Download Chapter 3 Digraphs and Tournaments Example Directed graph

Example
Chapter 3 Digraphs and
Tournaments
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Directed graph
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Vertices and Arcs
u
1. A directed graph can be obtained from a
graph by giving every edge a direction.
a
2. A directed graph is usually called a digraph.
b
v
1. A digraph consists of vertices and arcs.
2. An arc is an ordered pair of vertices, e.g, a = (u,v) and
b = (v, u).
3. a and b are different arcs.
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Formal definition
4
Arcs
u
A directed graph D = (V, A) consists of two finite
sets V and A, where
v
a
If a = (u,v) is an arc of D, then
(i) V is the vertex set, and
(i) a is said to join u to v,
(ii) A is the arc set, where each arc is an
(ii)u is called the original, or the initial vertex, or
the tail of a,
ordered pair of vertices.
(iii)v is called the terminus, or the terminal vertex,
or the head of a.
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Example
The underlying graph
1. Given a digraph D, we can get a graph G
from D by removing all the directions.
2. G is called the underlying graph of D.
Terminus, terminal
vertex or head
Original, initial vertex
or tail
D
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G
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Orientation of a graph
8
Example
1. Given a graph G, we can obtain a digraph D
from G by specifying a direction for each
edge of G.
2. Such a digraph D is called an orientation of
G.
One orientation of K3.
How many orientations are there in K3?
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Directed walks
3.2
Connectivity of digraphs
1. Let D be a digraph.
2. A directed walk in D is a finite sequence
W = v0a1v1a2v2...akvk.
a1
v0
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a2
v1
…
v2
v3
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ak
vk
12
Example
Directed trail, path and cycle.
Find three different direct walks from v4 to v3.
• A directed trail is a directed walk which does
not repeat any arc.
• A directed path is a directed walk which does
not repeat any vertex, except the two ends.
• A directed cycle is a directed path in which the
two ends are the same vertex.
v3
v4
a1
a6
a2
a7
a5
v1
v2
a4
a3
1 . v 4 a 7 v2 a 6 v3
2 . v 4 a 7 v 2 a 4 v1 a 3 v 2 a 6 v 3
3 . v 4 a 7 v 2 a 4 v1 a 5 v 2 a 6 v133
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Exercise
• Is it true that
if there is a directed walk in D from x to y,
then there is a directed path from x to y?
True.
Why?
v3
v4
a6
a2
a7
a5
v1
v2
a4
14
Question
Find all directed paths from v4 to v3.
a1
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a3
1. v4a7v2a6v3
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Weakly connected graph
• Let u and v be vertices in D.
• v is said to be reachable from u if there is a
directed path in D from u to v.
• D is said to be unilaterally connected if for
every pair of vertices u and v, either u is
reachable from v or v is reachable from u.
v3
v4
a6
a2
a7
a5
v1
a4
16
Unilaterally connected
• A digraph D is said to be weakly connected or
connected if its underlying graph is connected.
a1
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v2
a3
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Exercise
Theorem
• A digraph D is unilaterally connected if and
only if D has a spanning directed walk.
• The sufficiency is obvious. Why?
• Assume that P is a spanning directed walk in
D.
• For any two vertices u and v in D, they are
both on P and so there is a directed walk
from u to v or from v to u.
• Thus D is unilaterally connected.
• Draw a digraph D of order 5 such that D is
weakly connected but not unilaterally
connected.
u
v
x
w
y
Take u and w. u is not reachable from w and w is
not reachable from u.
Hence this digraph is not unilaterally connected.
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Proof of Necessity
• Suppose that D has no spanning directed
walk.
• Let P be a directed walk in D which contains
the maximum number of vertices in D.
• Denote the walk P by x1x2x3…xk.
• D has a vertex, say w, which is not on P.
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• Since D is unilaterally connected, for every
i=1,2,…, k, there is a directed walk P(xi,w)
from xi to w or P(w,xi) from w to xi.
w
x1
x2
x21k
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• There is no path P(w,x1); otherwise, D has a
directed walk which contains more vertices
than P, a contradiction.
• So there is a directed walk P(x1,w).
• Similarly, there is a walk P(w, xk).
w
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xk
x2
P(w,xi)
xk
xi
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• Let s be the maximum number in
{1,2,…,k} such that there is a directed walk
P(xs,w) from xs to w.
• We have 1 ≤ s < k. Why?
• Then there is a directed walk P(w,xs+1) from
w to xs+1.
P(w,xk)
P(w,x1)
x1
x1
P(xi, w)
xi
w
w
x117/08/05 x2
20
x2
xk
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w
P(xs,w)
x1
xs
x2
Strongly connected digraph
P(w,xs+1)
xs+1
xk
• A digraph is said to be strongly connected if
for every pair of vertices u and v in D, u is
reachable from v and v is reachable from u.
P2
P1
• Let P1 denote the directed walk on P from x1 to xs,
let P2 denote the directed walk on P from xs+1 to xk.
• Then P1,P(xs,w), P(w,xs+1) and P2 form a directed
walk in D which contains more vertices than P, a
contradiction.
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•Hence the necessity holds.
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Example
1. If u and v are in a directed closed walk,
then v is reachable from u and u is
reachable from v.
2. If v is reachable from u and u is
reachable from v, then u and v are in the
a directed closed walk.
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Example
v
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Theorem
• A digraph is strongly connected if and only if
it has a spanning closed walk.
• The sufficiency is obvious.
u
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Observations
A strongly connected digraph
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Example
Homework
Is this digraph strongly connected?
• Prove this theorem.
• Hint: apply the result for unilaterally
connected digraphs:
• Since D is strongly connected, D is also
unilaterally connected and so D has a
spanning directed walk.
V2
V1
V10
V3
V4
V5
V7
V9
V8
V6
• It is not strongly connected.
• It can be split into strongly connected subdigraphs.
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Strong component
Exercise
• A strong component of a digraph D is a
subdigraph of D which is strongly
connected and
is not a proper subdigraph of any other
strongly connected subdigraph of D.
Find all strong components of the following
digraph.
2
1
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Observation
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a2
a4
a3
a7
a1
5
a5
a6
3
4
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3.3 Matrix representations
• If two vertices are in a directed closed
cycle, then the two vertices are in the
same strong component.
• If u is reachable from v and v is reachable
from u , then u and v are in the same
strong component.
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35
1. Adjacency matrix
2. Incidence matrix
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Definition
Adjacency and incidence
• Let u and v be vertices in a digraph.
• If u and v are joined by an arc e, then u and v
are said to be adjacent.
• If the arc e is directed from u to v, then e is
said to be incident from u and incident to v.
e is incident from u and
incident to v
e
u
v
u and v are adjacent
e
u
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v
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Degrees
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Example
v
• Let D be a digraph and u be a vertex in D.
• The out-degree of u is the number of arcs incident
from u, and is denoted by od(u).
• The in-degree of u is the number of arcs incident
to u, and is denoted by id(u).
w
u
x
y
z
od(u)=1, od(v)=3, od(w)=2, od(x)=0, od(y)=1, od(z)=2
id(u)=0, id(v)=1, id(w)=1, id(x)=0, id(y)=5, id(z)=2
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Observation
1
∑ id ( x) = ∑ od ( x) = number of arcs in D
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Col. Col. Col. Col.
1 2 3
4
2
x∈V
Row 1
• Why?
• Since each arc has two ends, it must contribute
exactly 1 to the sum of the out-degrees and
exactly 1 to the sum of in-degrees. The result
follows immediately.
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Adjacency Matrix
• In any digraph,
x∈V
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3
4
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Row 2
Row 3
Row 4
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⎛0
⎜
⎜0
⎜0
⎜
⎜0
⎝
1
0
1
0
0
0
0
1
1⎞
⎟
2⎟
0⎟
⎟
0 ⎟⎠
42
Adjacency Matrix
Exercise
• Let D be a digraph with n vertices labeled
1,2,…,n.
Find the adjacency matrix A(D) of the
following digraph.
• The adjacency matrix A(D) is the n×n matrix:
2
A ( D ) = (a i , j )n × n
5
1
ai,j = the number of arcs from i to j.
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3
43
4
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0
0
1
1
1
1
1
1
2⎞
⎟
3⎟
0⎟
⎟
0 ⎟⎠
• Let D be a digraph and A(D)=(ai,j)n×n.
• Then
∑a
4
1≤ j ≤ n
i, j
∑a
1≤ i ≤ n
3
2
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= the out - degree of the i th vertex
i, j
∑a
1≤ i ≤ n
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1≤ j ≤ n
45
= the in - degree of j th vertex
i, j
= the number of arcs in D .
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1
2
• Let D be a digraph with vertices v1,v2,…,vn,
and A(D) be the adjacency matrix of D.
• If
( A( D) )
=
(a )
k
i , j n×n
,
( A ( D ) )2
k
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⎛0
⎜
⎜0
A(D ) = ⎜
0
⎜
⎜0
⎝
3
4
then a i , j is the number of directed walks of length k
from vi to vj for all i,j: 1≤i≤j≤n.
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Example
Theorem
k
44
Numbers of arcs
Find a digraph D such that its adjacency matrix is
the following matrix.
1
0 0 0 0⎞
⎟
0 0 2 0⎟
1 0 1 0⎟
⎟
0 0 0 1⎟
0 1 0 0 ⎟⎠
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Exercise
⎛1
⎜
⎜0
⎜0
⎜
⎜0
⎝
⎛0
⎜
⎜1
A(D) = ⎜ 0
⎜
⎜0
⎜0
⎝
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⎛0
⎜
⎜0
=⎜
0
⎜
⎜0
⎝
1
0
1
0
0
0
0
1
1⎞
⎟
2⎟
0⎟
⎟
0 ⎟⎠
0
0
0
1
2
0
1
0
2⎞
⎟
0⎟
2⎟
⎟
0 ⎟⎠
48
By induction,
Proof of Theorem
a 1i, t = number of arcs from v i to v t
a st, j = number of walks of length s from v t to v j
• It holds if k=1, by the definition of A(D).
• Assume that it holds when k≤s, where
s≥1. Now let k=s+1.
s +1
s
• Observe that ( A( D) ) = A( D) × ( A( D) ) ,
(a )
s +1
i , j n×n
i.e.,
( ) × (a )
= ai1, j
n×n
s
i , j n×n
Thus
n
ais,+j1 = ∑ ( number of arcs from v i to v t )
t =1
.
× (number of walks of length s from v t to v j )
= the number of walks of length s + 1 from vi to v j .
n
ais,+j1 = ∑ ai1,t × ats, j .
So
t =1
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Hence
for NIE
k=s+1.
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Incidence Matrix
v1
a1
a5
a4
a6
v4
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a3
a2
v3
Incidence Matrix
Col. Col. Col. Col. Col. Col.
3
5 6
4
2
1
v2
Row 1 ⎛ 1
0 0 1
⎜
Row 2 ⎜−1 −1 0 0
Row 3 ⎜ 0 1 −1 0
⎜
⎜
Row 4 ⎝ 0 0 1 −1
0⎞
⎟
1⎟
0 0⎟
⎟
−1 −1⎟⎠
0
1
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• The incidence matrix involves the incidence of
vertices and arcs.
• Let D be a digraph without loops, with n
vertices labeled v1, v2,…,vn, and m arcs labeled
a1, a2,… ,am.
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if arc a j is incident from vertex vi ;
if arc a j is incident to vertex vi ;
otherwise.
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Find the incidence matrix I(D) of the following
digraph D.
2
a2
5
a5
a4
a3
a7
a6
1
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Exercise
• The incidence matrix I(D) is the n×m
matrix in which the entry in row i and
column j is
⎧1,
⎪
⎨- 1,
⎪
⎩0,
50
53
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a1
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54
Exercise
0
0 −1 0 0
0⎞
⎛ 0
⎜
⎟
1
1
1 0 0 − 1⎟
⎜ 0
⎜ 1
0
0
0 −1 0
1⎟
⎜ −1 −1 −1 0 0 1
0⎟
⎜⎜
⎟⎟
0
1
0
0
0
0
−
1
⎝
⎠
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Find a digraph D such that its incidence matrix is
the following matrix.
⎛ 1
⎜
⎜−1
⎜ 0
⎜
⎜ 0
⎝
55
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3.4 Acyclic digraph
a1
v4
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a2
v3
57
Two cases:
y
z
…
x
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y
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1
0
−1
0
2
4
3
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• If z is on P(x,y), then one directed cycle C
can be formed by the arc yz and the path
P’(z,y) from z to y which is on the path
P(x,y), a contradiction.
• Hence Z is not on P(x,y).
• Thus the P(x,y) can be extended to a
longer path from x to z: P(x,y)z,
contradicting the assumption that P(x,y) is
a longest path in D.
• Therefore od(y)=0.
…
x
1
1
• Proof. Let P(x,y) be a longest path in D,
which is from x to y.
• It is clear that x≠y.
• We shall show that od(y) = 0.
• Suppose that od(y)>0. Let yz be an arc in
D.
v2
a6
0
0 ⎞
⎟
0 ⎟
1 ⎟
⎟
− 1 ⎟⎠
• An acyclic digraph D has at least one
vertex of out-degree 0.
a5
a4
0
Theorem
• A digraph is called acyclic if it contains no
directed cycles.
• The following digraph is acyclic.
v1
−1
59
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Theorem
Example
• An acyclic digraph D has at least one
vertex of in-degree 0.
v1
a1
v2
a5
a4
v4
od (v4 ) = 0
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a5
x4
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⎛0
⎜
⎜0
⎜0
⎜
⎜0
⎝
0
1
0
1
0
0
0
0
1⎞
⎟
1⎟
2⎟
⎟
0 ⎟⎠
63
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• Let D’=D-x.
• It is clear that D’ is acyclic.
• As the order of D’ is n-1 =k, by induction,
there is an ordering x2,x3,…,xn of vertices in
D’ such that the adjacency matrix M’ of D’ is
upper triangular.
• Since id(x1)=0, the adjacency matrix M of D
is
⎛ 0 a1, 2 a1,3 ... a1,n ⎞
• We shall prove by induction on the order n
of D.
• If n=1, the result holds.
• Assume that the result holds if n=k, where
k≥1.
• Now let n = k+1.
• Since D is acyclic, D has a vertex x with
id(x)=0.
• Let x1=x.
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The following statements are equivalent:
(1) D is acyclic;
(2) Every walk is a path;
(3) It is possible to order the vertices in D
such that A(D) is upper triangular.
Prove that (1) ⇒ (3).
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id (v1 ) = id (v3 ) = 0.
Theorem
• The following digraph D is acyclic .
• Can you find an order of the vertices in D
such that A(D) is upper-triangular? How?
x3
v3
a3
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Example
x1
a2
a6
⎜
0
M =⎜
⎜M
⎜
⎝0
65
M'
Hence
M is upper triangular.
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⎟
⎟
⎟
⎟
⎠
66
Algorithm
Algorithm
• Let D=(V,A) be an acyclic digraph. We shall
find an ordering of vertices in D such that its
adjacency matrix is upper-triangular.
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• D =(V, A) is an acyclic digraph.
•
•
•
•
•
•
Step 1: S := ∅ and s:=0.
Step 2: H:=D-S.
Step 3: Choose a vertex w in H with idH(w)=0.
Step 4: s:=s+1 and xs :=w;
Step 5: If s=|V|, then stop.
Step 6: S:=S∪{xs}. Go to step 2.
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3.5 Tournaments
Theorem 3.5.1
• A tournament is a digraph whose underlying
graph is a complete graph.
• The following digraph is a tournament on 4
vertices.
• Let D be a tournament and x be a vertex in
D which has the maximum out-degree.
• Then for any vertex y in D, there is a
directed path from x to y of length at most
2.
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Proof
• Let x1,x2,…,xk be the vertices in D which x is
adjacent to. So od(x) = k.
• If y is some vertex xi, then xy is a directed
path.
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…
y
x
x1
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x
…
x2
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• Now assume that yx is an arc.
• There must be some xi such that xiy is an arc in
D.
• Otherwise, yxi is an arc in D for all i=1,2,…,k,
implying that od(y)>k, contradicting the
assumption that x has the maximum out-degree.
• Thus xxiy is a path of length 2.
…
y
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xk
x1
71
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…
x2
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xk
72
Theorem 3.5.2
Theorem 3.5.2
• Any tournament D has a spanning directed
path.
• Proof.
• For every pair of vertices x and y, either xy
or yx is an arc.
• Thus D is unilaterally connected.
• Hence D has a spanning directed path.
• If D is a strongly connected tournament of
order n, then D contains a directed cycle
of length k for all k=3,4,…,n.
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• First show that D has a directed cycle of
order 3.
• Let x be any vertex in D.
• Let A = {w∈V(D): (w,x) is an arc in D}.
• Let B = {w∈V(D): (x,w) is an arc in D}.
• Since D is strongly connected, both A and B
are not empty.
x
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A
B
75
• Now we show that if D has a directed cycle
of order k, where 3≤k≤n-1, then D has a
directed cycle of order k+1.
74
• Since D is strongly connected, there must be
one arc (y,z), where y∈B and z∈C.
• Thus xyzx is a directed cycle of order 3.
y
…
z
…
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…
…
A
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x
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B
76
• Case 1: D contains a vertex w not on C
such that
( w, x ) ∈ A( D ), ( y , w) ∈ A( D ) for some x, y on C
• Denote the cycle C by x1x2…xkx1 with the
direction x1→x2 → …→xk→ x1.
• Assume that (x1,w) is an arc in D.
C
x1
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W
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78
• Since (w,x) is an arc in D for some vertex x
in C, there exists the minimum integer in
{1,2,…,k} such that (w,xi) is an arc in D.
• It is clear i>1. Then (xi-1,w) is an arc in D.
• Thus we have a cycle of order k+1:
• Case 2: for every vertex w not on C, either
(w,x) is an arc for all vertices x on C, or (x,w) is
an arc for all vertices x on C. Let
A1 = {w ∈V ( D) / V (C) : (w, x) ∈ A( D) for all x in V (C)}.
A2 = {w ∈V ( D) / V (C) : ( x, w) ∈ A( D) for all x in V (C)}.
x1 L x i −1 wx i L x k x1 .
C
C
xi-1
xi
x1
W
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• Since D is strongly connected, both A1 and A2
are not empty.
• Again since D is strongly connected, there
must be an arc (y,z) in D, where y∈A2 and
z∈A1.
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• Then the following cycle is of order k+1:
x1 yzx 3 L x k x1 .
• This completes this proof.
C
A1
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y
z
A2
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Theorem 3.5.3
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82
End
• A tournament of order n is strongly
connected if and only if D has a directed
cycle of order n, where n>2.
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