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Discrete Mathematics
Chapter 9
Graphs
By courtesy of Prof. Cheng-Chia Chen
Transparency No. 9-1
Discrete Mathematics
Chapter 9 Graphs
Chapter 8. Graphs
Old subject with many modern applications
Introduced by L. Euler, who used it to solve the famous
Konisberg bridge problem.
Example applications:
determine if a circuit can be implemented on a planer
circuit board (planer graph).
distinguish b/t two chemical compounds with the
same molecular formula but with different structures.
determine if two computers are connected in a
network.
schedule (with task precedence information given,
find an execution plan.)
..
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Discrete Mathematics
Chapter 9 Graphs
Table of contents
Introduction
Graph terminology
Representing graph and graph isomorphism
Connectivity
Euler and Hamilton paths
Shortest path problems
Planer graphs
Graph coloring
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Discrete Mathematics
Chapter 9 Graphs
8.1 Introduction
Graph: discrete structure consisting of vertices and edges
that connects these vertices.
Types of graphs:
Simple graph: G = (V,E) where E is a set of unordered
pairs of distinct elements of V.
Multi-graph: allow multiple edges b/t two distinct vertices.
Pseudo-graph: allow self-loops and multiple edges.
Digraph (directed graph): edges are ordered pairs of
vertices.
Directed multi-graph: allow multiple edges b/t any two
(possibly identical) vertices.
Hybrid graph: allow both directed and undirected edges.
Weighted (or labeled) graph: each edge is associated with
a weight (or a label).
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Discrete Mathematics
Chapter 9 Graphs
Formal definitions
Def 2: (multigraph) G = (V,E,f) where
f: E-> {{u,v} | u,v V and u v} with f(e) containing the two
vertices the edge e is connected to.
If f(e1) = f(e2) ==> e1 and e2 are parallel or multiple edges.
Def 3: A pseudo graph G = (V,E,f) where
f: E -> {{u,v} | u,v V}.
Note: if f(e) = {u} ==> e is a self-loop on u.
Note: If there is no worry of confusion, we usually use {u,v} to
identify the edge e with f(e) = {u,v}.
Def 4: A digraph G = (V,E) where E is a subset of V2.
Def 5: A directed multigraph G = (V,E,f) where f: E -> V2.
If f(e1) = f(e2) ==> e1 and e2 are parallel or multiple edges.
Note: f(e1) = (u,v) /\ f(e2) = (v,u) /\ u v => e1 and e2 are not
parallel edges (opposite directions).
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Discrete Mathematics
Chapter 9 Graphs
Labeled (or weighted) grapghs
A labeled graph G = (V,E,f,L,l) where
(V,E,f) is a graph of any kind (without labeling)
L is the set of labels, and
l: E->L is the labeling function. if f(e) = w, we say e is an w-
labeled edge or simply a w-edge.
Hyper graph: edges can connect more than 2 vertices:
e.g hyper k-graph G = (V,E,f) where f: E-> Vk.
Summary:
type directed edge? multiedge? loop?
simple no no no
multigraph no yes no
pseudo no yes yes
digraph yes no yes
directed- yes yes yes
-multigraph Transparency No. 9-6
Discrete Mathematics
Chapter 9 Graphs
Graph models
Ex1: Niche overlap graphs in ecology:
G = (V,E) is a simple graph where
V is the set of species.
{u,v} E iff u and v compete for food.
Ex2: V: Persons
(u,v) E iff u can influence v. => Influence graph: (digraph)
(u,v) E iff u, v know each other => Acquaintance graph
(undirected graph)
Ex3: Round-robin tournaments
One where each team plays each other team exactly once.
V: the set for all teams A B
(u,v) in E iff team u beats team v.
Ex: 4-team round-robin tournament C
D
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Discrete Mathematics
Chapter 9 Graphs
Graph model (cont'd)
Ex 4: (precedence graph and concurrent processing)
eg: 1. x := y + 1;
2. z := f(x)
=> (1) must be executed before (2) no matter which
processors execute (1) or (2), since (2) must use the
execution result of (1).
The executed-before relation in a program can be
represented by a digraph G = (V,E) where
V = the set of statements in the program
(u,v) E means u must be executed before v.
Note: if (u,v) and (v,w) E, ==> (u,w) E.
Hence E is transitive.
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Discrete Mathematics
Chapter 9 Graphs
Precedence graph of a program
Ex 4:
s1: a := 0; s1 s2
s2: b := 1;
s3: a := a+1; s3 s4
s4: d := b+a;
s5: e := d+1;
s5 s6
s6: e := c+d
==> What is the precedence
graph of this program?
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Discrete Mathematics
Chapter 9 Graphs
Precedence graph of a program
Ex 4:
s1: a := 0;
s2: b := 1; s1 s2
s3: a := a+1;
s4: d := b+a;
s5: e := d+1;
s3 s4
s6: e := c+d
==>Problems:
1. Minimum time needed to complete s5 s6
the program using multiple
processors? (Depth)
=> length of longest paths + 1
2. Minimum # of processors needed to
complete the program in minimum
time? (Width)
=> p1: [s1,s3,s4,s6], p2: [s2,-,-,s5]
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Discrete Mathematics
Chapter 9 Graphs
Graph terminology
Def 1: G = (V,E): undirected graph
if u --e-- v ==>
u and v are adjacent (or neighbors)
Edge e is incident with u (& v). D C B
e connects u and v.
G
u and v are end points of e. E F A
Def 2: G = (V,E): undirected graph; v V ==>
[Degree of v:] deg(v) = #edges incident with it =
|{e | v is an end point of e and e is an edge}|.
If deg(v) = 0 ==> v is isolated.
If deg(v) = 1 ==> v is pendant (or hang).
Ex: In Fig. 1, E has degree 3, D is pendant and G is
isolated.
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Discrete Mathematics
Chapter 9 Graphs
Handshaking theorem
Theorem 1: G = (V,E): an undirected graph
==> S v V deg(v) = 2 |E|. // 插孔數 = 插頭數
Pf: Simple induction on the number of edges |E|.
(Intuition: vertex deg = socket, edge end = plug)
Ex: G = (V,E) with |V| = 10 & each vertex with degree 6.
==> |E| =?
Corollary: The sum of the degrees of the vertices of an
undirected graph is even.
Theorem 2: The number of vertices in a graph with odd degree is
even.
Pf:
Sv in V deg(v) = Sv in V /\ deg(v) is odd deg(v) +Sv in V /\ deg(v) is even deg(v)
==> Sv in V /\ deg(v) is odd deg(v) is even ==> |{v in V | deg(v) is odd}| is
even. QED
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Discrete Mathematics
Chapter 9 Graphs
Terminology for digraphs
Def 3: G = (V,E): a digraph
if u --e--> v then
u is adjacent to v
v is adjacent from u
u is the initial (or starting) vertex of e
v is the end (or terminal) vertex of e
e is an edge from u to v.
Def. 4: [in-degree; out-degree]
In-deg(v) = #edge coming into v = |{e E | v is the terminal
vertex of e}|.
out-deg(v) = #edge going out from v = |{e E | v is the
starting vertex of e}|.
Theorem 3: Sv in V in-deg(v) = Sv in V out-deg(v) = |E|.
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Discrete Mathematics
Chapter 9 Graphs
Underlying undirected graph
G = (V,E,f): a directed graph
==> Its underlying undirected graph is a pseudo graph
G' = (V,E,f') where f': E -> {{u,v} | u,v V} s.t.
f'(e) = {a,b} iff f(e) = (a,b).
Note:
1. Both G and G' have the same number of edges (=
|E|).
2. If both (u,v) and (v,u) are edges in G ==> (u,v) and
(v,u) are two distinct edges in G'.
Ex: G = ({1,2,3}, {(1,2),(1,2),(1,3),(2,3),(3,2)})
==> G' =?
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Discrete Mathematics
Chapter 9 Graphs
Special simple graphs
Ex 4: complete graphs Kn = (V,E) where
|V| = n and E = {{x,y} | x and y are distinct vertices of V}.
Ex: K1,..., K5 =?
Ex 5: cycles: Cn (n > 2) = (V,E) where V = {1,2,...,n} and
E = {{i,i+1} | i = 1,...,n-1} U {{n,1}}.
Ex: C3,C4,C5,C6 =?
Ex 6: wheels: (n > 2) Wn = (V,E) where V = {0,1,...,n} and
E = {{0,i} | i = 1,...,n} U {{i,i+1} | i = 1,...,n-1} U {{n,1}}.
W3,W4,W5 =?
Ex7: n-cubes Qn = (V,E) where V = {0,1}n is the set of all bit
strings of length n and E ={{x,y} | x and y differ at exactly one
position}. Ex: 11011 and 11111 is connected.
Ex: Q1, Q2, Q3, Q4 =?
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Discrete Mathematics
Chapter 9 Graphs
Bipartite Graphs
Def 5: G = (V,E) is bipartite iff there is a binary partition {V1,V2}
of V s.t. for all e E, one end point of E is in V1 and the other
end point is in V2.
Ex8: Is C6 bipartite? (yes! why?) g a b
Ex9: Is K3 bipartite? (no! why?) f c
Ex10: Which of Fig G and H is bipartite? e
d
G
Ex11: Km,n (complete bipartite graph) a b
Km,n = ({u1,...,um} U {v1,...,vn}, E} , where f
c
E = {{ui, vj} | i=1..m v = 1..n} e d
H
eg: K2,3, K3,3 = ?
Problem: Km,n has ? edges
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Discrete Mathematics
Chapter 9 Graphs
Ex 13: interconnection network for processors
Types of connections
Linear array
Mesh
Hyper cube (m-cube)
Main measures: 1. #links 2. longest distance
linear array n-1 n-1
mesh 2sqrt(n)(sqrt(n)-1) 2(sqrt(n) -1)
m-cube (n=2m) nm/2 = O(n lg n) m = lg n
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Discrete Mathematics
Chapter 9 Graphs
New graphs from old
Def 6: [subgraph] G = (V,E): a graph
H = (V',E') is a subgraph of G iff
V‘ V and E' E and
H is a graph (i.e., if e E' is an edge connecting u and v in
V, then u, v must belong to V'.)
Ex: G = K5; find a subgraph H of K5.
Def 7: [union of simple graphs] G1 = (V1,E1); G2 = (V2,E2): two
graphs ==> G1 U G2 =def (V1 U V2, E1 U E2)
Note: Not only V1 and V2, but also E1 and E2 may overlap.
Ex: G1 = ({a,b,c,d,e}, {ab,bc,ce,be,ad,de})
G2 = ({a,b,c,d,f}, {ab,bc,bd,bf}}
==> G1 U G2 =?
Def 7': Disjoint union of graphs: (skip)
Disjoin union of G1 and G2 =?
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Discrete Mathematics
Chapter 9 Graphs
Graph representation and graph isomorphism
Graph representation: [suitable only for finite graphs]
adjacent list (for graphs w/o multiple edges): O(|E|+|V|)
adjacent matrices: O(|V|2)
incident matrices (or list): O(|V|x|E|).
adjacent list: for each vertex v, a[v] is the list of all adjacent
vertices of v.
adjacent matrix: M[u,v]=1 iff {u,v} (or (u,v) if directed) is an
edge.
incident matrix: I[u,e] = 1 iff u in f(e), or (for directed graph)
i[u,e]= 1 (or -1) iff u is the starting (or ending) vertex of e.
Ex: (p558 fig2)
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Discrete Mathematics
Chapter 9 Graphs
Graph isomorphism
Def . 1: G1 = (V1,E1), G2 = (V1,E2): two simple or digraphs
G1 and G2 are isomorphic iff $ a bijection h: V1->V2 s.t.
for all u,v in V1,
{u,v} (or (u,v)) E1 iff {h(u),h(v)} (or (h(u),h(v))) E2.
Such h is called an isomorphism (b/t G1 and G2).
Ex 8: Are G = ({1234},{12,23,34,41}) and H =
({abcd},{ad,ac,bc,bd}) isomorphic?
Fact: If G1 and G2 are isomorphic ==> |V1| = |V2| & |E1| = |E2|,
and G1 and G2 have the same degree type.
Def: For each G = (V,E), type(G) is the multiset {deg(v) | v V}.
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Discrete Mathematics
Chapter 9 Graphs
Connectivity
Def 1: G = (V,E,f): an undirected [multi-]graph
A path a of length n > 0 from u to v is a sequence
of edges e1,e2,...,en s.t. f(e1) = {u,x1}, f(e2) =
{x1,x2},…,f(en) = {xn-1,v}.
e1 e2 e3 e4 e5 e6 en-1 en
u=x0----x1-----x2----x3----x4----x5---- ….----xn-1----Xn=v
If G is a simple graph ==> a can be identified by
the sequence of vertices: x0 = u, x1,..., xn-1, xn = v.
a is a circuit (or cycle) iff x0 = xn.
a is simple if all edges in the path are distinct (i.e.,
for all 0≤ i<j≤ n, ei ej ).
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Discrete Mathematics
Chapter 9 Graphs
Connectivity (cont'd)
Def. 2: G = (V,E,f): a directed multigraph
A path a of length n > 0 from u to v is a
sequence of edges e1, e2, ...,en s.t. f(e1) = (u,x1),
f(e2) = (x1,x2),...f(en) = (xn-1,v).
If G is a digraph ==> a can be identified by the
sequence of vertices: x0=u, x1,..., xn-1, xn = v.
a is a circuit (or cycle) iff x0 = xn.
a is simple if all edges in the path are distinct
(i.e., for all 0<i<j<n+1, ei ej ).
Def 3: An undirected graph is connected if there is a
path between every pair of distinct vertices.
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Discrete Mathematics
Chapter 9 Graphs
Theorem about simple paths
Theorem 1: There is a simple path between every pair of distinct
vertices of a connected undirected [multi-]graph.
Pf: (u,v): any two distinct vertices of G.
Since G is connected, there exist paths from u to v.
Let a = e1,e2,...,en be any path of least length from u to v.
Then a must be a simple path. If it were not, then there would be
0 i <j n s.t. ei = ej.
==> the path b = e1,..,ei-1,ej+1,..,en or e1,..,ei,ej+1,..,en is another
path from u to v with length < |a|, a contradiction!. QED
join xi-1 and xj if x i-1 = xj
e1 e2 ei ej ej+1
..... ....... ......
u x1 x2 xi-1 xi xj-1 xj xj+1 v
join xi and xj if xi = xj
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Discrete Mathematics
Chapter 9 Graphs
xj
xi
ej+1
ei
e1 e2 xj+1
..... ej ......
u x1 x2 xi-1 v
xj-1
xj-1
xi
ei
e1 e2 xj+1
..... ej ......
u x1 x2 xi-1 v
xj ej+1
Note: Theorem 1 does not hold for circuit (cycle)
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Discrete Mathematics
Chapter 9 Graphs
A theorem about simple circuits
G = (V,E): an undirected multi-graph, u,v: two vertices in G.
Theorem: if there are two distinct simple paths from u to v, then
there is a simple circuit in G. (skip)
Pf: Let
a1: x0 --(e1)-- x1 --(e2)-- x2 --(e3)--…--(et)--xt--(et+1)--xt+1--…---xn=v,
and
a2: x0 --(e1)-- x1 --(e2)-- x2 --(e3)--…--(et)--xt--(e’t+1)--yt+1--…---ym=v,
be two distinct simple paths fro u to v in G
If either a1 or a2 contains a (simple) circuit, then we are done.
Otherwise let et+1, e’t+1 the first edges with et+1 ≠ e’t+1.
let J be the least number in {t+1,…,m} such that yJ = xs where s
is any vertex occurring in path a1 (i.e., yJ=xs ∈ {x0,…,xn}).
Note since ym = xn= v, such J must exist.
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Discrete Mathematics
Chapter 9 Graphs
A theorem about simple circuits
Now it is easy to see that:
1. If s ≤ t then { xs --(es+1)-- xs+1--…--(et)—xt } U
{ xt--(e’t+1)--yt+1--…--(e’J-1)-yJ } is a circuit. This is not
possible since all edges of the circuit are part of a2.
2. If s > t then {xt--(et+1)--xt+1--…-(es-1)--xs} U
{ xt--(e’t+1)--yt+1--…--(e’J-1)-yJ } ---(*) is a simple circuit.
Note: {xt,xt+1,…,xs} ⋂ {xt,yt+1,…,yJ } = {xt, yJ=xs}
If there were ea = e’b => f(ea) = {xt, xt+1=xs} = {xt,yt+1=yJ}
=> a=b=t+1 => e t+1 = e’t+1 a contradiction!
o/w, by def, (*) is a simple circuit.
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Discrete Mathematics
Chapter 9 Graphs
Xt
et+1
xs
u yJ e’t+1
v
es
xs
e’J yJ
e’J
yJ-1
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Discrete Mathematics
Chapter 9 Graphs
Connected components
G = (V,E): an undirected graph.
1. Any maximal connected subgraph of G is called a
connected component of G.
(i.e., G‘ = (V',E') is a connected component of G iff
1. G' is a connected subgraph of G and
2. There is no connected subgraph of G properly
covering G'.)
Ex:
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Discrete Mathematics
Chapter 9 Graphs
Connected components (cont'd)
Note 1. Every two distinct connected components are
disjoint.
Pf: G1 = (V1,E1), G2 = (V2,E2): two distinct connected
components.
If G1 and G2 overlap (i.e., V1 V2 ).
==> $v ∈ V1V2 ==> G3 = (V1 ⋃ V2,E1 U E2) is a
connected subgraph larger than G1 and G2, a
contradiction! QED
Note 2. Every connected subgraph G' of G must be
contained in some connected component of G. (skip)
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Discrete Mathematics
Chapter 9 Graphs
Connected components (cont'd)
Pf: Let i = 0 and Gi = G'. If Gi is maximal then we are
done.
o/w, there is connected Gi+1 contains Gi.
If Gi+1 is maximal, then we are done; o/t let i = i+1
and repeat the same process, we will eventually (if
G is finite) get a maximal graph containing G'.
Note 3. Let G1,G2,...,Gk be the set of all connected
components of G. Then G = Ui = 1..k Gi
Pf: 1. Ui=1,k Gi G since every Gi G.
2. G Ui=1,kGi since every edge and every vertex
must belong to some connected component.
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Discrete Mathematics
Chapter 9 Graphs
Connected components (cont'd)
Note 4: Every undirected graph G has a unique set of
connected components. (skip)
Pf: Let G = (V,E).
For each vertex u in V, let Gu = (Vu,Eu) where
Vu = {v | there is a path from u to v} V, and
Eu = {e | f(e) Vu} E.
It is easy to show that Gu is a connected component of G.
Moreover, we have
1. for all u, v ∈ V: Gu = Gv iff Vu=Vv /\ Eu=Ev and
2. for all e ∈ E: if f(e) = {u,v} ==> Gu=Gv and e Eu.
Hence the set of connected components of G = {Gu | u V}.
Note: Connected components in graphs play a role
like equivalence classes in equivalence relations.
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Discrete Mathematics
Chapter 9 Graphs
The connectivity relation in a graph (skip)
G = (V,E): an undirected graph
Let ~ be the connectivity relation induced by G, i.e., for all u,v
in V, u ~ v iff either u = v or u and v are connected in G.
Theorem:
1. ~ is an equivalence relation on V. (Hence V/~ is a partition of
V)
2. For all u,v in V, u and v are connected iff u and v are in the
same block of the partition.
3. For each u V, Gu = (Vu,Eu) = ([u]~, E|[u]) where
E|S is the set of edges restricted to the vertex set S, i.e.,
{e E | f(e) S}.
4. The set of all connected components of G =
{ (S, E|S) : S V/~}.
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Discrete Mathematics
Chapter 9 Graphs
Cut vertices and cut edges
A vertex in a graph is called a cut vertex (or articulation
point) if the removal of this vertex and all edges incident
with it will result in more connected components than in the
original graph.
Corollary: The removal of a cut vertex (and all edges incident
with it) produces a graph that is not connected.
Cut edges (bridges)
The removal of it will result in graph with more connected
components.
a d f g
Ex 4: Determine all cut vertices
and all bridges in the right graph.
cut vertices =? b h
c e
bridges =?
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Discrete Mathematics
Chapter 9 Graphs
Strongly connected digraphs
Def 4: G = (V,E): a directed graph
G is strongly connected iff there are paths from u to v and
from v to u for every pair of distinct vertices u and v in G.
Def 5: G is weakly connected iff there is a path between every
pair of distinct vertices in the underlying graph of G.
Ex 5:
G H
G is strongly connected.
H is weakly connected.
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Discrete Mathematics
Chapter 9 Graphs
Path and isomorphism
P(-) : a property on graphs
Ex: P(G) = "G has a simple cycle of length > 2"
P(G) = " G has an even number of vertices and edges".
P(G) = ...
P is said to be an isomorphic invariant if P is invariant under all
isomorphic graphs, i.e., for all graph G1 and G2,
if G1 and G2 are isomorphic then P(G1) P(G2).
Ex1: Pm,n(G) = " G has m vertices and n edges ", where m and n
are some constant numbers, is an isomorphic invariant.
Corollary: G1,G2: two graphs; P: an isomorphic invariant;
If P(G1) and P(G2) have distinct truth value, then G1 and G2 are
not isomorphic.
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Discrete Mathematics
Chapter 9 Graphs
More on graph isomorphism
Ex2: P2(G) = "G has a simple circuit of length k", where k is a
number > 2, is an isomorphic invariant.
Pf: G1 = (V1,E1), G2 = (V2,E2): two simple graphs.
Let h: V1->V2 be any isomorphism from G1 to G2.
Then if x0 --(e1)--> x2--(e2)--> x3--.... -->(en)-->xn is a simple path
on G1, then the sequence
h(x0) -->(h(e1))-->h(x2)--(h(e2))-->h(x3)--...-->((h(en))-->h(xn)
is also a simple path on G2.
Ex 6: G and H are not isomorphic since H has s simple cycle of
length 3 (1261), whereas G has no simple cycle of length 3.
2 3 2 3
1 4 1 4
6 5 6 5
G H
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Discrete Mathematics
Chapter 9 Graphs
Counting paths b/t vertices
G = (V,E) with V = {V1,...,Vn} : a simple graph with adjacent matrix M.
==> # different paths of length k from vi to vj = (Mk)ij, where scalar
multiplication are integer product (instead of boolean AND).
Pf: by math ind on k.
1. basis step: k = 1: => By def. Mij is the number of edges (= path of
length 1) from vi to vj.
2. Ind. step: assume (Mm)ij = #paths of length m from vi to vj for all m <
k and for all ui, uj ∈ V.
But #paths of length k from vi to vj =
#paths of length k-1 from ui to v1 x #paths of length 1 from v1 to vj +
#paths of length k-1 from ui to v2 x #paths of length 1 from v2 to vj +
... +
#paths of length k-1 from ui to vn x #paths of length 1 from vn to
vj
= St=1..n (Mk-1)it x Mtj = (MK)ij.
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Discrete Mathematics
Chapter 9 Graphs
Example
Ex 8: G = ({1,2,3,4}, {12,13,24,34})
==> M = [0110;1001;1001;0110]
==> M4 = [8008;0880;0880;8008]
==> There are 8 different paths of length 4 from a to d.
Notes:
1. The length of the shortest path from vi to vj is the least k s.t.
(MK)ij != 0.
2. G is connected if (Sk=1..n-1 Mk )ij != 0 for all 1≤ i<j ≤ n.
Pf: G is connected =>
for any i ≠ j, there is a simple path (of length t < n) from vi to vj
==> (Mt)ij > 0 ==> (Sk=1..n-1 Mk )ij != 0. QED
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Discrete Mathematics
Chapter 9 Graphs
Euler and Hamilton paths
The seven bridges problem: C
Problem: Is there a path A
D
passing through all bridges
w/o crossing any bridge twice? B
The town of Konigsburg
Multigraph model: C
Problem: Is there a simple
path of length 7? A D
B
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Discrete Mathematics
Chapter 9 Graphs
Eular paths and Eular cycles
Def. 1: A Eular path in a multigraph G is a simple path
containing all edges.
Def. 2: A Eular circuit in a multigraph G is a simple circuit
containing all edges.
Ex1: In G1, G2 and G3:
G1 has a Eular a b a b a b
circuit: a,e,c,d,e,b,a.
e e
G3 has a Eular path: d c d c c d e
bedbadca. G1 G2 G3
Note: If there is a Eular circuit beginning from a vertex v, then
there is a Eular path or circuit beginning from any other vertex.
Transparency No. 9-40
Discrete Mathematics
Chapter 9 Graphs
Necessary and sufficient conditions for Eular paths and Eular circuits
Theorem 1: A connected multigraph has a Eular circuit iff each of
its vertices has even degree.
Corollary: The seven-bridges problem has no Eular circuit.
Pf: "=>": G = (V,E): any multigraph.
Let a = x0 e1 x1 e2 x2 e2 x3 ... en xn=x0 be any Eular circuit.
For each v=xi in V ≠ x0, since ei-->xi --> ei+1 we have
deg(v) = 2 |{j | xj = v and 0<j<n}| and for x0 we have
deg(x0) = 2x|{j | xj = x0 and 0<j<n }| + 2.
Hence every vertex has even degree.
(<=): by induction on |E|. If |E|=0, by definition, it has a Eular ckt.
O/W Let a = x0 e1 x1 e2 x2 e3 ... en xn=x0 be any simple circuit.
(its existence will be shown later.)
If a is a Eular circuit, then we are done. O/W:
Transparency No. 9-41
Discrete Mathematics
Chapter 9 Graphs
Proof of Eular condition
Let G' = (V',E') be the resulting graph formed from G by removing all
e1,e2,...,en from E.
Let G1 = (V1,E1),...,Gk =( Vk,Ek) be all connected components of G'.
Since G is connected, {x0,...,xn} Vi != {} for all 0<i<k+1.
(o/w there is no path from x0 to vertices in Vi).
For each i let xti be any vertex in {x0,...,xn} Vi.
Since Gi = (Vi,Ei) and |Ei| < |E| and every vertex in Vi has even degree,
by ind. hyp. there is a Eular circuit ai = xti ->...-> xti in Gi.
Now join each ai with a at xti:
a1 ak
e1 e2 ei ej ej+1
..... ....... ......
x0 x1 x2 xt1 xi xtk xj xj+1 xn=x0
we can form a Eular circuit in G. QED.
Transparency No. 9-42
Discrete Mathematics
Chapter 9 Graphs
Example: (skip)
h g
a = a b c d a is a simple circuit in G.
removing all edges in a results in e f
three connected components: a d
G1, G2 and G3 intersecting with
G
a at {a,d}, {b} and {c} respectively. i
By ind. hyp., b
c j
$ Eular ckt :a1 = (a...a) h g
=(aedfghefa)
e f
a2: (b) and
a
a3: (c i j c) G1 d
==> join a and all ai, we obtain
i
a Eular ckt of G: G3
b G2
(aedfghefa)b(cijc)da. c j
Transparency No. 9-43
Discrete Mathematics
Chapter 9 Graphs
Eular condition for Eular paths:
Theorem 2: A connected multigraph has an Eular path but not
a Eular ckt iff it has exactly two vertices of odd degree.
Pf: (=>): G = (V,E): any multigraph.
Let a = x0 e1 x1 e2 x2 e2 x3 ... en xn != x0 be any Eular path.
For each v=xi in V != x0 or xn, since ei-->xi --> ei+1 we have
deg(v) = 2 |{j | xj = v and 0<j<n}| and for v = x0 or xn we have
deg(v) = 2x|{j | xj = v and 0<j<n } + 1.
Hence all vertices but x0 and xn have even degree.
(<=): Let G' = (V, EU{e}) where e is a new edge connecting a
and b, which are vertices of G with odd degree.
==> Every vertex of G has even degree. By theorem 1,
$ a Eular path a = a -> .... ->b->(e)->a.
==> removing e from a we get a Eular path of G. QED
Transparency No. 9-44
Discrete Mathematics
Chapter 9 Graphs
Existence of simple circuits
Lemma: G = (V,E): a multigraph s.t. E != {} and every vertex has
even degree. Then there exist a simple ckt in G from any
vertex of nonzero degree.
Pf: Let x0 be any vertex in G with nonzero degree.
We construct a sequece of simple paths a’s which eventually become a
simple circuit as follows:
0. Initially a1 = x0 --e1-- x1. G1 = (V, E \{e1}).
note: x1 in G1 has degree > 0 hence there is an edge leaving x1.
1. assume ai = x0...xi has been formed and Gi = (V, Ei= E\{e1,...,ei}).
If xi = xk for k<i then xk-ek- - … xi is a simple ckt and we are done.
o/w by ind. hyp. xi has odd degree in Gi, hence we can find an
edge e in Gi connecting xi and some other vertex u.
now let ai+1= ai -- e--u; xi+1 = u ; and i = i+1
2. goto 1.
Note: the procedure must terminate since every iteration of step 1 will
result in one edge removed from G, but G has only a finite number of
edges. So the prod. will exit from step 1 with a simple ckt returned. QED
Transparency No. 9-45
Discrete Mathematics
Chapter 9 Graphs
Hamilton paths and circuits
Def. 2: G = (V,E): a multigraph.
A path x0 e1 x1 e2 x2... en xn in G is a Hamilton path if {x0,...,xn}
= V and xi xj for all i j.
A ckt x0 e1 x1 e2 x2... en xn=x0 (n > 1) in G is a Hamilton ckt if
{x1,...,xn} = V and xi xj for all i j
Ex:
Problem: Is there a simple principle, like that of Eular ckt, to
determine whether a multigraph has a Hamilton ckt?
Ans: no ! In fact, there is no known polynominal time algorithm
which can test if a given input multigraph has a Hamilton ckt!
Theorem: [sufficient condition] G: a connected simple graph
with n 3 vertices. => If every vertex has degree n/2, then
G has Hamilton ckts.
Transparency No. 9-46
Discrete Mathematics
Chapter 9 Graphs
Grey code (skip)
Ex 8: label regions of a disc:
Problem: split the disc into 2n arcs of equal length and assign a
bit string of length n to each arc s.t. adjacent arcs are
assigned bit strings differing from neighbors by one bit only.
Problem:
How to find Gray code 111 000 100 000
of n-bit length? 110 001
101 001
Rule: Let Gn = (Vn,En) 101 010
100 011 111 011
= Qn be the n cube.
011 111 110 010
A cycle of the disc
(a grey code) corresponds 001 101
to an Hamilton ckt in Qn.
010 110
000 100
Transparency No. 9-47
Discrete Mathematics
Chapter 9 Graphs
Shortest path problems
A weighted graph is a graph G = (V,E) together with a
weighting function w:E->R+.
A shortest path from a to b is a path:
a e1 x1 e2 x2 ... en b s.t. Si=1,n w(ei) is minimized.
Problem: Given a graph and two vertices a, z, find the length
of a shortest path from a to z.
Alg: [Dijkstra's algorithm]
//L(v): the length of a shortest path from a to v//
1. L(a) := 0; L(v) = w(a,v) for all v ≠ a ; S := {a};
2. While( ~S ≠ { } ){ // i.e., S ≠ V
3. u = any vertex ∉ S with minimum L(u);
4. S = S U {u};
5. for( all v ∉ S)
6. L(v) = min(L(u) + w(u,v), L(v)) }
7. end /* L(z) = length of the shortest path from a to z}.
Transparency No. 9-48
Discrete Mathematics
Chapter 9 Graphs
Example:
Running Time: O(n2)
b 5 d
4 6
a 1 8 z
2
2 3
10 e
c
b,4 5 d
4 6
a,0 1 8 z
2
2 3
10 e
c,2
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Discrete Mathematics
Chapter 9 Graphs
b,3 5 d,10
4 6
a,0 1 8 z
2
2 3
10 e,12
c,2
b,3 5 d,8
4 6
a,0 1 8 z
2
2 3
10 e,12
c,2
Transparency No. 9-50
Discrete Mathematics
Chapter 9 Graphs
b,3 5 d,8
4 6
a,0 1 8 z,14
2
2 3
10 e,10
c,2
b,3 5 d,8
4 6
a,0 1 8 z,13
2
2 3
10 e,10
c,2
Transparency No. 9-51
Discrete Mathematics
Chapter 9 Graphs
Correctness (skipped!)
Let Sk = the value of S after the kth iteration of the for-loop.
uk = the vertex added to Sk at the kth iteration.
Lk(v) = the value of L(v) after kth iteration.
Notes
Sk = {u0=a,u1,…,uk}.
Lj(uk) = Lk-1(uk) for all j >k-1.
1. L(a) := 0; L(v) = w(a,v) for all v ≠ a ; S := {a};
2. While( ~S != { } ){
3. u = any vertex ∉ S with minimum L(u);
4. S = S U {u};
5. for( all v ∉ S)
6. L(v) = min(L(u) + w(u,v), L(v)) }
7. end /* L(z) = length of the shortest path from a to z}.
Transparency No. 9-52
Discrete Mathematics
Chapter 9 Graphs
Correctness of Dijkstra alg is a direct app. of the following lemma: (skipped)
Lemma: for all iterations k and all vertices v:
1. v Sk Lk(v) is the length of the shortest of all paths from a to v.
2. v ∉ Sk Lk(v) is the length of the shortest of all paths from a to v of
which all intermediate vertices must come from Sk.
pf: By induction on k.
k = 0: Then
S={a} , Lk (a) = 0 is the shortest length from a to a.
v ≠ a => Lk (v) = w(a,v) is the length of the shortest from a to v without passing
through any vertex in S.
k= j+1 > 0:
By step 3, Lj(uk ) = min { Lj(v) | v ~Sj }, and by step 4. Lk(uk) = Lj(uk).
Now we show that Lj(uk) (and hence Lk(uk) ) is the shortest length of all paths from
a to uk.
Assume it is not. Then there must exist a shorter path a from a to uk, but such path
must pass through a vertex not in Sj, since by ind.hyp, Lj(uk) is the least length of
all paths passing only through Sj.
Now let a = a --- u ----uk where u is the first vertex of a that are not in S and the
subpath a --- u contains only vertices from Sj.
Then the length |a| = | a---u | + |u --- uk|， but |a ---u| ≥ Lj(uk) Hence it is impossible
that |a| < Lj(uk). This shows (1) is true.
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Discrete Mathematics
Chapter 9 Graphs
(skipped)
Now consider (2). let v be any vertex ∉ Sk .
The shortest path from a to v with intermediate vertices among
Sk either contains uk or not.
1. If it does not, then it is also a shortest path from a to v
containing vertices form Sj, and by ind.hyp, its length is Lj(v).
2. If it conatins uk, then uk must be the last vertex before
reaching v, since if
a---uk,t---v is a shortest path, then the path a---t---v, where a--
-t is a shortest path from a to t, must be not longer than it.
As a result, it has distance Lj(uk) + w(uk,v), where by ind.hyp.,
Lj(uk) is the shortest distance of all paths from a to uk
passing through Sj.
Finally, Step 6 assigns the minimum of both cases to Lk(v)
and hence sastisfies lemma (2).
Transparency No. 9-54
Discrete Mathematics
Chapter 9 Graphs
Find the distances b/t all pairs of vertices
Floyd(G:(V,E,w))
1. for i = 1 to n
for j = 1 to n
d(i,j) = w(i,j);
2. for k = 1 to n ---- d(i,j) is the distance of minimum path from i
----- to j with interior points among {1,2,...,k}.
for i = 1 to n
for j = 1 to n
d(i,j) = min (d(i,k) + d(k,j), d(i,j));
end /* d(i,j) is the shortest distance b/t i and j */
running time : O(n3).
Note: we can also apply Dijkstra's alg to obtain a O(n3) alg for
distances of all pairs of vertices.
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Discrete Mathematics
Chapter 9 Graphs
Planar graphs
Def 1: A graph is planar iff it can be redrawn in the plane w/o
any edges crossing. Such a drawing is called a planar
representation of the graph.
Ex1: K4: ok
Ex2: Q3: ok
Ex3: K3,3: not planar.
Eular's formula:
A planar representation splits the plane into regions (including
an unbounded one. Eular showed a relation among #regions,
|E| and |V|).
Theorem 1: G=(V,E): a connected planar graph with e edges
and v vertices. Then r (#regions) = e - v + 2.
Pf: By induction on G =(V,E).
Let G0,...., Gn = G be a sequence of subgraphs of G.
Transparency No. 9-56
Discrete Mathematics
Chapter 9 Graphs
Planar graphs (cont'd)
G0 contains only one vertex.
Gi+1 is formed from Gi by adding one edge incident with one
vertices in Gi (and a vertex if needed)
Let ri, ei, vi are #regions, #edges, #vertices in Gi respectively.
Basis: e0 = 0; r0 = 1; v0 =1. Hence r0 = e0 - v0 + 2.
Ind. step: assume ri = ei - vi + 2.
Let (ai+1,bi+1) be the edges added to Gi to form Gi+1.
case 1: {ai+1,bi+1 } ∈ Vi:
==> ai+1 and bi+1 must be on the boundary of a common region.
(o/w crossing would occur!)
==> add {ai+1,bi+1} partition the region into 2 subregion
==> ri+1 = ri + 1; ei+1 = ei +1; vi+1 = vi. ==> ri+1 = ei+1-vi+1+2.
case 2: a i+1 ∈ Vn but b i+1 ∉ Vi => add {ai+1,bi+1 } does not
produce any region. => ri+1 = ri; ei+1=ei+1; vi+1 = vi+1
==> ri+1 = ei+1 -vi+1+ 2. QED
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Discrete Mathematics
Chapter 9 Graphs
More on planar graphs
Ex4: G has 20 vertices, each of degree 3.
==> #regions = ? r = e - v + 2 = 30 - 20 + 2 = 12.
Corollary: G: a connected planar simple graph with e edges
and v vertices 3. ==> e 3v - 6.
pf: for each region R:
define deg(R) = #edges on the boundary of R.
==> 2e = S deg(R) [each edge shared by 2 regions] =S deg(v)
Since deg(R) 3 (simple graph),
2e = S deg(R) 3 r. ==> 2e/3 r = e-v+2 ==> 3v- 6 e. QED
Ex5: K5 is not planar.
sol: K5 has C(5,2) = 10 edges and 5 vertices.
==> 3v - 6 = 15-6 = 9 < 10, violating the necessary condition !
Ex6: K3,3, though satisfying the condition: e 3v - 6, is not
planar.
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Discrete Mathematics
Chapter 9 Graphs
More on planar graph:
G: a connected planar simple graph of e edges and v > 2
vertices and no simple ckt of length 3 => e 2v - 4.
pf: no ckt of length 3 ==> every region has degree 4.
==> 2e = S deg(R) r. ==> e r = 2e-2v+4 ==> e 2v - 4.
Ex 6: K 3,3 is not planar.
K 3,3 has 9 edges and 6 vertices and has no cycle of length 3.
==> 2 x 6 - 4 = 8 < 9. Hence not planar.
General rule about planar graphs:
Def: elementary subdivision:
u--------------v ==> u------w------v.
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Discrete Mathematics
Chapter 9 Graphs
General rule about planar graphs
Def: G1 and G2 are homeomorphic iff they can be obtained
from the same graph by a sequence of elementary
subdivisions.
Ex12:
Fact: G in not planar if some of its subgraphs is not planar.
Theorem 2: [Kuratowski theorem]
A graph is nonplanar iff its contains a subgraph which is
homeomorphic to K5 or K3,3.
Transparency No. 9-60
Discrete Mathematics
Chapter 9 Graphs
Graph coloring
Problem: Given a map, determine at least how many colors
are needed to color the map s.t. neighbor regions (with a
common border) are assigned diff colors.
Ex:
B H
C I
G
A
D
E F
Transparency No. 9-61
Discrete Mathematics
Chapter 9 Graphs
Transform maps into graphs
Ideas:
regions ---> vertices
R1 and R2 share a border ---> {R1,R2} is an edges
Def: M: a map
GM = (V,E) where
V = the set of all regions
E={ {r1,r2} | r1 and r2 share a common border line in M}
GM is called the dual graph of the map M.
Notes:
1. Every dual graph of maps are planar graphs
2. Every planar graph is the dual graph of some map.
Transparency No. 9-62
Discrete Mathematics
Chapter 9 Graphs
Graph coloring
Def 1:G=(V,E): a graph.
A coloring of G is the assignment of a color to each vertex s.t.
no two adjacent vertices are assigned the same color.
Problem: What is the least number of colors necessary to color
a graph ?
Def 2: G=(V,E): a graph.
The chromatic number of G is the least number of colors
needed for a coloring of the graph G.
Problem: What is the maximum of all chromatic numbers of all
planar graphs ?
==>1. a problem studied for over 100 years ! [1850~1976,
solved by Appel&Haken]
2. 2000 cases generated and verified by computer programs.
3. issue: believable ? Transparency No. 9-63
Discrete Mathematics
Chapter 9 Graphs
The 4-color theorem
Theorem 1: The chromatic number of a planar graph is no
greater than 4.
Note: 1. Theorem 1 holds for planar graphs only.
2. Non-planar graphs can have any chromatic number > 4.
Ex: K5 has chromatic number 5.
In fact Kn has chromatic number n for any n.
Exs: #(Km ,n) = 2.
#(Cn) = 2 if n is even and 3 if n is odd.
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Discrete Mathematics
Chapter 9 Graphs
Applications (scheduling and assignments)
Ex 5: Problem: how to assign the final exams s.t. no student
has two exams at the same time.
Sol: G = (V,E) where
V = the set of all courses
{c1,c2} in E iff $ a student taking courses c1 and c2.
==> A schedule of the final examinations corresponds to a
coloring of the associated graph G.
1 I
IV Time courses
2 II I 1,6
7
==> II 2
III 3,5
3 III IV 4,7
I 6
III 5 4 IV
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