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2301-670 Graph theory 1.1 What is a graph? 1st semester 2550
1.1. What is a graph?

1.1.2. Definition. A graph G is a triple (V(G), E(G), ψG) consisting of V(G) of vertices, a set E(G),
disjoint from V(G), of edges, and an incidence function ψG that associates with each edge of G an
unordered pair of (not necessarily distinct) vertices of G.
If e is an edge and u and v are vertices such that ψG(e) = {u, v}(or is simply denoted by uv), then e
is said to join u and v; the vertices u and v are called endpoints of e.
Example. G = (V(G), E(G), ψG) where V(G) = {v1, v2, v3, v4, v5}, E(G) = {e1, e2, e3, e4, e5, e6, e7, e8}
and ψG is defined by ψG(e1) = v1v2, ψG(e2) = v2v3, ψG(e3) = v3v3, ψG(e4) = v3v4, ψG(e5) = v2v4,
ψG(e6) = v4v5, ψG(e7) = v2v5, ψG(e8) = v2v5. H = (V(H), E(H), ψH) where V(H) = {1, 2, 3, 4, 5}, E(H) = {a, b, c, d, e, f, g, h, k} and ψH is defined by ψH(a) = 12, ψH(b) = 15, ψH(c) = 13, ψH(d) = 34, ψH(e) = 24, ψH(f) = 23, Graphs are so named because they can be represented graphically, and it is this graphical representation which helps us understand many of their properties. Each vertex is indicated by a point, and each edge by a line joining the points which represent its ends. In such a drawing it is understood that no line intersects itself or passes through a point representing a vertex which is not an end of the corresponding edge, that is clearly always possible. The diagram itself is then referred to as a graph. Diagram of G and H are shown as follows:
1.1.4. Definition. A loop is an edge whose endpoints are equal.
Multiple edges are edges having the same pair of endpoints.
In the above diagram, e3 is a loop, e7 and e8 are multiple edges.
A simple graph is a graph having no loops or multiple edges, i.e. a simple graph G consists of a
vertex set V(G), an edge set E(G) where E(G) is a set of unordered pairs of vertices or a set of
2-elements subsets of V(G). In the above diagram, H is a simple graph.
When u and v are endpoints of an edge, they are adjacent and are neighbors. In the above diagram,
v3 and v4 are adjacent in G, v1 and v3 are not adjacent in G, 1 and 5 are neighbors in H.
A graph is finite if its vertex set and edge set are finite. We call a graph with just one vertex trivial
and all other graphs nontrivial.
1.1.6. Remark. The null graph is the graph whose vertex set and edge set are empty.
We emphasize finite simple graphs with a nonempty set of vertices.
Example. Consider the set S = {2, 3, 5, 8, 13, 21}. There are some pairs of distinct integers
belonging to S whose sum or difference(in absolute value) also belongs to S, namely, {2, 3}, {2, 5},
{3, 5}, {3, 8}, {5, 8}, {8, 13}, {8, 21}, and {13, 21}.
There is a more visual way of identifying these pairs, namely, by the graph G of the following
figure. In this case, V(G) = {2, 3, 5, 8, 13, 21}and E(G) = {{2, 3}, {2, 5}, {3, 5}, {3, 8}, {5, 8},
2301-670 Graph theory 1.1 What is a graph? 1st semester 2550 1.1.8. Definition. The complement G of a simple graph G is the simple graph with vertex set V(G)
defined by uv ∈ E( G ) if and only if uv ∉ E(G).
A clique in a graph is a set of pairwise adjacent vertices.
An independent set in a graph is a set of pairwise nonadjacent vertices.

Example
{a, b, c}is a clique in G, {a, i}is an independent set in G. {a, b, c}is an independent set in G , {i, d, b} is a clique in G . 1.1.10. Definition. A graph G is bipartite if V(G) is the union of two disjoint (possibly empty)
independent sets called partite sets of G.
A graph G is k-partite if V(G) is the union of k (possibly empty) independent sets.
Exercise 1.1.13. Let G be the graph whose vertex set is the set of k-tuples with coordinates in
{0,1}, with x adjacent to y when x and y differ in exactly one position.
Determine whether G is bipartite.
Solution. For example, k = 3, V(G) = {000, 001, 010, 011, 100, 101, 110, 111},
E(G) = {{000, 001}, {000, 010}, {000, 100}, {001, 011}, {001, 101}, {010, 011}, {010, 110},
{011, 111}, {100, 101}, {100, 110}{101, 111}, {110, 111}}.
Let X = {001, 010, 100, 111}, Y = {000, 011, 101, 110} be partite sets of G.
Then, adjacent vertices differ in exactly one position, no edges in X or Y, and G is a bipartite graph. In general, let X be the set of k-tuples with odd numbers of 1’s and Y be be the set of k-tuples with even numbers of 1’s. Then, adjacent vertices have opposite parity, no edges in X or Y and G is a bipartite graph.
1.1.12. Definition. The chromatic number of a graph G, written χ(G), is the minimum number of
colors needed to label the vertices so that adjacent vertices receive different colors.
Example.
2301-670 Graph theory 1.1 What is a graph? 1st semester 2550
1.1.15. Definition. A path is a simple graph whose vertices can be ordered so that two vertices are
adjacent if and only if they are consecutive in the list.
More formally, a path Pn (a path of n vertices) is a simple graph G with
V(G) = {v1, v2,…,vn} and E(G) = {v1v2, v2v3, …, vn-1vn}.
A cycle is a graph with an equal number of vertices and edges whose vertices can be placed around
a circle so that two vertices are adjacent if and only if they appear consecutively along the circle.
More formally, a cycle Cn (a cycle of n vertices) is a simple graph G with
V(G) = {v1, v2,…,vn} and E(G) = {v1v2, v2v3, …, vn-1vn, vnv1}.
1.1.16. Definition. A subgraph of a graph G is a graph H such that V(H) ⊆ V(G) and E(H) ⊆ E(G)
and the assignment of endpoint to edges in H is the same as in G.
We write H ⊆ G and say that “G contains H”.
A path in a graph G is a subgraph of G that is a path.
A graph G is connected if each pair of vertices in G belongs to a path;
otherwise, G is disconnected.

Example.
G is connected, while H is disconnected.
Exercise1.1.10. Prove or disprove: The complement of a simple disconnected graph G must be
connected.
Proof. Since G is disconnected, there exist 2 vertices x, y that do not belong to a path.
Thus, xy ∈ E( G ). Also x and y have no common neighbor in G, otherwise, that would yield a path connecting them. Every vertex not in {x, y} is adjacent in G to at least one of {x, y}. Hence every vertex can reach every other vertex in G using paths through {x, y}.
1.1.17. Definition. Let G be a loopless graph with vertex set V(G) = {v1, v2,…,vn}and edge set
E(G) = {e1, e2,…,em}.
The adjacency matrix of G, written A(G), is the n-by-n matrix in which entry aij is the number of
edges in G with endpoints {vi, vj}.
The incidence matrix of G, written M(G), is the n-by-m matrix in which entry mij is 1 if vi is an
endpoint of ej and otherwise is 0.
If vertex v is an endpoint of edge e, then v and e are incident.
The degree of vertex v(in a loopless graph), written d(v) is the number of incident edges
2301-670 Graph theory 1.1 What is a graph? 1st semester 2550 Example.
d e f A(G) = 1⎢ 0 1 0 0⎥ , M(G) = ⎢ d(u) = 3, d(v) = 4, d(w) = 2, d(x), d(y) = 1.
Exercise1.1.5. Prove or disprove: If every vertex of a simple graph G has degree 2, then G is a
cycle
Disproof: Such a graph G can be a disconnected graph with each component a cycle.

1.1.20. Definition. An isomorphism from a simple graph G to a simple graph H is a bijection f :
V(G) → V(H) such that uv ∈E(G) if and only if f(u)f(v) ∈E(H).
We say “G is isomorphic to H”, denoted by G ≅ H, if there is an isomorphism from G to H.
Remark. G ≅ H ↔ G ≅ H .
Proof. (→) Assume that G ≅ H. Let f be an isomorphism from V(G) to V(H). Then every two
adjacent vertices of G are mapped to adjacent vertices of H, also every two nonadjacent vertices of
G are mapped to nonadjacent vertices of H.
Since V( G ) = V(G) and V( H ) = V(H), the same function f : V( G ) →V( H ) also maps adjacent
vertices of G to adjacent vertices of H and nonadjacent vertices of G to nonadjacent vertices of H .
Therefore, G ≅ H .

Let G be any set of simple graphs. ≅ = {(G, H)∈ G x G : G is isomorphic to H}is a relation on G.
1.1.24. Proposition. The isomorphism relation (≅) is an equivalence relation on G.
Proof: Reflexive property. The identity permutation on V(G) is an isomorphism from G to itself.
Thus G ≅ G.
Symmetric property. If f : V(G) → V(H) is an isomorphism from G to H, then f-1 is an
isomorphism from H to G, because “uv ∈E(G) if and only if f(u)f(v) ∈E(H)” yields xy ∈E(H)
if and only if f-1(x)f-1(y) ∈E(G). Thus G ≅ H implies H ≅ G.
Transitive property. Suppose that f : V(F) → V(G) is an isomorphism from F to G and
g : V(G) → V(H) is an isomorphism from G to H.
We are given “uv ∈ E(F) if and only if f(u)f(v) ∈ E(G)” and “xy ∈ E(G) if and only if
g(x)g(y) ∈ E(H)”.
Since f is an isomorphism, for every xy ∈ E(G) we can find uv ∈ E(F) such that f(u) = x and
f(v) = y. This yields uv ∈ E(F) if and only if g(f(u))g(f(v)) ∈ E(H).
Thus the composition gÎf is an isomorphism from F to H.
We have prove that F ≅ G and G ≅ H together imply F ≅ H.

1.1.25. Definition. An isomorphic class of graphs is an equivalence class of graphs under the
isomorphism relation.
1.1.27. Definition. A complete graph is a simple graph whose vertices are pairwise adjacent; the
unlabeled complete graph with n vertices is denoted Kn.
A complete bipartite graph(biclique) is a simple bipartite graph such that two vertices are
adjacent if and only if they are in different partite sets.
When the sets have sizes r and s, the unlabeled complete bipartite graph is denoted Kr,s.
So the complete bipartite graph Km,n is a complete graph if and only if m = n = 1, i.e. K1,1 ≅ K2.
2301-670 Graph theory 1.1 What is a graph? 1st semester 2550 1.1.29. Remark. When we name a graph without naming its vertices, we often mean its isomorphic
class. H is a subgraph of G means that some subgraph of G is isomorphic to H and we say G
contains a copy of H.
1.1.30. Example.
Each graph has 6 vertices and 9 edges and is connected, but these graphs are not pairwise isomorphic. To prove that G1 ≅ G2, let f : V(G1) → V(G2) defined by f(u) = 1, f(v) = 3, f(w) = 5, f(x) = 2, f(y) = 4, f(z) = 6. Both G1 and G2 are bipartite. they are drawings of Κ3,3 as is G4. The graph G3 contains Κ3, so its vertices cannot be partitioned into 2 independent sets. Thus G3 is not isomorphic to the others. Sometimes we can test isomorphism quickly using the complements. Simple graphs G and H are isomorphic if and only if their complements are isomorphic. Hence G , G , G all consist of 2 disjoint 3-cycles and are not connected,
1.1.31. Example. When choosing 2 vertices from a set of size n, we can pick one and then the other
but don’t care about the order, the number of ways is ⎜ ⎟ . In a simple graph with n vertices, each vertex pair may form an edge or may not. Making the choice for each pair specifies the graph, so the number of n-vertex simple graphs is 2 . For example, there are 64 simple graphs on a fixed set of 4 vertices. These graphs form only 11 isomorphism classes. •
1.1.32. Definition. A graph is self-complementary if it is isomorphic to its complement.
A decomposition of a graph is a list of subgraphs such that each edge appears in exactly one
subgraph in the list.
Exercise1.1.6. Determine whether the graph below decomposes into copies of P4.
Solution. This graph decomposes into 3 copies of P4 as shown on the right.
2301-670 Graph theory 1.1 What is a graph? 1st semester 2550 1.1.33. Example. We can decompose Κ5 into 5-cycles, and thus the 5-cycle is self-complementary.
Any n-vertex graph and its complement decompose Κn. Also Κ1,n-1 and Κn-1 decompose Κn, even though one of these subgraphs omits a vertex. Below we show a decomposition of Κ4 using 3 copies of Ρ3. 1.1.34. Example. The question of which complete graphs decompose into copies of Κ3 is a
fundamental question in the theory of combinatorial designs.
On the left below we suggest a decomposition of Κ7 into copies of Κ3.
Rotating the triangle through 7 positions uses each edge exactly once.
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+
On the right we suggest a decomposition of Κ6 into copies of Ρ4. Placing one vertex in the center groups the edges into 3 types: the outer 5-cycle, the inner(crossing) 5-cycle on those vertices, and the edges involving the central vertex. Each 4-vertex path in the decomposition uses one edge of each type; we rotate the picture to get the next
Exercise1.1.7. Prove that a graph with more than 6 vertices of odd degree can not be decomposed
into 3 paths.
Proof. Since every vertex of odd degree must be the endpoint of some path in
a decomposition into paths and 3 paths need only 6 endpoints.

Exercise1.1.36. Prove that if Kn decomposes into triangles, then 6|(n-1) or 6|(n-3).
Proof. A decomposition of Kn into triangles requires the degree of each vertex is even and the
number of edges is divisible by 3. To have even degree, n must be odd.
If 3|n and n is odd, then 6|(n-3). If 3|(n-1) and n is odd, then 6|(n-1).
1.1.35. Example. The Graph Menagerie.
2301-670 Graph theory 1.1 What is a graph? 1st semester 2550 1.1.36. Definition. The Petersen graph is the simple graph whose vertices are the 2-element
subsets of a 5-element set and whose edges are the pairs of disjoint 2-element subsets.
1.1.37. Example. Structure of the Petersen graph.
Using [5] = {1, 2, 3, 4, 5}as our 5-element set, we write the pair {a, b}as ab or ba.
Since 12 and 34 are disjoint, they are adjacent vertices when we form the graph, but 12 and 23 are
not. For 2-set ab, there are 3 ways to pick a 2-set from the remaining 3 elements of [5], so every
vertex has degree 3.
The Petersen graph consists of 2 disjoint 5-cycles plus edges that pair up vertices on the two 5-
cycles.
The disjointness definition tells us that 12, 34, 25, 14, 35 in order are the vertices of a 5-cycle, and
similarly this holds for the remaining vertices 13, 24, 15, 23, 45.
Also 24 is adjacent to 35, and 15 is adjacent to 34, and so on.

1.1.38. Proposition. If 2 vertices are nonadjacent in the Petersen graph, then they have exactly 1
common neighbor.
Proof: Nonadjacent vertices are 2-sets sharing 1 element; their union S has size 3. A vertex adjacent
to both is a 2-set disjoint from both. Since the 2-sets are chosen from {1, 2, 3, 4, 5}, there is exactly
one

1.1.39. Definition. The girth of a graph with a cycle is the length of its shortest cycle.
A graph with no cycle has infinite girth.
1.1.40. Corollary. The Petersen graph has girth 5.
Proof: The graph is simple, so it has no 1-cycle or 2-cycle.
A 3-cycle would require 3 pairwise-disjoint 2-sets, which can’t occur among 5 elements.
A 4-cycle in the absence of 3-cycles would require nonadjacent vertices with 2 common neighbors,
which Proposition 1.1.38 forbids.
The vertices 12, 34, 25, 14, 35 yields a 5-cycle, so the girth of the Petersen graph is 5.

Exercise1.1.26. Let G be a graph with girth 4 in which every vertex has degree k.
Prove that G has at least 2k vertices. Determine all such graphs with exactly 2k vertices.
Proof. Since G has girth 4, thus G is simple and there are at least 4 edges in G, choose xy ∈ E(G)
then x, y has no common neighbors(why?). Thus, the neighborhoods N(x) and N(y) are disjoint sets
of size k, G must have at least 2 k vertices.
Kk,k is a k-regular graph with girth 4 and has exactly 2k vertices (why?)

1.1.41. Definition. An automorphism of a graph G is an isomorphism from G to G. A graph G is
vertex transitive if for every pair u, v ∈V(G) there is an automorphism that maps u to v.
1.1.42. Example. Let G be the P4 with vertex set {1, 2, 3, 4} and edge set {12, 23, 34}.
This graph has 2 automorphisms α1, α2 as follows:
1 : V(G) → V(G) defined by α1(v) = v for every vertex v of G. α2 : V(G) → V(G) defined by α2(1) = 4, α2(2) = 3, α2(3) = 2, α2(4) = 1. The function α3 : V(G)→V(G) defined by α3(1) = 2, α2(2) = 1, α2(3) = 3, α2(4) = 4 is not an automorphism of G, although G is isomorphic to the graph with vertex set {1,2,3,4} and edge 2301-670 Graph theory 1.1 What is a graph? 1st semester 2550 Example. Consider the following graph G :
There are 4 automorphisms α1, α2, α3, α4 of G as follows: α1 : V(G) → V(G) defined by α1(v) = v for every vertex v of G. 2 : V(G) → V(G) defined by α2(v) = ⎨v 3 : V(G) → V(G) defined by α3(v) = ⎨v 4 : V(G) → V(G) defined by α4(v) = ⎨v Remark. Since composition of functions is associative, the identity function is an automorphism,
the inverse of an automorphism is an automorphism., and the composition of 2 automorphisms is
an automorphism, it follows that the set of all automorphisms of a graph G from a group under the
operation of composition.
This group is denoted by Aut(G) and is called the automorphism group of G.
For the graph G above, Aut(G) = {α1, α2, α3, α4}.

Exercise 1.1.40. Count the automorphisms of Pn, Cn, and Kn.
The number of automorphisms of Pn is 2 since Pn can be left alone or flipped.
The number of automorphisms of Cn is 2n since Cn can be rotated or flipped.
The number of automorphisms of Kn is n! since Kn can be permuted arbitrarily.

Exercise 1.1.41. Construct a simple graph with 6 vertices that has only one automorphism.
Construct a simple graph that has only 3 automorphisms. Homework 1. 1.1.25, 1.1.34, 1.1.35, 1.1.38, 1.1.41 due on June 18.

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