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Table of content
1.
2.
Graph...................................................................................................................... 2
Definition 1.1 : of Graph........................................................................................ 2
Definition 1.2 : Loop .............................................................................................. 5
Definition 1.3 : Isomorphic graph ......................................................................... 6
Definition 1.4 : Degree of vertex........................................................................... 7
Theorem 1 : Handshaking Lemma for vertices ..................................................... 8
Definition1.5 : Path ............................................................................................... 9
Definition1.6 : Cycle .............................................................................................. 9
Definition 1.7 : Connected graph ........................................................................ 10
Planar graph ......................................................................................................... 11
Face ...................................................................................................................... 12
Definition 2.2 Degree of face .............................................................................. 13
Theorem 2.1: Handshaking Lemma for faces ..................................................... 13
Theorem 2.2: Euler’s formula of graph theory ................................................... 13
Corollary of Euler’s formula ................................................................................ 14
Theorem 2.3: Kuratowski’s Theorem. ................................................................. 16
3.Platonic Solid ............................................................................................................. 17
Theorem 3.1 : There are only 5 platonic solids ............................................... 18
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1. Graph
1.1 Definition of Graph
A Graph G is an order pair of G= (V, E), where V is a set of elements
called vertices and E is a set of 2-element subset of V called edges.
Undirected graph
The elements in E are not ordered pairs in, i.e. {A,B} = {B,A} in E
for any A, B ∈ 𝑉
Königsberg bridge problem:
Is it possible to cross every bridge exactly once for a walk?
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Does there exist a path that traverses every edge once and only once?
V={A,B,C,D} , E={{A,B},{A,C},{A,D},{B,C},{B,D},{C,D}}
Both graph above are the same even the edge intercept differently, since the connections between
vertices are the same.
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Directed graph / digraph
The elements in E are not ordered pairs in, i.e. {A,B} ≠ {B,A} in E for
any A, B ∈ 𝑉
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Problem: A thief wants to break into a room, but he does not have the password. The keypad
has only 0 and 1 digits. If the length of the password is 3 and the room will open as long as a
correct sequence is input e.g. if the password is 100, when you input 0100, the room will also
open. What is the minimum number of digits that required trying all the combinations of the
passwords?
In the figure, each vertex is a two-digit number. The label on the edge is the digit you type.
Every vertex has only two outward direction arrows which are also called outdegrees since
you can type either 1 or 0. If the right-most digit of a vertex 𝑉1 is the same as the left-most
digit of a vertex 𝑉2 . Then an ordered pair {𝑉1 , 𝑉2 } is an edge. If we start with 00, then go
through every edge once, the result consists all the combinations of length 3 digits.
E.g.
Start at “00”
Accumulated digit typed: 00
Go from “00” to “01”
Accumulated digit typed: 00 1
Go from “01” to “11”
Accumulated digit typed: 0011
Go from “11” to “11”
Accumulated digit typed: 00111
Go from “11” to “10”
Accumulated digit typed: 001110
Go from “10” to “01”
Accumulated digit typed: 0011101
Go from “01” to “10”
Accumulated digit typed: 00111010
Go from “10” to “00”
Accumulated digit typed: 001110100
Go from “00” to “00”
Accumulated digit typed: 0011101000
Other related questions:
The wolf, cabbage, goat and farmer cross the river. There is only one boat and the
boat capacity is two. If the wolf is left with the goat, goat will be eaten, if the
cabbage is left with the goat, the cabbage will be eaten. How should the farmer cross
the river so that all things are safe? (State graph: Vertices representation different
states on one chosen side.)
Definition 1.2 : Loop
Let E be a set of edges for some Graphs.
For any {A, B} ∈ 𝐸, 𝑖𝑓 𝐴 = 𝐵 then {A, B} is called a loop
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The edge {A,A} is a loop
A simple digraph
Not a simple graph since more than one edges
Vertex A has only one edge outward and one
between vertices A and B. This is also called
inward.
multigraph.
Definition 1.3 : Isomorphic graph
Two graphics 𝐺1 = (𝑉1 , 𝐸1 ), G2 (𝑉2 , 𝐸2 ) having the same structure
which is isomorphic if there exists a bijective function f: 𝑉1 → 𝑉2 ,
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{x, y} ∈ E1 ⟺ {f(x), f(y)} ∈ E2
Adjacent
Two vertices 𝑣1 , 𝑣2 are adjacent if {𝑣1 , 𝑣2 } ∈ 𝐸 .
For direction graph, if either {𝑣1 , 𝑣2 } ∈ 𝐸 𝑜𝑟 {𝑣2 , 𝑣1 } ∈ 𝐸 then
𝑣1 , 𝑣2 are adjacent.
Example:
A and B are adjacent. B and C are adjacent. B and D are adjacent.
But A and C are not adjacent though there is a “path” from A to C (A->B->C).
Definition 1.4 : Degree of vertex
The degree of a vertex 𝑣1 is the number of edge ends at 𝑣1 .
For digraph, there are indegree and outdegree, which are number of
edges that start and end at the vertex respectively.
Remark:
1. A vertex that has degree 0 is said to be isolated.
2. In this note the degree of vertex v is denoted by deg(v)
3. For a loop, the edge on a vertex has to be counted twice.
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Degree of A=3, outdegree of A=2, indegree of A=1
Example:
In the graph, the vertices are names of some children. There are edges between two
children if they are friends. Which children have the least number of friends? How
about the most?
Since Peter has the least edge, he has the least friends. And Ben, Tom and Chris have
the highest degree, and they have the most friends.
Theorem 1: Handshaking Lemma for vertices
Let G=(V,E) be a graph with 𝑉 = (𝑣1 , 𝑣2 , … , 𝑣𝑛 ), then
n
2|E| = ∑ deg(𝑣𝑖 )
i=1
Proof:
For each edge, it is incident to two vertices, so the sum all the degree of vertices will
double count the each edge twice. For loops, the degree will add two to the same
vertex so the equation still holds.
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Example:
In a party, assume everyone only shake hands with the same person once. If the
number of handshake for each person is an odd number, then let the number of
handshake be (2ai + 1) for i = 1,2, … , n , where n is number of people in the party.
Consider the people as a set of vertices and if two people have handshake, then they
have edge connected. Then by the handshaking lemma, we have
n
2|E| = ∑ deg(𝑣𝑖 ) = 2(𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 ) + 𝑛
i=1
Therefore, the number of people must be even.
Try it:
What can you say about the number of vertices that have odd degree?
1.5 Definition of Path
A Path P in a graph is a sequence of vertices
(𝑣0 , 𝑣1 , 𝑣2 , … , 𝑣𝑛 )
such that {𝑣𝑖 , 𝑣𝑖+1 } ∈ 𝐸 for i = 0,1,2, … , n − 1
A path that pass through all edges and vertices
Since {𝐴, 𝐶} ∉ 𝐸 , this is not a path
Path (A,B,E,D,C)
Red lines are paths, black lines are edges
Definition 1.6 Cycle
For a path (𝑣0 , 𝑣1 , … , 𝑣𝑛 ), if 𝑣0 = 𝑣𝑛 , then the path is called a cycle.
If there is no repeated edges and vertices in a path (except v1 and vn ),
then it is called a simple cycle or circuit
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The path (A,B,C,D,E,F,B,A) is a cycle
The path (B,C,D,E,F,B) is a cycle
1.7 Definition of connected graph
A graph 𝐺 = (𝑉, 𝐸) is connected
if ∀𝑣1 , 𝑣2 ∈ 𝑉 𝑎𝑛𝑑 𝑣1 ≠ 𝑣2 , 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑎𝑛 𝑒𝑑𝑔𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑣1 𝑎𝑛𝑑 𝑣2
A digraph is strongly connected if it is connected and there exits edges in
both direction for 𝑣1 𝑎𝑛𝑑 𝑣2
A digraph is weakly connected if it is connected but not strongly connected.
So for a connected graph, for any vertices 𝑣1 , 𝑣2 , there exists a path starts from 𝑣1
and ends at 𝑣2 .
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A connected graph
A disconnected graph: no path from
vertices A to D
Example:
Let 𝐺 = (𝑉, 𝐸)be a simple graph.
𝑉 = {𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓}, 𝐸 = {{𝑎, 𝑏}, {𝑏, 𝑐}, {𝑐, 𝑒}, {𝑒, 𝑓}}
Is G connected?
Since there is no edge in E on d, so there is no any path from vertices a to d.
Therefore, the graph is disconnected.
2.
Planar graph
Definition 2.1 :Planar graph is a graph that can be drawn on a plane
such that edges do not intercept with others.
Examples of planar graph
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Example:
Is the following a planar graph?
Yes. Since for this graph, we can redraw such that no edges intercepting, while the
graph representing the same vertices V and edges E.
Face
The region that is separated by the edges of a planar graph is called face.
The bounded face is called interior face. And the unbounded face is also
counted and called exterior face.
Only 1 face which is
There are two faces, one
Also two faces, the edge between vertices C
exterior
exterior (face2) and one
& E does not form a new face
interior (face1).
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One exterior face only.
Definition 2.2 Degree of face
The degree of face is the number of edges across in a boundary walk.
The degree of f1 = length of “AF-FH-HI-IO-ON-NO-OI-IB-BA” (red line), which is 9.
The degree of f2 = length of “AF-FH-HI-IM-MI-IB-BA” (blue line), which is 7.
Theorem 2.1: Handshaking Lemma for faces
Let G=(V,E) be a graph, denote all the faces by 𝑓1 , 𝑓2 , … , 𝑓𝑛 , then
2|E| = ∑ni=1 deg(𝑓𝑖 )
Proof:
For each edge, it is contact to exactly two faces. (For edge that contact to one face, it
will contribute two to the degree of face.) Therefore, the sum of degree of faces will
be the double of the number of edges.
Theorem 2.2: Euler’s formula of graph theory
For a connected planar graph G, we have
v−e+f=2
where v is the number of vertex, e is number of edge and f is number of face.
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Proof:
This will be proved by induction on the number of edges.
Since when adding a loop to the graph, it will increase the number of edges and face
each by 1. Then the formula v – (e+1) + (f+1) = v − e + f = 2 still holds. Therefore we
can assume there are no loops.
When the number of face is 1, it is an exterior face. For a connected planar graph
G= (V,E) , since it is connected, there is no simple cycle, and we have v vertices and
v-1 edges for some v ∈ ℕ. Therefore v – e + f = 1 + 1 = 2.
Assume the formula is true for some positive integer k.
For any graph that has k+1 faces, since number of face= k+1 ≥ 2, there is a simple
cycle. Then we can remove an edge that connects two faces to form a new graph G’.
The number of edges (let it be e’) and faces (let it be f’) of G’ will both decrease by
one. Since G’ has k face, by induction hypothesis, i.e. v – e’ + f’=2, G’ satisfied the
formula. Since e’=e-1 and f’=f-1, we have v - ( e – 1 ) + ( f – 1 ) = 2. Hence, v – e + f =2.
Corollary of Euler’s formula
Corollary 1: Let G be a connected planar graph, with v vertices, e edges and f faces. If
𝑣 ≥ 3 , then 𝑒 ≤ 3𝑣 − 6.
Proof:
Since 𝑣 ≥ 3, each face has at least degree 3. Then sum of degree of faces is at least
3f. So by the handshaking lemma, we have 2𝑒 ≥ 3𝑓. And from Euler’s formula, we
have 𝑓 = 𝑒 − 𝑣 + 2 , which implies 2𝑒 ≥ 3𝑒 − 3𝑣 + 6 ⇒ 𝑒 ≤ 3𝑣 − 6.
Corollary 2: Let G be a planar graph, then there exists a vertex with degree less than
6.
Proof:
For a connected planar graph with v vertices, e edges and f faces. If v < 3, it is
obviously true. So assume 𝑣 ≥ 3. Then by Corollary 1, we have 𝑒 ≤ 3𝑣 − 6, which
implies 2𝑒 ≤ 6𝑣 − 6. If all vertices have degree 6 or above, then by handshaking
lemma, we have 2𝑒 ≥ 6𝑣 together with Corollary 1 yields a contradiction.
Therefore, it is impossible for all vertices having degree 6 or above.
For disconnected graph, we can consider each subgraph that is connected. Therefore,
we can also find a vertex with degree less than 6.
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Corollary 3: Let G be a connected planar graph, with v vertices, e edges and f faces. If
𝑣 ≥ 3 and all simple cycles with length ≥ 4, then 𝑒 ≤ 2𝑣 − 4.
Proof:
Since each simple cycle has at least length 4, the degree of each face is also at least 4.
Then by handshaking lemma for faces, we have 2𝑒 ≥ 4𝑓. By Euler’s formula, we
have f = e – v +2. So 2𝑒 ≥ 4𝑒 − 4𝑣 + 8, then 𝑒 ≤ 2𝑣 − 4.
Corollary 4: Let G be a connected planar graph, with v vertices, e edges and f faces. If
𝑣 ≥ 3 and all simple cycles with length ≥ 4, then there exists a vertex with degree
less than 4.
Example 1:
K 5 is not a planar graph. K n means that there are n vertices and there are edges
between all distinct vertices. For K 5 , there are 5 vertices and 1+2+3+4=10 edges.
And it is a connected graph with vertices more than 3. But 3𝑣 − 6 = 9 < 10 = 𝑒, so
Corollary 1 implies K 5 is not a planar graph.
Example 2:
There are three household and they need water supply, electricity supply and tower
gas supply. Assume for each supply (water, electricity and tower gas), there are only
one supply plant. If you would like to use a graph to represent the connection, is it
possible that there is no edges intersect?
This is equivalent to determine whether the graph is a planar graph.
Denote the three households as A = {v1,v2,v3} and the three plants as B ={v4,v5,v6}.
There is an edge between two vertices only if the vertices are in different sets of A, B.
It is obvious that it is connected graph. Besides, for any simple cycle starting from 𝑣𝑖
(𝑣𝑖 , 𝑣𝑎1 , … , 𝑣𝑎𝑘 , 𝑣𝑖 ), assume 𝑣𝑖 ∈ 𝐴 then 𝑣𝑎1 ∈ 𝐵, then 𝑣𝑎2 ∈ 𝐴, since no repeated
edge for a simple cycle, , so 𝑣𝑎2 ≠ 𝑣𝑖 . Similarily, 𝑣𝑖 ≠ 𝑣𝑎3 ∈ 𝐵 . And so, the simple
cycle has at least length four. If the graph is planar, then by corollary 2, we have 𝑒 ≤
2𝑣 − 4. But in this graph there are 6 vertices and 3x3=9 edges, i.e. 2𝑣 − 4 = 8 <
9 = 𝑒. Therefore, this graph is not planar and cannot be drawn without edges
intersect.
Remark:
1. In example 2, it is a graph called K 3,3 which vertex set V has two subsets
(partition) of vertices 𝑉1 𝑎𝑛𝑑 𝑉2 , each having 3 elements such that there is an
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edge between two vertices only if the vertices are in different sets of 𝑉1 𝑎𝑛𝑑 𝑉2 .
Theorem 2.3: Kuratowski’s Theorem
A graph is not planar if and only if it contains a subgraph that is a subdivision of
𝐾5 or 𝐾3,3.
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3.Platonic Solid
A platonic solid is a regular convex polyhedron. The faces are regular polygons and
having the same number of edges meeting at every vertex.
Remark:
Convex: For any two points on the surfaces of the polyhedron, the line segment
joining them is inside the polyhedron.
Example:
(from Wikipedia)
(from Wikipedia)
Not convex which is not a platonic solid/not all
Tetrahedron
vertices have the same number of edges
intercepting
Converting a platonic solid onto a planar graph
Consider the solid is made of stretchy material. Then make a hole on one face and
pull the face very large, such that the solid becomes flat. Then it becomes a planar
graph with the exterior face being the original face having the hole.
E.g.
Other insight: Consider there is a sphere surround the solid, then you can project the
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vertices and edges to the sphere which forms a planar graph.
Theorem 3.1 :
There are only 5 platonic solids
Proof:
1. The degree of vertex is constant since every vertex has the same number (say d)
of edges intercepting for a convex polyhedron, then by handshaking lemma
2e = dv
where e is number of edges and v is number of vertices.
2. Since each face are the same polygon, the degree (say s) of faces is constant and
so by handshaking lemma for face, we have
2e = sf
where e is number of edges and f is number of faces.
3. Since it is a planar graph, by Euler’s formula, we have
v−e+f= 2
Then
2e
2e
−e+
=2
d
s
1 1 1 1
+ − =
d s 2 e
This implies if d and s are determined, e is also determined, and hence v and f. So (s,d)
⇒
as an order pair is sufficient to describe a platonic solid.
And for faces, the degree of faces (s) should be at least three, since it is a regular
polygon.
Since the solid is 3D. We must have at least 3 edges (d) intercepting one vertex.
When d=3,
1 1 1
− =
s 6 e
Since e must be positive,
1 1
− >0
s 6
⇒6>𝑠
⇒
So the value of s is at most 5.
Similarly, when d=4, s is at most 3, when d=5, s is also at most 3.
Since the formula is “symmetric” for d and s, so we have the same conclusion.
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Therefore, we have
d (deg(v))
s (deg(f))
v
e
F
3
3
4
6
4
3
4
8
12
6
3
5
20
30
12
4
3
6
12
8
5
3
12
30
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Picture(Wolfram
Mathworld)
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Concave polyhedron (Kepler-Poinsot Solid)
(from Wolfram Alpha)
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