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CSCI 4260MATH 4150

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So far we have been mostly concerned with 'undirected' graphs ... can find a sequence of teams a,b,c,d,... where a beat b, and b beat c, and so on... – PowerPoint PPT presentation

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Title: CSCI 4260MATH 4150


1
CSCI 4260-MATH 4150
  • Lecture 11

2
Directed Graphs
  • So far we have been mostly concerned with
    undirected graphs
  • Today, we will focus on directed graphs

(u,v) is an arc the order is important u is
adjacent to v v is adjacent from u
v
u
3
Indegree vs. Outdegree
  • Indegree of v id(v) u u is adjacent to
    v
  • Outdegree of v
  • od(v) u u is adjacent from v

4
Theorem
For any directed graph G (V,E)
5
Orienting an undirected graph, underlying graph
G is the underlying graph of H
H is an orientation of G
Note that H does not have the arcs (u,v) and
(v,u) simultaneously
6
Connectivity in directed graphs
  • Our paths are now directed.
  • Weakly connected the underlying (undirected)
    graph is connected
  • Strongly connected there is a directed u-v path
    AND a directed v-u path for all u,v
  • In this case, we say the graph is strong.

7
Some basic properties
  • In the next few theorems, we will see the
    directed version some earlier ones.

8
(Warm-up) Theorem
  • If a digraph D contains a u-v walk of length l,
    then D contains a u-v path of length at most l

9
Theorem
  • A digraph is strong iff it contains a closed
    spanning walk

10
Theorem
  • A non-trivial, connected graph D is Eulerian iff
    od(v) id(v) for every vertex v.

11
Eulerian ? od(v) id(v) ?v

12
od(v) id(v) ?v ? Eulerian
  • Pick an initial vertex u and find the longest
    trail T
  • T must be closed, i.e a circuit.
  • Suppose T does not contain all edges
  • Let D D T
  • There must be a vertex w on T that is incident to
    an edge not in T
  • Note that od(v) id(v) for all v in D
  • Apply the same argument, there must be a circuit
    starting from w
  • But then T can be made longer a contradiction!

13
Theorem
  • A non-trivial (undirected) graph G has a strong
    orientation iff G contains no bridge.

14
Strong orientation ?no bridge
  • Take the contrapositive
  • If there is bridge,

15
No bridge ? Strong orientation
  • Remember, no bridge means that every edge is on a
    cycle
  • The main idea is to cover the graph with cycle
    and to choose an orientation for each cycle
  • Lets prove this formally

16
Tournaments
  • A tournament is an orientation of a complete
    graph
  • Think of it as a real tournament
  • Vertices teams
  • All teams play with each other
  • u ? v means u won the game
  • A transitive tournament is defined as arc (u,v)
    and (v,w) ? arc (u,w)

17
Theorem
  • A tournament is transitive iff it has no cycles

18
Transitive ? no cycles
  • Suppose a transitive tournament has a cycle

vk
v1
19
No cycles ? Transitive
  • If (u,v) and (v,w) are arcs in T
  • (w,u) can not be an arc in T
  • This means (u,w) is an arc

20
Theorem
  • If u is a vertex of maximum outdegree in a
    tournament T, then d(u,v) ? 2 for every vertex

21
Proof
Max outdegree vertex, degree k
22
Interpretation
  • If A is the team that won most matches,
  • For any team B
  • Either A defeated B or
  • A defeated a team that defeated B

23
Theorem
  • Every tournament contains a hamiltonian path

24
Proof
  • Suppose not. Take the longest path. There must be
    some vertex not on it

Cant have these red edges
v
25
Proof
v
26
Interpretation
  • In every tournament, we can find a sequence of
    teams a,b,c,d, where a beat b, and b beat c, and
    so on

27
But who is the best?
Remember transitive ? no cycles
28
Theorem
  • Every vertex in a non-trivial strong tournament
    belongs to a triangle

29
W
U
  • U ??? and W ???Why?
  • There must be an arc from w to some vertex in U

w
30
Theorem
  • A nontrivial tournament T is strong iff T is
    Hamiltonian

31
Hamiltonian ? Strong
  • ?

32
Strong ? Hamiltonian
  • Let C be the maximum length cycle and for
    contradiction assume that there exists a vertex
    v that is not on C

33
Case I
w
u
Cant happen
v
34
Case II ?
v
u
So, its the same case
v
35
Case II
It must be that v is either adjacent to or
adjacent from all vertices on C.
v
36
Cycle C
U ??
W ??
Why must this edge exist?
Can you find a bigger cycle?
37
Recap
  • Transitive tournaments have an absolute winner
  • Strong tournaments have hamiltonian cycles
  • Hence are not transitive

38
Also
  • Read sections 7.3 and 7.4 in the book
  • Its fun
  • Project proposal deadline is coming soon!
  • Next Thursday there will be an overview lecture,
    given by your TA
  • Come prepared

39
Next few weeks
40
HW2 is out today
  • You can discuss the problems with a single
    classmate
  • You have to put her/his name on your solution
  • You have to write your solutions independently
  • The school is enforcing a very strict cheating
    policy.
  • Basically, any type of cheating is reported all
    the way up to the dean
  • So, please dont!
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