Graph Theory
Judith Stumpp
6. 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
22