GRAPH THEORY - PART 2 pptx

These trees are sometimes called construction trees, de-composition trees, factorization trees or grammatical trees.. Each edge ofT is a bridge, and therefore uandvbelong to different c

Connectivity of Graphs

2.1 Bipartite graphs and trees

In problems such as the shortest path problem we look for minimum solutions that satisfy the given requirements The solutions in these cases are usually subgraphs without cycles

Such connected graphs will be called trees, and they are used, e.g., in search algorithms for

databases For concrete applications in this respect, see

T.H CORMEN, C.E LEISERSON AND R.L RIVEST, “Introduction to Algorithms”, MIT Press, 1993

Certain structures with operations are representable as trees

These trees are sometimes called construction trees,

de-composition trees, factorization trees or grammatical trees.

Grammatical trees occur especially in linguistics, where

syn-tactic structures of sentences are analyzed On the right there

is a tree of operations for the arithmetic formulax
(y +z)+y

+

x +

y

z y

Bipartite graphs

DEFINITION A graphG is called bipartite, ifV has a partition to two subsets X and Y such that each edgeuv 2 E

Gconnects a vertex ofX and a vertex ofY In this case,(X; Y )

is a bipartition ofG, andGis(X; Y )-bipartite.

A bipartite graphG(as in the above) is a complete(m; k)

-bipartite graph, ifjXj = m,jY j = k, anduv 2 E

Gfor all

u 2 Xandv 2 Y

All complete (m; k)-bipartite graphs are isomorphic Let

K

m;k denote such a graph

A subsetX  V is stable, ifG[X ℄is a discrete graph

K 2;3 The following result is clear from the definitions

Theorem 2.1 A graphGis bipartite if and only ifV has a partition to two stable subsets.

Example 2.1 Thek-cubeQ

kof Example 1.5 is bipartite for allk Indeed, considerA = fu j

uhas an even number of1

0

sgandB = fu j uhas an odd number of1

0

sg:Clearly, these sets partitionB

k

, and they are stable inQ

k

Theorem 2.2 A graphGis bipartite if and only if it has no odd cycles.

Proof ()) LetGbe(X; Y )-bipartite For a cycleC : v

1

! : : ! v

k+1

= v

1 of lengthk, v

1

2 Ximpliesv

2

2 Y,v 3

2 X, ,v

2i

2 Y,v 2i+1

2 X Consequently,k + 1 = 2m + 1is odd, andk = jCjis even

(() Suppose that all cycles inGare even First, we observe that it suffices to show the claim for connected graphs Indeed, ifGis disconnected, then each cycle ofGis contained in one of the connected components, G

1

; : ; G

p, ofG IfG

i is(X i

; Y i )-bipartite, then(X

1 [ X

2

[

[ X

p

; Y

1 [ Y 2 [

[ Y

p is a bipartition ofG Assume thus thatGis connected Letv 2 Gbe a chosen vertex, and define

X = fx j d

G (v; x) is eveng; Y = fy j d

G (v; y) is oddg : SinceGis connected,V

G

= X [ Y Also, by the definition of distance,X \ Y = ; Letu; w 2 Gbe both inX or both inY, and letP : v

?

! uandQ: v

?

! wbe (among the) shortest paths fromvtouandw

Assume thatx is the last common vertex of P and Q:P = P

1 P

2, Q = Q

1 Q

2, where P

2

: x

?

! uandQ

2 : x

?

! ware independent SinceP andQare shortest paths,P

1andQ

1 are shortest pathsv

?

! x Consequently,jP

1

j = jQ 1

j

So jP

2

j and jQ

2

j have the same parity Therefore Q

1

2

P

2

: w

?

! u is an even path It follows that

u and w are not adjacent in G, since otherwise

Q

1

2

P

2

(uw)would be an odd cycle ThereforeG[X℄

and G[Y ℄are discrete induced subgraphs, and Gis

u

w

P 1

Q 1

P 2

Q 2 uw

Checking whether a graph is bipartite is easy

In-deed, this can be done by using two ‘opposite’

colours, say 1 and2 Start from any vertexv

1, and colour it by1 Then colour the neighbours ofv

1 by

2, and proceed by colouring all neighbours of an

al-ready coloured vertex by an opposite colour

If the whole graph can be coloured, then G is

(X; Y )-bipartite, whereXconsists of those vertices

with colour 1, and Y of those vertices with colour

2; otherwise, at some point one of the vertices gets

both colours, and in this case,Gis not bipartite

1 2

2

1

2 1 1 2 1

2

Theorem 2.3 (ERDÖS (1965)) Each graph G has a bipartite subgraph H  G such that

"

H



1

2

"

G.

Proof LetV

G

= X [ Y be a partition such that the number of edges betweenXandY is as large as possible Denote

F = E \ fuv j u 2 X; v 2 Y g ;

and letH = G[F ℄ ObviouslyHis a spanning subgraph, and it is bipartite By the maximum condition,

d H (v)  1

2 d G (v) ;

since, otherwise, v is on the wrong side (That is, if v 2 X, then the pair X

0

= X n fvg, Y

0

= Y [ fvgdoes better that the pairX; Y.) Now

"

H

= 1

2 X

v2H d H (v)  1

2 X

v2G 1

2 d G (v) = 1

2

"

G :

u

Bridges

DEFINITION An edgee 2 E

G is a bridge of the graphG,

if G e has more connected components than G, that is, if

e) >

In particular, and most importantly, an edgeein a connected

Gis a bridge if and only ifG eis disconnected On the right

the two horizontal lines are bridges The rest are not

Theorem 2.4 An edgee 2 E

Gis a bridge if and only ifeis not in any cycle ofG.

Proof First of all, note that e = uv is a bridge if and only ifu and v belong to different connected components ofG e

()) If there is a cycle inGcontaining e, then there is a cycleC = eP : u ! v

?

! u, whereP : v

?

! uis a path inG e, and soeis not a bridge

(() Assume thate = uvis not a bridge Henceuandvare in the same connected com-ponent ofG e IfP : v

?

! u is a path inG e, theneP : u ! v

?

! uis a cycle inGthat

Lemma 2.1 Letebe a bridge in a connected graphG.

(i) Then e) = 2.

(ii) LetHbe a connected component ofG e Iff 2 E

H is a bridge ofH, thenf is a bridge

ofG.

Proof For (i), let e = uv Since eis a bridge, the endsu andv are not connected inG e Letw 2 G SinceGis connected, there exists a pathP : w

?

! vinG This is a path ofG e, unlessP : w

?

! u ! vcontainse = uv, in which case the partw

?

! uis a path inG e For (ii), iff 2 E

H belongs to a cycleCofG, thenCdoes not containe(sinceeis in no cycle), and thereforeCis insideH, andf is not a bridge ofH u

DEFINITION A graph is called acyclic, if it has no cycles An acyclic graph is also called a

forest A tree is a connected acyclic graph.

By Theorem 2.4 and the definition of a tree, we have

Corollary 2.1 A connected graph is a tree if and only if all its edges are bridges.

Example 2.2 The following enumeration result for trees has many different proofs, the first

of which was given by CAYLEYin 1889: There aren

n 2

trees on a vertex setV ofnelements.

We omit the proof

On the other hand, there are only a few trees up to isomorphism:

trees 47 106 235 551 1301 3159 7741 19 320 The nonisomorphic trees of order6are:

Theorem 2.5 The following are equivalent for a graphT.

(i)T is a tree.

(ii) Any two vertices are connected inTby a unique path.

(iii)T is acyclic and"

T

=  T

1.

Proof Let

T

= n Ifn = 1, then the claim is trivial Suppose thus thatn  2 (i))(ii) LetT be a tree Assume the claim does not hold, and letP; Q: u

?

! vbe two different paths between the same verticesuandv Suppose thatjP j  jQj SinceP 6= Q, there exists an edgeewhich belongs toP but not toQ Each edge ofT is a bridge, and therefore

uandvbelong to different connected components ofT e Henceemust also belong toQ; a contradiction

(ii))(iii) We prove the claim by induction onn Clearly, the claim holds forn = 2, and suppose it holds for graphs of order less thann LetT be any graph of ordernsatisfying (ii)

In particular,T is connected, and it is clearly acyclic

LetP : u

?

! vbe a maximal path inT, that is, there are no edgese, for whichP eoreP is

a path Such paths exist, because is finite It follows that , since, by maximality,

if vw 2 E

T, thenw belongs to P; otherwiseP (vw) would be a longer path In this case,

P : u

?

! w ! v, where vw is the unique edge having an end v The subgraph T v is connected, and therefore it satisfies the condition (ii) By induction hypothesis,"

T v

= n 2, and so"

T

= "

T v

+ 1 = n 1, and the claim follows

(iii))(i) Assume (iii) holds for T We need to show that T is connected Indeed, let the connected components ofT beT

i

= (V i

; E i ), for i 2 [1; k℄ Since T is acyclic, so are the connected graphsT

i, and hence they are trees, for which we have proved thatjE

i

j = jV i

j 1 Now,

T

=

P

k

i=1

jV i

j, and"

T

= P k i=1 jE i

j Therefore,

n 1 = "

T

= k X

i=1 (jV i

j 1) =

k X

i=1 jV i

j k = n k ;

Example 2.3 Consider a cup tournament ofnteams If during a round there arek teams left

in the tournament, then these are divided intobk pairs, and from each pair only the winner continues Ifk is odd, then one of the teams goes to the next round without having to play How many plays are needed to determine the winner?

So if there are14teams, after the first round7teams continue, and after the second round

4teams continue, then2 So13plays are needed in this example

The answer to our problem is n 1, since the cup tournament is a tree, where a play corresponds to an edge of the tree

Spanning trees

Theorem 2.6 Each connected graph has a spanning tree, that is, a spanning graph that is a

tree.

Proof LetH  Gbe a minimal connected spanning subgraph, that is, a connected spanning subgraph ofGsuch thatH eis disconnected for alle 2 E

H Such a subgraph is obtained fromGby removing nonbridges:

 To start with, letH

0

= G

 Fori  0, let H

i+1

= H i e

i, wheree

i is a not a bridge ofH

i Sincee

iis not a bridge, H

i+1is a connected spanning subgraph ofH

iand thus ofG

 H = H

k, when only bridges are left

Corollary 2.2 For each connected graph G,"

G

  G

1 Moreover, a connected graphG

is a tree if and only if"

G

=  G

1.

Proof LetT be a spanning tree ofG Then"

G

 "

T

=  T

1 =  G

1 The second claim

Corollary 2.3 Each treeT with

T

 2has at least two leaves.

Proof Let`be the number of leaves ofT By Corollary 2.2 and the handshaking lemma,

2
 T

2 = 2
"

T

= X

v2T d T (v) =

X

d (v)>1

d T (v) + `

 2
(

T

`) + ` = 2


T

` ;

Example 2.4 In Shannon’s switching game a positive player P and a negative player N play on a graphGwith two special vertices: a sourcesand a sinkr.P andN alternate turns

so that P designates an edge by+, andN by Each edge can be designated at most once

It isP’s purpose to designate a paths

?

! r (that is, to designate all edges in one such path), andNtries to block all pathss

?

! r(that is, to designate at least one edge in each such path)

We say that a game(G; s; r)is

 positive, ifP has a winning strategy no matter who begins the game,

 negative, ifN has a winning strategy no matter who begins the game,

 neutral, if the winner depends on who begins the game.

The game on the right is neutral

s r

LEHMANproved in 1964 that Shannon’s switching game(G; s; r)is positive if and only if there existsH  Gsuch thatHcontainssandrandHhas two spanning trees with no edges

in common.

In the other direction the claim can be proved along the following lines Assume that there exists a subgraph H containing sand r and that has two spanning trees with no edges in common ThenP plays as follows IfN marks by an edge from one of the two trees, then

P marks by+an edge in the other tree such that this edge reconnects the broken tree In this way,P always has two spanning trees for the subgraph H with only edges marked by+in common

In converse the claim is considerably more difficult to prove

There remains the problem to characterize those Shannon’s switching games(G; s; r)that are neutral (negative, respectively)

The connector problem

To build a network connectingnnodes (towns, computers, chips in a computer) it is desirable

to decrease the cost of construction of the links to the minimum This is the connector

prob-lem In graph theoretical terms we wish to find an optimal spanning subgraph of a weighted

graph Such an optimal subgraph is clearly a spanning tree, for, otherwise a deletion of any nonbridge will reduce the total weight of the subgraph

Let thenG

be a graphGtogether with a weight function : E

G

! R + (positive reals)

on the edges Kruskal’s algorithm (also known as the greedy algorithm) provides a solution

to the connector problem

Kruskal’s algorithm: For a connected and weighted graphG

of ordern: (i) Lete

1be an edge of smallest weight, and setE

1

= fe 1

g (ii) For eachi = 2; 3; : : n 1in this order, choose an edgee

i

=

2 E

i 1 of smallest possible weight such thate

idoes not produce a cycle when added toG[E

i 1

℄, and letE

i

= E

i 1 [ fe

i

g

The final outcome isT = (V

G

; E

n 1

By the construction, T = (V

G

; E

n 1 is a spanning tree of G, because it contains no cycles, it is connected and has n 1 edges We now show that T has the minimum total weight among the spanning trees ofG

SupposeT

1 is any spanning tree ofG Lete

k be the first edge produced by the algorithm that is not inT

1 If we adde

k toT

1, then a cycle C containing e

k is created Also, C must contain an edge ethat is not inT When we replaceebye

kinT

1, we still have a spanning tree, sayT

2 However, by the construction,(e

k )  (e), and therefore(T

2

 (T

1 Note thatT

2 has more edges in common withT thanT

1 Repeating the above procedure, we can transformT

1toT by replacing edges, one by one, such that the total weight does not increase We deduce that(T )  (T

1 The outcome of Kruskal’s algorithm need not be unique Indeed, there may exist several optimal spanning trees (with the same weight, of course) for a graph

Example 2.5 When applied to the weighted

graph on the right, the algorithm produces the

sequence: e

1

= v

2 v

4, e 2

= v 4 v

5, e 3

= v 3 v

6, e

4

= v

2

v

3ande

5

= v 1 v

2 The total weight of the spanning tree is thus 9

Also, the selection e

1

= v 2 v

5,e 2

= v 4 v

5,e 3

= v

5

v

6,e

4

= v

3

v

6,e 5

= v 1 v

2gives another optimal solution (of weight 9)

v 1

v 2

v 3

v 4

v 5

v 6

3

4

2

1 1

2 2 2

3

Problem Consider treesT with weight functions : E

T

! N Each tree T of ordernhas exactly n

2

paths (Why is this so?) Does there exist a weighted treeT

of ordernsuch that the (total) weights of its paths are1; 2; : :

n 2

?

In such a weighted treeT

 different paths have differ-ent weights, and each i 2 [1;

n 2

℄is a weight of one path Also,must be injective

No solutions are known for anyn  7.

2 1 5 8 4

TAYLOR (1977) proved: ifT of ordernexists, then necessarily n = k

2

or n = k

2 + 2for somek  1

Example 2.6 A computer network can be presented as a graphG, where the vertices are the node computers, and the edges indicate the direct links Each computervhas an addressa(v),

a bit string (of zeros and ones) The length of an address is the number of its bits A message

that is sent tov is preceded by the address a(v) The Hamming distance h(a(v); a(u)) of two addresses of the same length is the number of places, where a(v) and a(u)differ For example,h(00010; 01100) = 3andh(10000; 00000) = 1

It would be a good way to address the vertices so that the Hamming distance of two vertices

is the same as their distance inG In particular, if two vertices were adjacent, their addresses should differ by one symbol This would make it easier for a node computer to forward a message

A graph G is said to be addressable, if it

has an addressing a such that d

G (u; v) = h(a(u); a(v))

000

100

010

110 111

We prove that every treeT is addressable Moreover, the addresses of the vertices ofTcan

be chosen to be of length

T

1.

The proof goes by induction If

T

 2, then the claim is obvious In the case

T

= 2, the addresses of the vertices are simply 0 and 1

Let thenV

T

= fv 1

; : ; v k+1

g, and assume that d

T (v 1

= 1 (a leaf) and v

1 v 2

2 E

T

By the induction hypothesis, we can address the tree T v

1 by addresses of length k 1

We change this addressing: let a

i be the address of v

i in T v

1, and change it to 0a

i Set the address ofv

1 to1a

2 It is now easy to see that we have obtained an addressing forT as required

The triangleK

3is not addressable In order to gain more generality, we modify the address-ing for general graphs by introducaddress-ing a special symbolin addition to 0 and 1 A star address

will be a sequence of these three symbols The Hamming distance remains as it was, that is, h(u; v) is the number of places, whereu andv have a different symbol 0 or 1 The special symboldoes not affecth(u; v) So,h(10  01; 0  101) = 1andh(1   ; 00  ) = 0

We still want to haveh(u; v) = d

G (u; v)

We star address this graph as follows:

a(v

1

= 0000, a(v

2

= 10  0, a(v

3

= 1  01, a(v

4

=   11 These addresses have length 4 Can you design a

star addressing with addresses of length 3?

v 3

v4

WINKLERproved in 1983 a rather unexpected result: The minimum star address length of

a graphGis at most

G

1.

For the proof of this, see VANLINT ANDWILSON, “A Course in Combinatorics”

2.2 Connectivity

Spanning trees are often optimal solutions to problems, where cost is the criterion We may also wish to construct graphs that are as simple as possible, but where two vertices are always connected by at least two independent paths These problems occur especially in different aspects of fault tolerance and reliability of networks, where one has to make sure that a break-down of one connection does not affect the functionality of the network Similarly, in a reliable network we require that a break-down of a node (computer) should not result in the inactivity

of the whole network

Separating sets

DEFINITION A vertex v 2 G is a cut vertex, if

v) > A subsetA  V is a separating

set, ifG Ais disconnected We also say thatA

sep-arates verticesuandv, ifuandvbelong to different

connected components ofG A

IfGis connected, thenvis a cut vertex if and only ifG vis disconnected, that is,fvgis

a separating set We remark also that ifA  V

Gseparatesuandv, then every pathP : u

?

! v visits a vertex ofA

Lemma 2.2 If a connected graphGhas no separating sets, then it is a complete graph.

Proof If

G

 2, then the claim is clear For

G

 3, assume thatGis not complete, and let

uv 2 = E

G NowV n fu; v is a separating set The claim follows from this u

DEFINITION The (vertex) connectivity number(G)ofGis defined as

(G) = minfk j k = jAj; G Adisconnected or trivial; A  V

G

g :

A graphGisk-connected, if(G)  k

In other words,

 (G) = 0, ifGis disconnected,

 (G) = 

G

1, ifGis a complete graph, and

 otherwise(G)equals the minimum size of a separating set ofG

Clearly, ifGis connected, then it is 1-connected

DEFINITION An edge cutF ofGconsists of edges so thatG F is disconnected Let

 0 (G) = minfk j k = jF j; G F disconnected; F  E

G

g :

For trivial graphs, let

0 (G) = 0 A graphGisk-edge connected, if

0 (G)  k A minimal edge cutF  E

Gis a bond (F n fegis not an edge cut for anye 2 F)

Example 2.7 Again, if G is disconnected, then



0

(G) = 0 On the right,(G) = 2and

0 (G) =

2 Notice that the minimum degree isÆ(G) = 3

Lemma 2.3 LetGbe connected Ife = uvis a bridge, then eitherG = K

2or one ofuorv

is a cut vertex.

Proof Assume thatG 6= K

2and thus that

G

 3, sinceGis connected LetG

u

= N



G e (u) and G

v

= N



G e

(v)be the connected components ofG econtaining u and v Now, either



Gu

 2(anduis a cut vertex) or

Gv

Lemma 2.4 IfF be a bond of a connected graphG, then F ) = 2.

Proof SinceG Fis disconnected, andF is minimal, the subgraphG (F n feg)is connected for eache 2 F Henceeis a bridge inG (F n feg) By Lemma 2.1,G F has exactly two

Theorem 2.7 (WHITNEY(1932)) For any graphG,

(G)  

0 (G)  Æ(G) :

Proof AssumeGis nontrivial Clearly,

0 (G)  Æ(G), since if we remove all edges with an endv, we disconnectG If

0 (G) = 0, thenGis disconnected, and in this case also(G) = 0

If

0

(G) = 1, thenGis connected and contains a bridge By Lemma 2.3, eitherG = K

2orG has a cut vertex In both of these cases, also(G) = 1

Assume then that 

0 (G)  2 Let F be an edge cut of G with jF j = 

0 (G), and let

e = uv 2 F ThenF is a bond, andG F has two connected components