
proof: Consider the shortest walk between u and v is W . Then W is a path. If not, W has a repeated
vertex W = ue0 v1 e1 . . . vi . . . vi . . . v , then W 0 = W1 W2 is a shorter u-v-walk.
| {z } | {z } | {z }
=:W1
=:W̃
=:W2
- If G has an odd closed walk (i.e. odd # edges) then G has an odd cycle.
proof: If there are no repeated vertices (except for first and last) ⇒ we have an odd cycle.
If there is a repeated vertex vi , W = v0 e0 v1 . . . vi . . . vi . . . vn = v0 .
| {z } | {z } | {z }
W1
1’st part of W2
2’nd part of W2
W is a union of two closed walks W1 and W2 . Either W1 or W2 is an odd closed walk
⇒ by induction it contains an odd cycle.
- If G has a closed walk with a non-repeated edge W = v0 e0 v1 . . . ei . . .
contains a cycle.
ei is unique, then G
proof: Induction on # vertices.
Basis:
Step: W = v0 e0 v1 . . . vi . . . vi . . . vn = v0
| {z } | {z } | {z }
1’st part of W2
W1
2’nd part of W2
(note, there is a repeated vertex vj , otherwise W is a cycle)
So, W is a union of two closed walks W1 and W2 and either W1 or W2 has a non-repeated edge.
By induction, that walk contains a cycle.
Definition:
An Eulerian tour is a closed walk containing all edges of a graph and repeating no edge.
e.g.: Eulerian tour v1 e1 v2 e2 . . . e8 v9 = v1 in
v2
e2
e5
v6 = v3
e6
v7
e1
v1 = v5 = v9
e4
e3
v4
e7
e8
v8
Theorem: A connected graph G has an Eulerian tour iff (i.e. if and only if ) each degree of vertex
in G is even.
6
proof:
„⇒“ : If there is an Eulerian tour then clearly the number of edges entering the vertex is the number
of edges leaving the vertex.
„⇐“ : Assume that each degree is even.
Consider a walk with longest number of edges and no repeated edge, W = v0 . . . vk . Thus,
there is no edge incident to v0 that is not in W . Since deg v0 is even, v0 must be vn , i.e. W
is a closed walk.
If all edges are in W , done. Otherwise, there is an edge e, not in W . Since G is connected,
there is such e incident to a vertex in W . Say e = vi u. Then W 0 = uevi W vi is a longer walk
with no repeated edges.
Other idea: all edges in G are even, δ(G) ≥ 2 ⇒ G has a cycle C. Delete C from G
(problem: G − C maybe isn’t connected).
G
G
C
C
Connectivity:
We say that a Graph G is vertex k-connected if |V (G)| > k and deleting any (k − 1) vertices does
not disconnect the graph.
Any connected graph is 1-connected. If a graph is 2-connected then there exists no cut-vertex which
is a vertex whose deletion disconnects a graph. Trees are not 2-connected.
If G is connected, X ⊆ V, G − X disconnected ⇒ X is called a cut-set.
κ(G) = max{k : G is k-connected}
v1
e.g.: κ(
v3
v2
v4
) = 1, κ(Cn ) = 2, κ(Kn,m ) = min{m, n}.
G is called l-edge connected if G 6= En and G does not become disconnected after deleting any (l − 1)
edges.
λ(G) (= κ0 (G)) = max{l : G is l-edge-connected}
e.g.: λ(tree) = 1, λ(Cn ) = 2.
If λ(G) = 1 there exists a so called bridge (cut edge)
.
K100
Clearly λ(G) ≤ δ(G). But it could be that 1 = λ(G) << δ(G) = 99
K100
.
Lemma: For any connected G: κ(G) ≤ λ(G) ≤ δ(G).
proof: Idea: want to find the set of at most λ := λ(G) vertices that disconnects the graph.
7
Let Ẽ be a set of λ edges disconnecting G. Then Ẽ is a cut, i.e. ∃S ⊆ V : ∀e ∈ Ẽ, one endpoint of
e is in S, another is in S̄ := V − S.
Ẽ
S̄
S
If in G there are all edges between S and S̄. λ = |Ẽ| = |S| · |S̄| ≥ |V (G)| − 1 ≥ κ(G).
Otherwise ∃x ∈ S, y ∈ S̄, x 6∼ y (i.e. xy 6∈ E(G)).
T := (N (x) ∩ S̄) ∪ ({z ∈ S : z ∼ S̄} − {x})
T is a vertex cut, in particular after deleting T , x and y are in different connected components.
We have |T | ≤ |Ẽ| = λ because
|N (x)| ≤ #(edges incident to x) and |{z ∈ S : z ∼ S̄} − {x}| ≤ #(edges incident to this set).
Definition:
A graph G is d-degenerate if there is a vertex order v1 , v2 , . . . , vn :
|N (vi ) ∩ {vi+1 , . . . , vn }| ≤ d.
I.e. we eliminate the graph by deleting a vertices sequence, s.t. at most d edges are gone at a time.
≤d
v1
≤d
v2
≤d
vn
v3
Let T be a graph. T is a tree if it is connected and acyclic.
• T is a tree iff T is connected and has |V (T )| − 1 edges.
• T is 1-degenerate.
• A leaf in a nontrivial tree is a vertex of degree 1.
• If G is a graph with δ(G) ≥ |V (T )| − 1 (T tree) then G contains T as a subgraph.
T
G
δ(G) ≥ 6
8
Lemma: A graph is bipartite if and only if it has no odd cycles.
proof:
„⇒“ : Let G be a bipartite graph, then any cycle has a form u1 v1 u2 v2 . . . uk vk u1 , where ui ∈ U, vi ∈
V, 1 ≤ i ≤ k, U, V are partite sets of G.
„⇐“ : Assume that G is connected and has no odd cycles. We shall prove that G is bipartite with
partite sets U, V defined as follows.
Fix x ∈ V (G).
Let U = {u : dist(x, u) is even}, V {v : dist(x, v) is odd}
We need to verify that G[U ], G[V ] are empty graphs.
Assume that u, u0 ∈ U and {u, u0 } ∈ E(G).
Consider a walk formed by shortest x-u-path, shortest x-u0 -path and u, u0 .
u0
u
This is an odd closed walk that contains an odd cycle, a contradiction.
Thus G[U ] is an empty graph.
Similarly G[V ] is an empty graph.
Matchings:
A matching is a graph that is a disjoint (vertex) union of edges.
Philip Hall (Apr. 1904 - Dec. 1982) Cambridge, UK
Recall that N (S) for a set S of vertices is a set of neighbors of vertices in S.
Hall’s matching theorem 1935: Let G be a bipartite graph with partite sets A, B. Then G has
a matching containing all vertices of A if and only if |N (S)| ≥ |S| for any S ⊆ A.
S
A
S
bad
N (S)
B
N (S)
proof:
9
S
A
„⇒“ : obvious
B
N (S)
„⇐“ : Assume that |N (S)| ≥ |S| for any S ⊆ A.
We shall proof that there is a matching containing all elements of A by induction on |A|.
If |A| = 1, clear.
Assume that |A| > 1
Case 1: |N (S)| ≥ |S| + 1, for any S ⊂ A, S 6= A.
Let {x, y} =: e ∈ E(G). Consider G0 = G − {x, y}.
|NG0 (S)| ≥ |NG (S)| − 1 ≥ |S| + 1 − 1 = |S|, for any S ⊆ A − {y}.
B
x
A
y
S
Thus, Hall’s condition is true for G0 , and there is a matching M 0 , containing all elements
of A − {y}, by induction.
So, M 0 ∪ {x, y} is a matching saturating A in G.
B
x
A
y
M0
Case 2: ∃S ⊂ A, S 6= A such that |N (S)| = |S|.
B
A
N (S)
S
N (S 0 )
S0
By induction, there is a matching containing all vertices of S. Let apply induction to
G[A − S, B − N (S)].
Assume that there is S 0 ⊆ A − S such that |N (S 0 ) ∩ (B − N (S))| < |S 0 |.
Then |N (S 0 ∪ S)| = |N (S) ∪ (N (S 0 ) ∩ (B − N (S)))|<6= |S| + |S 0 |.
A contradiction to Hall’s condition applied to S ∪ S 0 .
Thus for any S 0 ⊆ A−S, |N (S 0 )∩(B −N (S))| ≥ |S 0 |, and there is a matching saturating
A − S in G[A − S, B − N (S)]. Together with a matching between S and N (S), it gives
a matching saturating A.
10
Corollaries of Hall’s theorem:
1)
Let G be bipartite with partite sets A, B, such that |N (S)| ≥ |S| − d for all S ⊆ A, and some
fixed positive integer d.
Then G contains a matching of size at least |A| − d.
2)
A k-regular bipartite graph has a perfect matching, i.e. matching containing all vertices of a
graph. Here k-regular is a graph with all degrees equal to k.
G has partite sets A, B :
|E(G)| = #edges incident to A = |A| · k
= #edges incident to B = |B| · k
⇒ |A| = |B|
3)
A k regular bipartite graph has a proper k-edge coloring.
proof:
1)
Construct G0 .
C
B
A
|C| = d, add all edges between A and C.
In G0 |NG0 (S)| ≥ |NG (S)| + d ≥ |S| − d + d = |S|.
By Hall’s theorem, there is a matching in G0 saturating A, with at most d edges not in G.
2)
Let’s verify Hall’s condition.
Is it true that |N (G)| ≥ |S| for any S ⊆ A?
#edges from S to B is |S| · k = #edges between S and N (S) = q
#edges from N (S) to A is |N (S)| · k ≥ #edges between S and N (S) = q.
|N (S)|· 6 k ≥ q = |S|· 6 k ⇒ |N (S)| ≥ |S|.
11
Non-bipartite graphs:
A k-factor in a graph is a spanning (containing each vertex) subgraph in which each vertex has degree
k.
perfect matching = 1-factor
2-factor
Denes König (Sep. 1884 - Oct. 1944)
Gyula König (Dec. 1849 - Apr. 1913)
Let ν(G) be the size of largest matching in G and τ (G) be the size of smallest vertex cover, i.e. set
of vertices such that each edge is incident to some of this vertices, i.e. a set X of vertices such that
G − X is an empty graph.
X
König’s theorem ’31: If G is a bipartite graph, then ν(G) = τ (G).
Classical approach:
B
Given a maximal matching M and want to find a vertex cover of size |M |
A
alternating path: starts with an unmatched vertex of M (alternating one point in A and one in B).
Take the longest alternating path.
vertex cover: for any element of {a, b} ∈ E(M ), a ∈ A, b ∈ B pick b if there is an alternating path
ending in b, otherwise pick a.
proof: (by Romeo Rizzi ’2000)
We want to prove that τ (G) ≤ ν(G) (τ (G) ≥ ν(G) trivial).
Assume that G is the smallest counterexample (#edges, #vertices).
Observe that G is connected, not a path, not a cycle, i.e. ∃v : deg(v) ≥ 3
12
Let v : deg v ≥ 3. u ∈ N (v) ,
Case 1: ν(G\u) < ν(G):
Take a vertex cover X by König’s theorem of G − u of size ≤ ν(G) − 1. Then X ∪ {u} is the
vertex cover of G of size ≤ ν(G).
Case 2: ν(G\u) = ν(G):
Then, in G there is a maximal matching, M , not containing u. There is u0 ∈ N (v) − {u}, such
that f := {v, u0 } ∈
/ E(M ).
0
Let W be a cover of G − f of size ν(G − f ) = ν(G). Then W 0 does not contain u (W 0 contains
vertices of M only and u ∈
/ V (M )). Thus W 0 contains v. So, W 0 covers f too. Thus W 0 covers G.
v
v
2
1
f
w
u’
M
u
|M | = ν(G)
Tutte’s theorem
odd
S
odd
odd
|S| ≥ odd components of G − S
For a subset S of vertices of G, let q(S)=#odd components of G − S.
Theorem: (Bill Tutte May 1917- May 2002)
A graph G has a perfect matching (1-factor) if and only if ∀S ⊆ V (G) q(S) ≤ |S|.
proof:
„⇒“ : trivial.
„⇐“ : Consider G, such that ∀S ⊆ V (G), q(S) ≤ |S|, and assume that G has no 1-factor. Add edge
one-by-one, so the resulting graph G0 is no 1-factor.
We shall show that in G0 is a „bad“ set S, q(S) > |S|.
We shall show that S is also a bad set in G.
Observation: If M1 , M2 are perfect matchings in G, M1 4M2 = (M1 ∪ M2 ) − (M1 ∩ M2 ) are
only cycles.
13
Let S be a set of vertices of degree |V (G)| − 1. We shall show that S is bad in G0 .
Claim: All components of G0 − S are complete.
Assume not, i.e. there is a non-complete component in G0 − S.
c
c
b
S
b
a
d
a
Then there is an induced path a, b, c in this component. Since b ∈
/ S, deg b < |V (G0 )| − 1,
there is d ∈
/ {a, b, c}, such that b 6∼ d.
By maximality of G0 , there is a perfect matching M , in G0 ∪ {{a, c}}, there is a perfect
matching M2 in G0 ∪ {{b, d}}. Note ac ∈ E(M1 ), bd ∈ E(M2 ). We shall create a perfect
matching of G0 .
Consider M1 4M2 , ac, bd ∈ E(M1 4M2 ). If ac, bd belong to different cycles of M1 4M2 :
bd
ac
Take the edges of M2 in a component containing ac, take edges of M1 in a component with
bd, otherwise take edges of M1 .
If ac, bd belong to the same cycle of M1 4M2 , then
a
a
b
c
c
d
or
d
b
A contradiction, since G0 has no 1-factor, so all components of G0 − S are complete. Claim
odd
even
even
S
odd
odd
If S is not bad, i.e. |q(S)| ≤ |S|, we can construct a perfect matching, a contradiction to the
fact that G0 has no perfect matching. Thus S is bad in G0 .
14
even
odd
even
S
odd
odd
odd
G is obtained from G0 by deleting edges, so qG (S) ≥ qG0 (S) > |S|.
1-factor
2
3-factor
f-factor
1
4
3
k-factor - spanning subgraph,
all degrees = k
f -factor: If f : V → N, an f -factor is a spanning subgraph H of G such that degH (v) = f (v).
Let e(v)= deg(v) − f (v) ≥ 0 (excess).
e(v)
B(v)
A(v)
d(v)
Replace each vertex of G with
Ke(v), d(v)
For adjacent u and v, put an edge between A(u) and A(v), such that these edges form a matching.
An f -factor, in a graph G, for f : V (G) → N ∪ {0}, such that ∀v ∈ V f (v) ≤ deg(v), is a spanning
subgraph H of G such that degH (v) = f (v).
1-factor or matching ≈ f -factor, f ≡ 1.
B(v1 )
3 v1
1 v2
3
2
A(v1 )
B(v2 )
∅
A(v2 )
2
1
f (v1 ) = 3, f (v2 ) = 1.
For a graph G and a function f : V (G) → N ∪ {0}, construct an auxiliary graph T (G, f ) by replacing
each vertex v with vertex sets A(v) ∪ B(v), |A(v)| = deg(v), |B(v)| = deg(v) − f (v), and for adjacent
vertices u, v placing an edge between A(u) and A(v), so that these edges are disjoint, and placing a
15
complete bipartite graph between A(u)4B(u) for each vertex u.
Claim: G has an f -factor if and only if T (G, f ) has 1-factor.
proof:
• Assume that M is an f -factor of G, to create a 1-factor in T , take the edges corresponding to
M , and take missing edges between A(u) and B(u) ∀u ∈ V .
• Assume that M is a 1-factor in T , create an f -factor in G by deleting B(u), u ∈ V (G), contracting A(u) into a single vertex, u ∈ V (G).
H-factor: Given a graph G, and a graph H, such that |V (G)|:̇|V (H)| (:̇ = divisible). An H-factor of
G is a spanning subgraph of G that is a vertex-disjoint union of copies of H.
G=
H=
H = K2 H-factor ≈ perfect matching.
Hajnal & Szemerédi ’70: If G satisfies δ(G) ≥
Kk
Kk
Kk Kk
k−1
n,
k
n:̇k, then G has a Kk -factor.
n
Alon-Yuster ’95: If G satisfies δ(G) ≥ χ(H)−1
n. Then G contains at least (1 − o(1)) |V (H)|
(H is
χ(H)
fixed, G is large, n = |V (G)|) copies of H vertex-disjoint.
...
o(n)
χ(H)-chromatic number of a graph H := min #parts into which vertex sets can be partitioned, so
that no two adjacent vertices are in same part.
χ(G) := min # colors assigned to V (G) such that no two adjacent vertices get the same color.
1
3
G
2
χ(G) = 3
χ(Kk ) = k, χ(C3 ) = 3, χ(C4 ) = 2, χ(Km,n ) = 2, χ(C2k+1 ) = 3
There are graphs with large |V (G)| and small χ(G).
2
1
16
A, B ⊆ V (G), A-B-path P is a path v0 , v1 , . . . , vk such that V (P )∩A = {v0 }, V (P )∩
Connectivity:
B = {vk }.
C ⊆ V ∪ E, we say that X separates A and B if each A-B-path contains an element of X.
A
B
A
B
v ∈ A ∩ B ⇒ v is an A − B path
e1 v2 e2
e5
A u1 e3
u e4
v3
v1
u3 B
2
A − B sep. set : {v2 , u2 }
{e1 , e5 , e4 }
{e1 , u1 }
Note that a separating set must contain A ∩ B.
Note B 0 ⊇ B and X separates A and B 0 ⇒ X separates A and B.
A
B0
=
B
Menger’s theorem (1927): (Karl Menger Jan. 1902 - Oct. 1985)
Let G be a graph, A, B ⊆ V (G). Min #vertices separating A and B = Max #vertex-disjoint
A-B-paths.
proof: Assume that A ∩ B = ∅.
Let k = k(G; A, B) = min #vertices separating A and B, k(G; A, B) ≥ max # vertex-disjoint A-Bpath (easy).
We shall prove that max # vertex-disjoint A-B-path ≥ k(G; A, B) = k by stronger induction:
If P is any set of less than k disjoint A-B-paths then there is a set Q of disjoint A-B-paths that
includes the endpoints of P and |Q| = |P | + 1.
P
A
B
Lets prove this by induction on |V (G) − B − A|.
Basis: |V (G) − B − A| = 0.
P
A
B
<k
There is an edge between A and B, not adjacent to vertices of P , otherwise |V (P ) ∩ A| < k is
17
a vertex separating A and B.
Step: We have P , a set of less than k A-B-path, vertex disjoint.
There is an A-v-path for v ∈ B\(V (P )), otherwise V (P ) ∩ B is a set of less than k vertices
separating A and B, call it R.
A
P
x
B
R
v
Let x be the last vertex of R that also belongs to a path in P call it P .
Let B 0 = B ∪ (V (xP ) ∪ V (xR)).
P 0 = P \{P } ∪ {P x}.
note k(G; A, B 0 ) ≥ k(G; A, B).
By induction, there is a larger set of A-B 0 -paths, Q0 , |Q0 | ≥ |P 0 | + 1, Q0 contains endpoints of
P 0.
A
P
y
B
Let y be an endpoint of a path in Q0 in B 0 that is not an endpoint of P 0 .
Case : 1
Case: 2
Q
Case: 3
y
y
Q1
y
,
Case 1: y ∈ B:
Take Q = Q0 −
{Q}
|{z}
∪{Q ∪ xP }.
path containing x
Case 2: y ∈ xP :
Take Q = Q0 − {Q} ∪ {Q ∪ xR} −
{Q1 }
| {z }
∪{Q1 ∪ yP }.
path containing y
Case 3: y ∈ xR:
Take Q = Q0 − {Q} ∪ {Q ∪ xP } − {Q1 } ∪ {Q1 ∪ yR}.
If G = (V, E) a graph, then a line graph L(G) of G is a graph L(G) = (E, E 0 ),
E 0 = {{e, ẽ} : e, ẽ ∈ E and e, ẽ are adjacent}.
18
e2
v2
e2
e1
v3
e1
e3
e3 e4 e5
v1
e5
v4
e4
Corollary 1: If a, b ∈ V (G), {a, b} ∈
/ E(G).
min #vertices separating a and b = max #independent a-b-paths
(here independent means that they share only a and b)
b
a
B
A
Apply Menger’s theorem to A = N (a) and B = N (b).
Corollary 2: (Global version of Menger’s theorem)
Any graph G is k-connected if and only if for any two vertices a, b there are k independent paths
between a and b.
outline of proof:
Suppose G contains k independent paths between any two vertices, thus we need ≥ k vertices to
separate G. So κ(G) ≥ k.
Let κ(G) = k, in particular |(G)| > k.
Assume that a and b are not connected by k independent paths. By corollary 1 a adjacent to b.
Let G0 = G − {a, b}, then G0 contains ≤ (k − 2) independent a-b-paths.
≤k−2
v
X
a
a
b
b
≤k−2
G0
By corollary 1, we can separate a and b in G0 by ≤ k − 2 vertices, X.
Since |V (G)| > k, there is v ∈
/ {a, b} and v ∈
/ component of a in G0 − X.
Observe that v and a are separated by X ∪ {b} in G.
So, v and a are separated by ≤ k − 1 vertices, a contradiction to the fact that κ(G) = k.
Edge-connectivity
1)
min #edges separating a and b in G = max #edge-disjoint a-b-paths.
A
B
a
b
19
Apply Menger’s theorem to L(G) with A = {edges incident to a}, B = {edges incident to b}.
2)
Global Menger’s theorem (edge-connected)
A graph is k-edge-connected if and only if there are k edge-disjoint paths between any two
vertices.
κ(G) = 1 blocks
block-cut-vertex tree.
A block - either a bridge or maximal 2-connected subgraph.
v1 v2
B1
B1
v3
B2 B3
B2
B4
...
B5
v4
v5
v1
B6
v2
Bi ∼ vj if vj ∈ V (Bi ).
Any two block intersect by at most 1 vertex.
Block-cut-vertex graph is a tree.
A block that is a leaf in a block-cut-vertex tree is a block leaf.
κ(G) ≥ 2 ⇔ G can be constructed using ear-decomposition
G is created using ear-decomposition if there is a sequence of graphs G0 ⊆ G1 ⊆ . . . G, such that G0
is a cycle, Gi+1 is created from Gi by adding a Gi -path (ear) (i.e. a path with endpoints in Gi and
no other vertices in Gi ).
Gi
outline of proof:
„⇒“ : κ(G) = 2: We have that G has a cycle. Consider the largest subgraph H of G that is built as
ear-decomposition.
Observe H ⊆ G. If u, v ∈ V (H), v 6∼H u, v ∼G u, then add uv as a ear. If H 6= G ⇒ ∃x ∈
ind
V (G) − V (H), such that x is adjacent to a vertex w ∈ V (H).
G − w is connected, so in G − w there is a path from x to H, call it P , call the first vertex
of P in H, w1 .
So wx ∪ xP w is an H-ear. A contradiction to maximality of H, so G = H.
20
G
w1
H
w
x
„⇐“ : Show that an ear-decomposition is 2-connected . . .
κ(G) = 3 : |V (G)| ≥ 5.
Observation:
If κ(G) = 3 then there is an edge e of G such that κ(G ◦ e) ≥ 3.
Let e = {x, y} ∈ E(G), G ◦ e is obtained from G by identifying x and y, removing (if necessary)
loops and multiple edges.
v3
v3
v4
y
v2
vxy
v2
v4
x
v1
v5
v1
v5
Tutte’s theorem 1961: A graph G is 3-connected if and only if it exists a sequence of graphs
G0 , G1 , . . . , Gn , such that G0 = K4 , Gn = G, Gi+1 is obtained from Gi :
Gi+1 has two vertices x, y of degree ≥ 3, x ∼ y and Gi = Gi+1 ◦ {x, y}.
x0
x
x00
y 00
y
y0
Lemma: If G is 3-connected, then there exists an edge e such that G ◦ e is 3-connected.
(without proof )
proof: We want to prove hat if Gi is 3-connected, then Gi+1 is also 3-connected. Assume not, i.e.
Gi = Gi+1 ◦ {x, y} and Gi+1 is not 3-connected, i.e. there exists a cut-set S with |S| ≤ 2.
G1
S
G2
Let G1 and G2 be connected components of Gi+1 − S. Observe, {x, y} 6= S, otherwise Gi is not
3-connected. But {x, y} ∩ S 6= ∅, otherwise Gi is not 3-connected (disconnected by S). So, w.l.o.g.
(without loss of generality) x ∈ S, y ∈ V (G2 ).
21
G1
S
w
G2
x
y
v
|Gi+1 | > |Gi |
Assume that there exists a vertex v ∈ V (G2 )\{y}, then in Gi {w, vxy } separates v from V (G1 ),
a contradiction. So V (G2 ) = {y}, so deg(y) ≤ 2, a contradiction.
A graph G is k-linked, if for any distinct vertices s1 , s2 , . . . , sk , t1 , t2 , . . . , tk , there are vertex-disjoint
si -ti -paths, i = 1, . . . , k.
t1
s1
t01 = t2
s1
s2
t2
s2
t02 = t3
s3
s3
t3
t03 = t1
G is k-linked ⇒ G is k-connected (Menger’s theorem)
G is f (k)-connected ⇒ G is k-linked. (Bollobás-Thomason ’96)
|{z}
22k
Plane graphs and planar graphs
plane graphs: subset of the real plane.
planar graphs: graph (V, E) reresentable a plane graph.
Line segment {p + λ(q − p) : 0 ≤ λ ≤ 1, p, q ∈ R2 }.
Polygon - union of finitely many line segments, homeomorphic to a cycle.
Homeomorphism - bijection, continuous, −1-continuous,
Arc - homeomorphic image of a straight line segment
(Distel - arc is a polygonal homeomorphic image of line segment.)
Being linked by an arc is an equivalence relation on R2 . The equivalence classes are called regions in
a subset of R2 .
A plane graph G is a subset of R2 , G = V sup E, V is a set of vertices, E is a set of edges,
(i)
E is a set of arcs between vertices.
(ii)
different edges have different endpoints.
(iii)
the interior of an edge contains no vertex and no points of other edge. (no crossings)
A graph G = (V, E) is called planar if there is a plane graph G0 = (V 0 , E 0 ), such that there is a
bijection f : V → V 0 such that {u, v} ∈ E if and only if {f (u), f (v)} ∈ E 0 .
22
5-color theorem,
list-colorings,
characterisation (Kuratowski theorem: G planar ⇔ G has no K5 , K3,3 or their minors)
Eulers formula (v − e + f = 2).
Here:
Definition and small results:
• Frontier of a region - boundary
• Jordan curve theorem: For every polygon P ⊆ R2 , R2 \P has two regions, one bounded,
one unbounded. Each of these regions has P as frontier.
P1
P2
P3
• Lemma 1.2:
P1 , P2 , P3 - internally disjoint arcs sharing endpoints. Then
1)
R2 \(P1 ∪ P2 ∪ P3 ) has exactly 3 regions with boundaries P1 ∪ P2 , P1 ∪ P3 , P2 ∪ P3 .
2)
Let P be an arc. From P (i.e. interior of P (= P without his endpoints)), to P 3 , whose
◦
◦
◦
◦
◦
interior is in the region of R2 \(P1 ∪ P3 ) containing P 2 . Then P ∩ P 2 6= ∅.
P
X2 Let X , X ⊆ R2 , X ∩ X = ∅.
X1
• Lemma 1.3:
1
2
1
2
Xi - union of vertices and arcs, finitely many.
Let P be an arc between a vertex in X1 and a vertex in X2 with interior in the region O of
◦
R2 \(X1 ∪ X2 ), then O\P is a region of R2 \(X1 ∪ X2 ∪ P ).
• Lemma 1.4: Given two spheres S 2 in R3 , and circle C1 and C2 on them, then a homeomorphism from C1 to C2 can be extended to a homeomorphism of disks with boundarier C1
and C2 .
• Let G = (V, E) be a plane graph (⊆ R2 ). Then the regions of R2 \G are called faces (exactly
one unbounded face) (G-finite). Notation: F (G) is the set of faces.
• Lemma 2.1: Let G be a plane graph and e an edge.
1)
2)
3)
◦
If X is the frontier of a face, then either e ⊆ X or X ∩ e = ∅.
If e is on a cycle C ⊆ G, then e is on frontier of exactly two faces.
If e is on no cycle, then e is on the frontier of exactly one face.
Corollaries:
• Frontier of a face f is a point set of a subgraph, G[f ].
• Forest has exactly one face.
• If a plane graph has different faces with the same boundary, then the plane is a cycle.
23
A plane graph is maximally plane if one can not add edges and obtain a plane graph.
Lemma: A graph is maximally plane if and only if each face is a triangle (cycle with 3 vertices).
Such graph is called a plane triangulation.
proof: Let G be a plane triangulation. We can add edges only in the faces of G. Impossible without
creating multiple arcs. Assume that G is maximally plane. Observe first that G does not have a
cut-vertex.
By taking a vertex on the faces of different resulting connected components, we can join them by an
arc. So, G is 2-connected, i.e. each face is bounded by a cycle. Assume that some face is bounded by
a cycle of length ≥ 4, call this face f .
Claim: the vertices of f include a complete graph in G, otherwise we could add a missing edge
incident the face f .
Let the vertices on the cycle be v1 , v2 , . . . , vk , k ≥ 4. We have an arc connecting v1 an v3 , and an arc
connecting v2 and v4 .
v1
v4
v2
v3
A contradiction to Lemma 1.2, applied to three internally disjoint arcs between v1 and v3 .
Euler’s formula: Let G be a plane graph with v vertices, e edges and f faces, then v − e + f = 2.
proof 2: (Induction)
Basis: e = v − 1, G is a tree, by lemma ..., f = 1. Then v − e + f = v − (v − 1) + 1 = 2. X
Step: e > v − 1, G has an edge e on a cycle. Then e belongs to the boundary of exactly two faces
f1 , f2 .
◦
◦
Idea: delete e, this wil create a face f1,2 = f1 ∪ f2 \e, in a new graph G0 , v = v, e(G0 ) =
e(G) − 1, f (G0 ) = f (G) − 1. By induction, in G0 v(G0 ) − e(G0 ) + f (G0 ) = 2. Thus v(G) −
(e(G) − 1) + (f (G) − 1) = 2.
◦
f1,2 = f1 ∪ f2 \e.
First we shall prove Claim: F (G0 )\f1,2 = F (G)\{f1 , f2 }. (without assuming that f1,2 is a face of G0 ).
formalities:
◦
Using this claim, observe e is in the face of G0 , not in the face of G, it must be f1,2 , so Idea could be
applied.
proof of the claim:
24
◦
• Let f ∈ F (G)\{f1 , f2 }. By lemmma 2.1(i) G[f ] ⊆ G\e = G0 , so f is a face of G0 (not equal to
f1,2 ).
• Let f 0 ∈ F (G0 ) not equal to f1,2 . Any two points of f1,2 are linked by an arc in R2 \G0 , so
f1,2 ⊆ f 00 , a face of G0 . Since f 0 6= f1,2 , f 0 6⊆ f1,2 because otherwise f 0 ⊂ f1,2 ⊆ f 00 (can not be
a face being a subset of another face). Let x be a point in f 0 \f1,2 , x belongs to a face f of G,
f 6= f1 , f2 . So f is a face of G0 , so x ∈ f ∩ f 0 , so f = f 0 , f 0 ∈ F (G)\{f1 , f2 }.
proof 2: (outline) (Dual graph)
Definition: For a plane graph G with vertex set V , edge set E and face set F , let D(G), the dual of
G, be a plane graph (multigraph) with a single vertex in each face of G, with two vertices connected
by an arc if and only if the corresponding faces of G share an edge, on this boundary. In this case
the arc in D(G) must cross e.
G
D(G)
Proof of Euler’s formula using dual graphs.
Let T be a spanning tree of G. Consider edges of D(G) crossing edges of G, not in T , call the resulting
graph T 0 .
T and T 0 are trees and T and T 0 have e(G) edges. v(T ) = |v(G)|, v(T 0 ) = |F (G)|, e(T ) = |v(T )| −
1, e(T 0 ) = |F (G)| − 1.
e(T ) + e(T 0 ) = e(G) = |v(G)| − 1 + |F (G)| − 1.
proof 2: (Pick’s formula)
area = #internal points
+# 12 bound. points −1
Let G be a 2-connected plane graph, embedded in the grid, such that each arc is a straight line
segment. We’ll calculate the area of big polygon using Rick’s formula and using areas of smaller
polygons.
25
Let vext
vint
I
eint
eext
=
=
=
=
=
#exterior vertices of G,
#interior vertices of G,
#other points,
#gridpoints of interior edges but not vertices,
#gridpoints of exterior edges but not vertices.
vext + eext
+ (I + eint + vint ) − 1
Area =
2X
=
area(F )
F - bounded face
= I + eint +
X
d(v)
+
2
v - interior
|vertex of G
X
v - exterior
vertex of G
{z
(∗)
d(v) − 1 eext
+
− 1(|F | − 1)
2
2
}
X d(v)
X d(v) vext
(∗)
+
−
=
2
2
2
v - ext
v - int
X
vext
d(v) vext
+ vint − 1 =
−
− |F | + 1
2
2
2
v - vertex of G
X
d(v) vext
−
2
2
v - vertex of G
vext + vint −2 + |F | = e(G)
| {z }
v
so |V (G)| − 2 + |F (G)| = |E(G)|. (|V (G)| = v, |E(G)| = e)
Outerplane graph is a graph that has a face containing all the vertices.
Corollary: (to Euler’s formula)
Let G be a planar graph, then |E(G)| ≤ 3|V (G)| − 6 (equality for triangulation).
proof: Consider a plane embedding of a triangulation with v vertices, e edges and f faces. Each
face has 3 edges on its boundary, so e = 23 f . Euler’s formula gives us v − e + f = v − e + 23 e = 2.
Thus 3v − e = 6 and so e = 3v − 6. If G is not a triangulation, G is a subgraph of a triangulation,
so |E(G)| ≤ 3|V (G)| − 6.
Corollary: Let G be a planar graph, where each face is not a triangle. Then |E(G)| ≤ 2|V (G)|−4.
proof: |E(G)| ≥ 4·#faces
(f ≤ 2e ). Euler’s formula:
2
2 = v − e + f ≤ v − e + 2e = v − 2e , thus e ≤ 2v − 4.
Planar graphs are characterized by forbidden subgraphs, and by poset (partially ordered set) dimentions.
26
Given two graphs, G and X, we say that X is a minor of G, write G = M X. Say also
that G contains X as a minor if X can be obtained from G by edge deletions and edge contractions.
(delete isolated vertices)
Minors:
G
G
X
G = MX
X
G = MX
G contains X as a minor, if a subgraph of G can be vertex-partitioned into sets V1 , V2 , . . . , V|V (X)| ,
such that G[Vi ] is connected and Vi ∼ Vj if and only if the corresponding vertices vi , vj ∈ V (X) are
adjacent.
G = MX
X
For two graphs G and X, we say that G contains X as a topological minor, we
write G ≥ T X, if a subgraph of G is a subdivision (german: Unterteilung) of X.
Topological minor:
H 0 is obtained from H by a single edge-subdivision, if V (H 0 ) = V (H)∪{v}, and E(H 0 ) = E(H)\{x, y}∪
{x, v} ∪ {v, y}. H 0 is a subdivision of H if H 0 is obtained from H by series of single edge-subdivisions.
H
subdivision of H
examples
G
G ≥ TX
X
G = T K4
In the definition of minor, V1 , V2 , . . . , V|V (X)| are branch sets. In the definition of topological minor,
the vertices of T X, corresponding to the vertices of X are called branch vertices.
Kuratowski 1930, Wagner 1937: The following are equivalent:
1)
G is planar,
2)
G has no T K5 or T K3,3 and
3)
G has no M K5 or M K3,3 .
Another characterization: Given a graph G = (V, E), let P , partially ordered set (poset) be
P = (V
∪ E}, „R“ = „ < “ ), v < e for v ∈ V, e ∈ E, if and only if v is incident to e.
| {z
|
{z
}
X
relation R⊂X×X
27
v2
e2
e1
e2
e1
e3
v3
v1
e3
v4
v1
v2
v4
v3
Schnyder (1989): G is planar if and only if dim(P ) ≤ 3, where P is an incidence poset of P .
For a poset P , dim(P ) is the smallest number of total orders (chains), such that P is the intersection
of these total orders.
dim
=1
dim( a
3
6
3
6
2
5
2
1
5
1
4
4
b
)=2
a
b
b
a
a
b
= ∩
6
∩
5
4
3
2
1
Corollary: (Euler’s formula).
A planar graph has neither K5 nor K3,3 as a topological minor.
Lemma 4.4.2: G contains K5 or K3,3 as a minor if and only if G contains K5 or K3,3 as a
topological minor.
Lemma 4.4.3: Every 3-connected graph without K5 or K3,3 as a minor is planar.
Lemma 4.4.5: |G| > 4, G is edge maximal with respect to not containing T K5 and T K3,3 , then
G is 3-connected.
Kuratowski’s theorem:
1)
G is planar,
2)
G as no T K5 or T K3,3 and
3)
G has no M K5 or M K3,3 .
proof:
(1)⇒(2): Corollary
(2)⇔(3): Lemma 4.4.2
(2)&(3)⇒(1): Lemma 4.4.3 and Lemma 4.4.5
proof of corollary:
v(K5 ) = 5 e(K5 ) = 10
28
We know that e ≤ 3v − 6. Here 10 6≤ 15 − 6.
v(T K5 ) = 5 + α e(T K5 ) = 10 + α
The Euler’s formula does not hold for K5 .
Let v 0 = v(T K5 ), e0 = e(T K5 ) and f 0 = f (T K5 ).
If T K5 is planar, then v 0 − e0 + f 0 = 2, but
v 0 − e0 + f 0 = 5 + α − (10 + α) + f = v(K5 ) − e(K5 ) + f (K5 ) = 5 − 10 + f 6= 2
v(K3,3 ) = 6, e(K3,3 ) = 9
Assume that K3,3 is planar. K3,3 is triangle-free, so e ≤ 2v−4 must hold. But it doesn’t 9 6≤ 2v−4 = 8,
a contradiction.
Trees with 3 leaves:
G = T K1,3
Trees with 4 leaves:
T
T K1,4
proof of Lemma 4.4.2:
If G contains X as a topological minor, then G contains X as a minor.
Assume that G contains K3,3 as a minor.
Consider the branch sets of this minor.
29
connected
connected
connected
connected
connected
connected
Consider a spanning tree in a branch set and three edges leaving the branch set.
T K1,3
connected
connected
connected
connected
connected
connected
This is a topological minor of K3,3
Assume that G cotains a minor of K5 . Consider smallest such minor, so each branch set together
with 4 edges leaving it to another branch set is tree with 4 leaves.
type 1
type 2
If each branch set is of type 1, then G has a T K5 . If some branch set is of type 2, then
1
2
1
4
6
3
5
2
4
3
5
6
So G contains T K3,3 .
30
proof of Lemma 4.4.3: Induction on |V (G)|.
|V (G)| = 4, done. Assume that |V (G)| > 4. Then there is an edge {x, y}, such that G ◦ {x, y} is
3-connected. Lets call G0 = G ◦ {x, y}, call the vertex vx,y obtained by contracting x and y. We have
that G0 is K5 , K3,3 -minor-free, |v(G0 )| < |V (G)| and G0 is 3-connected, by induction G0 is planar.
Consider its embedding. let x1 , x2 , . . . , xk in cyclic order be the neighbours of x in G in a cycle C,
bounding of the face containing vx,y in G0 − vx,y .
(Note: G0 − vxy is 2-connected ⇒ each face is bounded by a cycle.)
x1
C
xk
vxy
G0
C
x2
vxy
x3
Let xi -xi+1 -path in C be Pi , i = 1, . . . k − 1 and xk -x1 -path in C be Pk . And let "Y be the set of
neightbours of y in C.
C
Pk
xk
x1
P1
x2
vxy
P2
x3
Case 1:
,
Y ⊆ Pi for 1 ≤ i ≤ k − 1:
xi
x y
xi+1
Case 2:
|Y ∩ N (x) ∩ V (C)| ≥ 3:
w1 ∈ Y
w3 ∈ Y
y
xi+1
w2 ∈ Y
Then w1 , w2 , w3 , x, y correspond to branch vertices of T K5 , where w1 , w2 , w3 ∈ Y ∩ N (x) ∩ V (C).
Case 3:
◦
|Y ∩ Pi | =
6 0, |Y ∩ V (C\Pi )| =
6 0 for some 1 ≤ i ≤ k.:
31
xi
x
w1
y
xi+1
w2
G contains T K3,3 .
Tutte’s theorem: Let G be a 3-connected planar graph with convex embedding. Then each bounded
face of G is a convex polygon.
Lemma4.4.4: Let G be a graph with κ(G) ≤ 2 and G = G1 ∪ G2 with |G1 ∩ G2 | ≤ 2. If G
is edge-maximal with respect to not containing T X, X is 3-connected, then also G1 and G2 are
3-connected, and if G1 ∩ G2 = {x, y}, then x ∼ y.
proof: (Outline)
V (G1 ) ∩ V (G2 ) = S. We have ∀v ∈ S v ∼ G1 \S, v ∼ G2 \S.
• S = ∅:
e
G1
G2
Add e between G1 and G2 . Contradiction since G ∪ e ⊇ T X, e ∈ T X, e is a bridge in T X, but
T X is 3-connected.
• |S| = 1:
G1
e
v2
v1
G2
v
Add edge e = {v1 , v2 }. In G ∪ e ⊇ T X, the branch vertices of T X are all, w.l.o.g. in G1 . So, e
is on the path P between the branch vertices. Replace P with vv1 , so T X ⊆ G1 .
e
G1
v1
P
v2
G2
v
• |S| = 2:
x
G1
y
P
G2
32
Assume that x 6∼ y. Add e = {x, y}, G ∪ e ⊇ T X. All branch vertices of T X are in, w.l.o.g.,
G1 . Replace e with a x-y-path P in G2 , obtaining T X ⊆ G, a contradiction.
e1
x
P
y
G1
G2
Add e1 to G1 , T X ⊆ G ∪ e1 . All branch vertices in one of G1 and G2 . Replace xP y with a
{x, y}, thus G1 ∪ e1 ⊇ T X.
proof of Lemma 4.4.5:
Assume not, there is a cut S = {x, y}, x ∼ y (lemma 4.4.4). G = G1 ∪ G2 , S = V (G1 ) ∩ V (G2 ).
T K5 , T K3,3 ⊆ G1 , G2
⇒
G1 , G2 3-connected ⇒ by lemma 4.4.3 and corollary ⇒ G1 , G2 3lemma 4.4.4
connectod and has no M K5 , M K3,3 ⇒ G1 , G2 are planar.
Embed G1 and G2 such that {x, y} lies on the unbounded face.
z2
z1
x
P
G2
G1
y
G ∪ {z1 , z2 } ⊇ T K5 or T K3,3 . ,
Case 1: G ∪ {z1 , z2 } ⊇ T K5 :
All branch vertices of T K5 are w.l.o.g. in G1 . Replace P with z1 , y.
z1
G1
x
y
G1 ∪ {z1 , y} ⊇ T K5 it is planar
Case 2: G ∪ {z1 , z2 } ⊇ T K3,3 :
If all branch vertices are in G1 , done as in case 1. Otherwise branch vertex in G2
z1
z2
x
z1
x
y
y
Replace those path with edges.
T K3,3 ⊆ G1 ∪ {z1 , v} ∪ {x, v} ∪ {y, v} and it is planar
4-color theorem:
„solution“ Alfred Kempe 1879 in 1890 Heawood found a mistake. (proved 5-color
theorem)
33
(Heawood Sep. 1861 - Jan. 1955, alfred Kempe Jul. 1849 - Apr. 1922 (multiset discovery))
5-color theorem: Let G be a planar graph. Then χ(G) ≤ 5 (χ(G) is the smallest number of
colors you need to color G (chromatic number)).
proof: Induction on |V (G)|.
Claim: ∃v ∈ V (G) with deg(v) ≤ 5. Otherwise, deg(v) ≥ 6∀v ∈ V (G)., |E(G)| ≥
(contradiction to Euler’s formula).
6·|V (G)|
2
= 3|V (G)|
Consider an embedding of G in the plane. Lets order neighbors v1 , . . . , v5 of v cyclically according to
the cyclic order of edges vv1 , vv2 , . . . , vv5 in the neighborhood of v. Note that if |N (v)| < 5, χ(G−v) ≤
5, there are at most 4 colors used on neighbors of v, so v can be colored with 5th color.
v1
v5
v
v4
v2
G
v3
Color G − v with at most 5 colors. If #colors on v1 , . . . , v5 is at most 4, we can use the 5th color on v.
Thus, wlog, color of vi is i, i = 1, . . . , 5. Let Gij (x) be the maximal subgraph of Gv , connected, colored
with i, j, and containing x. Note, that if we flip colors i and j in Gij (x) then we get proper coloring.
If V (G13 (v1 )) 63 v3 , then we flip colors 1 and 3 in G13 (v1 ), color v with 1. Thus v3 ∈ V (G13 (v1 )),
similary v4 ∈ V (G24 (v2 )). So we have a v1 -v3 -path and v2 -v4 -path. These two paths do not share
vertices and cross, contradiction to planarity.
v1
v5
v4
v
v2
v4 , v2 not connected
colorflip
v3
Heawood formula (Weisstein): G is embedded on Sγ (torus in R with γ holes), γ > 0,
√
7 + 1 + 48γ
χ(G) ≤
.
2
3
For γ = 1, χ(G) ≤ 7.
is planar then χ(G) ≤ 4. G is toroidal then χ(G) ≤ 7 sharp.
Extremal example: Sizilassi
7
6
34
G
each face shares an edge with every other face.
The polytope where each faces shares an edge with every other face: Tetrahedron, Szilasi.
General chromatic theory: χ(G) =?
χ(G) ≤ ∆(G) + 1 color greedily (use smallest available color)
≤∆
1
...
2
1
≤ ∆ colors are not usable
use (∆ + 1)st color.
Brook’s theorem: χ(G) ≤ ∆(G), unless G = K∆+1 or G = C2k+1 (Cycle of any odd degree).
∆
∆=2
3 colors needed
Intuition: color greedily
≤∆−1
∆−1
∆
1 1
Lemma 1: Let G be a connected graph, v ∈ V (G), then there is an ordering of vertices of G,
v1 , . . . , vn , such that vn = v and vi ∼ vj for each i ≤ n − 1, some j > i.
proof: Pick a spanning tree, T , of G. Let v1 be a leaf of T , v1 6= v. Apply induction to G − v1
(connected as we deleted a leaf).
...
v1 v2
v1
v7
v2
vn = v
v3
v4
v5
v6
v
v8
Observe that we could choose any vertex v = vn and so if G is not regular, then χ(G) ≤ ∆. Choose
v = vn to be a vertex of degree ≤ ∆ − 1, apply a greedy coloring to this order.
Lemma 2: Let G be 2-connected, ∆(G) ≥ 3, G 6= Kk for any k. Then ∃x, y, z ∈ V (G), x ∼ y, x ∼
z, y 6∼ z, G − {y, z} = G[V − {y, z}] is connected.
proof: ,
Case 1:
∃v, deg(v) ≤ |V (G)| − 2, such that G − v is 2-connected:
x
z
Take z ∈ V (G) at distance 2 from v. Let x be a common neighbor of v and z. v
Let y = v. We want to show that G − {y, z} is connected. We have that G − y = G − v is 2connected, so G − y has no cut-vertex, so G − y−(any vertex) is connected, particulary G − y − z
35
is connected.
Case 2:
There is v ∈ V (G), such that G − v has a cut-vertex:
deg(v) ≥ 3
deg(v) = 2
y
y
z
z
v=x
v=x
We have that v is adjacent to non-cut-vertices of each leaf of Block G−v (since G is 2-connected).
If deg(v) ≥ 3 take x = v, z, y ∈ N (v), z, y are in the different leaf-blocks of G − v.
If deg(v) = 2, so G − v has a block-cut-vertex-tree isomorphic to a path. If all blocks are K2 ’s
v
z
x
is impossible since ∆(G) ≥ 3. So, there is a nontrivial block in G − v.
y
v
proof of Brook’s theorem:
If G has a cut-vertex, v, then G = G1 ∪ G2 , V (G1 ) ∩ V (G2 ) = {v}.
G2
G1
K∆+1
By induction on #vertices, χ(G1 ) ≤ ∆(G1 ) ≤ ∆(G) and χ(G2 ) ≤ ∆(G2 ) ≤ ∆(G) unless G1 or G2
are K∆+1 or (C2k+1 and ∆ = 2). Nether G1 nor G2 is K∆+1 (otherwise deg(v) > ∆). neither G1 nor
G2 is (C2k+1 and ∆ = 2) since otherwise deg(v) ≥ 3.
G1
G2
v1 v2
Color G1 and G2 with at most ∆ colors, permute colors if necessary so that the color of v is the same
in G1 and in G2 .
x
z G − {y, z} is connected. Let
If G is 2-connected, let x, y, z be as in Lemma 2. v
v1 = y, v2 = z, vn = x; to order vertices, apply Lemma 1 to G − {y, z} with vn = x, such that
∀vi , i = 1, . . . , n − 1 ∃j > i : vi ∼ vj .
36
1
1
y = v1 z = v2
1
2
2
x = vn
1
∆=4
Apply greedy coloring to this ordering of V (G). Since vn has ∆ colored neighbors, but two neighbors
have the same color, vn can be colored using the ∆th color.
ω(G) = #vertices in a largest complete subgraph of G (clique) (clique number)
α(G) = #vertices in a largest induced empty subgraph of G (co-clique, independent set) (independence number)
ω(G) ≤ χ(G),
n
α
≤ χ(G)
...
1
3
2
χ(G)
each color class is an independent set, i.e. has size ≤ α(G).
χ(G) 6= ω(G).
Theorem (Erdo’s, Hajanal?) ∀k, l > 0 positive integers there is a graph G: χ(G) ≥
k, girth(G) ≥ l.
Some other bounds:
• χ(G) ≤
1
2
+
q
2m + 14 , m = |E(G)| ≥
...
1
2
3
χ(G)
2
.
χ(G)
Note in χ(G)-coloring, there is an edge between two color classes, otherwise take a union of
these two color classes.
• χ(G) ≤ max{δ(H) : H ⊆ G} +1 =
|
{z
}
=:δ
≤δ
vn−1
≤δ
vn
Idea: color vertices v1 , . . . , vn such that deg(Vn ) = δ(G), deg(vn−1 )G−vn = δ(G−vn ), . . . , deg(vi )G−{vm ,...,vi+1 } =
δ(G − {vm , . . . , vi+1 }). Color greedily.
• Every graph has a subgraph with minimum degree ≥ χ(G) − 1.
List-colorings: Usually: have a graph , each vertex has a palette (set of available colors) {1, 2, 3, 4, 5}.
We need to pick a color for each vertex from a palette.
37
{1, 2, 3}
{1, 2, 3}
{1, 2, 3}
{1, 2, 3}
Given a set of palettes, lists assigned to vertices L : V → 2N , we say that G = (V, E) is L-colorable
if ∃c : V → N, c is proper, c(v) ∈ L(v) (say H is list-colorable from L).
{2, 3}
{1, 3}
{1, 2}
{1, 2}
{3, 1}
G is not L-colorable
{2, 3}
{1, 2}
{1, 2}
{3, 1}
{2, 3}
χch (G) is the choosability of G, list chromatic number of G
= min{k ∈ N : ∀L : V → 2N , |L(v)| = k ∀v ∈ V (G) G is L-colorable}.
χch (G) ≤ 5 (G planar), χch (K3,3 ) ≥ 3, χch (Kn,n ) ≥∼ log(n).
Es gibt planare Graphen mit χch (G) > 4 (Maryam Mirzakhani)
|V (G)| = 63, L = {1, 2, 3, 4}
124
134
124
123
234
134
123
124
134
234
123
234
not colorable
is not colorable
2345
1235
2345
2354
1235
1234
1235
1345
1234
1245
1245
2345
1234
1234
2345
1235
1345
1235
1345
1324
1235
1235
1234
4523
2354
1524
1345
1254
1345
4523
1345
1342
3452
1523
1453
1245
1352
1253
1245
1243
1245
4523
1243
1453
1245
1345
1542
1243
1245
1542
1352
1235
1354
3452
1345
1354
1254
1234
1245
1342
1234
1253
1453
L : V → 2N lists, c : V → N.
c(v) ∈ L(v) ∀v ∈ V, c(v) 6= c(u) for ny two adjacent u an v.
χch (G) = χlist (G) = ch(G) = min{k : ∀L : V → 2N , |L(v)| = k, G is L-colorable}.
(χch (G) is the chromatic number or choosability of G)
There is a graph G χlist (G) > χ(G). (e.g. χch (K3,3 ) > 2 = χ(K3,3 ).
38
χch (Kn,n ) ≥ log n.
Consider Kn,n , where n =
[2k−1]
.
k
Example:
{1, 2, 3}
2k−1
k
, with partite sets A, B. Each vertex in A gets a distinct list from
k = 3, [2k − 1] = {1, 2, 3, 4, 5}
{1, 2, 4}
{1, 2, 5} . . .
{3, 4, 5}
...
{1, 2, 3}
{1, 2, 4}
Use same lists in B.
...
{1, 2, 5}
{3, 4, 5}
2k−1
k
2k−1
k
Suppose there is a coloring.
We must just ≥ k colors in B, otherwise if only colors 1, 2, . . . , k − 1 are used, neither of these colors
are from {k, . . . , 2k − 1} = L(v), so v could not have been colored.
{1, 2, ..., k}
v
{k, k + 1, k + 2, ..., 2k − 1}
1
2
k
So, in B we must use ≥ k colors.
Then a vertex u ∈ A with L(u) = {1, 2, . . . , k} can not be colored.
χlist (Kn,n ) >> χ(Kn,n ) = 2.
Theorem: (5-list-coloring) Casten Thomassen: χlist (G) ≤ 5 for any planar graph G.
proof: Stronger inductive statement: If G is a planar graph, F a set of vertices on a face in some
embedding of G, and given lists L, such
|L(v)| = 5, v 6∈ F
|L(x)| = |L(y)| = 1, x, y adjacent in F
|L(u)| = 3, u ∈ F \{x, y}, L(x) 6= L(y).
39
3
1
x
3
y
3
1
3
3
Then G is L-colorable.
1 {a}
Basis:
3
{a, b, c}
1 {b}
Assume that G is 2-connected (easy).
Step: ,
Case 1:
There is a chord, e, that splits G into G1 an G2 with G = G1 ∪ G2 , G1 ∩ G2 = e:
3
x 1
G1
G2 y
3
1
3
Let, wlog, G2 contain x and y. Apply induction to G2 to color it from these list with
coloring c. Let e = {x1 , x2 }. Apply induction to G, with x1 , x2 playing a role of x and y,
L(x1 ) = {c(x1 )}, L(x2 ) = {c(x2 )}.
Case 2: There does not exist such a chord:
xk
xk−1
x
y
Apply induction to G − xk with new lists L0 .
(
L(v),
v 6∈ N (xk ) − {x, xk−1 }
L0 (v) =
L(v) − {α, β}, v ∈ N (xk ) − {x, xk−1 }, where {α, β} ⊆ L(xk ) − L(x).
By induction we can color G − xk from L0 , with coloring c.
xx−1
{α, β}
xk {α, β}
{α, β} {α, β}
x color 6= α, β
40
If c(xk−1 ) = α, let c(xk ) = β and if c(xk−1 ) = β, let c(xk ) = α.
A graph G with the property, for any induced subgraph H of G, χ(H) = ω(H)
holds, is called a perfect graph.
Perfect graphs:
Bipartite graphs are perfect χ = 2, ω = 2.
Perfect graph theorem: (Lorász) G is perfect ⇔ G is perfect.
Strong perfect graph theorem: (Chudnowsky, Seymour, Robertsin, Thosmas) 2002:
G is perfect ⇔ it has no odd hole (odd cycle on at least 5 vertices) or antihole (complement of
odd hole) as induced subgraph.
Recall χ0 (G) is the edge chromatic number.
Coloring edges:
2
1
3
1
χ0 (G) = 3
König’s theorem 1916: If G is bipartite, then χ0 (X) = ∆(G).
proof: Since a vertex of degree ∆ = ∆(G) is incident to ∆ edges, χ0 (G) ≥ ∆.
To prove the upper bound χ0 (G) ≥ ∆, use induction on |E(G)|.
Let e ∈ E(G), color G − e with colors from {1, 2, . . . , ∆} by induction.
x
no β but α
≤ ∆−1
≤ ∆−1
no α but β
y
There is a color α not incident to x, and a color β not incident to y. If α = β, let c(e) = α. Otherwise,
wlog α is incident to y, β is incident to x.
x
y
α
β
x
colorflip
y
Consider an α-β-subgraph, Gαβ , it is a disjoint union of pahs and even cycles. If a component of
Gαβ containing x does not contain y, flip colors α and β in this component, color e with β. If the
component of Gα,β containing x also contains y, it is a path of even length with end-edges of different
colors, a contradiction.
41
Vizing theorem (1964): For any G ∆ ≤ χ0 (G) ≤ ∆ + 1.
proof: Lower bound is easy. For the upper bound we use induction on number of edges. Let e =
{x, y} ∈ E(G). Let c be a (∆ + 1)-coloring of G0 = G − e.
Let Mis(v) be the set of colors not used at v in c (missing colors).
Note, if Mis(x) ∩ Mis(y) 3 s, color xy with s. Mis(v) 6= ∅ ∀v ∈ V (G) because total number of colors
is ∆ + 1, that is less than ∆.
Let y1 = y, y2 , . . . , yh+1 be vertices in N (x) and t1 , t2 , . . . , th+1 be colors such that ti ∈ Mis(yi ), c(xyi+1 ) =
ti . Let s ∈ Mis(x)
y = y1 no t = t1
t1
t2
no S
x
y2 no t2
y3 no t3
th
yh+1
Such that ti 6= s, i = 1, . . . , h, the sequence is largest, ti ’s are distance.
If ti = s (take smallest i)
no t = t1
y = y1
t1
t2
no S
x
If t3 = s
y2 no t2
t1
x
y3 no t3
y1
t2
s
y2
y3
Since the graph is finite, the sequence will stop if th+1 = tj , j < h + 1.
y1
no S
x
tj−1
yj
tj
S
no tj
yj+1
tj
Try to recolor G0 such that s is not incident to yj+1 . Do this by taking a path colored tj and s, and
flip colors. If this path does not end in yj , done.
y1
no S
x
t2
tj
yj
no tj
yj+1
yh+1
no th+1 = tj
The recoloring path between yh+1 & yj and yj+1 & yj can not exist simultaneously.
Edge colorings of multigraphs (allow multiple edges but no loops).
42
Theorem: (Shannon 1949)
χ0 (G) ≤ 32 ∆
χ0 (G) ≤ 32 ∆ for a multigraph G.
k
k
3k colors needed
multigraph
∆ = 2k
k
Main results in colorings:
• χ(G) ≤ ∆ + 1
• χ(G) ≤ ∆ if G 6= K∆+1 , C2k+1 (Brook’s theorem)
• ∆ ≤ χ0 (G) ≤ ∆ + 1 (Vizing’s theorem)
• χlist (G) ≤ 5, G planar ⇒ χ(G) ≤ 5, G planar
• ω(G) = χ(G) for perfect graphs (not induced C2k+1 , C2k+1 , k ≥ 2).
• There exists graph G with:
– χ0 (G) = ∆ G bipartite
– χ(G) <<< ∆(G)
– χlist (G) >> χ(G)
– ω(G) ≤ χ(G)
– ω(G) <<< χ(G) for some graphs
– χlist 0 (G) =?
Mycielski’s construction:
G1
G2
G1 , G2 , . . . , Gk triangle-free, ω(Gi ) ≤ 2 for i = 1, . . . , k and χ(Gi ) = i
G3
Let Gk be constructed, build Gk+1 . Let Gk = (V, E),
Let V (Gk+1 ) = V ∪ U ∪ {w}, V ∩ U = ∅, |V | = |U |, V = {v1 , . . . , vn }, U = {u1 , . . . , un }
n
S
E(Gk+1 ) = {wui : i = 1, . . . , k} ∪ E(Gk ) ∪ {ui v : v ∈ NGk (vi )}.
i=1
43
v1
u1
v2
v3
u2
u3
u4
v4
v5
w
u5
U
V
Observe that χ(Gk+1 ) ≤ k + 1 (Color the copy of Gk on V with k colors, mimic these colors in U ,
c(vi) = c(ui ), color w with (k + 1)).
Assume that χ(Gk+1 ) = k, let c be a k-coloring using 1, 2, . . . , k.
A
vi
k
Gk
Assume that c(w) = k, let A ⊆ V, c(v) = k for v ∈ A.
Note A 6= ∅, otherwise Gk was colored with ≤ k + 1 colors, which is a contradiction to χ(Gk ) = k.
Recolor vertices in A: if vi ∈ A, let c0 (vi ) = c(ui ), vi 6∈ A, let c0 (vi ) = c(vi ).
Claim: c0 is a proper (k − 1)-coloring of Gk .
Since c(ui ) 6= k, c0 does not use color k. Take two vertices vi , vj :
• vi , vj 6∈ A c(vi ) = c0 (vi ) 6= c0 (vj ) = c(vj )
• vi , vj ∈ A vi 6∼ vj
• vi ∈ A, vj 6∈ A vi ∼ vj
vi
ui
vj
Assume that c0 (vi ) = c0 (vj ). c0 (vi ) = c(ui ) ui ∼ vj vj ∈ N (vi ) c(ui ) = c(vj ), contradiction since
c is a proper coloring.
χlist 0 (G) edge choosability.
Conjecture:
∀G χlist 0 (G) = χ0 (G)
44
Theorem: (Galvin 1995) χlist 0 (G) = χ0 (G) for bipartite graph G.
Lemma: Let H be a graph, L : V → 2N a list assignment. If H has an orientation of the edges
such that d+ (v) < |L(v)|, and ever induced subgraph has a kernel, then G is L-colorable.
orientation d+ (v) = 4
A kernel is an independent set S in a graph H = (V, E), such that
s
v
∀v ∈ V \S ∃s ∈ S
S
proof: Let H(α) be set of vertices whose list contain α. Color the vertices of kernel of H(α) with
α, remove color α from other lists. Apply induction to uncolored part.
H(α)
α
α
α
α
α
α
α
α
For v ∈ V (H(α)) the list size decreased by 1, outdegree decreased by 1.
outline (proof of Galvin’s theorem):
Let c be a coloring of E(G) with k colors 1, . . . , k, partite sets X and Y . For two edges e, e0
e0 if c(e) < c(e0) and meet in Y , or c(e) > c(e0) and meet in X.
e
e 7
2
e0
3 e
5
X
e0
Y
X
Y
Rest apply Lemma.
Extremal graph theory
Given fixed graph H, find
ex(n, H) := max{|E(G)| : H 6⊆ G, |V (G)| = n}
45
(H 6⊆ G means that H ist not a subgraph of G), the largest number of edges in H - free graph on n
vertices.
H is fixed and n-large.
An extremal graph EX(n, H) is a graph on n vertices, ex(n, H) edges, no subgraph isomorphic to H.
Examples:
H = K2
H = P3
ex(n, K2 ) = 0, EX(n, K2 ) = En
jnk
jnk
ex(n, P3 ) =
, EX(n, P3 ) =
K2 .
2
2
H=
EX(n, H)
Here:
H ∈ {Kt , Pt , Ct , Kt,s }, χ(H) ≥ 3
...
Turán graph T (n, r), r ≤ n is a complete r-partite graph with partite sets of almost equal sizes, n
vertices.
T (n, 5)
T (5, 2)
T (6, 2)
T (7, 3)
|E(T (n, r))| =t(n, r)
Note:
Claim:
χ(T (n, r)) = r
The largest number of edges in a bipartite graph is
Kq,n−q
≈
n2
4
n n
. The densest bipartite graph is
2
2
n − 1 edges
edges
f (q) := |E(Kq,n−q ))| = q(n − q)
n2
max f (q) ∼
=n
i≤q≤n−1
q= 2 4
Combinatorial argument:
46
G
A
B
q
n−q q <n−q−1
Move a vertex from B to A, get a new graph G0 .
|E(G0 )| = |E(G)| − q + (n − q − 1) = |E(G)| − 2q + n − 1 > |E(G)|.
Lemma: Among all r-partite graphs on n vertices (r-chromatic graphs) T (n, r) has largest number
of edges.
proof: Let G be such a graph. Then G is a complete r-partite graph with parts V1 , V2 , . . . , Vr . If
||Vi | − |Vj || ≥ 2, then shifting a vertex from larger part to smaller gives a graph with larger number
of edges.
Lemma: t(n, r − 1) = t(n − (r − 1), r − 1) + (n − r + 1)(r − 2) + r−1
.
2
X2
X1
T (n − (r − 1), r − 1)
(r − 1) blobs
Xr−1
proof:
|E(T (n − (r − 1), r − 1))| = t(n − (r − 1), r − 1)
(Number of edges between{x1 , x2 , . . . , xr−1 } and V − {x1 , x2 , . . . , xr−1 })
= |V − {x1 , x2 , . . . , xr−1 }| · (r − 2) = (n − r + 1)(r − 2)
r−1
(Number of edges betweenx1 , x2 , . . . , xr−1 ) =
.
2
Lemma: lim
n→∞
t(n,r−1)
(n2 )
= (1 −
1
)
r−1
.
proof:
n n
1
n
n
n
r−1
∼
t(n, r − 1) =
− (r − 1) ·
− (r − 1)
=
2
2
2
2 r−1
r−1
2
2
2
n
1
n ∼ n
1
n
n
∼
−
(1 −
) (
≈
for large n)
=
=
2
r−1 2
2
r−1
2
2
47
Theorem (Turàn 1941): For all positive intgers r, n, r ≥ 3:
EX(n, Kr ) = {T (n, r − 1)}
n
1
ex(n, Kr ) = t(n, r − 1) ≈
(1 −
).
2
r−1
Theorem (Erdös-Stone): ex(n, H)
1
2
2
ex(n,
1
2
).
n
2
3
Theorem: ex(n, Pk+1 ) &
1
3−1 )
(1 −
∼
=
n
2
χ(H)=r
(1 −
1
)
r−1
= ex(n, Kr ).
n 1
2 2
=
k−1
n.
2
proof: Let G be a disjoint union of cliques on k vertices.
k
k
k
k n
2 k
...
k
=
k(k−1) n
2
k
=
(k−1)n
2
For the upper bound, use induction on n.
Basis: n = k, Pk+1 6⊆ Kk
|E(Kk )| = k(k−1)
= n(k−1)
.
2
2
Step: Observe that if Pk+1 6⊆ G, then δ(G) ≤
k−1
.
2
forbidden
longest path
graph is connected (otherwise apply induction)
∃ longer path
v
u
v is not adjacent to these
If u has t neighbours, v has ≤ |V (P )| − t neightbors.
Together u and v has at most |V (P )| neighbours, or ≥ 2δ neightbours. If δ(G) ≥ k−1
+ 1, then
2
|V (P )| ≥ k + 1.
48
Let v be a vertex of minimum degree, deg(v) ≤
|E(G)| ≤ |E(G − v)| +
k−1
2
k−1
.
2
by induction
≤
Then
(n − 1)
k−1 k−1
k−1
+
=n·
2
2
2
Theorem: k ≥ 2, let G be a graph where each cycle has length at most k, then |E(G)| ≤ k2 (n − 1).
(without proof )
outline (proof of Turán’s theorem):
Basis: n ≤ r − 1, then Kn does not contain Kr−1 , Kn = T (n, r − 1).
Step: n ≥ r, consider a graph G with no Kr , has largest number of edges and n vertices. We have
that Kr−1 ⊆ G, on vertex set K. Also, we have that |E(G)| ≥ t(n, r − 1).
≤ r−2
Kr−1
n − (r − 1)
≤ r−2
t(n, r − 1) ≤ |E(G)| ≤ |E(G[K])| + |E(K, V − K)| + |E(G[V − K])
(1)
r−1
≤ |
+ (n − r − 1)| · (r − 2) + t(n − (r − 1), r − 1)
|
{z
}
2
(2)
by induction since G[V −K]6⊇Kr .
1
= t(n, r − 1)
⇒ |E(G)| = t(n, r − 1)
Since equality holds in (1) and (2), |E(K, V − K)| = (n − (r − 1)) · (r − 2), |E(G[V − K])| =
t(n − (r − 1), r − 1).
We have ∀v ∈ V − K, v sends exactly (r − 2) edges to K, by induction G[V − K] = T (n − (r −
1), r − 1). We have that the number of nonedges between K and V − K is n − (r − 1)
(why: total number of possible edges is (n−(r−1))·(r−1)− number of edges (n−(r−1))·(r−2))
If , for some v ∈ K, v is adjacent to each part V1 , . . . , Vr−1 of T (n − (r − 1), r − 1), then Kr ⊆ G,
a contradiction. Thus ∀v ∈ K there is Vj such that v 6∼ Vj .
• If vi 6∼ Vi , i = 1, . . . , r − 1, vi is adjacent to all vertices in V − K − Vi (count number of
nonedges), also vi is adjacent to all of K − vi , G = T (n, r − 1).
49
vr−1
v1
v3
V −K
K = {v1 , . . . , vr−1 }
v2
• If vi 6∼ Vi , vj 6∼ Vi , then vi , vj and (r − 2) vertices from distinct {V1 , . . . , Vr−1 } − {Vi } form
a Kr , a contradiction.
vi
vj
ex(n, K3 )
ex(n, K4 )
ex(n, K5 )
t(n, r − 1)
ex(10, K4 )
n
1
∼
=
2 2
1
n
∼
= (1 − )
3 2
1 n
∼
= (1 − )
4 2
n
1
)
= (1 −
r−1 2
= 9 + 12 + 12 = 33.
Forbid Krs = T (rs, r)
If s is largest enough, then any H, χ(H) = r is contained in Krs .
If H 6⊆ G ⇒ Krs 6⊆ G.
Note: T (n, r − 1) 6⊇ H.
Densest (r − 1)-chromatic graph is T (n, r − 1).
ex(n, Kr ) ≈ ex(n, H) for any H : χ(H) = r ≥ 3.
50
Theorem (Erdös-Stone
1946, Chvatàl-Szemerédi 1981): ∀ > 0 ∀r ∈ N, r ≥ 3, if |E(G)| ≥
n
1
(1 − r−1 + ) 2 , |V (G)| = n, then
Krt ⊆ G, t =
There is a graph G on (1 −
is amost the best).
1+ n
)
r−1 2
log n
.
500 · log( 1 )
edges, n vertices and no Krt for t =
5 log n
log( 1 )
(this tells us that t
In particular H ⊆ G for any H : χ(H) ≤ r.
log n
r=3
log n
log n
ex(n, H) = 1 −
1
χ(H)−1
n
2
(1 + o(1))
∀ > 0 ∀t ∈ Z+ ∃n = n(, t), E(G) ≥
1
ex(n, K310 ) = n2 (1 −
+ )
2
n
1
10
ex(n, K3 ) = 2 (1 + o(1)).
Corollary: lim
n→∞
ex(n,
).
ex(n,H)
(n2 )
= 1−
1
r−1
n
2
1−
=
r=χ(H)
1
r−1
1−
+ , |V (G)| = n ⇒ Krt ⊆ G.
=
1
2−1
1
χ(H)−1
χ(H)−2
χ(H)−1
2 n
3 2 What if χ(H) = 2?
ex(n, H)
lim
=
n
n→∞
2
1−
= 0.
Conjecture (Erdös): For any H ex(n, H) ∼
= cnr , r ∈ Q, 0 ≤ r ≤ 2. ∀r ∈ Q, 0 ≤ r ≤ s ∃H :
ex(n, H) = cnr .
5
3
known: n2 , n 2 , n 3
Forbidding Ks,t in subgraphs of Km,n
z(m, n; s, t):= max number of edges in a bipartite graph with parts X, Y of sizes m, n, respectively,
containing no Ks,t with s vertices in X, t vertices in Y . (Zarankiewitz)
51
m
X
s
forbidden
n
Y
t
1
1
z(m, n; s, t) ≤ (s − 1) t (n − t + 1)m1− t + (t − 1)m
1
1
Corollary: z(n, n; t, t) ≤ c n · n1− t + cn ≈ cn2− t
t, s constants.
proof: Consider G ⊆ Km,n , Ks,t 6⊆ G.
X v s
d(v)
Y
t=4
t
Let T be the number of stars with t leaves centered in X. T ≥
P d(v)
≤ (s − 1) stars ⇒ T ≤ nt (s − 1) ⇒
≤ nt (s − 1).
t
P
v∈X
d(v)
t
, each t-subset of Y gives
v∈X
Jensen’s inequality: ϕ-convex function ϕ : R → R, ai > 0, ai ∈ R
P
P
ai ϕ(xi )
ai x i
.
≤ P
ϕ P
ai
ai
We have ϕ(x) = xt , ai = 1
P d(v) X d(v)
|x|
|x|
≤
.
t
t
v∈X
Why is xt convex for x ≥ t?
Idea: f (x) = x · g(x) and g(x) convex (g 0 (x) > 0, x > 0) ⇒ f (x) convex. f 0 = xg 0 + g, f 00 =
xg 00 + g 0 + g 0 > 0
P
P e
P d(v)
e m
With e :=
d(v) it follows m mt =
≤
≤ nt (s − 1)
t
t
v∈X
x∈X
Jensen’s inequality v∈X
52
e
n
m m
≤ (s − 1)
m
t
e
nt
m m t ≤ (s − 1)
6 t!
6 t!
e
− t + 1)t ≤ (s − 1)(n − t + 1)t
m
1
1
1
e
(s − 1)(n − t + 1)t t
= (s − 1) t (n − t + 1) · m− t
( − t + 1) ≤
m
m
m·(
1
1
e ≤ (t − 1)m + (s − 1) t (n − t + 1)m · m− t .
1
Corollary: ex(n, Ks,t ) ≤ cn2− s , s ≤ t
2
2
Lemma: ex(n, Kr,r ) ≥ c · n2− (r+1)
1
(⇒ c · n2− (r+1) ≤ ex(n, Kr,r ) ≤ cn2− r )
proof: Construct a graph G on n vertices, no Kr,r subgraph with many edges. Pick an edge with
probability p, get G.
G
2r
r
r
(EXP # means expected
number)
EXP(#E(G)) = p n2 .
2r r2
n
EXP(#copies of Kr,r in G) = 2r
p .
r
0
Construct a graph G from G by deleing an edge from each copy of Kr,r in G. So, G0 6⊇ Kr,r .
2
EXP #deleted edges= EXP(#copies of Kr,r in G) < 12 EXP(#E(G)) if p = cn− r+1 .
Verify:
n
2r r2 ? 1 n
p
<
p
2r
r
2 2
−1
1 n
n
2r
n2
r2 −1
∼
p
<
= C 2r = Cn2−2r
2 2
2r
r
n
p < C2 n
2(1−r)
r 2 −1
2
= C2 n− r+1 .
3
Lemma: ex(n, K2,2 ) = ex(n, C4 ) ∼
= c · n2 .
proof: Upper bound: follows from z(n, n; 2, 2)
Lower bound: construct G on n vertices, no C4 . V (G) = Zp \{0} × Zp , prime p.
(a, b) ∼ (c, d) ⇔ ac = b + d(mod p) (Lazebnik)
53
p=3
{1, 2} × {0, 1, 2}
(1, 0)
(2, 0)
(1, 1)
(2, 1)
(1, 2)
(2, 2)
1·2=0+2
Claim 1: G is (p − 1)-regular.
Fix a vertex (a, b), count #(x, y) ax = b+y. We can choose x in (p−1) ways, then y is uniquely
determined.
Claim 2: G has no C4 .
Fix (a, b), (c, d) ∈ V (G), we’ll show that these vertices have at most 1 common neighbor.
(a, b)
(c, d)
(x, y)
Let (x, y) ∈ V (G). If (x, y) ∼ (a, b) and (x, y) ∼ (c, d) then
(
(
ax = y + b
x = (b − d)(a − c)−1
⇒
cx = y + d
2y = x(a + c) − b − d
Either a 6= c or b 6= d ⇒ unique solution or not, then there are no solution.
1
1
|V (G)| · (p − 1) = (p − 1)p · (p − 1)
2
2
2
p
3
p3 ∼
= |V (G)| 2
|E(G)| =
|V (G)| ∼
=
|E(G)| ∼
=
n
∼
= cn2 if χ(H) ≥ 3.
ex(n, H) =? = o(n2 ) if χ(H) = 2.
ex(n, H) =
2
1−
1
χ(H)−1
repetition:
ex(n, H) = max #edges, in an n-vertex H-free graph.
n2
ex(n, K3 ) ∼
=
4
1
n2
ex(n, Kr ) ≈ (1 −
)( )
r−1 2 1
n
ex(n, H) ∼
)
, χ(H) ≥ 3
= (1 −
χ(H) − 1 2
ex(n, H) = ? for H-bipartite
2
2− t+1
c1
3
1
≤ ex(n, Kt,t ) ≤ c · n2− t
3
c1 n 2 ≤ ex(n, K2,2 ) = ex(n, C4 ) ≤ c · n 2
ex(n, H) ∼
= cnr (Erdös Conjecture, r ∈ Q)
54
For the lower bound constructed a graph on ≈ p2 vertices and p3 edges for a prime p.
(We know that for any n ∈ Z+ there exists a prime p such that n ≤ p ≤ 2n (Bertrand’s postulate).
Erdös: ∀k ∈ Z+ ∃N ∈ N : ∀n > N there are at least k primes between n and 2n.
Pierre Dusart (2010): For x ≥ 396748 ∃prime between x and x +
x
.
25 ln2 x
√
Oppermann’s conjecture (1817-1883): There is a prime between x2 (n) and x2 + x (n + n).
For Zarankiewics problem, look at subgraph of Km,n
For extremal problem, look at subgraph of Kn
3
3
z(n, n; t, t) ≤ cn 2 ⇒ ex(n, Kt,t ) ≤ cn 2
Any graph G can be vertex split into X, Y , #edges(X, Y ) ≥ 12 #edges of G.
1st Way:
G
Y
X
3
G 6⊇ C4 , then bipartite G(X, Y ) does no have C4 , has a Zarankiewics bound e(G(X, Y )) ≤ cn 2 , thus
3
e(G) ≤ 2e(G(X, Y )) ≤ 2cn 2
2nd Way:
Consider G on n vertices and no C4 .
G
n
2
n
2
G
n
2
3
For any spanning bipartite subgraph of G, it has at most cn 2 edges.
X
n−2
e(G) ≤ c
e(H) n
−1
2
H-balance spanning
bipartite subgraph of G
n−2
n
≤ cn n
n
−1
2
2
3 n(n − 1)
= cn 2
nn
3
2
2 2
≈ c1 n
Szemerédi’s regularity lemma:
3
2
Given a graph G, X, Y ⊆ V (G), X ∩ Y = ∅
kX, Y k = #edges between X and Y.
d(X, Y ) =
55
kX, Y k
|X||Y |
−1
n−2
the density of a pair (X, Y ), 0 ≤ d(X, Y ) ≤ 1
(X, Y ) is an -regular pair if ∀A ⊆ X, |A| ≥ |X|, ∀B ⊆ Y, |B| ≥ |Y | :
|d(X, Y ) − d(A, B)| ≤ X
A
B Y
˙ 1 ∪V
˙ 2 ∪˙ . . . ∪V
˙ k:
An -regular partition of G = (V, E) is a partition of vertex set V = V0 ∪V
(i)
|V0 | ≤ |V |
(ii)
|V1 | = |V2 | = . . . = |Vk |
(iii)
All but at most k 2 pairs (Vi , Vj ) are -regular 1 ≤ i < j ≤ k.
vk
v0
vi
vj
Regularity Lemma of Szemerédi (about 1970): ∀ > 0 ∀m ≥ 1, m ∈ Z, ∃M ∈ Z+ , ∀graph
G of order ≥ m, there is an -regular partition of G with at most M parts.
Note:
M -constant M does not depend on |V (G)|.
proof: Let G be given, let V0 , V1 , . . . , Vk be -regular partition, let d ∈ (0, 1]. Construct R, reduced
subgraph R = R(G) on v1 , . . . , vk so that vi ∼ vj iff (Vi , Vj ) ist -regular of density ≥ d. Construct
R = Rs (s ∈ Z+ ) - blow up of R with parameter s (each vertex s replaced by s vertices and each edge
is replace by Ks,s between the corresponding sets).
v0
v4
v3
G
v1
v2
v1
Rs = R3
R
v4
v2
v3
K3 ⊆ G
Embedding Lemma: If H ⊆ Rs ⇒ H ⊆ G.
(∀d ∈ (0, 1], ∆ ≥ 1 ∃0 > 0 ∀G ∀H ∆(H) ≤ ∆, s ∈ N, R = RG ( ≤ 0 , e = |Vi | ≥
proof of Erdös-Stone theorem):
56
K3 ⊆ Rs
s
, d)
0
− ...
1
|E(G)| ≥ t(n, r − 1) + γn2 = (1 − r−1
+ 2γ) n2 . Want to show Krs ⊆ G (s-blow-up of Kr ).
⇒
Krs ⊆ G.
Apply Regularity Lemma to G, get R, Rs . If Kr ⊆ R, then Krs ⊆ Rs
Embedding Lemma
To show that Kr ⊆ R, count the number of -regular pairs of positive density, apply Turán theorem
to R.
Fix > 0 and d ∈ [1, 0), d very small.
A good pair (vi , vj ) is a -regular one with density ≥ d.
v0
vk
v1
v2
Total number of edges which are not in good pairs =
= #edges in G[vi ] + #edges in not -regular pairs + #edges in -regular pairs of density < d
n n
n
2 n n
k
+
≤ k
+k ( · ) + d
2
2
k k
2
|
{z
}
edges inside the blobs
2
2 2
∼ n
=
n large
2k
+ n + n2 + dn2
1
n
2
2
= ( + + + d) n choose , k, d such that this constant is < γ
2
| 2k
{z
}
<γ
n2
1
n2
n2
1 n2
≥ (1 −
+ γ) − γ
≥ (1 −
) .
2
r−1
2
2
r−1 2
1 n2 k k
1 k2
(1 −
) ·
= (1 −
) .
n n
r−1 2 nn
r−1 2
edges
k k
#edges in good pairs ≥ e(G) − γ
|E(R)|
&
each pair has at most
Since |V (R)| = k, by Turán’s theorem R ⊇ Kr .
Unavoidable induced subgraphs. When do we have induced C5 ?
We say that a graph H has binary chromatic number r if ∀0 ≤ c ≤ r V (H) can be split into c
cliques and r − c co-cliques (independent sets) and r is the smallest such number.
Binary cromatic number of C5 is 3.
ex(n, C5 )
c=0
c=2
c=1
c=3
57
(Vi , Vj ) is -regular pair of density 0 < d < 1 ⇒ C5 ⊆ G.
H ⊆ G where H has binary chromatic number 4.
So far:
many edges → subgraph H is unavoidable
#edges c · n2 in G →H ⊆ H, χ(H) ≥ 3
1
|E(G)| > (1 − r−1
) n2 ⇒ Kr ⊆ G.
What do we need to force M Kr or T K3 ?
recall if ∆(H) ≤ 3 M H ⊇ T H.
average degree is high → M Kr , T Kr
?
high chromatic number → M Kr , T Kr
high chromatic number 6→ Kr (Recal Mycielsky construction ∀k ∃G : χ(G) ≥ k, ω(G) = 2)
Theorem 1996 (Bollobás-Thomason, Komlós-Szemerédi):
∃c ∈ R ∀r ∈ N ∀G : average degree d(G) ≥ cr2 , T Kr ⊆ G.
e(G) ≥ Cr2 n (sharp up to c)
G ⊇ Kr if e(G) ≥ cr n2
G ⊇ T Kr if e(G) ≥ c0r n
Theorem (Kostochka ’82, Thomason
’84):
√
∃c ∈ R ∀r ∈ N ∀G : d(G) ≥ cr log r, M Kr ⊆ G (sharp up to c)
p
e(G) = Cr log r n
| {z }
cr
?
χ(G) high → M Kr ⊆ G T Kr ⊆ G
Hadwiger’s conjecture 1943: ∀r ∈ Z+ ∀G
χ(G) ≥ r ⇒ M Kr ⊆ G.
trivial for r = 2, 3, easy for r = 4, equivalent to 4-color-theorem for r = 5, r = 6, r ≥ 7 open.
Hajos conjecture: ∀r ∈ Z+ ∀G
χ(G) ≥ r ⇒ G ⊇ T Kr
true for k ≤ 4, open for k = 5, 6, false for k ≥ 7.
Bollobás, Thomason c = 1116.
58
Theorem (Thomassen ’83): ∀k ∈ Z+ ∀G girth(G) ≥ 4k − 3, δ(G) ≥ 3 : G ⊇ M H, for some
H with minimum degree k.
proof: If k = 2, δ(G) = 3 ⇒ G contains a cycle (that is a minor of H, χ(H) = 2).
Let k ≥ 3, assume that G is connected. Let V1 , V2 , . . . , V, ⊆ V (G) such that they are disjoint, G[Vi ]
connected, |Vi | ≥ 2k − 2, m is as large as possible. (we’ll show that these Vi ’s are branch sets of M H)
Claim 1: G[Vi ] is a tree, i = 1, . . . , m.
Assume not, i.e., G[Vi ] has a cycle C, length(C) ≥ 4k − 3
≥ 2k − 2
v4k−3
v1
v2k−1
v2k−2
≥ 2k − 2
Let Vi0 = {v1 , . . . , v2k−2 }, Vi00 = {v2k−1 , . . . , v4k−3 }. Then G[Vi0 ], G[Vi00 ] conneted, have ≥ (2k −2)
vertices, thus m was not maximal, a contradiction.
Claim 2: Between any two parts Vi , Vj there are at most 2 edges.
Assume not, i.e. ∃ three internally dijoint u-v-paths, u ∈ Vi , v ∈ Vj .
Vi
P1
u
≥ 2k − 2
≥ 2k − 2
≥ 2k − 2
Vj
P2
P3
v
We have l(P1 )+l(P2 ) ≥ 4k −3, l(P1 )+l(P3 ) ≥ 4k −3, l(P2 )+l(P3 ) ≥ 4k −3. Thus ∃i ∈ {1, 2, 3}
◦
◦
wlog i = 1 : l(Pi ) ≥ 4k−3
= 2k − 1. We have |V (P1 )| = 2k, so |v(uP1 v)| ≥ 2k − 2. We have
2
that P2 ∪ P3 is a cycle of length ≥ 4k − 3, lets split its vertex set into two sets of ≥ 2k − 2
◦
◦
vertices inducing paths, call them Vi0 , Vi00 . Replace Vi , Vj with Vi0 , Vi00 , V (uP1 v).
≥ 2k
≥ 2k − 2
How many edges are incident to Vi ? We have δ(G) ≥ 3, |E(G[Vi ])| = |Vi | − 1, #edges leaving Vi is
≥ 3|Vi | − (2|Vi | − 2) = |Vi | + 2 ≥ (2k − 2) + 2 = 2k. This Vi is adjacent to ≥ 2k
= k other Vj ’s. Since
2
this the Vi ’s form branch sets of M H, δ(H) ≥ k.
Structure:
how does a graph without M Kr look like? (Hadwiser’s conjecture χ(G) < r)
We say that a graph G is obtained from G1 and G2 by pasting along S if G = G1 ∪ G2 , S = G1 ∩ G2 .
59
S
S
G1
Pasting
G2
S
S
along
S = K3
G = G1 ∪ G2
Pasting
S
along
S = K2
• A graph on ≥ 3 vertices, edge-maximal without M K4 iff it is constructed from triangles via
pasting along K2 ’s.
• Wagner 1937 (Klaus Wagner): A graph G is edge-maximal without M K5 (has ≥ 4 vertices)
then G can be constructed recursively from plane trianguations and Wagner graph via pasting
along K2 ’s and K3 ’s.
Wagner’s graph
Ramsey theory
Frank Ramsey
Color E(Kn )
in 2 colors or
≈ log n
≈ logn
Example: Any 2-colored K6 (edges are colored) has a monochromatic triangle (i.e. K3 whose edges
have the same color).
v
v
Wlog
done
don’t know
know
v
done
60
Lemma: Let E(Kn ) be colored with 2 colors, n > 4s + 1, ten there is a monochromatic Ks .
v
blue neighborhood of v
S
yellow neighborhood of v
proof:
Define a sequence of vertices. Let v1 ∈ V . For any vertex v ∈ V , any S ⊆ V , we say that S 0 ⊆ S is a
yellow neighborhood of v in S if all edges between v and S 0 are yellow and all edges between S − S 0
and v are blue (S − S 0 is blue neightborhood of v in S).
bor of v1 , pick V1 to be yellow neighborhood of v1 , otherwise let V1 be blue neighborhood of v1 .
Continue recursively.
v3
v1
v2
repeat log2 4s = log2 22s = 2s.
Ramsey theorem: ∀k ∃N (k) in any yellow-blue coloring of E(Kn ), n ≥ N there is a monochromatic Kk . Smallest such N is denoted R(k), called Ramsay number.
R(3) = 6
√ k
2 ≤ R(k) ≤ 4k
1
lim R(k) k exists? =? $100, $250.
k→∞
proof: Upper bound: Let c be a 2-coloring of E(Kn ), n ≥ 4k . For a vertex x, let
Nyellow (x) = {y ∈ V : c(xy) = yellow}, Nblue (x) = {y ∈ V : c(xy) = blue}.
Ex:
=K
ˆ 3 yellow
Pick v1 ∈ V , pick V1 := {Nyellow : |Nyellow (vi )| > |Nblue (v1 )|}, V1 := Nblue (v1 ) otherwise.
Assume that v1 , . . . , vi , V1 , . . . , Vi are constructed. Pick vi+1 ∈ Vi , let Vi+1 = {Nyellow (vi+1 ) ∩ Vi : If
|Nyellow (vi+1 ) ∩ Vi | ≥ |Nblue (vi+1 ) ∩ Vi |}, Vi+1 = Nblue (vi+1 ) ∩ Vi otherwise.
So |Vi+1 | ≥ |V2i | and at least log2 n ≥ log2 4k = 2k vertices can be chosen, and these vertices give us a
monochromatic Kk .
Upper bound: Need to construct a coloring of E(Kk ) with no monochromatic Kk . Color edges
randomly independently, with probability 21 yellow and probability 12 blue. If prob(there is no mono61
chromatic Kk )> 0, there is a coloring with no monochromatic Kk .
Let S ⊆ V, |S| = k. Prob(S induces a monochromatic Kk ) = Prob(G[S] yellow ∨G[S] blue) ≤
(k)
k
2Prob(G[S] yellow) = 2 · 21 2 = 21−(2)
P
k
Prob(Some S ⊆ V, |S| = k is monochromatic)≤
Prob(G[S] is monochromatic)≤ nk 21−(2) < 1.
k
If n ≤ k2 2
√1
2k
k
+ o(1) , in particular if n ≤ 2 2
S⊂V
|S|=k
Non symmetric Ramsay number:
R(k, l) := min{n : any yellow-blue coloring of E(Kn ) has yellow Kk or blue Kk }
R(2, l) = l, R(k, 2) = k.
Lemma: R(k, l) ≤ k+l−2
k−1
te: R(k, k) ≤ 2k−2
≤ 4k .
k−1
No-
proof: Consider a coloring of E(Kn ) with no yellow Kk and no blue Kl . We’ll show n ≤ k+l−2
− 1.
k−1
The yellow neighborhood has no yellow Kk−1 and no blue Kl , the blus neighborhood has no blue
Kl−1 and no yellow Kk . Then
k−1+l−2
k+l−1−2
k+l−2
n ≤ 1+(R(k−1, l)−1)+(R(k, l−1)−1) = −1+
+
= −1+
.
k−2
k−1
k−1
no yellow Kk−1 no yellow Kk
no blue Kl−1
no blue Kl
Graph Ramsey numbers:
H, G graphs
R(G, H) := min{n : any yellow-blue coloring of E(Kn ) has either yellow G or blue H}
Lemma: R(sK2 , tK2 ) = 2s + t − 1, s ≥ t ≥ 1.
2s − 1
t−1
proof: Lower bound:
Upper bound: We shall show that R((s + 1)K2 , (t + 1)K2 ) ≤ R(sK2 , tK2 ) + 3. Let c be a coloring
of a graph on R(sK2 , tK2 ) + 3 vertices. We shall show that it has yellow (s + 1)K2 or blue (t + 1)K2 .
62
In c:
s
s+1
t
t+1
R(sk2 , tK2 )
R(sK2 , tK2 ) ≤ R((s − 1)K2 , (t − 1)K2 ) + 3
≤ R((s − 2)K2 , (t − 2)K2 ) + 3 · 2
. . . (Repeat until w reach 2)
≤ R([s − (t − 1)]K2 , (t − 1)K2 ) + 3(t − 1)
= 2(s − t + 1) + 3t − 3 = 2s + t − 1
Ramsay theory is a generalization of Pigeonhole principle, Dirichlet’s principle, Schubfachprinzip
(1834)
Pigeonhole principle: If n objects placed in k boxes ⇒ ∃ box with ≥ nk objects.
Erdös-Szekerés (1935) theorem: For any list of n2 distinct numbers, there is a sublist of > n
numbers that s increasing or decreasing
1 , 7 , 2 , 5 , 3 , 4 , 0 , 10 , 6
proof: Given a1 , a2 , . . . , an2 of distinct numbers, let (ui , di ) be a pair, where ui is the length of
longest increasing subequence ending with ai and di is the length of longest decreasing subsequence
ending with ai . Assume that each increaing, decreasing subsequence has length < n. Then we have
< n2 labels (ui , di ). By pigeonhole principle, the same lavel (u, d) appear on different elts, i.e.,
(ui , di ) = (uj , dj ) i 6= j, i < j
a1 , . . . , ai , . . . , aj , . . .
(ui ,di )
This is possible since if aj > ai , uj > ui or if aj < ai , dj > di .
Ramsay theory for set system (hypergraphs):
63
(uj ,dj )
V
V
V
so far
pdgeonhole princ.
color V
graph Ramsey
color V2
color
c:
V
K
V
k
−set of k-element subsets of V
→ {yellow, blue}
n
V
Rr (l) = R(l, r) = min n : for any yellow/blue coloring of
, |V | = n,
r
0
V
0
0
there is either V ⊆ V, |V | = l,
is yellow,
r
00 o
V
00
00
is blue
or ∃V ⊆ V, |V | = l,
r
l
l
Ramsey proved that this exists.
n
V
Rr (l1 , l2 , . . . lk ) = min n : for any coloring of
into k colors 1, 2, . . . , k, |V | = n
r
V1
∃V1 ⊆ V, |V1 | = l1 ,
has color 1, or
r
o
V1
∃V2 ⊆ V, |V2 | = l2 ,
has color 2, . . .
r
Lemma:
Rr (p, q) ≤ Rr−1 (Rr (p − 1, q), Rr (p, q − 1)) + 1
(two colors, color r-element subsets, force p-clique in red, q-clique in blue).
proof: Let V be given, |V | = Rr−1 (Rr (p − 1, q), Rr (p, q − 1)) + 1, c : Vr → {red, blue} is given.
0
00 We want to show that either ∃V 0 ⊂ V, |V 0 | = p, Vr is red, or ∃V 00 ⊂ V, |V 00 | = q, Vr is blue.
64
x r
V0
r−1
Rr (p − 1, q)
r−1
Rr (p, q − 1)
Let V = x ∪ {X}. We shall color
c0 :
X
r−1
.
(
X
red , if c(X 0 ∪ {x}) = red
→ {red, blue}, c0 (X 0 ) =
r−1
blue , if c(X 0 ∪ {x}) = blue.
We have |X| = Rr−1 (Rr (p − 1, q), Rr (p, q − 1)).
{z
} |
{z
}
|
red
blue
V0
Thus ∃V 0 ⊆ X, r−1
is red under c0 , |V 0 | = Rr (p − 1, q)
0
V
or ∃V 0 ⊆ X, r−1
is blue under c0 , |V 0 | = Rr (p − 1, q).
0
00 V0
Assume r−1
is red under c0 . Then, look at Vr under c. So, we habe either V 00 ⊂ V 0 with Vr is
00 00
blue, |V 00 | = q, we are done; or V 00 ⊂ V 0 wih Vr is red, |V 00 | = p − 1, then V ∪{x}
is red and
r
00
|V ∪ {x}| = p.
Applications:
1)
Erdös-Szerés 1935: ∀m ∈ Z≥3 ∃N (m) ∀set of N (m) points in R2 (general position no 3 on a
line) contains a convex m-gon.
proof(1): N := R4 (m, 5) (use red/blue, color 4-element subsets, force red
[m]
4
, or blue
[5]
4
).
Let V ⊆ R2 in general position is given, |V | = N. Lets color V4 with c. Let c(X) be red,
if X is in convex position, and let c(X) be blue if X is not in convex position.
0
0
V
V
0
0
So, either ∃V ⊂ V, |V | = m,
is red or ∃V ⊂ V, |V | = 5,
is blue.
4
4
|
{z
}
|
{z
}
(a)
(b)
Claim: (b) is impossible
65
|V 0 | = 5, convex hull of V 0 is
V0 :
, otherwise
-red
- contradiction
So (a) holds. Then V 0 must form a convex m-gon. Indeed, if not, and a convex hull of V 0
has less than m endpoints
a contradiction.
proof(2): due to Tarjan.
[m]
Let N = R3 (m, m) (color triples, red/blue, force red [m]
or
blue
).
3
3
Given {x1 , x2 , . . . , xN } ⊆ R2 in a general position. Color c({xi , xj , xk }) =red if i < j < k
and xi , xj , xk appear clockwise.
xi
xi
otherwise blue
xk
xj
xj
xk
V0
Let, wlog, V 0 ⊆ {x1 , x2 , . . . , xN }, 3 red, |V 0 | = m.
Then V 0 forms a convex m-gon. Indeed, otherwise
x2
impossible
x4
x5
x3
2m − 5
m−2
2
− 1 ≤ N (m)
≤
+2
m−2
Erdös
Toth-Valtr 1988
2)
Schur theorem 1916: If c : N → {1, 2, . . . , r} then ∃x, y, z ∈ N : c(x) = c(y) = c(z) and
x + y = z.
proof: Let c : {1, 2, . . . , n} → {1, 2, . . . , r}, where n = R2 (3, 3, 3, . . . , 3) (color pairs, r-colors,
|
{z
}
r times
force monochromatic triangles.)
Lets color the edges of Kn on vertex set {1, 2, . . . , n}, c0 ({i, j}) = c(|i − j|).
66
c
1
c0
3
2
4
1
2
5
5
3
4
There is a monochromatic triangle on vertex set {i, j, k}, i < j < k. Let x = j − i, y =
k − j, z = k − i.
We have c(x) = c0 ({i, j}) = c0 ({k, j}) = c(y) = c(z). x + y = j − i + k − j = k − i = z. Known:
k!(e −
1
) ≥ R2 (3, 3, . . . , 3) ≥ c · (3.17)k
| {z }
12
k
Claim: c · 2k ≤ R2 (3, 3, . . . , 3)c1 · k!
| {z }
k
proof: Lower bound: construct a coloring of E(Kn ), n = 2k , in k colors, with no monochromatic
K3 .
Upper bound: Suppose we are given a k-coloring of E(Kn ), n = c1 · k!. We need to show that there
is a monochromatic triangle.
Ni (v)
v
≥
n−1
k
Let v ∈ V (Kn ), there is a color i, v is incident to at least
Ni (v). We have Ni (v) does not induce edges of color i.
n−1
k
edges of color i, call their endpoint set
R(3, 3, . . . , 3) ≤ k · R(3, 3, . . . , 3) + 1 ≤ k · (k − 1) · . . . · (3)R(3, 3) + o(k!)
| {z }
| {z }
k
k−1
= k · (k − 1) · . . . · 3 · 6 + o(k!) = k! · 3 + o(k!).
monochromatic solutions to x + y = z.
67
Question:
Do we always have a monochromatic solution to a1 x1 + a2 x2 + . . . + ar xr = 0, variables
xi ’s?
We say that a linear equation S is r-regular on A ⊆ Z, if, for any r-coloring of A, there is a
monochromatic solution of S. For example x + y = z is r-regular on N for any r. A linear equation
S is regular on A ⊆ Z, if, for any r, S is r-regular. (e.g.: x + y = z is regular on N).
Theorem(Rado): S : a1 x1 + a2P
x2 + . . . + ar xr = 0, ai ∈ Z. We have that S is regular on N if
∃subset I ⊆ {1, . . . , m} such that
ai = 0.
i∈I
Example:
2x1 + x3 − 4x4 + x2 = 0 is regular, because 2 + 1 − 4 + 1 = 0.
Column condition for systems:


c11 x1 + c12 x2 + . . . + c1n xn = 0



c21 x1 + c22 x2 + . . . + c2n xn = 0
..

.



c x + c x + . . . + c x = 0
m1 1
m2 2
mn n
Given variables x1 , . . . , xn .


c1i
, (c̄i :=  ... )
cmi
⇔ c̄1 x1 + c̄2 x2 + . . . + c̄n xn = 0̄ has a column condition if up to recordering the columns the following
holds:
A1
A2
A3
c̄1 , c̄2
, c̄n
sum = 0̄
sum =
linear combination of previous colums
∃k0 = 1 ≤ k1 < k2 < . . . < kt = n if Ai =
ki
P
c̄j
j=ki−1 +1
Ai = 0̄, for 2 ≤ i ≤ t, Ai is a linear combination of c̄1 , c̄2 , . . . , c̄ki −1 .
If a system of equations has a column condition, then for any r ∈ Z≥1 and any r-coloring of N, there
is a monochromatic solution of the system.
Some graph-Ramsay numbers:
Recall R(H, G) = min{n : any red/blue coloring of E(Kn ) has red
G or blue H}.
Lemma: R(sK3 , tK3 ) = 3s + 2t, s ≥ t ≥ 1, s ≥ 2.
proof: (s=red,
ˆ
t=blue)
ˆ
Lower bound: construction:
68
3s − 1
2t − 1
˙ 1,2t−1 .
Red subgraph is isomorphic to K3s−1 ∪K
Blue subgraph is complement of red.
#vertices 3s − 1 + 2t − 1 + 1 = 3s + 2t − 1
Upper bound: Consider a coloring of K3s+2t in red and blue, show there are either s red triangles
or t blue triangles. We shall show R((s + 1)K3 , (t + 1)K3 ) ≤ R(sK3 , tK3 ) + 5.
If have this fact, then
R((s + 1)K3 , (t + 1)K3 ) ≤
≤
R(K3 , K3 ) = 6, R(2K3 , K3 ) ≤ 8 (∗),
≤
R(sK3 , tK3 ) + 5 ≤ R((s − 1)K3 , (t − 1)K3 ) + 2 · 5
. . . ≤ R((s − t + 1)K3 , K3 ) + t · 5
R(qK, K3 ) ≤ 3 + R((q − 1)K3 , K3 ) = 3 + 3(q − 1) + 2 = 3q + 2
3(s − t + 1) + 2 + 5t = 3(s + 1) + 2(t + 1)
R(2K3 , K3 ) ≤ 8.
proof: Consider red/blue coloring of E(K8 ), need to show that ∃blue triangle (∆) or ∃red double
triangle (∆∆).
(∗) Claim:
Case 1: ∃vertex v incident to ≥ 6 edges. Either Nblue (v) has a blue edge, and we have blue triangle,
or Nblue (v) induces monochromatic (red) K≥6 , that contains two red triangles.
done
Case 2: ∃vertex v, |Nblue (v)| = 5.
done
u
w
N (v)
blue
u
w
we ave that u sets at most one red edge to Nblue (v). Otherwise:
69
uw is blue, otherwise:
Since u, w send each ≥ 4 blue edges to Nblue (v), ∃x ∈ Nblue (v), such that ux, wx are blue. So,
x, u, w form a blue triangle.
Case 3: ∃vertex of blue degree 4.
Case 4: ∃vertex of blue degree 3. (almost the same as Case 3)
Case 5: all blue degrees are ≤ 2, then the blue graph s a vertex disjoint union of cycles of length
≥ 4 and paths.
If blue cycle of length:
≥6
5
4
If there are no cycles, the blue graph is a union of paths, so red graph contains wo red triangles.
• We shall prove that there are two adjacent triangles - one red, one blue. Note, we have red
70
triangle, otherwise # small, similary here is a blue triangle.
- done
9 edges between, at east 5 of these, wlog, red. There are
≥ 2 yellow edges from a vertex of blue triangle.
≥ R(sK3 , tK3 )
•
5
Know:
√ n
2 ≤ R(Kn , Kn )
?
R(H, H) ≤ f (|V (H)|) (small)
R(H, H) ≤ c(∆(H)) · |V (H)|
Question:
?
≤ R(H, G)
Theorem (Ramsey for bounded degree graphs): (Chvátal, Rödl, Szemerédi„ Trotter ’83)
∀positive integer ∆ there exists c = c(∆) such that R(H, H) ≤ c · |V (H)| for ∆(H) = ∆.
Corollary: For a n-vertex graph H of maximum degree 3 R(H, H) ≤ c · n (compare R(Kn , Kn ) ≥
n
22)
proof: (Outline)
Recall Szemerédi’s Regularity Lemma. Given graph G there exists partition
vk
v1
− regular
v2
k
1
2
k-constand # blobs
if corresponding pair Vi Vj
− regular density > 0
R-reduced graph
if Kt ⊆ R ⇒ Kts ⊆ G.
Ex. ∆ ⊆ R ⇒
⊆ G, in particular G0 ⊆ G for any 3-chromatic graph.
Given red/blue coloring of E(Kcn ). We need to show that ∃monochromatic H.
et a red graph be G, blue graph be Ḡ.
Let’s apply the Regularity Lemma to G.
71
v1
vk
v2
3 types of pairs (Vi , Vj ):
1)
not -regular (≤ k 2 of such)
2)
-regular of density ≥
1
2
(majority - red)
3)
-regular of density ≥
1
2
(majority - blue)
v1
Ri
reduced
v2
vk
grey
Color vi vj grey if it is of type 1, red if it is of type 2 andblue if it is of type 3.
#grey edges ≤ k 2 ⇒ #non-greq edges ≥ k2 − k 2 ≈ k2 (1 − 2)
⇒ non-grey graph contains Kr = KR(∆+1,∆+1) ⇒ R contains red K∆+1 or blue K∆+1 , assume that
R contains red K∆+1 .
s
s
⇒ G ⊇ K∆+1
⇒ G ⊇ K∆+1
⊇ H since χ(H) = ∆ + 1. (recall ∆(H) = ∆ ⇒ χ(H) ≤ ∆ + 1 by greedy
coloring.)
Theorem (Lower bound on graph Ramsey numbers): (Chvátal, Harary ’72)
R(H, G) ≥ (χ(G) − 1)(c(H) − 1) + 1, c(H) is the cardinality of the largest connected component
of H.
KC(h)−1
χ(G) − 1 < χ(G)
proof:
There is no red H because the largest connected component of a red subgraph has order c(H) − 1,
that is strictly maller than the order of largest connected component of H.
There is no blue G since the blue graph has chromatic numer χ(G) − 1 that is strictly less then the
chromatic number of G.
G:
H:
χ(G) = 4
χ(G) = 4
72
Induced Ramsey number:
Rind (H, G) = min{n : ∃F on n vertices, such that any blue/red coloring of E(F)
contains induces H that is blue, or induced G that is red.}
Theorem: (Deuber, Derdös, Hajnal, Pósa, Rödl ’73)
Rind (H, G) exists for any graphs H and G.
In this class, we’ll prove thet Rind (H, H) exits for bipartite graphs H. We say that a bipartite graph
H = (A ∪ B, E) is embedded int a bipartite graph G = (A0 ∪ B 0 , E 0 ), if H ⊆ G and A ⊆ A0 , B ⊆ B 0 .
ind
A0
A
G
induced
B
B0
Lemma: Any bipartite graph can be embedded into the incidence graph ((X,
{xY : x ∈ X, Y ∈ Xq , x ∈ Y }
Example:
1
12
((X,
X
2
2
), E), X = {1, 2, 3, 4},
3
13 14
23
4
24
X
34
x
2
X
2
X
q
), E), where E =
= {{12}, {13}, . . . , {34}}
We’ll prove induced Ramsey theorem for incidence graph.
Induced Ramsey theory
Rind (H, G) = min{n : ∃F any red/blue coloring of E(F)
has induced red H or induced blue G}
Rind (Kk , Kl ) = R(Kk , Kl )
Lemma: Every bipartite graph is embedded in ((X,
Comment: ((X,
X
q
X
q
), E) for some X, q.
), E) is an universal graph for all bipartite graph of certain size.
73
proof: Let P be a given bipartie graph with parts {a1 , . . . , an }, {b1 , . . . , bm }.
Let X := {x1 , . . . , xn , y1 , . . . , yn , z1 , . . . , zm }, q = n + 1.
yn z1
xn y1
x1
zn
an
a1
ϕ
bm
b1
Let ϕ(ai ) = xi , 1 ≤ i ≤ n,
ϕ(bi ) = Y , such that Y ∩ {x1 , . . . , xn } = {xi1 , . . . , xik }, where N (bi ) = {ai1 , ai2 , . . . , aik }. Y ∩
{z1 , . . . , zm } = {zi }, |Y ∩ {y1 , . . . , yn }| = n + 1 − (1 + k).
a1 a2 a3
a4 a5
Ex:
b1
X
X
6
x1
x2
x3 x4
b2
b3
x5 y1 y2 y3 y4 y5 z1 z2
z3
{x1 , x2 , x5 , z1 , y1 , y2 } {x2 , x3 , z2 , y1 , y2 , y3 } {x3 , x4 , x5 , z3 , y4 , y5 }
≈ N (b1 )
dummy
dummy ≈ N (b2 )
≈ N (b3 )
dummy
Lemma: ∀bipartite P ∃bipartite P 0 ∀2-coloring of E(P 0 ) there is a monochromatic P embedded
in P 0 .
X0
proof: We shall prove the result for P = ((X, Xk ), E). Let P 0 = ((X 0 , 2k−1
), E 0 ), where
|X| = R2k−1 (k|x| + k − 1, 2
size of hyperedges
sets colored
2k−1
k
)
2k − 1
# colors
size of a set s.t. all its (2k − 1)-subsets
are of the same color
k|x| + k − 1
Consider a red/blue coloring c of E(P 0 ) = E 0 . Assume that X, X 0 subsets of integers.
Definition:
A = {1, 8, 11}
0
σ (A0 ) = {1, 4, 5}
c0 (A0 ) = (blue, {1, 4, 5})
σ(A0 ) = blue
1
3
7
8
11
X0
k=3
A0
x0
2k−1
{1, 3, 7, 8, 11}
74
X0
X0
We shall corot the elements of 2k−1
. Let A0 ∈ 2k−1
, there is a color such that the majority of
0
0
edges incident to A is that color, call it σ(A ). There is a set A of k vertices A ⊆ A0 , such that
∀x ∈ A : c(x, A0 ) = σ(A0 ). Let σ 0 (A0 ) be a set of positions of A in A0 .
Define c0 (A0 ) = (σ(A0 ), σ 0 (A0 )). The number of colors in c0 is 2 · 2k−1
.
k
0
X
By Ramsey theorm applied to 2k−1
with c0 , there is a subset W ⊆ X 0 , |W | = k|X| + k − 1, such
W
that 2k−1
have the same color under c0 . Assume, wlog, W = {1, 2, . . . , k|X| + k − 1}.
ϕ
Pick X̃ = {k, 2k, 3k, . . . , |X|k} ⊆ W , see |X̃| = |X|. Map X −→ X̃, map X̃k into (2k − 1) element
subsets of W .
W
ϕ
0
A = {i1 k, i2 k, . . . , ik k} −→ A ∈
, such that A ⊆ A0 , (A0 − A) ∩ X̃ = ∅.
2k − 1
This gives a monochromatic P .
Example:
X
3
), E)
k = 3, |X| = 4,
((X,
b
c
abc
abd
acd
8
11
a
d
bcd
W
1
3
7
X0
X0
(blue, {1, 4, 5})
{1, 3, 7, 8, 11}
1 2 3
X0
5
c0
x0
5
W
5
{3, 6, 12, 4, 5}
-monochromatic
of color (blue,{1, 4, 5})
W
4 5 6 7 8 9 10 11 12 13 14
{3, 6, 9, 4, 5}
{6, 9, 12, 7, 8}
W
5
|W | = k|X| + k − 1 =
= 3 · 4 + 3 − 1 = 14
FLOWS
Let G = (V, E) a graph.
→
E = {(x, y) : {x, y} ∈ E}
→
We say that pair G, f is a circulation if f : E → H, H is an abelian semigroup and
(F1 ) f (x, y) = −f (y, x),
∀{x, y} ∈ E
75
(F2 ) f (v, V ) :=
P
f (v, x) = 0 ∀v ∈ V
{x,v}∈E
P
(f (X, Y ) =
f (x, y))
x∈X,y∈Y
{x,y}∈E
Example:
x1
x1
1
1
x4
1
1
−1
1
x2
x4
1
1
x3
x2
x3
f (x1 , x2 ) = 1, f (x2 , x3 ) = 1, f (x3 , x4 ) = 1, f (x4 , x1 ) = 1,
f (x2 , x1 ) = −1, f (x3 , x2 ) = −1, f (x4 , x3 ) = −1, f (x1 , x4 ) = −1
f (x2 , V ) = f (x2 , x1 ) + f (x2 , x3 ) = −1 + 1 = 0.
• f satisfies (F1 ) ⇒ f (X, X) = 0 ∀X ⊆ V
• f satisfies (F2 ) ⇒ f (X, V ) = 0.
• f -circulation ⇒ f (X, X̄) = f (X, V − X) = 0 ∀X ⊆ V
(f (X, X̄) = f (X, V ) − f (X, X) = 0 − 0 = 0 )
Netork flow:
→
Given G = (V, E), s, t ∈ V , s - sourse, t - sink, c : E → Z≥0 - capacity function. A
→
network flow on (G, s, t, c) is f : E → R such that
(F 1) f (x, y) = −f (y, x)
(F 2) f (x, V ) = 0 for x ∈ V − {s, t}
(F 3) f (x, y) ≤ c(x, y)
f1 = 2
f =2
x1
f1 = 2
f =2
5
2
f =0
s
1 f1 = 0
7
f =0
f1 = 3
3
x2
Definition:
f =0
f1 = 3
t
c(s, x1 ) = 2 c(x1 , s) = 0
c(x1 , t) = 5 c(t, x1 ) = 0
..
..
.
.
A cut (S, S̄) is a pair of sets S ⊆ V, s ∈ S, S̄ = V − S, t ∈ S̄.
76
2
5
s
t
1
7
Proposition: If f is a network flow, then for any cut (S, S̄)
f (S, S̄) = f (s, V )
proof: recall: f (S, S̄) =
P
f (x, y)
x∈S,y∈S̄
f (S, S̄ ) = f (S, V ) − f (S, S) = f ( S , V )
V −S
(F 1)
s∪(S−{s})
X
= f (s, V ) +
f (v, V ) = f (s, V ).
v∈S−{s}
(F 2)
The value of the flow f , denoted |f | := f (s, V ) = f (S, S̄) = f (V − t, t) = f (V, t) for any cut (S, S̄).
|f | = f (S, S̄) ≤ c(S, S̄)
Note:
(F 3)
Question?
∃f : |f | = min c(S, S̄)
(S,S̄) cut
Theorem (Ford-Fulkerson 1956): In any network, max value of a flow = min capacity of a
cut.
proof: Define the sequence of integral flows f0 , f1 , . . . , such that |f0 | < |f1 | < . . . = min c(S, S̄).
(S,S̄)-cut
Let f0 ≡ 0. Assume that f1 , . . . , fn−1 were constructed. Lets build fn .
Case 1: ∃path x0 = s, x1 , x2 , . . . , xm = t such that fn−1 (xi , xi+1 ) < c(xi , xi+1 ),
i = 0, . . . , m − 1.
x1 c = 7 x2
x
c=5
c=2 3 c=4
fn−1 = 3 fn−1 = 1 fn−1 = 0 fn−1 = 1
fn = 3
fn = 5
fn = 3
fn = 2
s
Let =
min
i=0,...,m−1
t
c(xi , xi+1 ) − fn−1 (xi , xi+1 ), let fn (xi , xi+1 ) = fn−1 (xi , xi+1 ) + .
Case 2: Opposite of Case 1 (no such path exists)
Let X = {v, ∈ V : ∃path s = x0 , x1 , . . . , xm = v such that fn−1 (xi , xi+1 ) < c(xi , xi+1 ), i =
77
0, . . . , m − 1}, t 6∈ X.
X
fn−1 = c
s
∈ X
∈ X
∈ X
t
fn = c
So, (s ∪ X, V − (s ∪ X)) is a cut.
For all x ∈ X ∪ s, y 6∈ X ∪ s, x, y ∈ E, fn−1 (x, y) = c(x, y).
Then |fn−1 | = f (s ∪ X, s ∪ X) = c(s ∪ X, s ∪ X).
1
Ex:
2
s
0
3
3
2
2
3
0
0
t
3
0
0
s
0
1
1
1
1
t
0
0
1
f1
0
1
1+1
t
s
s
t
1
1
1
f2
stop |f2 | = 2 = c(S, S̄), done
Group-valuet flows:
Major assumption: bridgeless!!!
→
Given G = (V, E), f : E → H, H-abelian group, f -circulation, f is callen H-valued / group valued
flow (H-flow), it is no-where zero flow if f (x, y) 6= 0.
1
1
1
1
2
1∈R
1
1
1 ∈ Z2
2
2
1
Z3
1
1
1 no Z2 flow
1
Theorem (Tutte 1954): ∀multigraph G ∃polynomial P such that for any finite abelian group
H, the number of H-flows (no-where zero) is P (|H| − 1).
78
Corollary: If H-flow exists ⇒ H1 -flow exists for all H1 , |H1 | = |H|.
Z4 -flow exist ⇔ Z2 × Z2 -flow exist.
Definition:
Defk-flow is a Z-flow such that 0 < |f (x, y)| < k. Flow number of a graph G,
ϕ(G) = min{k : G has k-flow}.
Theorem (Tutte 1950): A multigraph admits a k-flow if and only if it admits a Zk -flow.
Therorem (Tutte 1954): For each pair of dual planar graphs G, G∗ , χ(G) = ϕ(G∗ ).
Lemma: A graph has a 2-floor if and only if all degrees are even.
proof: 2-flow exists ⇔ Z2 -flow exists
Tutte’s flow conjecture: Every bridgeless multigraph has flow number at most 5. ϕ(Petersen’s
graph)= 5.
Seymore 1981 ϕ(G) ≤ 6, G bridgeless.
Lemma: A cubic graph has a 3-flow if and only if it is bipartite.
proof: By Tutte’s theorem, G has a 3-flow iff G has a Z3 -flow.
→
Let f : E → Z3 flow, i.e. f (x, y) = 1 or 2; note iff f (x, y) = 1, then f (y, x) = 2; G is cubic. We need
to prve that G is bipartite by showing that there are no odd cycles.
Consider a cycle x1 , x2 , . . . , xm , x1 .
x2
x1
1
1
x3
x4
2
1
≈
2
≈
2
2
1
1
1
2
2
Observe that if f (xi , xi+1 ) = 1 ⇒ f (xi+1 , xi+1 ) 6= 1.
Thus the values of f on the edges of a cycle iterate between 1 and 2, so the cycle is even.
Now, assume that G is cubic and bipartite with parts A and B, show that ∃Z3 -flow on G.
A
2
2
2
1 1 1
B
Let f (a, b) = 1, a ∈ A, bP
∈ B, i.e. f (b, a) = 2, b ∈ B, a ∈ A.
Then for any a ∈ A :
f (a, y) = 1 + 1 + 1 = 0 (mod 3),
y∈N (a)
79
∀b ∈ B :
P
f (b, x) = 2 + 2 + 2 = 0 (mod 3).
x∈N (b)
Random graphs
G(n, p) Erdös-Renyi random graph model
choose with probability p
(Alon & Spencer „Probabilistic method“)
fix the degree, choose gaphs randomly from this set (internet graph)
(Fan Chung)
Given fixed graph, use probability method to show some properties or show existence results.
(Alon & Spencer „Probabilistic method“)
G(n, p) = set of graphs on vertices x1 , . . . , xn such that P (xi , xj ∈ E(G)) = p (edges are chosen
independently).
n
For a given graph on e edges P (G) = p · p · . . . · p (1 − p) · (1 − p) · . . . · (1 − p) = p(1 − p)( 2 )−e .
| {z } |
{z
}
e
(n2 )−e
Ex:
G(5, 13 )
P(
2 10
3
1 t
3
)=
1
3
2 9
3
X
·
2 6
3
2 9
3
1
3
G∈G(n,p)-labeled
graph on nvertices
P (G) =
1 2
3
(n2 ) X
X
i=0
2 8
3
P (G)
G∈G(n,p)
|E(G)|=i
(n2 ) n
X
n
2
=
p(1 − p)( 2 )−i
i
i=0
n
= (p + (1 − p))( 2 ) = 1.
80
1 10
3
Lemma 1: ∀ integers n, k, n ≥ k ≥ 2:
k
n
Prob(G ∈ G(n, p),
α(G)
≥ k) ≤
(1 − p)(2) (1 = 1 − p).
k
independent number
k
proof: Let S ⊆ V, |S| = k, Prob(G[S] idependent) = (1 − p)(2) .


 [

Prob(some set on k vertices induces an independent set) = Prob 
G[S]-independent
S∈(Vk )
X
≤
Prob(G[S]-ind.)
V
S∈( k )
k
n
=
(1 − p)(2) .
k
Lemma 2: ∀ integer k, n, n ≥ k ≥ 2:
n (k2)
p .
Prob(G ∈ G(n, p), ω(G) ≥ k) ≤
k
clique number
k
Theorem (Erdös, 1947): R(k, k) ≥ 2 2 .
proof: Let G ∈ G(n, 12 ).
Prob(α(G) ≥ k ∨ ω(G) ≥ k) ≤ Prob(α(G) ≥ k) + Prob(ω(G) ≥ k)
(k2)
n
1
≤
· 2 < 1.
k
2
k
if n = 2 2
Prob(α(G) < k & ω(G) < k) > 0 ⇒ ∃G : α(G) < k & ω(G) < k.
X : G(n, p) → [n, ∞) random variable Exp(X) = E(X) =
Markov inequality P (X ≥ a) ≤
E(X)
.
a
Lemma 3: E(#k-cycles in G ∈ G(n, p)) =
(n)k
2k
P
G∈G(n,p)
Prob(G) · X(G).
· pk . Here (n)k := n · (n − 1) · . . . · (n − k + 1).
(
1 ,C⊆G
proof: For a k-cycle C in Kn , let XC : G →
.
0 , otherwise.
81
Let X =
P
XC ,
C-k-cycles in Kn
E(X) =
X
X
E(XC ) =
C
=
X
C
pk =
C
(1 · Prob(Xc = 1) + 0 · Prob(XC = 0))
(n)k k
·p .
2k
Corollary: P (#k cycles in G ∈ G(n, p) ≥ n2 )
Lemma 4: k > 0, p = p(n), p ≥
6k ln n
, n-large,
n
≤
E(#k-cycles)
Markov’s
n
2
=
(n)k k
·p
2k
n
2
.
then
lim P (α(G) ≥
n→∞
1n
) = 0.
2k
proof:
Prob(α(G) ≥ r)
≤
Lemma 1
=
plug r ≥
1n
,p
2k
=
r
r
n
(1 − p)(2) ≤ nr (1 − p)(2)
r
p(r−1) r
r−1 r
− 2
2
n(1 − p)
≤
ne
−→ 0
n→∞
1−p≤e−p
1k ln n
.
n
Theorem (Erdös 1959): ∀k ∃graph H : girth(H) ≥ k and χ(H) ≥ k.
proof: Let k ≥ 3. Fix : 0 < < k1 , choose p = n−1 , let G ∈ G(n, p).
Let χ = #cycles of length ≤ k in G.
Exp(X) =
k
X
Exp(#cycles of length k)
i=3
(∗) Exp(#k-cycles in G ∈ G(n, p)) =
=
(∗)
≤
k
X
(n)i
i=3
k
X
i=3
2i
pi
k
ni pi
1X
≤
(np)i
2i
2 i=3
1
≤
(k − 2)(np)k .
2
82
(n)k k p
2k
(since np ≥ 1, indeed n · n−1 = n ≥ 1)
n
Prob(X ≥ )
2
⇒ Prob(X ≥
Prob(α ≥
(∗∗) p ≥
6k ln n
n
E(X)
≤
n
2
Makov
≤
1
(k
2
n
2
=
(k − 2)nk−1 pk
=
(k − 2)nk−1
n
1
)<
2
2
− 2)(np)k
=
plug p
(k − 2)nk−1 n(−1)k
−→
n→∞
k−1< 1 k−1<0
k
0
for n ≥ n0 , for some n0 .
1n
1
1n
) −→
)<
for n ≥ n00 , for some n00 .
0 ⇒ Prob(α ≥
n→∞
2 k (∗∗)
2k
2
⇒ lim Prob(α(G) ≥
n→∞
1n
)
2k
=0
Prob(X <
n
1n
and α <
)>0
2
2k
There is a graph G on n vertices with < n2 short cycles of length ≤ k and α(G) < 12 nk . Let G0 be
obtained from G by deleting a vertex from each short cycle.
n
(G0 )|
2
We have: G0 has girth > k, |V (G0 )| > n2 , α(G0 ) ≤ α(G) < 12 nk , χ(G0 ) ≥ |Vα(G
0 ) ≥ 1 n = k.
2 k
S, χ(G0 ) ≥ k, girth(G0 ) ≥ k.
Properties of almost all graphs
-set of graphs
A property P of being all k-connected ≈ P = {G : G-k-connected}.
We say that almost all graphs in G(n, p) have property P if Prob(G ∈ P, G ∈ G(n, p)) −→ 1, we
A property
write a.a.have P. (a.s.- almost surely) (a.a.- almost always, almost all)
n→∞
Proposition: ∀p - constant p ∈ (0, 1), ∀graph H a.a.G ∈ G(n, p) contain H as an induced
subgraph.
proof: Let k := |V (H)|, U ⊆ V (G), |U | = k.
Prob(G[U ] ≈ H) = f (p, k) (independent of n)
˙ 2 ∪˙ . . . ∪U
˙ n , where |U i| = k.
Let V ⊇ U1 ∪U
bkc
83
U
k
k
U1
k
k
U2
k
n
U3
Prob(G[Ui ] 6≈ H)
=
Prob(G 6⊆ G)
⊆
ind
=
=
1 − f (p, k)


\

Prob 
G[U ] 6≈ H 

Prob 
|U |=k
U ⊆V
bn
c
k
\
G[Ui ]
i=1 independent events
n
(1 − f (p, k))b k c −→ 0
n→∞
| {z }

6≈ H 
<1
>0
Prob(H ⊆ G) −→ 1
ind
n→∞
Proposition: ∀p - constant p ∈ (0, 1), ∀ > 0 a.a.G ∈ G(n, p):


1
log 1−p
 n .

χ(G) >
2+
log n
84
proof:
Prob(α(G) ≥ k)
≤
=
k
k
n
(1 − p)(2) ≤ nk (1 − p)(2)
k
log n
1
(1 − p)k log(1−p) + 2 k(k−1)
|
{z
}
nk
=
=
(1+)
log n
1 )
log( 1−p
(1 − p)
k
(−2 log 1n +k−1)
2
log( 1−p )
k
k
(1 − p) 2 (−2 (2+) +k−1)
>0
=
=
Prob(α(G) < k)
n
Prob(χ(G) ≥ )
k
1
n(log( 1−p
))
i.e. Prob(χ(G) ≥
)
2 + log n
k
(2+)
(1 − p)
1 − p −→ 0
n→∞
−→
1
−→
1,
−→
1.
n→∞
n→∞
n→∞
}|
{
z
(−2k + 2k + k − 1)
Evolution of random graphs
0 ≤ p1 < p2 < . . . < pi−1 < pi < . . . < pk ≤ 1 probabilities.
¬P
a.a.
¬P
a.a.
P
a.a.
¬P
a.a.
P
a.a.
P - property, example P - beging connected.
A funcion f (n) is a threshold function for a property P if
0
for any p0 : f p(n) −→ 0 a.a. G(n, p0 ) has no P
00
for any p :
n→∞
p00
−→
f (n) n→∞
∞ a.a. G(n, p00 ) has P
p <<√n−2 G a.a. has no edges,
p = 2n−2 G a.a. has a component with > 2 vertices,
p = n−1 G a.a. has a cycle,
p = logn n G a.a. connected,
p = (1 + ) logn n G a.a. has a Hamiltonia cycle.
−2
p = n k−1 - threshold function for containing Kk , i.e. for example
p0 =
−2
n k−1
log n
⇒ a.a. G(n, p0 ) 6⊇ Kk
−2
p00 = log log log log log log n k−1 ⇒ a.a. G(n, p00 ) ⊇ Kk .
v
p = n− e - threshols function for containing a balanced v-vertex, e-edged graph, here H is balanced
0)
e(H)
if ∀H 0 ⊆ H : Ve(H
≤ v(H)
, e = |E|, v = |V |.
(H 0 )
85
Lovász
nLocal
Lemma: Let A1 , . . . , An be events in some probabilistic space.
T
Prob
Āi > 0 if Prob(Ai ) ≤ p ∀i and each Ai is independent to all but at most d other events
i=1
Ai , i = 1, . . . , n and ep(d + 1) ≤ 1.
A1
A2
An
A3
d
Van der Waerden’s number W (k) = min{n : any coloring 1, . . . , n in red/blue ∃monochromatic arithmetic progression of length k}.
1 2 3 4 5 6 7 8
Lemma: W (k) ≥
2k−1
ek2
k−1
proof: We need to show that ∃coloring of [n], n = 2ek2 in 2 colors with no monochromatic AP (k).
Color [n] in 2 colors randomly, probability of a given color assigned to a number is 21 . Let AS be an
event hat arithmetic progression S of length k is monochromatic, for all such AP (k).
n
1
Prob
T
ĀS
S−AP (k)
Prob(AS ) =
S!
1 k
2
S1
> 0.
· 2 = p, d ≤ nk 2
so ep(d + 1) = e · 21−k · (nk 2 ) ≤!
1 if n =
T
Thus by LLL Prob
ĀS > 0.
k·k·n
2k−1
.
ek2
,
S−AP (k)
(LLL application - Wikipedia)
Given: coloring of vertices on C11 n in n different colors, 11 vertices of each color.
Prove that one can choose n ertices of different colors, no two adjacent.
Example:
i
n=3
For a fixed color, pick a vertex of that color with probability
adjacent vertices.
Ai = {vi , vi+1 were choosen}, add (mod (11n)).
86
1
.
11
A bad event is that we choose two
Prob(Ai ) =
1
1 1
·
=
.
11 11
121
} 22 Adjacent pairs containing
} a vertex of fixed color.
d ≤ 42
1
Apply LLL: ep(d + 1) ≤ e |2|
· 43 ≤ 1
T
⇒ Prob( Āi ) > 0.
87
Hamiltonian Cycles - spanning cycle
Hamilton, W.R. invented a game in 1857.
(On a graph travel through all vertices exactly only once and back to the original point)
(If G = (V, E)-given, w : E → R, finding a Hamiltonian cycle of smallest weight - traveling salesperson problem.)
Necessary condition for the existence of Hamitonian cycle
If Hamiltonian cycle exists ⇒ ∀S ⊆ V, S 6= ∅, G − S has ≤ |S| components.
v
u
S = {u, v}
G − S has 3 components →
G is not Hamiltonian
Theorem (Dirac 1952): Every graph with n ≥ 3 vertices and δ(G) ≥
cycle.
δ=
n
Note: can not do better K 2
K n2
n
2
n
2
contains a Hamiltonian
−1
proof: Note that G is connected, otherwise there is a component on ≤ n2 vertices with all vertices
in thos component of degree ≤ n2 − 1. Let P be the longest path in G, P = x0 , . . . , xk .
x0
xk
x4 x5
4∈I
4∈J
N (x0 ), N (xk ) ⊆ V (P ). Let I = {i : x0 ∼ xi+1 }, J = {i : xk ∼ xi }, lvertI|, |J| ≥ n2 , k 6∈ I ∩ J,
I, J ⊆ {0, . . . , k − 1}, |{0, . . . , k − 1}| ≤ n − 1, thus ∃i ∈ I ∩ J.
Thus
v
xi
xi−1
xk
, then x0 . . . xi xk xk−1 . . . xi+1 x0 is a cycle. Either this is a Hamilto-
nian cycle, or ∃v ∈ G not in this cycle but adjacent to it. In this case there is a longer path v
Theorem: |V (G)| ≥ 3, k = k(G) ≥ α(G) ⇒ G has a Hamiltonian cycle.
proof: Let C be the longest cycle, so if C is not Hamiltonian, pick v ∈ V (G) − V (C).
88
x11
xm−1
x13
x3 x2
v
x0
x1
I = {0, 2, 11, 13}
y Menger’s theorem, there are ≥ k (V, C), internally disjoint paths. Let I = {i : xi -endpoint of one
of there paths}.
Observe that if i ∈ I, i + 1 6∈ I (addition mod m), otherwise there is a longer cyle if i, j ∈ I, then
xi+1 6∼ xj+1 , otherwise, there is a longer cycle.
xi
xi+1
v
xi
xi+1
v
xj
xj+1
We have |I| ≥ k, |{xi+1 : i ∈ I}| = |I| ≥ k, so {xi : i ∈ I} ∩ {v} is also an independent set of size
≥ k + 1, a contradicion.
Theorem (Tutte 1956): Every 4-connected planar graph is Hamiltonian (this means is has a
Hamiltonian cycle).
Theorem (Fleischner 1974): If G is 2-connected, then G2 is Hamiltonian, where G2 =
(V, E 0 ), G = (V, E), E 0 = {{u, v} : u, v ∈ V : distG (u, v) ≤ 2}.
G
Theorem (Chrátal 1972): Let (a1 , . . . , an ), 0 ≤ ai ≤ . . . ≤ an < n an integer sequence, n ≥ 3,
∀i <
n
(ai ≤ i ⇒ an−i ≥ n − i) ⇒ any graph with degree sequence a1 , . . . , an is Hamiltonian.
2
(⇐ is true too)
89
Line graphs
2
1
H
4
2
3
1
2
1
3
3
L(H) 4
1
2
3
4
4
not a line graph
L(H1 ) =? there is no H1
Theorem (Beineke 1968): A simple graph is a line graph of another simple graph ⇔ it has
none of F as induced subgraph
Last
time:
Hamiltonicity:
δ(G) ≥ n2 , n ≥ 3
K(G) ≥ α(G), n ≥ 3
K(G) ≥ 4 and G planar
G = H 2 , where k(H) ≥ 2
G has degree sequence d1 ≤ d2 ≤ . . . ,
di + dn − i ≥ n ∀i < n2
−→
G-Hamiltonian
−→
−→
G-Hamiltonian
G-Hamiltonian
Dirac
Tutte
−→
G-Hamiltonian
−→
G-Hamiltonian
chrátal
−→
∀S ⊆ V (G), S 6= ∅,
G − Shas at most |S| components.
WQO (structural graph theory)
Quasi-ordering - relation that is reflexive and transitive. We say a set X is well-quasi-ordered by
quasi-ordering ≤ if for any infinite sequence x1 , x2 , . . . , xi ∈ X, ∃i < j, xi ≤ xj .
X Y for graphs X and Y if Y = M X, i.e. X can be obtained from Y by edge-deletions, edgecontractions and vertex-deletions.
90
X
XY
Y
Graph minor theorem (Neil Robertson, Paul Seymour 1990): „“-minor containment is
a well-quasi-ordering.
I.e., for any infnite set of graph there are two graphs such that one is a minor of the other.
A famiy F of graph is closed under taking minors if
Y ∈ F, X : M X = Y, ⇒ X ∈ F,
i.e. if Y ∈ F and X is obtained from Y by edge-deletion, contraction, vertex-deletion, then X ∈ F.
(Diestel hereditary)
PLanar graph, graphs on surfaces.
Example:
Corollary (Graph minor theorem): If F-minor-closed family, then ∃finite set of graphs
H1 , . . . , Hk such that
F = F orb(H1 , . . . , Hk ) = {G : Hi 6 G, i = 1, . . . , k}.
Planar graph = F orb(K3,3 , K5 )
Lemma: Let ≤ be a well-quasi-ordering of X, we write A≤B for A, B ⊆ X if ∃injective map
f : A → B such that a ≤ f (a) ∀a ∈ A. Then ≤ is a well-quasi-ordering of X <ω - finite subsets of
X.
Rooted tree is a pair (T, r), r ∈ V (T ). In a rooted tree x ≤ y if x is on r − ypath in T .
Define ≤∗ on the set of rooted trees (T, r) ≤∗ (T − 1, r1 ) if T1 contains a subdivision of T and there
is a map ϕ : V (T ) → V (T1 ) such that x ≤ y ⇒ ϕ(x) ≤ ϕ(y).
(T,r)
T
T1
(T1 ,r1 )
ϕ(r)
r
r1
Graph minor theorem for trees (Kruska 1960): ≤∗ is well-quasi-ordering.
proof: Assume not, then there is a „bad“ sequence of rooted trees such that no two (T, r), (T 0 , r0 )
satisfy (T, r) ≤∗ (T 0 , r0 ). We say a sequence is „good“ otherwise. Fix n ∈ N, then there are trees
T0 , T1 , T2 , . . . , Tn−1 such that some bad sequence starts with these. Choose min order tree Tn , such
that some bad sequence starts with T0 , T1 , . . . , Tn . Let root of Ti be ri , i = 0, . . .. Let An be a set of
rooted trees obtained as comploments of Tn − {rn }.
91
Tn
An =
,
,
rn
Claim: A =
∞
S
i=1
An is well-quasi-ordered under ≤∗ .
proof of claim: Let (T k )k=1,... be any sequence of trees in A. For each k, let n(k) be smallest
index such that T k ∈ An(k) . Then T0 , T1 , . . . , Tn(k)−1 , T k , T k+1 , . . . is a good sequence since T0 , T1 , . . . , Tn(k)−1 , Tn(k) , . . . is bad with smallest Tn(k) and T k is smaller than Tn(k) since T k ∈ An(k) . Let (T, T 0 ) be a good pair, T ≤ 1 ∗ T 0 in T0 , T1 , . . . , Tn(k)−1 , T k , T k+1 , . . ..
If T, T 0 ∈ {T k , T k+1 , . . .}, done, so (T, T 0 ) is a good pair in (T k )k=1,... . Thus, assume that
T ∈ {T0 , . . . , Tn(k)−1 }. So, T 0 6∈ {T0 , . . . , Tn(k)−1 }. So, T ≤∗ T 0 = T i ≤∗ Tn(i) , (T = Tj , j ∈
{0, . . . , n(k) − 1}), so (T, Tn(i) ) is a good pair in the original sequence, contradicion.
Recall T0 , T1 , . . . bad sequence A = A0 ∪ A1 ∪ A2 .
A1
A0
A2
A = A0 ∪ A1 ∪ A2 . . .
By lemma [A]<ω is well-quasi-ordered, so A0 , A1 , A2 , . . . contains a good pair (Ai , Aj ), i.e., ∃injective
f : Ai → Aj ∀T ∈ Ai T ≤∗ f (T ).
⊆A ⊆A
Aj
Ai
Ti
Tj
⇒ Ti ≤∗ Tj .
92

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