Connected Graphs and Connectivity

Connectivity 1May 2014

Connected Graphs and Connectivity

prepared and Instructed by Shmuel Wimer

Eng. Faculty, Bar-Ilan University

Connectivity 2

The Friendship Theorem

May 2014

Theorem. (Erdös et. al. 1966) Let be a simple, -vertex graph, in which any two vertices (people) have exactly one common neighbor (friends). Then has a vertex of degree (everyone’s friend).

Proof. Suppose in contrary that . We first show that is regular. Consider two non adjacent vertices and , and assume w.l.o.g that .

Since and have a single common neighbor , has no subgraph. 𝑥 𝑦

𝑧1

𝑧 2


We establish a one-to-one mapping of to .

For each denote by the common neighbor of and .

𝑥𝑁 (𝑥 ) 𝑦 𝑁 ( 𝑦 )𝑧

𝑣𝑓 (𝑣 )𝑣

𝑓 (𝑣 )impossible

is therefore one-to-one mapping from to , hence .

There is , hence for any non adjacent vertices of .


Any other vertex is not a neighbor of at least one of , say ( has no ). By the same argument as above for and there is .

What about ? It could be the friend of all, but by the contrary assumption there is .

It has therefore one non adjacent vertex , and by same argument as above there is .

In conclusion . is therefore -regular.

We next look for a relation between and by counting the number of 2-edge paths in .


By the theorem hypothesis, any two vertices have a unique common adjacent.

Picking two vertices yields a total of distinct 2-edge paths.

All in all there is .

For there are distinct paths, yielding a total of distinct 2-edge paths.

To investigate the possible values of , we examine the vertex adjacency matrix of .

Since is -regular, each row and column of has 1s.


Let us consider .

Each row and column of has 1s. The diagonal of has therefore elements .

Since two vertices have one and only one common neighbor , there is .

is therefore , where is the matrix of all s and is the identity matrix.

the rank of is so it has an eigenvalue with multiplicity and eigenvalues with multiplicity .

has therefore one eigenvalue and eigenvalues .


has therefore eigenvalues with total multiplicity and eigenvalue with multiplicity .

Since is simple, ‘s diagonal entries are all , so its trace is , and so is the sum of the eigenvalues.

Consequently, there is some integer (follows from the ) such that .

The only integers and solving is , implying , hence , contradicting the supposition . ■

The eigenvalues of are the square root of . By 2-edge path counting argument there is .

Connectivity 8

Euler Tours

May 2014

A tour of a connected graph is a closed walk traversing each edge of a graph at least once.

Let be Eulerian and an Euler tour with initial and terminal vertex . Each time an internal vertex occurs, two edges are accounted.

It is called Euler tour if each edge is traversed exactly once. is Eulerian if it admits an Euler tour.

is therefore even for all , and also for .

Eulerian graph is therefore necessarily even.


Proof. Let be a maximal trail but not closed. Since is open, the terminal edge has odd incident ’s edges.

But then has another non traversed incident edge which contradicts that is maximal. ■

Theorem. A finite graph (parallel edges and loops are allowed) is Eulerian iff it is connected and all its vertex degrees are even.

Proof. Necessity was shown. For sufficiency, Let be a maximal trail in (must be closed by the lemma).

Lemma. Every maximal trail in an even graph is closed.


𝐺 ′𝑇𝑮

𝑣𝑒

Let be a maximal trail in , starting at along . By the lemma is closed.

If , let . is an even graph. There must be an edge incident to at a vertex .

Once the traversal of reached , we could switch to , consume its edges and return to v, a contradiction to being maximal trail at . ■

𝑇 ′


Theorem. For a connected nontrivial graph with odd vertices (why even?), the minimum number of pairwise edge disjoint trails covering the edges is .

Proof. The internal nodes of trails contribute even degree, and their terminals odd degree.

A trail has two terminals at most (zero if it is a tour) hence trails at least are required.

One trail at least is required since . It was shown also that for one trail (a tour) suffices, so at least max is required for .


Add an edge connecting each paired odd vertices. is connected and each vertex is even, hence an Euler tour exists.

Traversing the Euler tour, a new trail starts each time an edge of is traversed, yielding a total of edge-disjoint trail cover of . ■ G

𝐺 ′

To see that edge-disjoint trails cover , pair up the odd vertices of arbitrarily.


Algorithm. (Fleury 1883, Eulerian trail construction)

Input: A connected graph with at most two odd vertices.

Initialization: Start at an odd vertex if exists, otherwise start arbitrarily at any vertex.

Iteration: Traverse from the current vertex any non cut-edge, unless there is no other alternative.

Theorem. If has one non trivial component and at most two odd vertices, then Fleury’s algorithm constructs an Eulerian trail.

Proof. By induction on . Immediate if .


If is even, it has no cut-edge. Otherwise, the removal of that edge would leave two separate components, each with a single odd degree vertex, which is impossible. (why?)

Suppose the construction claim holds for .

Consider . turn to odd degree vertices. Starting from , by the induction hypothesis a Eulerian trail exists. Then close to a tour along .

Suppose has two odd vertices . If , starting from , the rest graph has by induction an Eulerian trail from to .

Connectivity 15

?

May 2014

Since is connected, there is a path . Since is odd, there is an edge not on .

Assume first . Remove from . Is connected?

So let .

Yes, otherwise would have been a single odd vertex of its component, which is impossible. (why?)

𝑃𝑢

𝑥

𝑣𝑮


has two odd vertices and .

By induction there is an Eulerian trail, which can then be extended by to an Eulerian trail of .

If then is even.

By induction there is an Eulerian tour with starting (and terminating) at . It is then extended into an Eulerian trail by . ■

Layout of CMOS Compound Gates

May 2014 Connectivity 17

Connectivity 18

The Chinese Postman Problem

May 2014

(Guan Meigu 1962) A postman has to traverse all the roads of a town (a graph ), where every road (an edge) has a positive weight (e.g. length, time, biting dogs ). The postman starts and ends at the same vertex.

If is even, an Eulerian tour is optimal. Otherwise, edges must be repeated (multigarph, parallel edges).Edges are duplicates to produce an even graph. The problem is therefore to minimize the total weight of edge duplicates producing an even graph.

Find a closed walk of minimum weight that traverses all the edges.


Edges need not be multiplied more than once. (why?)

If an edge is used three or more times, two duplicates can be removed while stays even.

4 4

4 4

7

2

2

7

1 2 2 13

3

3

3

1 1

1 1

Cost: .

Cost: .

Better solution:.

Duplicated edge connecting odd and even vertices switch their evenness.


Edge addition must proceed until an odd vertex is met.

(Edmonds and Johnson 1973). If there were only two odd vertices, a shortest path connecting those solves the problem.

Given odd vertices, a weighted graph is defined. An edge weight is the length of the shortest path in G connecting the corresponding vertices.

The problem turns into finding a minimum weight perfect matching in the above weighted , for which a polynomial algorithm exists. ■

Connectivity 21

Connection in Digraphs

May 2014

A directed walk in a digraph is an alternating sequence of vertices and arcs , such that and are the tail and head vertices of the arc , respectively, .

is the portion of starting at and ending at .

𝝏+¿ (𝑿 )¿

𝑿 is the outgoing arcs connected to .


Theorem. Given digraph , let . is reachable from iff , there is .

Proof. Let be a directed path from to . Consider any such that .

Let be reachable from and let be the last vertex on and its successor. Then and hence .

Conversely, suppose that is not reachable from , and let be the set of vertices reachable from .

There is . Since no vertex of is reachable from , the out-cut . ■


Definition. A digraph is strongly connected if ordered vertex pair there is path in .

An Eulerian tour in implies .

This is also sufficient for the edges of to belong to a strongly connected component. The proofs are similar to those of undirected graphs.

Algorithm. (Directed Eulerian Tour)

Input. A digraph that is an orientation of a connected graph, satisfying .


Step 1. Choose a vertex . Derive by reverting the edge directions of .

Construct a BFS spanning tree of rooted at . (It is possible since is strongly connected. Proven in next Theorem).

Step 2. Let be the reversal of . Designate the arc of .

Step 3. Construct an Eulerian tour from , where leaving from a vertex is on edge of only if all other outgoing arcs have already been used. ■

Since , a path exists for each , hence is strongly connected.

Connectivity 25

𝑮

𝑻

𝑣

May 2014

𝑻 ′

𝑮 𝑮 ′𝑣


Theorem. If is multi digraph with one non trivial component and , then the algorithm constructs an Eulerian tour of .

Proof. The construction of by BFS (oriented) must reach all . If it did not, let be the set of those reached and .

An arc within contributes one to the in-degree and one to the out-degree of ’s vertices.

is connected to only by entering arcs, otherwise could be expanded. Such arcs contribute only to the in-degree of ’s vertices.


The total in-degree of ’s vertices is therefore greater than their total out-degree, which is impossible since . Hence .

The algorithm starts traversal from . We show that it must terminate at and consume .

Notice that all entering arcs of belong to .

The traversal leaves a vertex along an edge only after all the other out-arcs are consumed.

Since , it implies that all ’s in-arcs are also consumed, in particular, all those of .


Finally, spans , so all the arcs must be traversed. ■

Therefore, entering into with an arc () implies that all the vertices and incident arcs of the sub tree rooted at are consumed.

Application. Testing the position of a rotating drum.

Is there a cyclic arrangement of binary digits, such that the -bit words obtained by successively sliding -bit window are all distinct?

By encoding the currently read -bit word (with a mounted sensor) the position of the drum is known.


0 00

01

11

111

0

0

01

1

0

solves the problem for .

The problem can be solved using Eulerian digraph.

Associate the distinct ()-bit words with the vertices of a digraph .


101 111

011

110

0 00

01

11

111

00

01

1

0Place an arc from sequence to sequence if the LSBs of a agree with the MSBs of .

000 010

001

100

1

1 1 1

1

11

10

000

0

00 0

Label an arc with the LSB of .

110

100

101

011

111

0

0

0

1


For each sequence there are two out-going arcs labeled 0 and 1.

There are also two in-coming arcs labeled with the vertex’s code LSB. Hence is Eulerian.

head vertex

tail vertex

The -bit code obtained by appending the arc bit (LSB) to the bits at a tail vertex, agrees with the bits at the head vertex. ■


Application. Street-Sweeping Problem.

Curbs of a city are described by a digraph .

Curbs are swept in the traffic direction. A two-way street implies two parallel oppositely directed arcs.

A one-way street implies two parallel arcs of same direction.

In NYC parking is prohibited form some street curbs each weekday to allow for street sweeping.

This defines a sub-graph of , consisting of the arcs available for sweeping on that day.


The problem is how to sweep while minimizing deadheading time (no sweeping).

Each has a deadheading time .

If is Eulerian no deadheading time is needed.

Otherwise, arcs of are duplicated or arcs of are added (not being swept).

Let satisfy , , and satisfy , .

Set and . There is


Since in the super-digraph the in and out degrees must be balanced, the additions are paths in from to , whose cost is the shortest path length.

The Eulerian super-digraph must add arcs whose tails are in and arcs whose heads are in .

The cost of shipment of one unit from to is (shortest path length), with .

This turns into a Transportation Problem with supplies for and demands for .

We have solved a similar problem of farms and factories in the graph matching discussion. ■

Connectivity 35

Cuts and Connectivity

May 2014

It is desired to preserve network service when some nodes or connections break.

For expensive connections it is desired to maintain connectivity preservation with as few edges as possible.

Graphs and digraphs are assumed loopless.

Definitions. A set of a graph is a separating set or vertex cut if has more than one component.

is -connected if for every such there is .

The connectivity is the smallest vertex cut size.


Example. Though has no separating set, we define .

What is ?

Every induced subgraph of having at least one and is connected.

Hence either or must be included in a separating set.

Since and are separating sets by themselves, there is . ■


A -d cube has vertices, obtained from two copies by connecting matched vertices.

is -regular. Deletion of the neighbors of a vertex separates and hence .

Example. What is the connectivity of a -dimensional cube ?

To prove that we show by induction that every vertex cut has at least vertices.


is obtained by matching the corresponding vertices of two copies and .

Let be any vertex cut of .

If and are connected then is connected too, unless contains one end vertex of each of the matching edges.

But then for , hence we may assume that is disconnected.

has vertices, hence by induction , hence has vertices in .


If would not have vertices in then would be connected and all the vertices of have neighbors in (by the matching edges).

would therefore be connected unless has at least one vertex in , yielding . ■

When is not a clique, deleting all the neighbors of a vertex disconnects , so , but equality does not necessarily hold.

Connectivity 40

Edge Connectivity

May 2014

Perhaps that the transceivers (vertices) of a network are so reliable that more than never fail, hence communication is guaranteed.It is desired that the links (edges) are also designed so it is hard to separate by edge deletion.

Definitions. A disconnecting set of edges is a set such that has more than one component.

Given , denotes the edges having one vertex in and one in . An edge cut is of the form , where .


is -edge connected if every disconnecting set has at least edges.The edge-connectivity of is the minimum size of a disconnecting set.Every edge cut is a disconnecting set since there is no path from to .The converse is false. The 3 edges of are disconnecting set, but not an edge cut. Still, there isProposition. Every minimal disconnecting set is an edge cut.Let and have more than one component.


there must be some component whose outgoing edges are deleted.Hence contains the edge cut . is not a minimal disconnecting set unless . ■

Deleting one endpoint of each edge of disconnects .

It is therefore expected that will always hold, unless a vertex deletion eliminates a component of , producing a connected subgraph. Theorem. .Proof. follows immediately since the deletion of the incident edges of a vertex disconnects .


To show let be a minimum edge cut.

If every vertex of is adjacent to every vertex of then .

We therefore assume that there exist , but an edge does not exist.

Let T be a vertex set of .

𝑥

𝑦𝑆 𝑆

𝑇𝑇


Since and belong to different components of , is a separating set.

The vertices of can be associated with distinct edges connecting to , hence . ■

Connectivity 45

-connected Graphs

May 2014

A communication network is fault-tolerant if there are alternative paths between vertices.

The more vertex disjoint paths (except ends) the better.

Lemma. A graph G is connected iff for every non trivial partition , , there is where and .

Proof. Suppose is connected, and let and be a partition.

Since is connected, there is a path between every and .


Let be the last vertex of in and its successor. is the desired edge.Conversely, if is disconnected, let be a component of . Then and is a partition.

There cannot exist an edge between and , otherwise would not be a component. ■

The above lemma shows that each pair of vertices is connected with a path iff is -edge-connected.

We subsequently generalize this characterization to -edge-connected graphs and to -connected graphs.


Theorem. (Whitney 1932) A graph , , is -connected iff each are connected with a pair of internally-disjoint paths (disjoint vertices except and ).

Proof. Suppose that any two vertices are connected with a pair of internally-disjoint paths.

Deletion of one vertex cannot disconnect those vertices, and at least two vertex deletion is required, hence G is 2-connected.Conversely, suppose that is -connected. We prove by induction on (shortest path) that two internally-disjoint -paths exist.


For the base . is still connected, since ( is -connected).

A -path is internally-disjoint from , which being an edge, has no internal vertices, hence two disjoint paths exist.

Assume by induction that has a pair of internally-disjoint -paths for .

Let and let be the vertex before on the -path.

, and by the induction hypothesis there are vertex-disjoint paths and .


Since is connected, there is a -paths , and .

If is vertex-disjoint of or we are done, since and either of or are edge disjoint paths. .

Otherwise, let intersect both and .

Assume w.l.o.g that its last common vertex is on .

Combining the -path of with -path of yields a vertex-disjoint path to path. ■

Rx

u vw

PQ


Theorem. (-connected graph characterization)If , the following conditions are equivalent.• is connected and has no cut-vertex.• For all , there are internally-disjoint -paths.• For all , there is a cycle through and .• , and every pair of edges of is on a common cycle.


Let be a graph with two distinguished vertices.

The local connectivity is the maximum number of pairwise internally disjoint . denotes the smallest vertex cut separating and .

Theorem. (Menger 1927, Göring’s proof 2000)Let be non adjacent. Then .

Proof. Let

There is , because any family of internally disjoint meets any -vertex-cut in at least distinct vertices.


We subsequently show by induction on that .

We can assume that there is an edge incident neither to nor to .

Otherwise, every is of lengths , so the interim vertices of the -paths define an -vertex-cut and the Theorem’s conclusion follows immediately.

Set . Because is a subgraph of there is

By induction there is .


We may assume that .

Otherwise, , hence and the Theorem’s conclusion follows.

Thus, in particular, .

Let be a minimum -vertex-cut in .

Since every -vertex-cut of together with either end of is an -vertex-cut of G , there is , yielding

.

Connectivity 54

𝑥 𝑦

May 2014

𝑢 𝑣𝑒

𝑋 𝑌𝑆

Let the set of vertices reachable from in .

Since , cannot be an -vertex-cut of ().

There must exist therefore an -path in , that includes , where w.l.o.g and .


Let us contract into a single vertex , and denote the outcome by .

Likewise, let us contract into a single vertex , and denote the outcome by .

Every -vertex-cut in is necessarily so in . Otherwise, there was an -path in avoiding .

𝑥 𝑦𝑢

𝑋 𝑆 𝑮 /𝒀𝑦𝑥

𝑣

𝑌𝑆𝑮 /𝑿


The subgraph of would then contain an -path avoiding , impossible since is an -vertex-cut in .

Consequently ().

On the other hand, which is an -vertex-cut of , is also such in , and therefore .

Consequently, , and is a minimum -vertex-cut in .

By the induction hypothesis (), there are internally disjoint xy-paths in , and each must lie on one and only of them.


Assume w.l.o.g that , and .

Likewise, there are internally disjoint -paths in , obtained by contracting to , such that , and .

𝑥𝑢

𝑦𝑣

𝑌𝑆𝑋

𝑄1 ,…,𝑄𝑘𝑃1 ,…,𝑃𝑘


It follows that there are internally disjoint -paths in , , and .

Consequently . ■

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Connected Graphs and Connectivity