Graphs

1
Chapter 9

• Graphs

By courtesy of Prof. Cheng-Chia Chen
2
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.)
• ..

3
Table of contents

• Introduction
• Graph terminology
• Representing graph and graph isomorphism
• Connectivity
• Euler and Hamilton paths
• Shortest path problems
• Planer graphs
• Graph coloring

4
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).

5
Formal definitions

• Def 2 (multigraph) G (V,E,f) where
• f E-gt 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) gt e1 and e2 are parallel or
  multiple edges.
• Def 3 A pseudo graph G (V,E,f) where
• f E -gt u,v u,v ? V.
• Note if f(e) u gt 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 -gt V2.
• If f(e1) f(e2) gt e1 and e2 are parallel
  or multiple edges.
• Note f(e1) (u,v) /\ f(e2) (v,u) /\ u ¹ v
  gt e1 and e2 are not parallel edges (opposite
  directions).

6
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-gtL 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-gt 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

7
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. gt
  Influence graph (digraph)
• (u,v) ? E iff u, v know each other gt
  Acquaintance graph
• 
  (undirected graph)
• Ex3 Round-robin tournaments
• One where each team plays each other team
  exactly once.
• V the set for all teams
• (u,v) in E iff team u beats team v.
• Ex 4-team round-robin tournament

8
Graph model (cont'd)

• Ex 4 (precedence graph and concurrent
  processing)
• eg 1. x y 1
• 2. z f(x)
• gt (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, gt (u,w) ? E.
• Hence E is transitive.

9
Precedence graph of a program

• Ex 4
• s1 a 0
• s2 b 1
• s3 a a1
• s4 d ba
• s5 e d1
• s6 e cd
• gt What is the precedence graph of this program?

s1
s2
s3
s4
s6
s5
10
Precedence graph of a program

• Ex 4
• s1 a 0
• s2 b 1
• s3 a a1
• s4 d ba
• s5 e d1
• s6 e cd
• gtProblems
• 1. Minimum time needed to complete the program
  using multiple processors? (Depth)
• gt length of longest paths 1
• 2. Minimum of processors needed to complete
  the program in minimum time? (Width)
• gt p1 s1,s3,s4,s6, p2 s2,-,-,s5

s1
s2
s3
s4
s6
s5
11
Graph terminology

• Def 1 G (V,E) undirected graph
• if u --e-- v gt
• u and v are adjacent (or neighbors)
• Edge e is incident with u ( v).
• e connects u and v.
• u and v are end points of e.
• Def 2 G (V,E) undirected graph v ? V gt
• 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 gt v is isolated.
• If deg(v) 1 gt v is pendant (or hang).
• Ex In Fig. 1, E has degree 3, D is pendant and G
  is isolated.

12
Handshaking theorem

• Theorem 1 G (V,E) an undirected graph
• gt S v ? V deg(v) 2 E. // ??? ???
• 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.
• gt 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)
• gt Sv in V /\ deg(v) is odd deg(v) is even gt
  v in V deg(v) is odd is even. QED

13
Terminology for digraphs

• Def 3 G (V,E) a digraph
• if u --e--gt 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.

14
Underlying undirected graph

• G (V,E,f) a directed graph
• gt Its underlying undirected graph is a pseudo
  graph G' (V,E,f') where f' E -gt 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 gt
  (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))
  
• gt G' ?

15
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 gt 2) (V,E) where V
  1,2,...,n and
• E i,i1 i 1,...,n-1 U n,1.
• Ex C3,C4,C5,C6 ?
• Ex 6 wheels (n gt 2) Wn (V,E) where V
  0,1,...,n and
• E 0,i i 1,...,n U i,i1 i
  1,...,n-1 U n,1.
• W3,W4,W5 ?
• Ex7 n-cubes Qn (V,E) where V 0,1n 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 ?

16
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?)
• Ex9 Is K3 bipartite? (no! why?)
• Ex10 Which of Fig G and H is bipartite?
• Ex11 Km,n (complete bipartite graph)
• Km,n (u1,...,um U v1,...,vn, E , where
• E ui, vj i1..m v 1..n
• eg K2,3, K3,3 ?
• Problem Km,n has ? edges

17
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 (n2m) nm/2 O(n lg n) m lg n

18
New graphs from old

• Def 6 subgraph G (V,E) a graph
• H (V',E') is a subgraph of G iff
• V Í V and E' Í E and
• H is a graph (i.e., if e Î 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 gt 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
• gt G1 U G2 ?
• Def 7' Disjoint union of graphs (skip)
• Disjoin union of G1 and G2 ?

19
Graph representation and graph isomorphism

• Graph representation suitable only for finite
  graphs
• adjacent list (for graphs w/o multiple edges)
  O(EV)
• adjacent matrices O(V2)
• incident matrices (or list) O(VxE).
• adjacent list for each vertex v, av is the
  list of all adjacent vertices of v.
• adjacent matrix Mu,v1 iff u,v (or (u,v) if
  directed) is an edge.
• incident matrix Iu,e 1 iff u in f(e), or
  (for directed graph)
• iu,e 1 (or -1) iff u is the
  starting (or ending) vertex of e.
• Ex (p558 fig2)

20
Graph isomorphism

• Def . 1 G1 (V1,E1), G2 (V1,E2) two simple
  or digraphs
• G1 and G2 are isomorphic iff a bijection h
  V1-gtV2 s.t.
• for all u,v in V1,
• u,v (or (u,v)) Î E1 iff h(u),h(v)
  (or (h(u),h(v))) Î 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 gt 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.

21
Connectivity

• Def 1 G (V,E,f) an undirected multi-graph
• A path a of length n gt 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
• ux0----x1-----x2----x3----x4----x5----
  .----xn-1----Xnv
• If G is a simple graph gt 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 iltj n, ei ¹ ej ).

22
Connectivity (cont'd)

• Def. 2 G (V,E,f) a directed multigraph
• A path a of length n gt 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 gt a can be identified by the
  sequence of vertices x0u, 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 0ltiltjltn1, ei ¹ ej ).
• Def 3 An undirected graph is connected if there
  is a path between every pair of distinct vertices.

23
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 ltj n s.t. ei ej.
• gt the path b e1,..,ei-1,ej1,..,en or
  e1,..,ei,ej1,..,en is another path from u to v
  with length lt a, a contradiction!. QED

join xi and xj if xi xj
24
Note Theorem 1 does not hold for circuit (cycle)
25
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
  --(et1)--xt1-----xnv,
• and
• a2 x0 --(e1)-- x1 --(e2)-- x2 --(e3)----(et)--xt
  --(et1)--yt1-----ymv,
• 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 et1, et1 the first edges with
  et1 ? et1.
• let J be the least number in t1,,m such that
  yJ xs where s is any vertex occurring in path
  a1 (i.e., yJxs ? x0,,xn).
• Note since ym xn v, such J must exist.

26
A theorem about simple circuits

• Now it is easy to see that
• 1. If s t then xs --(es1)--
  xs1----(et)xt U
• xt--(et1)--yt1----(eJ-1)-yJ is a
  circuit. This is not possible since all edges of
  the circuit are part of a2.
• 2. If s gt t then xt--(et1)--xt1---(es-1)--xs
  U
• xt--(et1)--yt1----(eJ-1)-yJ ---() is
  a simple circuit.
• Note xt,xt1,,xs ? xt,yt1,,yJ xt,
  yJxs
• If there were ea eb gt f(ea) xt, xt1xs
  xt,yt1yJ
• gt abt1 gt e t1 et1 a
  contradiction!
• o/w, by def, () is a simple circuit.

27
Xt
et1
xs
u
yJ
et1
v
es
xs
eJ
yJ
eJ
yJ-1
28
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

29
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., V1Ç V2 ¹ Æ ).
• gt v ? V1ÇV2 gt 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)

30
Connected components (cont'd)

• Pf Let i 0 and Gi G'. If Gi is maximal
  then we are done.
• o/w, there is connected Gi1 contains Gi.
• If Gi1 is maximal, then we are done o/t let
  i i1 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. Ui1,k Gi Í G since every Gi Í G.
• 2. G Í Ui1,kGi since every edge and every
  vertex
• must belong to some connected
  component.

31
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 Í V, and
• Eu e f(e) Í Vu Í 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 VuVv /\ EuEv
  and
• 2. for all e ? E if f(e) u,v gt GuGv and
  e Î Eu.
• Hence the set of connected components of G Gu
  u Î V.
• Note Connected components in graphs play a role
  like equivalence classes in equivalence relations.

32
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 Î V, Gu (Vu,Eu) (u, Eu)
  where
• ES is the set of edges restricted to the
  vertex set S, i.e.,
• e Î E f(e) Í S.
• 4. The set of all connected components of G
• (S, ES) S Î V/.

33
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.
• Ex 4 Determine all cut vertices
• and all bridges in the right graph.
• cut vertices ?
• bridges ?

34
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.
35
Path and isomorphism

• P(-) a property on graphs
• Ex P(G) "G has a simple cycle of length gt 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.

36
More on graph isomorphism

• Ex2 P2(G) "G has a simple circuit of length
  k", where k is a number gt 2, is an isomorphic
  invariant.
• Pf G1 (V1,E1), G2 (V2,E2) two simple
  graphs.
• Let h V1-gtV2 be any isomorphism from G1 to G2.
• Then if x0 --(e1)--gt x2--(e2)--gt x3--....
  --gt(en)--gtxn is a simple path on G1, then the
  sequence
• h(x0) --gt(h(e1))--gth(x2)--(h(e2))--gth(x3)--...--gt(
  (h(en))--gth(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.

G
H
37
Counting paths b/t vertices

• G (V,E) with V V1,...,Vn a simple graph
  with adjacent matrix M.
• gt 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 gt 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 lt 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
• St1..n (Mk-1)it x Mtj (MK)ij.

38
Example

• Ex 8 G (1,2,3,4, 12,13,24,34)
• gt M 0110100110010110
• gt M4 8008088008808008
• gt 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 (Sk1..n-1 Mk )ij ! 0 for
  all 1 iltj n.
• Pf G is connected gt
• for any i ? j, there is a simple path (of length
  t lt n) from vi to vj
• gt (Mt)ij gt 0 gt (Sk1..n-1 Mk )ij ! 0. QED

39
Euler and Hamilton paths
C

• The seven bridges problem
• Problem Is there a path
• passing through all bridges
• w/o crossing any bridge twice?
• Multigraph model
• Problem Is there a simple
• path of length 7?

A
D
B
The town of Konigsburg
B
40
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
• circuit a,e,c,d,e,b,a.
• G3 has a Eular path
• bedbadca.
• 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.

41
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 "gt" G (V,E) any multigraph.
• Let a x0 e1 x1 e2 x2 e2 x3 ... en xnx0 be any
  Eular circuit.
• For each vxi in V ? x0, since ei--gtxi --gt ei1
  we have
• deg(v) 2 j xj v and 0ltjltn and for x0
  we have
• deg(x0) 2xj xj x0 and 0ltjltn 2.
• Hence every vertex has even degree.
• (lt) by induction on E. If E0, by
  definition, it has a Eular ckt.
• O/W Let a x0 e1 x1 e2 x2 e3 ... en xnx0 be
  any simple circuit.
• (its existence will be shown later.)
• If a is a Eular circuit, then we are done. O/W

42
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 ÇVi ! for
  all 0ltiltk1.
• (o/w there is no path from x0 to vertices in
  Vi).
• For each i let xti be any vertex in x0,...,xn
  Ç Vi.
• Since Gi (Vi,Ei) and Ei lt E and every
  vertex in Vi has even degree, by ind. hyp. there
  is a Eular circuit ai xti -gt...-gt xti in Gi.
• Now join each ai with a at xti
• we can form a Eular circuit in G. QED.

43
Example (skip)
g
h

• a a b c d a is a simple circuit in G.
• removing all edges in a results in
• three connected components
• G1, G2 and G3 intersecting with
• a at a,d, b and c respectively.
• By ind. hyp.,
• Eular ckt a1 (a...a)
• (aedfghefa)
• a2 (b) and
• a3 (c i j c)
• gt join a and all ai, we obtain
• a Eular ckt of G
• (aedfghefa)b(cijc)da.

f
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44
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 (gt) G (V,E) any multigraph.
• Let a x0 e1 x1 e2 x2 e2 x3 ... en xn ! x0 be
  any Eular path.
• For each vxi in V ! x0 or xn, since ei--gtxi
  --gt ei1 we have
• deg(v) 2 j xj v and 0ltjltn and for v
  x0 or xn we have
• deg(v) 2xj xj v and 0ltjltn 1.
• Hence all vertices but x0 and xn have even
  degree.
• (lt) Let G' (V, EUe) where e is a new edge
  connecting a and b, which are vertices of G with
  odd degree.
• gt Every vertex of G has even degree. By
  theorem 1,
• a Eular path a a -gt .... -gtb-gt(e)-gta.
• gt removing e from a we get a Eular path of G.
  QED

45
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 as which
  eventually become a simple circuit as follows
• 0. Initially a1 x0 --e1-- x1. G1 (V, E
  \e1).
• note x1 in G1 has degree gt 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 klti 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 ai1 ai -- e--u xi1 u and i
  i1
• 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

46
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 xnx0 (n gt 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 3 vertices. gt If every
  vertex has degree ³ n/2, then G has Hamilton ckts.

47
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
• of n-bit length?
• Rule Let Gn (Vn,En)
• Qn be the n cube.
• A cycle of the disc
• (a grey code) corresponds
• to an Hamilton ckt in Qn.

48
Shortest path problems

• A weighted graph is a graph G (V,E) together
  with a weighting function wE-gtR.
• A shortest path from a to b is a path
• a e1 x1 e2 x2 ... en b s.t. Si1,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.

49
Example

• Running Time O(n2)

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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 u0a,u1,,uk.
• Lj(uk) Lk-1(uk) for all j gtk-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.

53
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
• Sa , Lk (a) 0 is the shortest length from
  a to a.
• v ? a gt Lk (v) w(a,v) is the length of the
  shortest from a to v without passing through any
  vertex in S.
• k j1 gt 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 lt Lj(uk). This shows (1) is true.

54
(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).

55
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.

56
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.

57
Planar graphs (cont'd)

• G0 contains only one vertex.
• Gi1 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 (ai1,bi1) be the edges added to Gi to form
  Gi1.
• case 1 ai1,bi1 ? Vi
• gt ai1 and bi1 must be on the boundary of a
  common region. (o/w crossing would occur!)
• gt add ai1,bi1 partition the region into 2
  subregion
• gt ri1 ri 1 ei1 ei 1 vi1 vi. gt
  ri1 ei1-vi12.
• case 2 a i1 ? Vn but b i1 ? Vi gt add
  ai1,bi1 does not produce any region. gt ri1
  ri ei1ei1 vi1 vi1
• gt ri1 ei1 -vi1 2. QED

58
More on planar graphs

• Ex4 G has 20 vertices, each of degree 3.
• gt regions ? r e - v 2 30 - 20
  2 12.
• Corollary G a connected planar simple graph
  with e edges and v vertices ³ 3. gt e 3v - 6.
• pf for each region R
• define deg(R) edges on the boundary of R.
• gt 2e S deg(R) each edge shared by 2
  regions S deg(v)
• Since deg(R) ³ 3 (simple graph),
• 2e S deg(R) ³ 3 r. gt 2e/3 ³ r e-v2 gt
  3v- 6 ³ e. QED
• Ex5 K5 is not planar.
• sol K5 has C(5,2) 10 edges and 5 vertices.
• gt 3v - 6 15-6 9 lt 10, violating the
  necessary condition !
• Ex6 K3,3, though satisfying the condition e
  3v - 6, is not planar.

59
More on planar graph

• G a connected planar simple graph of e edges and
  v gt 2 vertices and no simple ckt of length 3 gt e
  2v - 4.
• pf no ckt of length 3 gt every region has
  degree ³ 4.
• gt 2e S deg(R) ³ 4 r. gt e ³ 2r 2e-2v4
  gt 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.
• gt 2 x 6 - 4 8 lt 9. Hence not planar.
• General rule about planar graphs
• Def elementary subdivision
• u--------------v gt u------w------v.

60
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.

61
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

62
Transform maps into graphs

• Ideas
• regions ---gt vertices
• R1 and R2 share a border ---gt 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.

63
Graph coloring

• Def 1G(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 ?
• gt1. a problem studied for over 100 years !
  18501976,
• solved by AppelHaken
• 2. 2000 cases generated and verified by
  computer programs.
• 3. issue believable ?

64
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 gt 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.

65
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.
• gt A schedule of the final examinations
  corresponds to a
• coloring of the associated graph G.

1
I
Time courses I
1,6 gt II 2 III 3,5 IV 4,7
IV
II
2
7
3
III
I
6
IV
4
III
5