Graph Theory - KIT · Bei dem folgenden Skript handelt es sich um einen Mitschrieb der Vorlesung...
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Graph Theory Judith Stumpp 7. November 2013
Transcript of Graph Theory - KIT · Bei dem folgenden Skript handelt es sich um einen Mitschrieb der Vorlesung...
Graph Theory
Judith Stumpp
7. November 2013
Bei dem folgenden Skript handelt es sich um einen Mitschrieb der Vorlesung Graph Theory vom Winter-semester 2011/2012. Sie wurde gehalten von Prof. Maria Axenovich Ph.D. . Der Mitschrieb erhebt wederAnspruch auf Vollständigkeit, noch auf Richtigkeit!
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Kapitel 1
Definitions
The graph is a pair V,E. V is a finite set and E ⊆(V2
)a pair of elements in V . V is called the set of vertices
and E the set of edges.
Visualize: G = (V,E), V = {1, 2, 3, 4, 5}, E = {{1, 2}, {1, 3}, {2, 4}}
1 2
3
4
5
History: word: Sylvester (1814-1897) and Cayley (1821-1895)Euler - developed graph theory
Königsberg bridges (today Kaliningrad in Russia):
A
B
D C
A
D
B
C
Problem: Travel through each bridge once, come back to the original point.Impossible!
Notations:
• Kn= (V,(V2
)) - complete graph on n vertices |V | = n
K5 K3
v1v2
v3
v4v5
v6
C6 = v1v2v3v4v5v6
• Cn - cycle on n verticesV = {v1, v2, . . . , vn}, E = {{v1, v2}, {v2, v3}, . . . , {vn−1, vn}, {vn, v1}}
• Pn - path on n vertices (Note: Pn . path on n edges (Diestel))V = {v1, v2, . . . , vn}, E = {{v1, v2}, {v2, v3}, . . . , {vn−1, vn}}
2
• Let P be a path from v1 to vn. The subpath of P from vi to vj is viPvj and the subpath from vi+1 tovj is
◦viPvj .
• En= (V, ∅), |V | = n isolated vertices
E10
m
n
Km,n
• Kn,m= (A ∪B,A×B), A ∩B = ∅ complete bipartite graph
• Peterson graph: V =({1,2,3,4,5}
2
), E = {{{i, j}, {k, l} : {i, j} ∩ {k, l} = ∅}
{1, 2}
{3, 4}{3, 5}{4, 5}
{1, 3}
{2, 3}
{1, 4} {2, 4}
{1, 5}
{2, 5}
• Kneser Graph K(n, k)= ((Vk
), E)
|V | = n, E = {{A,B} : A,B ∈(Vk
)and A ∩B = ∅}.(
Vk
)is the set of k-element subsets of V, |
(Vk
)| =
(|V |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 Q2 Q3
V = {∅, {1}} = {(0), (1)} V = {(0, 0), (0, 1), (1, 0), (1, 1)} V = {(0, 0, 0), . . . , (1, 1, 1)}
1
0
(1,1)
(1,0)(0,1)
(0,0)
(1,1,1)
(1,1,0) (1,0,1) (0,1,1)
(1,0,0)(0,1,0)
(0,0,1)
(0,0,0)
(1, . . . , 1)
(0, . . . , 0)
weight 1 (# 1’s in a binary tuple)weight 2
weight n-1
n
(n2
)(n3
)
(nn2
)
(n
n−1
)
Qn−1
Qn−1
001001
01
Qn :
3
Parameter: Let G = (V,E) be a graph. The order of G ist the number of vertices (|V |) and the size of Gis 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 G = ({v1, . . . , vn}, E) which is called the adjacent matrix.
For
v1
v2v3 v4 A =
0 1 1 01 0 1 11 1 0 00 1 0 0
.
Subgraph: H ⊆ G, H = (V ′, E′), G = (V,E), V ′ ⊆ V, E′ ⊆ Ev1
v2v3 v4
v1
v2v3
H ⊆ind
H is an induced subgraph of G if H ⊆ G and for v1, v2 ∈ V (H): {v1, v2} ∈ E(H)⇔ {v1, v2} ∈ E(G).
In the upper example it is no induced subgraph.
An induced subgraph is obtained from G by deleting vertices. E.g.:
v1
v2v3 v4
v1
v2v3
v1
v2 v4 v3 v4v2v3 v4
v1
Let G = (V,E) and G′ = (V ′, E′) be graphs. Then we define G ∪ G′ := (V ∪ C ′, E ∪ E′) and G ∩ G′ :=(V ∩ C ′, E ∩ E′).G[X] := (X, {{x, y} : x, y ∈ X, {x, y} ∈ E(G)}) is called the subgraph of G induced by a vertex setX ⊆ V (G). E.g.:
4 1
2
3 5 4 1
2
3
G G[{1, 2, 3, 4}]
A degree d(v) = deg v of a vertex is the number of edges incident to that vertex.v1
v2v3 v4 deg v1 = 2, deg v2 = 3, deg v3 = 2, deg v4 = 1In this example the degree sequence is (2, 3, 2, 1), the minimum degree δ(G) is 1 and the maximum degree∆(G) is 3.
Apparently |E(G)| = 12
n∑i=1
deg vi is true.
Thusn∑i=1
deg vi is even and therefore the number of vertices with odd degree is even.
d(G) := 1n
n∑i=1
deg vi is called the average degree of G.
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Extremal graph theorem: We’ll prove that if G has n vertices and >⌈n2
4
⌉edges ⇒ G has a
triangle.
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 agraph is a connected component.
A connected graph without cycles is called a tree. A graph without cycles (acyclic graph) is called aforest.
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) anda 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 aneighbor of x if {x, y} ∈ E). In particular k ≥ δ(G).Let i = min{j ∈ {0, . . . , k} : {xk, xj} ∈ E}. Then xixkxk−1 . . . xi is a cycle of length at least δ + 1.
x0xixk
≥ δ
�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 XAssume it holds for all trees of order < n.
Let v be a leaf of T , T ′ := T − v.⇒ |T ′| = n− 1 < n.T ′ is acyclic.Let v′, w ∈ T ′. ∃P v′ = v0, v1, . . . , vn = w ⊆ T .
To show: vi 6= v for all i = 0, . . . , n
v0, vn 6= v because v0, vn ∈ T ′, v 6∈ T ′vi 6= v (i = 1, . . . , n− 1) because dT (vi) ≥ 2, vi is not a leaf.
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⇒ P ⊆ T ′ connecting v0 and w ⇒ T ′ connected ⇒ T is a tree.
With induction hypothesis T ′ has (n− 1)− 1 edges. Thus T has (n− 1)− 1 + 2 = n− 1 edges. �A walk is an alternating sequence v0e0v1e1 . . . vn of vertices and edges so that ei = vivi+1 for alln = 0, . . . , n − 1. Compared to a path it is allowed to pass edges and vertices more than once. Ifv0 = 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 isW . ThenW is a path. If not,W has a repeatedvertex W = ue0v1e1 . . .︸ ︷︷ ︸
=:W1
vi . . .︸ ︷︷ ︸=:W
vi . . . v︸ ︷︷ ︸=:W2
, then W ′ = W1W2 is a shorter u-v-walk. �
- 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 = v0e0v1 . . .︸ ︷︷ ︸1’st part of W2
vi . . . vi︸ ︷︷ ︸W1
. . . vn = v0︸ ︷︷ ︸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 = v0e0v1 . . . ei . . . ei is unique, then Gcontains a cycle.
proof: Induction on # vertices.
Basis:
Step: W = v0e0v1 . . .︸ ︷︷ ︸1’st part of W2
vi . . . vi︸ ︷︷ ︸W1
. . . vn = v0︸ ︷︷ ︸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 v1e1v2e2 . . . e8v9 = v1 in
e1
v1 = v5 = v9
v2e2
e3 e4
e5
e6
e7
e8v4
v6 = v3
v7 v8
Theorem: A connected graph G has an Eulerian tour iff (i.e. if and only if) each degree of vertexin G is even.
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proof:
„⇒“ : If there is an Eulerian tour then clearly the number of edges entering the vertex is the numberof 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. Wis 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 = viu. Then W ′ = ueviWvi is a longer walkwith 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).
C C
G G
�
Connectivity:We say that a Graph G is vertex k-connected if |V (G)| > k and deleting any (k − 1) vertices doesnot disconnect the graph.
Any connected graph is 1-connected. If a graph is 2-connected then there exists no cut-vertex whichis 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}
e.g.: κ(v1
v2v3 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) (= κ′(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) .
Clearly λ(G) ≤ δ(G). But it could be that 1 = λ(G) << δ(G) = 99K100K100
.
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.
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Let E be a set of λ edges disconnecting G. Then E is a cut, i.e. ∃S ⊆ V : ∀e ∈ E, one endpoint ofe is in S, another is in S := V − S.
E
SS
If in G there are all edges between S and S. λ = |E| = |S| · |S| ≥ |V (G)| − 1 ≥ κ(G).
Otherwise ∃x ∈ S, y ∈ S, x 6∼ y (i.e. xy 6∈ E(G)).
T := (N(x) ∩ S) ∪ ({z ∈ S : z ∼ S} − {x})
T is a vertex cut, in particular after deleting T , x and y are in different connected components.We have |T | ≤ |E| = λ because
|N(x)| ≤ #(edges incident to x) and |{z ∈ S : z ∼ S} − {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.
v1 v2 vnv3
≤ d ≤ d ≤ d
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.
G T
δ(G) ≥ 6
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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 u1v1u2v2 . . . ukvku1, 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 withpartite 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, u′ ∈ U and {u, u′} ∈ E(G).Consider a walk formed by shortest x-u-path, shortest x-u′-path and u, u′.
u 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 hasa matching containing all vertices of A if and only if |N(S)| ≥ |S| for any S ⊆ A.
S
N(S)
S
N(S)
A
B
bad
proof:
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„⇒“ : obvious
S
N(S)
A
B
„⇐“ : 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 G′ = G− {x, y}.|NG′(S)| ≥ |NG(S)| − 1 ≥ |S|+ 1− 1 = |S|, for any S ⊆ A− {y}.
S
A
B x
y
Thus, Hall’s condition is true for G′, and there is a matchingM ′, containing all elementsof A− {y}, by induction.So, M ′ ∪ {x, y} is a matching saturating A in G.
A
B x
yM ′
Case 2: ∃S ⊂ A, S 6= A such that |N(S)| = |S|.
S
N(S)
A
B
S′
N(S′)
By induction, there is a matching containing all vertices of S. Let apply induction toG[A− S,B −N(S)].Assume that there is S ′ ⊆ A− S such that |N(S ′) ∩ (B −N(S))| < |S ′|.Then |N(S ′ ∪ S)| = |N(S) ∪ (N(S ′) ∩ (B −N(S)))|<6= |S|+ |S ′|.A contradiction to Hall’s condition applied to S ∪ S ′.Thus for any S ′ ⊆ A−S, |N(S ′)∩(B−N(S))| ≥ |S ′|, and there is a matching saturatingA− S in G[A− S,B −N(S)]. Together with a matching between S and N(S), it givesa matching saturating A.
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�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 somefixed 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 agraph. 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 G′.
B
A
C
|C| = d, add all edges between A and C.In G′ |NG′(S)| ≥ |NG(S)|+ d ≥ |S| − d+ d = |S|.By Hall’s theorem, there is a matching in G′ 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|.
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�
Non-bipartite graphs:
A k-factor in a graph is a spanning (containing each vertex) subgraph in which each vertex has degreek.
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. setof vertices such that each edge is incident to some of this vertices, i.e. a set X of vertices such thatG−X is an empty graph.
X
König’s theorem ’31: If G is a bipartite graph, then ν(G) = τ(G).
Classical approach: Given a maximal matching M and want to find a vertex cover of size |M |B
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 pathending 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
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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 thevertex 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 u′ ∈ N(v) − {u}, suchthat f := {v, u′} /∈ E(M).Let W ′ be a cover of G− f of size ν(G− f) = ν(G). Then W ′ does not contain u (W ′ containsvertices of M only and u /∈ V (M)). Thus W ′ contains v. So, W ′ covers f too. Thus W ′ covers G.
v
w
1
v
u
2
u’
f
M |M | = ν(G) �
Tutte’s theorem
S
odd
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 edgeone-by-one, so the resulting graph G′ is no 1-factor.We shall show that in G′ 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, M14M2 = (M1 ∪M2)− (M1 ∩M2) areonly cycles.
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Let S be a set of vertices of degree |V (G)| − 1. We shall show that S is bad in G′.Claim: All components of G′ − S are complete.Assume not, i.e. there is a non-complete component in G′ − S.
S
c
b
aa
c
bd
Then there is an induced path a, b, c in this component. Since b /∈ S, deg b < |V (G′)| − 1,there is d /∈ {a, b, c}, such that b 6∼ d.By maximality of G′, there is a perfect matching M , in G′ ∪ {{a, c}}, there is a perfectmatching M2 in G′ ∪ {{b, d}}. Note ac ∈ E(M1), bd ∈ E(M2). We shall create a perfectmatching of G′.Consider M14M2, ac, bd ∈ E(M14M2). If ac, bd belong to different cycles of M14M2:ac
bd
Take the edges of M2 in a component containing ac, take edges of M1 in a component withbd, otherwise take edges of M1.If ac, bd belong to the same cycle of M14M2, then
a
cb
d
ac
b
dor
A contradiction, since G′ has no 1-factor, so all components of G′− S are complete.� Claim
S
odd
odd
odd
even
even
If S is not bad, i.e. |q(S)| ≤ |S|, we can construct a perfect matching, a contradiction to thefact that G′ has no perfect matching. Thus S is bad in G′.
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S
odd
odd
odd
odd
even
even
G is obtained from G′ by deleting edges, so qG(S) ≥ qG′(S) > |S|.
�
3-factor
f-factor12
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1-factor
k-factor - spanning subgraph,all degrees = kf -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).
Replace each vertex of G withA(v)
B(v) e(v)
d(v)
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 spanningsubgraph H of G such that degH(v) = f(v).
1-factor or matching ≈ f -factor, f ≡ 1.
1 v2
1
3 v1
2
3 2
∅B(v1)
A(v1) B(v2)
A(v2)
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 replacingeach vertex v with vertex sets A(v)∪B(v), |A(v)| = deg(v), |B(v)| = deg(v)−f(v), and for adjacentvertices 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 toM , 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), con-tracting 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 ofG is a spanning subgraph of G that is a vertex-disjoint union of copies of H.
H = G =
H = K2 H-factor ≈ perfect matching.
Hajnal & Szemeredi ’70: If G satisfies δ(G) ≥ k−1kn, n:k, then G has a Kk-factor.
Kk Kk KkKk
Alon-Yuster ’95: If G satisfies δ(G) ≥ χ(H)−1χ(H)
n. Then G contains at least (1− o(1)) n|V (H)| (H is
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, sothat 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
2
3
2
1
G
χ(G) = 3χ(Kk) = k, χ(C3) = 3, χ(C4) = 2, χ(Km,n) = 2, χ(C2k+1) = 3There are graphs with large |V (G)| and small χ(G).
16
Connectivity: A,B ⊆ V (G), A-B-path P is a path v0, v1, . . . , vk such that V (P )∩A = {v0}, V (P )∩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
A B
v3v2
v1
u1u2
u3
A−B sep. set : {v2, u2}{e1, e5, e4}{e1, u1}
e1 e2
e3 e4
e5
Note that a separating set must contain A ∩B.Note B′ ⊇ B and X separates A and B′ ⇒ X separates A and B.
AB
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-disjointA-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-B-path (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 thatincludes the endpoints of P and |Q| = |P |+ 1.
BA
P
Lets prove this by induction on |V (G)−B − A|.
Basis: |V (G)−B − A| = 0.
A B
P
< 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 verticesseparating A and B, call it R.
A B
v
xR
P
Let x be the last vertex of R that also belongs to a path in P call it P .Let B′ = B ∪ (V (xP ) ∪ V (xR)).P ′ = P\{P} ∪ {Px}.note k(G;A,B′) ≥ k(G;A,B).By induction, there is a larger set of A-B′-paths, Q′, |Q′| ≥ |P ′| + 1, Q′ contains endpoints ofP ′.
A BP
y
Let y be an endpoint of a path in Q′ in B′ that is not an endpoint of P ′.
y
y
y
Case : 1 Case: 2 Case: 3
Q1
Q
,
Case 1: y ∈ B:Take Q = Q′ − {Q}︸︷︷︸
path containing x
∪{Q ∪ xP}.
Case 2: y ∈ xP :Take Q = Q′ − {Q} ∪ {Q ∪ xR} − {Q1}︸ ︷︷ ︸
path containing y
∪{Q1 ∪ yP}.
Case 3: y ∈ xR:Take Q = Q′ − {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 ′),E ′ = {{e, e} : e, e ∈ E and e, e are adjacent}.
18
v2v3
v4
v1
e2e1
e3 e4 e5
e1
e2
e3
e5e4
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)
a b
A B
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 pathsbetween a and b.
outline of proof:Suppose G contains k independent paths between any two vertices, thus we need ≥ k vertices toseparate 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 G′ = G− {a, b}, then G′ contains ≤ (k − 2) independent a-b-paths.
ab
≤ k − 2
a b
≤ k − 2v
X
G′
By corollary 1, we can separate a and b in G′ by ≤ k − 2 vertices, X.Since |V (G)| > k, there is v /∈ {a, b} and v /∈ component of a in G′ −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
AB
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 twovertices.
κ(G) = 1 blocks block-cut-vertex tree.A block - either a bridge or maximal 2-connected subgraph.
B1 B2B3 B4
B5
B6
v1 v2v3
v4
v5
B1 B2
v1 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 andno 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 asear-decomposition.Observe H ⊆
indG. If u, v ∈ V (H), v 6∼H u, v ∼G u, then add uv as a ear. If H 6= G ⇒ ∃x ∈
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 vertexof P in H, w1.So wx ∪ xPw is an H-ear. A contradiction to maximality of H, so G = H.
20
H
G
w x
w1
„⇐“ : 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.
x
y
v3
v2
v1
v4
v5
vxyv2
v1
v3
v4
v5
Tutte’s theorem 1961: A graph G is 3-connected if and only if it exists a sequence of graphsG0, 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}.
x
x′
x′′
yy′
y′′
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.
G1G2S
Let G1 and G2 be connected components of Gi+1 − S. Observe, {x, y} 6= S, otherwise Gi is not3-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
G2S
x yv
w
|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-disjointsi-ti-paths, i = 1, . . . , k.s1s2
s3
t1
t2
t3
s1s2
s3
t′1 = t2
t′2 = t3
t′3 = t1
G is k-linked ⇒ G is k-connected (Menger’s theorem)
G is f(k)︸︷︷︸22k
-connected ⇒ G is k-linked. (Bollobas-Thomason ’96)
Plane graphs and planar graphsplane 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 ina subset of R2.
A plane graph G is a subset of R2, G = V supE, 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 G′ = (V ′, E ′), such that there is abijection f : V → V ′ such that {u, v} ∈ E if and only if {f(u), f(v)} ∈ E ′.
22
Here: 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).
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.
• Lemma 1.2:
P1P2
P3
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= ∅.
• Lemma 1.3: X1X2
P
Let X1, X2 ⊆ R2, X1 ∩X2 = ∅.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 ofR2\(X1 ∪X2), then O\
◦P is a region of R2\(X1 ∪X2 ∪ P ).
• Lemma 1.4: Given two spheres S2 in R3, and circle C1 and C2 on them, then a homeo-morphism 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 (exactlyone 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) If X is the frontier of a face, then either e ⊆ X or X ∩ ◦e = ∅.2) If e is on a cycle C ⊆ G, then e is on frontier of exactly two faces.3) 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 withoutcreating multiple arcs. Assume that G is maximally plane. Observe first that G does not have acut-vertex.
By taking a vertex on the faces of different resulting connected components, we can join them by anarc. So, G is 2-connected, i.e. each face is bounded by a cycle. Assume that some face is bounded bya 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 edgeincident 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 arcconnecting v2 and v4.
v1 v2
v3v4
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 facesf1, f2.
Idea: delete ◦e, this wil create a face f1,2 = f1 ∪ f2\◦e, in a new graph G′, v = v, e(G′) =
e(G) − 1, f(G′) = f(G) − 1. By induction, in G′ v(G′) − e(G′) + f(G′) = 2. Thus v(G) −(e(G)− 1) + (f(G)− 1) = 2.
formalities: f1,2 = f1 ∪ f2\◦e.
First we shall prove Claim: F (G′)\f1,2 = F (G)\{f1, f2}. (without assuming that f1,2 is a face of G′).
Using this claim, observe ◦e is in the face of G′, not in the face of G, it must be f1,2, so Idea could beapplied.
proof of the claim:
24
• Let f ∈ F (G)\{f1, f2}. By lemmma 2.1(i) G[f ] ⊆ G\◦e = G′, so f is a face of G′ (not equal tof1,2).
• Let f ′ ∈ F (G′) not equal to f1,2. Any two points of f1,2 are linked by an arc in R2\G′, sof1,2 ⊆ f ′′, a face of G′. Since f ′ 6= f1,2, f
′ 6⊆ f1,2 because otherwise f ′ ⊂ f1,2 ⊆ f ′′ (can not bea face being a subset of another face). Let x be a point in f ′\f1,2, x belongs to a face f of G,f 6= f1, f2. So f is a face of G′, so x ∈ f ∩ f ′, so f = f ′, f ′ ∈ 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 ofG, be a plane graph (multigraph) with a single vertex in each face of G, with two vertices connectedby an arc if and only if the corresponding faces of G share an edge, on this boundary. In this casethe 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 resultinggraph T ′.
T and T ′ are trees and T and T ′ have e(G) edges. v(T ) = |v(G)|, v(T ′) = |F (G)|, e(T ) = |v(T )| −1, e(T ′) = |F (G)| − 1.
e(T ) + e(T ′) = 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 linesegment. We’ll calculate the area of big polygon using Rick’s formula and using areas of smallerpolygons.
25
Let vext = #exterior vertices of G,vint = #interior vertices of G,I = #other points,
eint = #gridpoints of interior edges but not vertices,eext = #gridpoints of exterior edges but not vertices.
Area =vext + eext
2+ (I + eint + vint)− 1
=∑
F - bounded face
area(F )
= I + eint +∑
v - interiorvertex of G
d(v)
2+
∑
v - exteriorvertex of G
d(v)− 1
2
︸ ︷︷ ︸(∗)
+eext
2− 1(|F | − 1)
(∗)∑
v - int
d(v)
2+∑
v - ext
d(v)
2− vext
2=
∑
v - vertex of G
d(v)
2− vext
2
vext
2+ vint − 1 =
∑
v - vertex of G
d(v)
2− vext
2− |F |+ 1
vext + vint︸ ︷︷ ︸v
−2 + |F | = e(G)
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. Eachface has 3 edges on its boundary, so e = 3
2f . Euler’s formula gives us v − e + f = v − e + 2
3e = 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·#faces2
(f ≤ e2). Euler’s formula:
2 = v − e+ f ≤ v − e+ e2
= v − e2, thus e ≤ 2v − 4. �
Planar graphs are characterized by forbidden subgraphs, and by poset (partially ordered set) dimen-tions.
26
Minors: Given two graphs, G and X, we say that X is a minor of G, write G = MX. Say alsothat G contains X as a minor if X can be obtained from G by edge deletions and edge contractions.(delete isolated vertices)
G
X
G = MX
G 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) areadjacent.
X
G = MX
Topological minor: For two graphs G and X, we say that G contains X as a topological minor, wewrite G ≥ TX, if a subgraph of G is a subdivision (german: Unterteilung) of X.
H ′ is obtained fromH by a single edge-subdivision, if V (H ′) = V (H)∪{v}, andE(H ′) = E(H)\{x, y}∪{x, v}∪{v, y}. H ′ is a subdivision of H if H ′ is obtained from H by series of single edge-subdivisions.
H subdivision of H
GX
G ≥ TX
G = TK4
examples
In the definition of minor, V1, V2, . . . , V|V (X)| are branch sets. In the definition of topological minor,the vertices of TX, 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 TK5 or TK3,3 and
3) G has no MK5 or MK3,3.
Another characterization: Given a graph G = (V,E), let P , partially ordered set (poset) beP = (V ∪ E︸ ︷︷ ︸
X
, „R“ = „ < “︸ ︷︷ ︸relation R⊂X×X
), v < e for v ∈ V, e ∈ E, if and only if v is incident to e.
27
v1
v2
v3
v4
e1e2
e3
e1 e2 e3
v1 v2 v3 v4
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 intersectionof these total orders.
dim = 1 dim( ) = 2a b a b
b
ba
a
= ∩
1
2
3 6
5
4
36
215
4 12345
6
∩
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 atopological 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 TK5 and TK3,3, thenG is 3-connected.
Kuratowski’s theorem:
1) G is planar,
2) G as no TK5 or TK3,3 and
3) G has no MK5 or MK3,3.
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(TK5) = 5 + α e(TK5) = 10 + α
The Euler’s formula does not hold for K5.Let v′ = v(TK5), e′ = e(TK5) and f ′ = f(TK5).If TK5 is planar, then v′ − e′ + f ′ = 2, but
v′ − e′ + f ′ = 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 thatK3,3 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 = TK1,3
Trees with 4 leaves:
T TK1,4
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
connectedconnected connected
Consider a spanning tree in a branch set and three edges leaving the branch set.
TK1,3
connected connected connected
connectedconnected 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 togetherwith 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 TK5. If some branch set is of type 2, then
1 2
3 4
5 6
1
3
4
5
6
2
So G contains TK3,3. �
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} is3-connected. Lets call G′ = G ◦ {x, y}, call the vertex vx,y obtained by contracting x and y. We havethat G′ is K5, K3,3-minor-free, |v(G′)| < |V (G)| and G′ is 3-connected, by induction G′ 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 G′ − vx,y.(Note: G′ − vxy is 2-connected ⇒ each face is bounded by a cycle.)
G′ vxy vxy
x1
x2
x3
xkC
C
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 ofneightbours of y in C.
vxy
x1
x2
x3
xk
CP1
P2
Pk
,
Case 1: Y ⊆ Pi for 1 ≤ i ≤ k − 1:
x
xi
xi+1
y
Case 2: |Y ∩N(x) ∩ V (C)| ≥ 3:
xi+1
y
w1 ∈ Y
w2 ∈ Y
w3 ∈ Y
Then w1, w2, w3, x, y correspond to branch vertices of TK5, 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
x
xi
xi+1
y
w1
w2
G contains TK3,3. �
Tutte’s theorem: Let G be a 3-connected planar graph with convex embedding. Then each boundedface 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 Gis edge-maximal with respect to not containing TX, X is 3-connected, then also G1 and G2 are3-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 = ∅:
G1 G2
e
Add e between G1 and G2. Contradiction since G ∪ e ⊇ TX, e ∈ TX, e is a bridge in TX, butTX is 3-connected.
• |S| = 1:
G1 G2
ev2v1
vAdd edge e = {v1, v2}. In G ∪ e ⊇ TX, the branch vertices of TX are all, w.l.o.g. in G1. So, eis on the path P between the branch vertices. Replace P with vv1, so TX ⊆ G1.
G1 G2
ev2v1
v
P
• |S| = 2:
x
yG1G2
P
32
Assume that x 6∼ y. Add e = {x, y}, G ∪ e ⊇ TX. All branch vertices of TX are in, w.l.o.g.,G1. Replace e with a x-y-path P in G2, obtaining TX ⊆ G, a contradiction.
x
yG1G2
e1P
Add e1 to G1, TX ⊆ G ∪ e1. All branch vertices in one of G1 and G2. Replace xPy with a{x, y}, thus G1 ∪ e1 ⊇ TX.
�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).TK5, TK3,3 ⊆ G1, G2 ⇒
lemma 4.4.4G1, G2 3-connected ⇒ by lemma 4.4.3 and corollary ⇒ G1, G2 3-
connectod and has no MK5,MK3,3 ⇒ G1, G2 are planar.Embed G1 and G2 such that {x, y} lies on the unbounded face.
x
yG1
G2
z1 z2
P
G ∪ {z1, z2} ⊇ TK5 or TK3,3. ,
Case 1: G ∪ {z1, z2} ⊇ TK5:All branch vertices of TK5 are w.l.o.g. in G1. Replace P with z1, y.
G1
z1 xy
G1 ∪ {z1, y} ⊇ TK5 it is planar
Case 2: G ∪ {z1, z2} ⊇ TK3,3:If all branch vertices are in G1, done as in case 1. Otherwise branch vertex in G2
x
y
z1 z2z1
x
y
Replace those path with edges.TK3,3 ⊆ 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-colortheorem)
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 ofcolors 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)| ≥ 6·|V (G)|2
= 3|V (G)|(contradiction to Euler’s formula).
Consider an embedding of G in the plane. Lets order neighbors v1, . . . , v5 of v cyclically according tothe 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.
v
v1
v2
v3
v4
v5
G
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, coloredwith 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 sharevertices and cross, contradiction to planarity.
v
v1
v2
v3
v4
v5
v4, v2 not connected
colorflip
�Heawood formula (Weisstein): G is embedded on Sγ (torus in R3 with γ holes), γ > 0,
χ(G) ≤⌊
7 +√
1 + 48γ
2
⌋.
For γ = 1, χ(G) ≤ 7.
G
is planar then χ(G) ≤ 4. G is toroidal then χ(G) ≤ 7 sharp.
Extremal example: Sizilassi
6
7
34
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
∆
11
∆ − 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
. . .vn = v
v1
v2
v3
v4
v5 v6
vv7
v8
Observe that we could choose any vertex v = vn and so if G is not regular, then χ(G) ≤ ∆. Choosev = 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:
Take z ∈ V (G) at distance 2 from v. Let x be a common neighbor of v and z.
x
v z
Let y = v. We want to show that G − {y, z} is connected. We have that G − y = G − v is 2-connected, 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:
yz
v = x
y z
v = x
deg(v) ≥ 3deg(v) = 2
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
vis impossible since ∆(G) ≥ 3. So, there is a nontrivial block in G− v.
v
z xy
�proof of Brook’s theorem:If G has a cut-vertex, v, then G = G1 ∪G2, V (G1) ∩ V (G2) = {v}.
G1G2
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 norG2 is (C2k+1 and ∆ = 2) since otherwise deg(v) ≥ 3.
G1G2
v1 v2
Color G1 and G2 with at most ∆ colors, permute colors if necessary so that the color of v is the samein G1 and in G2.
If G is 2-connected, let x, y, z be as in Lemma 2.
x
v z G − {y, z} is connected. Letv1 = 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
x = vn
∆ = 4
y = v1 z = v22 2 1
11 1
Apply greedy coloring to this ordering of V (G). Since vn has ∆ colored neighbors, but two neighborshave 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) (indepen-dence number)
ω(G) ≤ χ(G), nα≤ χ(G)
1 2 3
. . .
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) ≤ 12
+√
2m+ 14,m = |E(G)| ≥
(χ(G)
2
).
1 2 3
. . .
χ(G)
Note in χ(G)-coloring, there is an edge between two color classes, otherwise take a union ofthese two color classes.
• χ(G) ≤ max{δ(H) : H ⊆ G}︸ ︷︷ ︸=:δ
+1 =
≤ δ
vnvn−1
≤ δ
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-colorableif ∃c : V → N, c is proper, c(v) ∈ L(v) (say H is list-colorable from L).
{1, 2}
{2, 3}{3, 1}
{1, 2}
{1, 2}
{3, 1}{1, 2} {2, 3}
{2, 3}{1, 3}
G is not L-colorable
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
124
124123
123
123
134
134
134234
234
234
not colorable
is not colorable
1245 2345 1254
1234
1234
1234
1234
1234 1235 1235
1235
1235
1235 1345 1345
1345
1345
1345 1245 1245
1245
1245
1245
1235 1345
1235 2345
123523451245
1245 1345
1345
1254
2354
2354
1354
1354
1234
1234
1324
1524
1342
1342
1542
1542
3452
3452
1352
1352
4523
1523
4523
4523
1253
1253
1243
1243
12431453
1453
1453
2345
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−1k
), with partite sets A,B. Each vertex in A gets a distinct list from(
[2k−1]k
).
Example: k = 3, [2k − 1] = {1, 2, 3, 4, 5}{1, 2, 3} {1, 2, 4} {1, 2, 5} {3, 4, 5}
{1, 2, 3} {1, 2, 4} {1, 2, 5} {3, 4, 5}
. . .
. . .
. . .
Use same lists in B. (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 colorsare from {k, . . . , 2k − 1} = L(v), so v could not have been colored.
{k, k + 1, k + 2, ..., 2k − 1}
v
{1, 2, ..., k}
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 someembedding 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
x
y
1
1
3
3
3
3
3
Then G is L-colorable.
Basis:3 1
1
{a, b, c}
{a}
{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
y
G1
G2
1
1
33
Let, wlog, G2 contain x and y. Apply induction to G2 to color it from these list withcoloring 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−1
xk
x
y
Apply induction to G− xk with new lists L′.
L′(v) =
{L(v), v 6∈ N(xk)− {x, xk−1}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 L′, with coloring c.
xx−1
xk {α, β}
{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.
�
Perfect graphs: A graph G with the property, for any induced subgraph H of G, χ(H) = ω(H)holds, is called a perfect graph.
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 ofodd hole) as induced subgraph.
Coloring edges: Recall χ′(G) is the edge chromatic number.
12
31
χ′(G) = 3
König’s theorem 1916: If G is bipartite, then χ′(X) = ∆(G).
proof: Since a vertex of degree ∆ = ∆(G) is incident to ∆ edges, χ′(G) ≥ ∆.
To prove the upper bound χ′(G) ≥ ∆, use induction on |E(G)|.Let e ∈ E(G), color G− e with colors from {1, 2, . . . ,∆} by induction.
x
y
≤ ∆ − 1≤ ∆ − 1
no β but α
no α but β
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
αβ
x
ycolorflip
Consider an α-β-subgraph, Gαβ, it is a disjoint union of pahs and even cycles. If a component ofGαβ containing x does not contain y, flip colors α and β in this component, color e with β. If thecomponent of Gα,β containing x also contains y, it is a path of even length with end-edges of differentcolors, a contradiction. �
41
Vizing theorem (1964): For any G ∆ ≤ χ′(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 G′ = 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 colorsis ∆ + 1, that is less than ∆.
Let y1 = y, y2, . . . , yh+1 be vertices inN(x) and t1, t2, . . . , th+1 be colors such that ti ∈ Mis(yi), c(xyi+1) =ti. Let s ∈ Mis(x)
x
y = y1
y2
y3
no t = t1
no t2
no t3
t1t2
no S
yh+1
th
Such that ti 6= s, i = 1, . . . , h, the sequence is largest, ti’s are distance.If ti = s (take smallest i)
x
y = y1
y2
y3
no t = t1
no t2
no t3
t1t2
no S
If t3 = s
xy2
y3
t1t2
y1
s
Since the graph is finite, the sequence will stop if th+1 = tj, j < h+ 1.
x
y1
yj
tj−1no S
yj+1
tj no tjS
tj
Try to recolor G′ such that s is not incident to yj+1. Do this by taking a path colored tj and s, andflip colors. If this path does not end in yj, done.
x
y1
yjt2
no S
yj+1
tj no tj
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) χ′(G) ≤ 32∆ for a multigraph G.
k k
k
3k colors needed
∆ = 2k
χ′(G) ≤ 32∆
multigraph
Main results in colorings:
• χ(G) ≤ ∆ + 1
• χ(G) ≤ ∆ if G 6= K∆+1, C2k+1 (Brook’s theorem)
• ∆ ≤ χ′(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:
– χ′(G) = ∆ G bipartite– χ(G) <<< ∆(G)
– χlist(G) >> χ(G)
– ω(G) ≤ χ(G)
– ω(G) <<< χ(G) for some graphs– χlist
′(G) =?
Mycielski’s construction: G1, G2, . . . , Gk triangle-free, ω(Gi) ≤ 2 for i = 1, . . . , k and χ(Gi) = i
G1 G2 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}E(Gk+1) = {wui : i = 1, . . . , k} ∪ E(Gk) ∪
n⋃i=1
{uiv : v ∈ NGk(vi)}.
43
v1
v2v3
v4v5
V U
u1
u2
u3
u4
u5
w
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.
k
Gk
vi
A
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 c′(vi) = c(ui), vi 6∈ A, let c′(vi) = c(vi).
Claim: c′ is a proper (k − 1)-coloring of Gk.
Since c(ui) 6= k, c′ does not use color k. Take two vertices vi, vj:
• vi, vj 6∈ A c(vi) = c′(vi) 6= c′(vj) = c(vj)
• vi, vj ∈ A vi 6∼ vj
• vi ∈ A, vj 6∈ A vi ∼ vj
vi
vj
ui
Assume that c′(vi) = c′(vj). c′(vi) = c(ui) ui ∼ vj vj ∈ N(vi) c(ui) = c(vj), contradiction sincec is a proper coloring. �
χlist′(G) edge choosability.
Conjecture: ∀G χlist′(G) = χ′(G)
44
Theorem: (Galvin 1995) χlist′(G) = χ′(G) for bipartite graph G.
Lemma: Let H be a graph, L : V → 2N a list assignment. If H has an orientation of the edgessuch 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 \S ∃s ∈ Sv s
proof: Let H(α) be set of vertices whose list contain α. Color the vertices of kernel of H(α) withα, remove color α from other lists. Apply induction to uncolored part.
ααα
αα
ααα
H(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, e′
e e′ if c(e) < c(e′) and meet in Y , or c(e) > c(e′) and meet in X.
X Y X Y
e
e′
e
e′
3
5
7
2
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 nvertices.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 ex(n,K2) = 0, EX(n,K2) = En
H = P3 ex(n, P3) =⌊n
2
⌋, EX(n, P3) =
⌊n2
⌋K2.
. . .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, nvertices.
T (n, 5)
T (5, 2) T (6, 2) T (7, 3)
|E(T (n, r))| =t(n, r)
Note: χ(T (n, r)) = r
Claim: The largest number of edges in a bipartite graph is⌊n2
⌋ ⌈n2
⌉. The densest bipartite graph is
Kq,n−q
≈ n2
4 edges n− 1 edges
f(q) := |E(Kq,n−q))| = q(n− q)
maxi≤q≤n−1
f(q) ∼=q=n
2
n2
4
Combinatorial argument:
46
A
B
q
n− q
G
q < n− q − 1
Move a vertex from B to A, get a new graph G′.
|E(G′)| = |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 numberof 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 numberof edges. �
Lemma: t(n, r − 1) = t(n− (r − 1), r − 1) + (n− r + 1)(r − 2) +(r−1
2
).
proof:
(r − 1) blobs
T (n− (r − 1), r − 1)X1 X2
Xr−1
|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)
(Number of edges betweenx1, x2, . . . , xr−1) =
(r − 1
2
).
�
Lemma: limn→∞
t(n,r−1)
(n2)= (1− 1
r−1) .
proof:
t(n, r − 1) =
(n
2
)− (r − 1) ·
(nr−1
2
)∼=(n
2
)− (r − 1)
1
2
(n
r − 1
)(n
r − 1
)
∼= n2
2−(
1
r − 1
)n2
2∼=(n
2
)(1− 1
r − 1) (
(n
2
)≈ n2
2for large n)
47
�Theorem (Turàn 1941): For all positive intgers r, n, r ≥ 3:
EX(n,Kr) = {T (n, r − 1)}
ex(n,Kr) = t(n, r − 1) ≈(n
2
)(1− 1
r − 1).
Theorem (Erdös-Stone): ex(n,H) ∼=χ(H)=r
(n2
)(1− 1
r−1) = ex(n,Kr).
ex(n,1
21
2
2
3
) .(n2
)(1− 1
3−1 ) =(n2
)12
Theorem: ex(n, Pk+1) & k−12n.
proof: Let G be a disjoint union of cliques on k vertices.
k k k
. . .
k
(k2
)nk = k(k−1)
2nk = (k−1)n
2
For the upper bound, use induction on n.
Basis: n = k, Pk+1 6⊆ Kk
|E(Kk)| = k(k−1)2
= n(k−1)2
.
Step: Observe that if Pk+1 6⊆ G, then δ(G) ≤ k−12.
forbidden
longest path
graph is connected (otherwise apply induction)
∃ longer path
u v
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
2+ 1, then
|V (P )| ≥ k + 1.
48
Let v be a vertex of minimum degree, deg(v) ≤ k−12. Then
|E(G)| ≤ |E(G− v)|+ k − 1
2
by induction≤ (n− 1)
k − 1
2+k − 1
2= n · k − 1
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 havethat Kr−1 ⊆ G, on vertex set K. Also, we have that |E(G)| ≥ t(n, r − 1).
Kr−1
≤ r − 2
≤ r − 2
n− (r − 1)
t(n, r − 1) ≤ |E(G)| ≤(1)|E(G[K])|+ |E(K,V −K)|+ |E(G[V −K])
≤(2)|(r − 1
2
)+ (n− r − 1)| · (r − 2) + t(n− (r − 1), r − 1)︸ ︷︷ ︸
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 ofnonedges), also vi is adjacent to all of K − vi, G = T (n, r − 1).
49
K = {v1, . . . , vr−1}V −K
vr−1
v1
v2
v3
• If vi 6∼ Vi, vj 6∼ Vi, then vi, vj and (r− 2) vertices from distinct {V1, . . . , Vr−1} − {Vi} forma Kr, a contradiction.vi vj
�
ex(n,K3) ∼= 1
2
(n
2
)
ex(n,K4) ∼= (1− 1
3)
(n
2
)
ex(n,K5) ∼= (1− 1
4)
(n
2
)
t(n, r − 1) = (1− 1
r − 1)
(n
2
)
ex(10, K4) = 9 + 12 + 12 = 33.
Forbid Ksr= T (rs, r)
If s is largest enough, then any H, χ(H) = r is contained in Ksr .
If H 6⊆ G ⇒ Ksr 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-Szemeredi 1981): ∀ε > 0∀r ∈ N, r ≥ 3, if |E(G)| ≥(1− 1
r−1+ ε)
(n2
), |V (G)| = n, then
Ktr ⊆ G, t =
log n
500 · log(1ε).
There is a graph G on (1− 1+εr−1
)(n2
)edges, n vertices and no Kt
r for t = 5 logn
log( 1ε)(this tells us that t
is amost the best).
In particular H ⊆ G for any H : χ(H) ≤ r.
r = 3log n
log n
log n
ex(n,H) =(
1− 1χ(H)−1
) (n2
)(1 + o(1))
∀ε > 0∀t ∈ Z+ ∃n = n(ε, t), E(G) ≥(n2
) (1− 1
r−1+ ε), |V (G)| = n ⇒ Kt
r ⊆ G.
ex(n,K103 ) =
(n2
)(1− 1
2+ ε)
ex(n,K103 ) =
(n2
) (1ε
)(1 + o(1)).
Corollary: limn→∞
ex(n,H)
(n2)=(1− 1
r−1
)=
r=χ(H)
(1− 1
χ(H)−1
)=(χ(H)−2χ(H)−1
)
ex(n, ) . 23
(n2
)What if χ(H) = 2?
limn→∞
ex(n,H)(n2
) =
(1− 1
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.known: n2, n
32 , n
53
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
X
Y
m
n
s
t
forbidden
z(m,n; s, t) ≤ (s− 1)1t (n− t+ 1)m1− 1
t + (t− 1)m
Corollary: z(n, n; t, t) ≤ c n · n1− 1t + cn ≈ cn2− 1
t t, s constants.
proof: Consider G ⊆ Km,n, Ks,t 6⊆ G.
X
Y
s
t
t = 4
v
d(v)
Let T be the number of stars with t leaves centered in X. T ≥ ∑v∈X
(d(v)t
), each t-subset of Y gives
≤ (s− 1) stars ⇒ T ≤(nt
)(s− 1)⇒ ∑
v∈X
(d(v)t
)≤(nt
)(s− 1).
Jensen’s inequality: ϕ-convex function ϕ : R→ R, ai > 0, ai ∈ R
ϕ
(∑aixi∑ai
)≤∑aiϕ(xi)∑ai
.
We have ϕ(x) =(xt
), ai = 1
|x|(∑
d(v)|x|t
)≤∑
v∈X
(d(v)
t
).
Why is(xt
)convex for x ≥ t?
Idea: f(x) = x · g(x) and g(x) convex (g′(x) > 0, x > 0) ⇒ f(x) convex. f ′ = xg′ + g, f ′′ =xg′′ + g′ + g′ > 0
With e :=∑v∈X
d(v) it follows m( emt
)=∑x∈X
( emt
)≤
Jensen’s inequality
∑v∈X
(d(v)t
)≤(nt
)(s− 1)
52
m
(em
m
)≤ (s− 1)
(n
t
)
m
(em
)t
6 t! ≤ (s− 1)nt6 t!
m · ( em− t+ 1)t ≤ (s− 1)(n− t+ 1)t
(e
m− t+ 1) ≤
((s− 1)(n− t+ 1)t
m
) 1t
= (s− 1)1t (n− t+ 1) ·m− 1
t
e ≤ (t− 1)m+ (s− 1)1t (n− t+ 1)m ·m− 1
t .
�
Corollary: ex(n,Ks,t) ≤ cn2− 1s , s ≤ t
Lemma: ex(n,Kr,r) ≥ c · n2− 2(r+1) (⇒ c · n2− 2
(r+1) ≤ ex(n,Kr,r) ≤ cn2− 1r )
proof: Construct a graph G on n vertices, no Kr,r subgraph with many edges. Pick an edge withprobability p, get G.
G2r r r
(EXP # means expected number)EXP(#E(G)) = p
(n2
).
EXP(#copies of Kr,r in G) =(n2r
)(2rr
)pr
2 .Construct a graph G′ from G by deleing an edge from each copy of Kr,r in G. So, G′ 6⊇ Kr,r.EXP #deleted edges= EXP(#copies of Kr,r in G) < 1
2EXP(#E(G)) if p = cn−
2r+1 .
Verify:(n
2r
)(2r
r
)pr
2 ?<
1
2p
(n
2
)
pr2−1 <
1
2
(n
2
)((n
2r
)(2r
r
))−1
∼= Cn2
n2r= Cn2−2r
p < C2n2(1−r)r2−1 = C2n
− 2r+1 .
�
Lemma: ex(n,K2,2) = ex(n,C4) ∼= c · n 32 .
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
(1, 0)
(1, 1)
(1, 2)
(2, 0)
(2, 1)
(2, 2)
p = 3 {1, 2} × {0, 1, 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 uniquelydetermined.
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.
(x, y)
(a, b) (c, d)
Let (x, y) ∈ V (G). If (x, y) ∼ (a, b) and (x, y) ∼ (c, d) then{ax = y + b
cx = y + d⇒{x = (b− d)(a− c)−1
2y = x(a+ c)− b− d
Either a 6= c or b 6= d⇒ unique solution or not, then there are no solution.
|E(G)| =1
2|V (G)| · (p− 1) =
1
2(p− 1)p · (p− 1)
|V (G)| ∼= p2
|E(G)| ∼= p3 ∼= |V (G)| 32
�ex(n,H) =
(n2
) (1− 1
χ(H)−1
)∼= cn2 if χ(H) ≥ 3.
ex(n,H) =? = o(n2) if χ(H) = 2.
repetition:
ex(n,H) = max #edges, in an n-vertex H-free graph.
ex(n,K3) ∼= n2
4
ex(n,Kr) ≈ (1− 1
r − 1)(n2
2)
ex(n,H) ∼= (1− 1
χ(H)− 1)
(n
2
), χ(H) ≥ 3
ex(n,H) = ? for H-bipartite
c2− 2
t+1
1 ≤ ex(n,Kt,t) ≤ c · n2− 1t
c1n32 ≤ ex(n,K2,2) = ex(n,C4) ≤ c · n 3
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+ x25 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
z(n, n; t, t) ≤ cn32 ⇒ ex(n,Kt,t) ≤ cn
32
1st Way: Any graph G can be vertex split into X, Y , #edges(X, Y ) ≥ 12#edges of G.
G X Y
G 6⊇ C4, then bipartite G(X, Y ) does no have C4, has a Zarankiewics bound e(G(X, Y )) ≤ cn32 , thus
e(G) ≤ 2e(G(X, Y )) ≤ 2cn32
2nd Way: Consider G on n vertices and no C4.
Gn2
n2
For any spanning bipartite subgraph ofG, it has at most cn32 edges.
G
n2 − 1
n− 2
e(G) ≤ c∑
H-balance spanningbipartite subgraph of G
e(H)
(n− 2n2− 1
)
≤ c n32
(nn2
)(n− 2n2− 1
)
= cn32n(n− 1)
n2n2
≈ c1n32
Szemeredi’s regularity lemma: Given a graph G, X, Y ⊆ V (G), X ∩ Y = ∅
‖X, Y ‖ = #edges between X and Y.
d(X, Y ) =‖X, Y ‖|X||Y |
55
the density of a pair (X, Y ), 0 ≤ d(X, Y ) ≤ 1
(X, Y ) is an ε-regular pair if ∀A ⊆ X, |A| ≥ ε|X|,∀B ⊆ Y, |B| ≥ ε|Y | :
|d(X, Y )− d(A,B)| ≤ ε
X YA B
An ε-regular partition of G = (V,E) is a partition of vertex set V = V0∪V1∪V2∪ . . . ∪Vk:
(i) |V0| ≤ ε|V |(ii) |V1| = |V2| = . . . = |Vk|(iii) All but at most εk2 pairs (Vi, Vj) are ε-regular 1 ≤ i < j ≤ k.
v0vk vi
vj
Regularity Lemma of Szemeredi (about 1970): ∀ε > 0 ∀m ≥ 1, m ∈ Z, ∃M ∈ Z+, ∀graphG of order ≥ m, there is an ε-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 ε-regular partition, let d ∈ (0, 1]. Construct R, reducedsubgraph R = R(G) on v1, . . . , vk so that vi ∼ vj iff (Vi, Vj) ist ε-regular of density ≥ d. ConstructR = Rs (s ∈ Z+) - blow up of R with parameter s (each vertex s replaced by s vertices and each edgeis replace by Ks,s between the corresponding sets).
v0
v1
v2
v3
v4 G
v4
v1
v2
v3
R Rs = R3
K3 ⊆ RsK3 ⊆ G
Embedding Lemma: If H ⊆ Rs ⇒ H ⊆ G.(∀d ∈ (0, 1],∆ ≥ 1∃ε0 > 0∀G∀H ∆(H) ≤ ∆, s ∈ N, R = RG(ε ≤ ε0, e = |Vi| ≥ s
ε0, d)− ... �
proof of Erdös-Stone theorem):
56
|E(G)| ≥ t(n, r − 1) + γn2 = (1− 1r−1
+ 2γ)(n2
). Want to show Ks
r ⊆ G (s-blow-up of Kr).Apply Regularity Lemma to G, get R,Rs. If Kr ⊆ R, then Ks
r ⊆ Rs ⇒Embedding Lemma
Ksr ⊆ G.
To show that Kr ⊆ R, count the number of ε-regular pairs of positive density, apply Turan theoremto R.Fix ε > 0 and d ∈ [1, 0), d very small.A good pair (vi, vj) is a ε-regular one with density ≥ d.
v0vk
v1
v2
Total number of edges which are not in good pairs =
= #edges in G[vi] + #edges in not ε-regular pairs + #edges in ε-regular pairs of density < d
≤ k
(nk
2
)+
(εn
2
)
︸ ︷︷ ︸edges inside the blobs
+εk2(n
k· nk
) + d
(n
2
)
∼=n large
n2
2k+ ε2n2 + εn2 + dn2
= (1
2k+ ε2 + ε+ d)
︸ ︷︷ ︸<γ
n2choose ε, k, d such that this constant is < γ
(n
2
)
#edges in good pairs ≥ e(G)− γn2
2≥ (1− 1
r − 1+ γ)
n2
2− γn
2
2≥ (1− 1
r − 1)n2
2.
|E(R)| &each pair has at most n
knk
edges(1− 1
r − 1)n2
2· kn
k
n= (1− 1
r − 1)k2
2.
Since |V (R)| = k, by Turan’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 ccliques 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 = 1
c = 2 c = 3
57
(Vi, Vj) is ε-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|E(G)| > (1− 1
r−1)(n2
)⇒ Kr ⊆ G.
What do we need to force MKr or TK3?
recall if ∆(H) ≤ 3 MH ⊇ TH.
average degree is high → MKr, TKr
high chromatic number ?→MKr, TKr
high chromatic number 6→ Kr (Recal Mycielsky construction ∀k ∃G : χ(G) ≥ k, ω(G) = 2)
Theorem 1996 (Bollobas-Thomason, Komlos-Szemeredi):∃c ∈ R ∀r ∈ N ∀G : average degree d(G) ≥ cr2, TKr ⊆ G.
e(G) ≥ Cr2n (sharp up to c)
G ⊇ Kr if e(G) ≥ crn2
G ⊇ TKr if e(G) ≥ c′rn
Theorem (Kostochka ’82, Thomason ’84):∃c ∈ R ∀r ∈ N ∀G : d(G) ≥ cr
√log r, MKr ⊆ G (sharp up to c)
e(G) = Cr√
log r︸ ︷︷ ︸cr
n
χ(G) high ?→MKr ⊆ G TKr ⊆ G
Hadwiger’s conjecture 1943: ∀r ∈ Z+ ∀G
χ(G) ≥ r ⇒MKr ⊆ 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+ ∀Gχ(G) ≥ r ⇒ G ⊇ TKr
true for k ≤ 4, open for k = 5, 6, false for k ≥ 7.
Bollobas, Thomason c = 1116.
58
Theorem (Thomassen ’83): ∀k ∈ Z+ ∀G girth(G) ≥ 4k − 3, δ(G) ≥ 3 : G ⊇ MH, for someH 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 ofMH)
Claim 1: G[Vi] is a tree, i = 1, . . . ,m.Assume not, i.e., G[Vi] has a cycle C, length(C) ≥ 4k − 3
v4k−3v1v2k−2
v2k−1
≥ 2k − 2
≥ 2k − 2
Let V ′i = {v1, . . . , v2k−2}, V ′′i = {v2k−1, . . . , v4k−3}. Then G[V ′i ], G[V ′′i ] 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.
P1
P2
P3 v
u
≥ 2k − 2
≥ 2k − 2
≥ 2k − 2
Vi Vj
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
2
⌉= 2k − 1. We have |V (P1)| = 2k, so |v(
◦uP1
◦v)| ≥ 2k − 2. We have
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 V ′i , V′′i . Replace Vi, Vj with V ′i , V ′′i , 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
2= k other Vj’s. Since
this the Vi’s form branch sets of MH, δ(H) ≥ k. �
Structure: how does a graph without MKr 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
Pasting
alongS = K3
G1 G2 G = G1 ∪G2
S SS
S SPasting
alongS = K2
• A graph on ≥ 3 vertices, edge-maximal without MK4 iff it is constructed from triangles viapasting along K2’s.
• Wagner 1937 (Klaus Wagner): A graph G is edge-maximal without MK5 (has ≥ 4 vertices)then G can be constructed recursively from plane trianguations and Wagner graph via pastingalong K2’s and K3’s.Wagner’s graph
Ramsey theory
Frank Ramsey
Color E(Kn)or
≈ log n
≈ logn
in 2 colors
Example: Any 2-colored K6 (edges are colored) has a monochromatic triangle (i.e. K3 whose edgeshave the same color).
vWlog
v
v
done
done
know
don’t know
60
Lemma: Let E(Kn) be colored with 2 colors, n > 4s + 1, ten there is a monochromatic Ks.
proof:
v
S
blue neighborhood of v
yellow neighborhood of v
Define a sequence of vertices. Let v1 ∈ V . For any vertex v ∈ V , any S ⊆ V , we say that S ′ ⊆ S is ayellow neighborhood of v in S if all edges between v and S ′ are yellow and all edges between S − S ′and v are blue (S − S ′ 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.
v1 v2
v3
repeat log2 4s = log2 22s = 2s. �
Ramsey theorem: ∀k ∃N(k) in any yellow-blue coloring of E(Kn), n ≥ N there is a monochro-matic Kk. Smallest such N is denoted R(k), called Ramsay number.
R(3) = 6√2k ≤ R(k) ≤ 4k
limk→∞
R(k)1k exists? =? $100, $250.
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}.
=K3 yellowEx:
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| ≥ |Vi|
2and 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 edgesrandomly independently, with probability 1
2yellow and probability 1
2blue. If prob(there is no mono-
61
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) ≤2Prob(G[S] yellow) = 2 ·
(12
)(k2) = 21−(k2)
Prob(Some S ⊆ V, |S| = k is monochromatic)≤ ∑S⊂V|S|=k
Prob(G[S] is monochromatic)≤(nk
)21−(k2) < 1.
If n ≤ k2k2
(1√2k
+ o(1)), in particular if n ≤ 2
k2 �
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−2k−1
)No-
te: R(k, k) ≤(
2k−2k−1
)≤ 4k.
proof: Consider a coloring of E(Kn) with no yellow Kk and no blue Kl. We’ll show n ≤(k+l−2k−1
)−1.
The yellow neighborhood has no yellow Kk−1 and no blue Kl, the blus neighborhood has no blueKl−1 and no yellow Kk. Then
n ≤ 1+(R(k−1, l)−1)+(R(k, l−1)−1) = −1+
(k − 1 + l − 2
k − 2
)+
(k + l − 1− 2
k − 1
)= −1+
(k + l − 2
k − 1
).
no yellow Kk−1no blue Kl
no yellow Kkno blue Kl−1
no yellow Kk−1no blue Kl
no yellow Kk−1no 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.
proof: Lower bound:
t− 12s− 1
Upper bound: We shall show that R((s+ 1)K2, (t+ 1)K2) ≤ R(sK2, tK2) + 3. Let c be a coloringof 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:
R(sk2, tK2)
t
s s+ 1
t+ 1
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-Szekeres (1935) theorem: For any list of n2 distinct numbers, there is a sublist of > nnumbers 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 oflongest increasing subequence ending with ai and di is the length of longest decreasing subsequenceending 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(ui,di)
, . . . , aj(uj ,dj)
, . . .
This is possible since if aj > ai, uj > ui or if aj < ai, dj > di. �
Ramsay theory for set system (hypergraphs):
63
so far
pdgeonhole princ.
color V
graph Ramsey
color(V2
)
V V V
color(VK
)−set of k-element subsets of V
c:(Vk
)→ {yellow, blue}
Rr(l) = R(l, r) = min{n : for any yellow/blue coloring of
(V
r
), |V | = n,
there is either V ′ ⊆ V, |V ′| = l,
(V ′
r
)is yellow,
or ∃V ′′ ⊆ V, |V ′′| = l,
(V ′′
r
)is blue
}
l l
Ramsey proved that this exists.
Rr(l1, l2, . . . lk) = min{n : for any coloring of
(V
r
)into k colors 1, 2, . . . , k, |V | = n
∃V1 ⊆ V, |V1| = l1,
(V1
r
)has color 1, or
∃V2 ⊆ V, |V2| = l2,
(V1
r
)has color 2, . . .
}
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.
We want to show that either ∃V ′ ⊂ V, |V ′| = p,(V ′
r
)is red, or ∃V ′′ ⊂ V, |V ′′| = q,
(V ′′
r
)is blue.
64
x r
r − 1
r − 1
Rr(p− 1, q)Rr(p, q − 1)
V ′
Let V = x ∪ {X}. We shall color(Xr−1
).
c′ :
(X
r − 1
)→ {red, blue}, c′(X ′) =
{red , if c(X ′ ∪ {x}) = redblue , if c(X ′ ∪ {x}) = blue.
We have |X| = Rr−1(Rr(p− 1, q)︸ ︷︷ ︸red
, Rr(p, q − 1)︸ ︷︷ ︸blue
).
Thus ∃V ′ ⊆ X,(V ′
r−1
)is red under c′, |V ′| = Rr(p− 1, q)
or ∃V ′ ⊆ X,(V ′
r−1
)is blue under c′, |V ′| = Rr(p− 1, q).
Assume(V ′
r−1
)is red under c′. Then, look at
(V ′
r
)under c. So, we habe either V ′′ ⊂ V ′ with
(V ′′
r
)is
blue, |V ′′| = q, we are done; or V ′′ ⊂ V ′ wih(V ′′
r
)is red, |V ′′| = p − 1, then
(V ′′∪{x}
r
)is red and
|V ′′ ∪ {x}| = p. �
Applications:
1) Erdös-Szeres 1935: ∀m ∈ Z≥3 ∃N(m)∀set of N(m) points in R2 (general position no 3 on aline) 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.
So, either ∃V ′ ⊂ V, |V | = m,
(V ′
4
)is red
︸ ︷︷ ︸(a)
or ∃V ′ ⊂ V, |V | = 5,
(V ′
4
)is blue
︸ ︷︷ ︸(b)
.
Claim: (b) is impossible
65
|V ′| = 5, convex hull of V ′ is , otherwise -red
V ′ : - contradiction
So (a) holds. Then V ′ must form a convex m-gon. Indeed, if not, and a convex hull of V ′has less than m endpoints
a contradiction. �proof(2): due to Tarjan.
Let N = R3(m,m) (color triples, red/blue, force red(
[m]3
)or blue
([m]3
)).
Given {x1, x2, . . . , xN} ⊆ R2 in a general position. Color c({xi, xj, xk}) =red if i < j < kand xi, xj, xk appear clockwise.
xi
xjxk
xi
xj xk
otherwise blue
Let, wlog, V ′ ⊆ {x1, x2, . . . , xN},(V ′
3
)red, |V ′| = m.
Then V ′ forms a convex m-gon. Indeed, otherwisex2
x3
x4
x5
impossible
�
2m−2 − 1 ≤Erdös
N(m) ≤Toth-Valtr 1988
(2m− 5
m− 2
)+ 2
2) Schur theorem 1916: If c : N → {1, 2, . . . , r} then ∃x, y, z ∈ N : c(x) = c(y) = c(z) andx+ y = z.
proof: Let c : {1, 2, . . . , n} → {1, 2, . . . , r}, where n = R2(3, 3, 3, . . . , 3︸ ︷︷ ︸r times
) (color pairs, r-colors,
force monochromatic triangles.)Lets color the edges of Kn on vertex set {1, 2, . . . , n}, c′({i, j}) = c(|i− j|).
66
1 2 3 4 5
cc′
1
2
34
5
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) = c′({i, j}) = c′({k, j}) = c(y) = c(z). x+ y = j − i+ k − j = k − i = z. �
Known:k!(e− 1
12) ≥ R2(3, 3, . . . , 3︸ ︷︷ ︸
k
) ≥ c · (3.17)k
Claim: c · 2k ≤ R2(3, 3, . . . , 3︸ ︷︷ ︸k
)c1 · k!
proof: Lower bound: construct a coloring of E(Kn), n = 2k, in k colors, with no monochromaticK3.
Upper bound: Suppose we are given a k-coloring of E(Kn), n = c1 · k!. We need to show that thereis a monochromatic triangle.
v
Ni(v)
≥ n−1k
Let v ∈ V (Kn), there is a color i, v is incident to at least n−1k
edges of color i, call their endpoint setNi(v). We have Ni(v) does not induce edges of color i.
R(3, 3, . . . , 3︸ ︷︷ ︸k
) ≤ k ·R(3, 3, . . . , 3︸ ︷︷ ︸k−1
) + 1 ≤ k · (k − 1) · . . . · (3)R(3, 3) + o(k!)
= 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 a1x1 + a2x2 + . . .+ arxr = 0, variablesxi’s?
We say that a linear equation S is r-regular on A ⊆ Z, if, for any r-coloring of A, there is amonochromatic solution of S. For example x + y = z is r-regular on N for any r. A linear equationS 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 : a1x1 + a2x2 + . . .+ arxr = 0, ai ∈ Z. We have that S is regular on N if∃subset I ⊆ {1, . . . ,m} such that
∑i∈Iai = 0.
Example: 2x1 + x3 − 4x4 + x2 = 0 is regular, because 2 + 1− 4 + 1 = 0.
Column condition for systems: Given variables x1, . . . , xn.
c11x1 + c12x2 + . . .+ c1nxn = 0
c21x1 + c22x2 + . . .+ c2nxn = 0...cm1x1 + cm2x2 + . . .+ cmnxn = 0
, (ci :=
c1i...cmi
)
⇔ c1x1 + c2x2 + . . .+ cnxn = 0 has a column condition if up to recordering the columns the followingholds:
c1 , c2 , cn
sum = 0
sum =
A1 A2 A3
linear combination of previous colums
∃k0 = 1 ≤ k1 < k2 < . . . < kt = n if Ai =ki∑
j=ki−1+1
cj
Ai = 0, for 2 ≤ i ≤ t, Ai is a linear combination of c1, c2, . . . , cki−1.
If a system of equations has a column condition, then for any r ∈ Z≥1 and any r-coloring of N, thereis a monochromatic solution of the system.
Some graph-Ramsay numbers: Recall R(H,G) = min{n : any red/blue coloring of E(Kn) has redG or blue H}.
Lemma: R(sK3, tK3) = 3s+ 2t, s ≥ t ≥ 1, s ≥ 2.
proof: (s=red, t=blue)Lower bound: construction:
68
3s− 1
2t− 1
Red subgraph is isomorphic to K3s−1∪K1,2t−1.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 trianglesor 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(sK3, tK3) + 5 ≤ R((s− 1)K3, (t− 1)K3) + 2 · 5≤ . . . ≤ R((s− t+ 1)K3, K3) + t · 5(
R(K3, K3) = 6, R(2K3, K3) ≤ 8 (∗), 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)
(∗) Claim: R(2K3, K3) ≤ 8.proof: Consider red/blue coloring of E(K8), need to show that ∃blue triangle (∆) or ∃red doubletriangle (∆∆).
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
N(v)blue
uw
uw
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.
- done9 edges between, at east 5 of these, wlog, red. There are
≥ 2 yellow edges from a vertex of blue triangle.
• 5
≥ R(sK3, tK3)
�
Know:√
2n ≤ R(Kn, Kn)
Question: R(H,H)?
≤ f(|V (H)|) (small)R(H,H) ≤ c(∆(H)) · |V (H)|?
≤ R(H,G)
Theorem (Ramsey for bounded degree graphs): (Chvatal, Rödl, Szemeredi„ 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) ≥2n2 )
proof: (Outline)Recall Szemeredi’s Regularity Lemma. Given graph G there exists partition
v1v2vk
ε− regular
k-constand # blobs
1
2
k
R-reduced graph
if corresponding pair ViVjε− regular density > 0
if Kt ⊆ R⇒ Kst ⊆ G.
Ex. ∆ ⊆ R⇒ ⊆ G, in particular G′ ⊆ 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 G.Let’s apply the Regularity Lemma to G.
71
v1v2vk
3 types of pairs (Vi, Vj):
1) not ε-regular (≤ εk2 of such)
2) ε-regular of density ≥ 12(majority - red)
3) ε-regular of density ≥ 12(majority - blue)
v1v2
vkRi
reduced
grey
Color vivj grey if it is of type 1, red if it is of type 2 and blue if it is of type 3.#grey edges ≤ εk2 ⇒ #non-greq edges ≥
(k2
)− εk2 ≈
(k2
)(1− 2ε)
⇒ non-grey graph contains Kr = KR(∆+1,∆+1) ⇒ R contains red K∆+1 or blue K∆+1, assume thatR contains red K∆+1.⇒ G ⊇ Ks
∆+1 ⇒ G ⊇ Ks∆+1 ⊇ H since χ(H) = ∆ + 1. (recall ∆(H) = ∆⇒ χ(H) ≤ ∆ + 1 by greedy
coloring.) �
Theorem (Lower bound on graph Ramsey numbers): (Chvatal, Harary ’72)R(H,G) ≥ (χ(G)− 1)(c(H)− 1) + 1, c(H) is the cardinality of the largest connected componentof H.
proof:KC(h)−1
χ(G)− 1 < χ(G)
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 thechromatic number of G.
H: G:
χ(G) = 4χ(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, Posa, 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 graphH = (A ∪B,E) is embedded int a bipartite graph G = (A′ ∪B′, E ′), if H ⊆
indG and A ⊆ A′, B ⊆ B′.
GA
B
A′
B′
induced
Lemma: Any bipartite graph can be embedded into the incidence graph ((X,(Xq
)), E), where E =
{xY : x ∈ X, Y ∈(Xq
), x ∈ Y }
Example: ((X,(X2
)), E), X = {1, 2, 3, 4},
(X2
)= {{12}, {13}, . . . , {34}}
1 2 3 4
12 13 14 23 24 34
X
(x2
)
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,(Xq
)), E) for some X, q.
Comment: ((X,(Xq
)), 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.
x1 xn y1 yn z1 zn a1 an
b1 bm
ϕ
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
b1 b2 b3
Ex:
x1 x2 x3 x4 x5 y1 y2 y3 y4 y5 z1 z2 z3X
(X6
){x1, x2, x5, z1, y1, y2} {x2, x3, z2, y1, y2, y3} {x3, x4, x5, z3, y4, y5}
≈ N(b1) ≈ N(b2) ≈ N(b3)dummy dummy dummy
Lemma: ∀bipartite P ∃bipartite P ′ ∀2-coloring of E(P ′) there is a monochromatic P embeddedin P ′.
proof: We shall prove the result for P = ((X,(Xk
)), E). Let P ′ = ((X ′,
(X′
2k−1
)), E ′), where
|X| = R2k−1(k|x|+ k − 1, 2(2k−1k
))
size of hyperedgessets colored
size of a set s.t. all its (2k − 1)-subsetsare of the same color
# colors
k|x|+ k − 1
2k − 1
Consider a red/blue coloring c of E(P ′) = E ′. Assume that X,X ′ subsets of integers.Definition:
1 3 7 8 11
X ′
{1, 3, 7, 8, 11}A′ (
x′
2k−1) k = 3
σ(A′) = blue
A = {1, 8, 11}σ′(A′) = {1, 4, 5}
c′(A′) = (blue, {1, 4, 5})
74
We shall corot the elements of(X′
2k−1
). Let A′ ∈
(X′
2k−1
), there is a color such that the majority of
edges incident to A′ is that color, call it σ(A′). There is a set A of k vertices A ⊆ A′, such that∀x ∈ A : c(x,A′) = σ(A′). Let σ′(A′) be a set of positions of A in A′.Define c′(A′) = (σ(A′), σ′(A′)). The number of colors in c′ is 2 ·
(2k−1k
).
By Ramsey theorm applied to(X′
2k−1
)with c′, there is a subset W ⊆ X ′, |W | = k|X| + k − 1, such
that(W
2k−1
)have the same color under c′. Assume, wlog, W = {1, 2, . . . , k|X|+ k − 1}.
Pick X = {k, 2k, 3k, . . . , |X|k} ⊆ W , see |X| = |X|. Map X ϕ−→ X, map(Xk
)into (2k − 1) element
subsets of W .
A = {i1k, i2k, . . . , ikk} ϕ−→ A′ ∈(
W
2k − 1
), such that A ⊆ A′, (A′ − A) ∩ X = ∅.
This gives a monochromatic P . �
Example: k = 3, |X| = 4, ((X,(X3
)), E)
1 3 7 8 11
X ′
{1, 3, 7, 8, 11}(X′
5
)(blue, {1, 4, 5})
a b c d
abc abd acd bcd
c′
X ′
(x′
5
)
W
(W5
)-monochromatic
of color (blue,{1, 4, 5})
1 2 3 4 5 6 7 8 9 10 11 12 13 14
{3, 6, 9, 4, 5} {3, 6, 12, 4, 5} {6, 9, 12, 7, 8} (W5
)
W
|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 ) :=∑
{x,v}∈Ef(v, x) = 0 ∀v ∈ V
(f(X, Y ) =∑
x∈X,y∈Y{x,y}∈E
f(x, y))
Example:x1
x2
x3
x4
−11
1 1
x1
x2
x3
x4
11
1 1
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) = −1f(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, X) = f(X, V −X) = 0 ∀X ⊆ V(f(X, X) = 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
(F1) f(x, y) = −f(y, x)
(F2) f(x, V ) = 0 for x ∈ V − {s, t}(F3) f(x, y) ≤ c(x, y)
2
7
1
3
5
x1
x2
s t
f = 2 f = 2
f = 0
f = 0
f = 0
f1 = 2f1 = 2
f1 = 3f1 = 3
f1 = 0c(s, x1) = 2 c(x1, s) = 0c(x1, t) = 5 c(t, x1) = 0
......
Definition: A cut (S, S) is a pair of sets S ⊆ V, s ∈ S, S = V − S, t ∈ S.
76
s t
2
5
7
1
Proposition: If f is a network flow, then for any cut (S, S)
f(S, S) = f(s, V )
proof: recall: f(S, S) =∑
x∈S,y∈Sf(x, y)
f(S, SV−S
) = f(S, V )− f(S, S) =(F1)
f( Ss∪(S−{s})
, V )
= f(s, V ) +∑
v∈S−{s}f(v, V ) =
(F2)f(s, V ).
�The value of the flow f , denoted |f | := f(s, V ) = f(S, S) = f(V − t, t) = f(V, t) for any cut (S, S).
Note: |f | = f(S, S) ≤(F3)
c(S, S)
Question? ∃f : |f | = min(S,S) cut
c(S, S)
Theorem (Ford-Fulkerson 1956): In any network, max value of a flow = min capacity of acut.
proof: Define the sequence of integral flows f0, f1, . . . , such that |f0| < |f1| < . . . = min(S,S)-cut
c(S, S).
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.
s tx1 x2 x3c = 5 c = 7 c = 2 c = 4
fn−1 = 3 fn−1 = 1 fn−1 = 0 fn−1 = 1
fn = 5 fn = 3 fn = 2 fn = 3
Let ε = mini=0,...,m−1
c(xi, xi+1)− fn−1(xi, xi+1), let fn(xi, xi+1) = fn−1(xi, xi+1) + ε.
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.
s
X
t∈ X
∈ X
∈ X
fn−1 = c
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).
�
Ex:s t
3
2
1
s t
1
3
3
3
2 20
0
00
0
00
0 0
s t1
1
1
s t1
1
11+
1
1 1
stop |f2| = 2 = c(S, S), done
f1
f2
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
11
1 ∈ R1 ∈ Z2
1 12
22
Z3
1
11 1
1
no Z2 flow
Theorem (Tutte 1954): ∀multigraph G ∃polynomial P such that for any finite abelian groupH, 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’sgraph)= 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.
x1
x2
x3
x4
≈
≈
1 1
1 1
1
12 2
222
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
B
1 1 122
2
Let f(a, b) = 1, a ∈ A, b ∈ B, i.e. f(b, a) = 2, b ∈ B, a ∈ A.Then for any a ∈ A :
∑y∈N(a)
f(a, y) = 1 + 1 + 1 = 0 (mod 3),
79
∀b ∈ B :∑
x∈N(b)
f(b, x) = 2 + 2 + 2 = 0 (mod 3). �
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 chosenindependently).For a given graph on e edges P (G) = p · p · . . . · p︸ ︷︷ ︸
e
(1− p) · (1− p) · . . . · (1− p)︸ ︷︷ ︸(n2)−e
= p(1− p)(n2)−e.
Ex: G(5, 13 )
P ( ) =(13
)t ·(23
)6
(23
)10 (13
) (23
)9 (13
) (23
)9 (13
)2 ( 23
)8 (13
)10
∑
G∈G(n,p)-labeledgraph on nvertices
P (G) =
(n2)∑
i=0
∑
G∈G(n,p)|E(G)|=i
P (G)
=
(n2)∑
i=0
((n2
)
i
)p(1− p)(n2)−i
= (p+ (1− p))(n2) = 1.
80
Lemma 1: ∀ integers n, k, n ≥ k ≥ 2:
Prob(G ∈ G(n, p), α(G)independent number
≥ k) ≤(n
k
)(1− p)(k2)(1 = 1− p).
proof: Let S ⊆ V, |S| = k, Prob(G[S] idependent) = (1− p)(k2).
Prob(some set on k vertices induces an independent set) = Prob
⋃
S∈(Vk)
G[S]-independent
≤∑
S∈(Vk)
Prob(G[S]-ind.)
=
(n
k
)(1− p)(k2).
�Lemma 2: ∀ integer k, n, n ≥ k ≥ 2:
Prob(G ∈ G(n, p), ω(G)clique number
≥ k) ≤(n
k
)p(
k2).
Theorem (Erdös, 1947): R(k, k) ≥ 2k2 .
proof: Let G ∈ G(n, 12).
Prob(α(G) ≥ k ∨ ω(G) ≥ k) ≤ Prob(α(G) ≥ k) + Prob(ω(G) ≥ k)
≤(n
k
)(1
2
)(k2)· 2 < 1.
if n = 2k2
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) =
∑G∈G(n,p)
Prob(G) ·X(G).
Markov inequality P (X ≥ a) ≤ E(X)a.
Lemma 3: E(#k-cycles in G ∈ G(n, p)) = (n)k2k· pk. Here (n)k := n · (n− 1) · . . . · (n− k + 1).
proof: For a k-cycle C in Kn, let XC : G→{
1 , C ⊆ G
0 , otherwise..
81
Let X =∑
C-k-cycles in KnXC ,
E(X) =∑
C
E(XC) =∑
C
(1 · Prob(Xc = 1) + 0 · Prob(XC = 0))
=∑
C
pk =(n)k2k· pk.
�
Corollary: P (#k cycles in G ∈ G(n, p) ≥ n2) ≤
Markov’s
E(#k-cycles)n2
=(n)k2k·pk
n2
.
Lemma 4: k > 0, p = p(n), p ≥ 6k lnnn
, n-large, then
limn→∞
P (α(G) ≥ 1
2
n
k) = 0.
proof:
Prob(α(G) ≥ r) ≤Lemma 1
(n
r
)(1− p)(r2) ≤ nr(1− p)(r2)
=(n(1− p) r−1
2
)r≤
1−p≤e−p
(ne−
p(r−1)2
)r−→n→∞
0
plug r ≥ 12nk, p = 1k lnn
n. �
Theorem (Erdös 1959): ∀k ∃graph H : girth(H) ≥ k and χ(H) ≥ k.
proof: Let k ≥ 3. Fix ε : 0 < ε < 1k, choose p = nε−1, let G ∈ G(n, p).
Let χ = #cycles of length ≤ k in G.
Exp(X) =k∑
i=3
Exp(#cycles of length k)
((∗) Exp(#k-cycles in G ∈ G(n, p)) =
(n)k2k
pk)
=(∗)
k∑
i=3
(n)i2i
pi
≤k∑
i=3
nipi
2i≤ 1
2
k∑
i=3
(np)i
≤ 1
2(k − 2)(np)k.
82
(since np ≥ 1, indeed n · nε−1 = nε ≥ 1)
Prob(X ≥ n
2) ≤
Makov
E(X)n2
≤12(k − 2)(np)k
n2
= (k − 2)nk−1pk =plug p
(k − 2)nk−1n(ε−1)k
= (k − 2)nkε−1 −→n→∞
kε−1< 1kk−1<0
0
⇒ Prob(X ≥ n
2) <
1
2for n ≥ n0, for some n0.
Prob(α ≥ 1
2
n
k) −→n→∞(∗∗)
0 ⇒ Prob(α ≥ 1
2
n
k) <
1
2for n ≥ n00, for some n00.
((∗∗) p ≥ 6k lnn
n⇒ lim
n→∞Prob(α(G) ≥ 1
2nk) = 0
)
Prob(X <n
2and α <
1
2
n
k) > 0
There is a graph G on n vertices with < n2short cycles of length ≤ k and α(G) < 1
2nk. Let G′ be
obtained from G by deleting a vertex from each short cycle.We have: G′ has girth > k, |V (G′)| > n
2, α(G′) ≤ α(G) < 1
2nk, χ(G′) ≥ |V (G′)|
α(G′) ≥n2
12nk
= k.
S, χ(G′) ≥ k, girth(G′) ≥ k. �
Properties of almost all graphs
A property -set of graphsA 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)) −→
n→∞1, we
write a.a.have P . (a.s.- almost surely) (a.a.- almost always, almost all)
Proposition: ∀p - constant p ∈ (0, 1), ∀graph H a.a.G ∈ G(n, p) contain H as an inducedsubgraph.
proof: Let k := |V (H)|, U ⊆ V (G), |U | = k.Prob(G[U ] ≈ H) = f(p, k) (independent of n)Let V ⊇ U1∪U2∪ . . . ∪Ubnkc, where |Ui| = k.
83
k
U
n
k
k
kk
U1
U2
U3
Prob(G[Ui] 6≈ H) = 1− f(p, k)
Prob(G 6⊆indG) ⊆ Prob
⋂
|U|=kU⊆V
G[U ] 6≈ H
= Prob
bnkc⋂
i=1
G[Ui]independent events
6≈ H
= (1− f(p, k)︸ ︷︷ ︸<1>0
)bnkc −→n→∞
0
Prob(H ⊆indG) −→
n→∞1
�Proposition: ∀p - constant p ∈ (0, 1), ∀ε > 0 a.a.G ∈ G(n, p):
χ(G) >
log(
11−p
)
2 + ε
n
log n.
84
proof:
Prob(α(G) ≥ k) ≤(n
k
)(1− p)(k2) ≤ nk(1− p)(k2)
= (1− p)klogn
log(1−p)+ 12k(k−1)
︸ ︷︷ ︸nk
= (1− p)k2
(−2 logn
log( 11−p )
+k−1)
=(1+ε) logn
log( 11−p )
(1− p) k2 (−2 k(2+ε)
+k−1)
= (1− p) k(2+ε)
>0︷ ︸︸ ︷(−2k + 2k + εk − 1)
= 1− p −→n→∞
0
Prob(α(G) < k) −→n→∞
1
Prob(χ(G) ≥ n
k) −→
n→∞1,
i.e. Prob(χ(G) ≥n(log( 1
1−p))
2 + ε log n) −→
n→∞1.
�
Evolution of random graphs
0 ≤ p1¬Pa.a.
< p2¬Pa.a.
< . . . < pi−1¬Pa.a.
< piPa.a.
< . . . < pkPa.a.
≤ 1 probabilities.
P - property, example P - beging connected.
A funcion f(n) is a threshold function for a property P iffor any p′ : p′
f(n)−→n→∞
0 a.a. G(n, p′) has no Pfor any p′′ : p′′
f(n)−→n→∞
∞ a.a. G(n, p′′) 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
nG a.a. connected,
p = (1 + ε) lognn
G a.a. has a Hamiltonia cycle.
p = n−2k−1 - threshold function for containing Kk, i.e. for example
p′ = n−2k−1
logn⇒ a.a. G(n, p′) 6⊇ Kk
p′′ = log log log log log log n−2k−1 ⇒ a.a. G(n, p′′) ⊇ Kk.
p = n−ve - threshols function for containing a balanced v-vertex, e-edged graph, here H is balanced
if ∀H ′ ⊆ H : e(H′)V (H′) ≤
e(H)v(H)
, e = |E|, v = |V |.
85
Lovasz Local Lemma: Let A1, . . . , An be events in some probabilistic space.
Prob
(n⋂i=1
Ai
)> 0 if Prob(Ai) ≤ p ∀i and each Ai is independent to all but at most d other events
Ai, i = 1, . . . , n and ep(d+ 1) ≤ 1.
An
A1 A2
A3d
Van der Waerden’s numberW (k) = min{n : any coloring 1, . . . , n in red/blue ∃monochromatic arith-metic progression of length k}.
1 2 3 4 5 6 7 8
Lemma: W (k) ≥ 2k−1
ek2
proof: We need to show that ∃coloring of [n], n = 2k−1
ek2in 2 colors with no monochromatic AP (k).
Color [n] in 2 colors randomly, probability of a given color assigned to a number is 12. Let AS be an
event hat arithmetic progression S of length k is monochromatic, for all such AP (k).1 n
S S1
Prob
(⋂
S−AP (k)
AS
)> 0.
Prob(AS) =(
12
)k · 2 = p, d ≤ nk2(
k · k · n),
so ep(d+ 1) = e · 21−k · (nk2) ≤ 1 if n = 2k−1
ek2.
Thus by LLL Prob
(⋂
S−AP (k)
AS
)> 0. �
Example: (LLL application - Wikipedia)Given: coloring of vertices on C11n in n different colors, 11 vertices of each color.Prove that one can choose n ertices of different colors, no two adjacent.
i
n = 3
For a fixed color, pick a vertex of that color with probability 111. A bad event is that we choose two
adjacent vertices.
Ai = {vi, vi+1 were choosen}, add (mod (11n)). Prob(Ai) =1
11· 1
11=
1
121.
86
}} 22 Adjacent pairs containinga vertex of fixed color.
d ≤ 42
Apply LLL: ep(d+ 1) ≤ e 1|2| · 43 ≤ 1
⇒ Prob(⋂Ai) > 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 sales-person problem.)
Necessary condition for the existence of Hamitonian cycle
If Hamiltonian cycle exists ⇒ ∀S ⊆ V, S 6= ∅, G− S has ≤ |S| components.
u vS = {u, v}G− S has 3 components →G is not Hamiltonian
Theorem (Dirac 1952): Every graph with n ≥ 3 vertices and δ(G) ≥ n2contains a Hamiltonian
cycle.
Note: can not do better Kn2
Kn2
δ = n2 − 1
proof: Note that G is connected, otherwise there is a component on ≤ n2vertices with all vertices
in thos component of degree ≤ n2− 1. Let P be the longest path in G, P = x0, . . . , xk.
x0 xkx4 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 .
Thusxkxi xi−1
v , then x0 . . . xixkxk−1 . . . xi+1x0 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
x0
x1
xm−1v
x2x3
x11x13
I = {0, 2, 11, 13}
y Menger’s theorem, there are ≥ k (V,C), internally disjoint paths. Let I = {i : xi-endpoint of oneof there paths}.Observe that if i ∈ I, i + 1 6∈ I (addition mod m), otherwise there is a longer cyle if i, j ∈ I, thenxi+1 6∼ xj+1, otherwise, there is a longer cycle.
v
xi
xi+1
xixi+1
xj
xj+1
v
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 aHamiltonian cycle).
Theorem (Fleischner 1974): If G is 2-connected, then G2 is Hamiltonian, where G2 =(V,E ′), G = (V,E), E ′ = {{u, v} : u, v ∈ V : distG(u, v) ≤ 2}.
G
Theorem (Chratal 1972): Let (a1, . . . , an), 0 ≤ ai ≤ . . . ≤ an < n an integer sequence, n ≥ 3,
∀i < n
2(ai ≤ i⇒ an−i ≥ n− i) ⇒ any graph with degree sequence a1, . . . , an is Hamiltonian.
(⇐ is true too)
89
Line graphs
1 23
4H
1
2
3
4L(H)
1
2
3
4
12
34
L(H1) =? there is no H1not a line graph
Theorem (Beineke 1968): A simple graph is a line graph of another simple graph ⇔ it hasnone of F as induced subgraph
Last
time:Hamiltonicity:
δ(G) ≥ n2, n ≥ 3 −→
DiracG-Hamiltonian −→ ∀S ⊆ V (G), S 6= ∅,
G− Shas at most |S| components.K(G) ≥ α(G), n ≥ 3 −→ G-HamiltonianK(G) ≥ 4 and G planar −→
TutteG-Hamiltonian
G = H2, where k(H) ≥ 2 −→ G-HamiltonianG has degree sequence d1 ≤ d2 ≤ . . . ,
di + dn− i ≥ n ∀i < n2
−→chratal
G-Hamiltonian
WQO (structural graph theory)
Quasi-ordering - relation that is reflexive and transitive. We say a set X is well-quasi-ordered byquasi-ordering ≤ if for any infinite sequence x1, x2, . . . , xi ∈ X, ∃i < j, xi ≤ xj.
X � Y for graphs X and Y if Y = MX, i.e. X can be obtained from Y by edge-deletions, edge-contractions and vertex-deletions.
90
X Y X � Y
Graph minor theorem (Neil Robertson, Paul Seymour 1990): „�“-minor containment isa 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 : MX = 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)
Example: PLanar graph, graphs on surfaces.
Corollary (Graph minor theorem): If F-minor-closed family, then ∃finite set of graphsH1, . . . , 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 mapf : A→ B such that a ≤ f(a) ∀a ∈ A. Then ≤ is a well-quasi-ordering of X<ω - finite subsets ofX.
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 thereis a map ϕ : V (T )→ V (T1) such that x ≤
(T,r)y ⇒ ϕ(x) ≤
(T1,r1)ϕ(y).
r
T T1
r1
ϕ(r)
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 ′, r′)satisfy (T, r) ≤∗ (T ′, r′). We say a sequence is „good“ otherwise. Fix n ∈ N, then there are treesT0, T1, T2, . . . , Tn−1 such that some bad sequence starts with these. Choose min order tree Tn, suchthat some bad sequence starts with T0, T1, . . . , Tn. Let root of Ti be ri, i = 0, . . .. Let An be a set ofrooted trees obtained as comploments of Tn − {rn}.
91
rn
Tn An =, ,
Claim: A =∞⋃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 smallestindex such that T k ∈ An(k). Then T0, T1, . . . , Tn(k)−1, T
k, T k+1, . . . is a good sequence sin-ce T0, T1, . . . , Tn(k)−1, Tn(k), . . . is bad with smallest Tn(k) and T k is smaller than Tn(k) sin-ce T k ∈ An(k). Let (T, T ′) be a good pair, T ≤ 1 ∗ T ′ in T0, T1, . . . , Tn(k)−1, T
k, T k+1, . . ..If T, T ′ ∈ {T k, T k+1, . . .}, done, so (T, T ′) is a good pair in (T k)k=1,.... Thus, assume thatT ∈ {T0, . . . , Tn(k)−1}. So, T ′ 6∈ {T0, . . . , Tn(k)−1}. So, T ≤∗ T ′ = 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
A2
A0 A = A0 ∪ A1 ∪ A2 . . .
By lemma [A]<ω is well-quasi-ordered, so A0⊆A, A1⊆A, A2, . . . contains a good pair (Ai, Aj), i.e., ∃injective
f : Ai → Aj ∀T ∈ Ai T ≤∗ f(T ).
Ai
Ti Tj
Aj
⇒ Ti ≤∗ Tj. �
92
![Automatic Structural Synthesis of Planar Mechanisms and ... · 1.2.2 Graph theory based methods Freudenstein [27] and Crossley [28] employed the graphs in graph theory [29] to represent](https://static.fdokument.com/doc/165x107/5f5339e685676e6bab74e680/automatic-structural-synthesis-of-planar-mechanisms-and-122-graph-theory-based.jpg)
