prepared and Instructed by Shmuel Wimer Eng Faculty BarIlan University The Friendship Theorem May 2014 Connectivity 2 Theorem Erdös et al 1966 Let be a simple vertex graph in which any two vertices people have exactly one common ID: 661401
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Slide1
May 2014
Connectivity
1
Connected Graphs and Connectivity
prepared and instructed by Shmuel WimerEng. Faculty, Bar-Ilan UniversitySlide2
The Friendship Theorem
May 2014
Connectivity
2Theorem. (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.
Slide3
May 2014
Connectivity
3
We establish a one-to-one mapping of to . For each
denote by the common neighbor of and .
impossible
is therefore one-to-one mapping from
to
, hence
.
There is
, hence
for any
non adjacent
vertices of
.
Slide4
May 2014
Connectivity
4
Any other vertex is not a neighbor of at least one of , say ( has no
). By the same one-to-one mapping for and . In conclusion
. must therefore be -regular
.
We next look for a relation between
and
by counting the number of 2-edge paths in
.
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 one-to-one mapping for
and
.
Slide5
May 2014
Connectivity
5
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.
Slide6
May 2014
Connectivity
6
Let us consider . Since each row and column of has 1s, there is
.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
.
Slide7
May 2014
Connectivity
7
has therefore eigenvalues with and one eigenvalue .
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
-edge path counting there is
.
Slide8
Euler Tours
May 2014
Connectivity
8A 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
.
Slide9
May 2014
Connectivity
9
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 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.Slide10
May 2014
Connectivity
10
If
, let
.
is even. 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
, a contradiction to
being maximal trail at
.
■
Let
be a maximal trail in
, starting at
along
. By the lemma
is closed.
Slide11
May 2014
Connectivity
11
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
is required for
.
Slide12
May 2014
Connectivity
12
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
To see that
edge-disjoint trails cover
, pair up the odd vertices of
arbitrarily.
Slide13
May 2014
Connectivity
13
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 .
Slide14
May 2014
Connectivity
14
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 induction Fleury algorithm finds Eulerian trail. Then close to a tour along
.
Suppose
has two odd vertices
. If
, starting from
, Fleury algorithm finds Eulerian trail from
to
.
Slide15
?
May 2014
Connectivity
15Since is connected, there is a path
. Since , 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?)
Slide16
May 2014
Connectivity
16
has two odd vertices and .By induction Fleury algorithm finds a Eulerian trail, extendable by to an Eulerian trail of
. If then is even. By induction Fleury algorithm finds Eulerian tour starting (and terminating) at , extendable by to an Eulerian trail of
. ■ Slide17
Application: Layout of CMOS Gates
May 2014
Connectivity
17
polysilicondiffusion
metal
contactSlide18
The Chinese Postman Problem
May 2014
Connectivity
18(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 duplicated 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. Slide19
May 2014
Connectivity
19
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
1
3
3
3
3
1
1
1
1
Cost:
.
Cost:
.
Better solution:
.
Duplicated edge connecting odd and even vertices switch their evenness.Slide20
May 2014
Connectivity
20
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. ■ Slide21
Connection in Digraphs
May 2014
Connectivity
21A 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
.
is the
out-cut
(outgoing arcs) connected to
.
Slide22
May 2014
Connectivity
22
Theorem. Given digraph , let . is reachable from iff
, there is .
Proof. Let be a directed path from to
. Consider any
such that
and
.
Let
be
reachable from
and let
be the last vertex on
and
its successor. Then
and hence
.
Slide23
May 2014
Connectivity
23
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
.
Slide24
May 2014
Connectivity
24
An Eulerian tour in implies
.This is also sufficient, if the edges of belong to one connected component. Proofs are similar to undirected graphs. Algorithm. (Directed Eulerian Tour)Input. A digraph that is an orientation of a connected graph, satisfying
.
Since
, a
path
exists for each
, hence
is
strongly
connected (proven later).
Slide25
May 2014
Connectivity
25
Step 1. Choose a vertex . Derive by reverting the edge directions of .
Find a spanning tree
of
rooted at
(BFS, other). It is possible since
is strongly connected (proven later).
Slide26
May 2014
Connectivity
26
Step 2. Let be the reversal of . Designate ’s arcs. 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. ■
Slide27
May 2014
Connectivity
27
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.
Slide28
May 2014
Connectivity
28
is connected to only by entering arcs, otherwise (and ) 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
.
Slide29
May 2014
Connectivity
29
The trail must terminate at , since 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 vertices, and hence 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.
Slide30
May 2014
Connectivity
30
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?
0
0
0
0
1
1
1
1
1
1
0
0
0
1
1
0Slide31
May 2014
Connectivity
31
0
0
0
0
1
1
1
1
1
1
0
0
0
1
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
.
By encoding the currently read
-bit word (with a mounted sensor) the position of the drum is known.
Slide32
May 2014
Connectivity
32
0
0
0
0
1
1
1
1
1
1
0
0
0
1
1
0
Place an arc from sequence
to sequence
if the
LSB
s of
agree with the
MSB
s of
.
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
Label an arc with the LSB of
.
11
0
10
0
10
1
011
111
0
0
0
1
tail vertex
head vertex
101
111
011
110
000
010
001
100Slide33
May 2014
Connectivity
33
For each -bit sequence (vertex) there are two out-going arcs labeled 0 and 1.There are also two in-coming arcs labeled with the vertex’s LSB. Hence is Eulerian.
tail vertex
head vertex
The
-bit
codes
obtained by
appending the arc bit (LSB) to the
bits at
a tail
vertex are all distinct. Two successive codes agree on the
bits. ■
Slide34
May 2014
Connectivity
34
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 from 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. Slide35
May 2014
Connectivity
35
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
Slide36
May 2014
Connectivity
36
Since in the super-digraph the in and out degrees must be balanced, the additions should be paths in from to , which cost is the shortest path length. The Eulerian super-digraph must add
arcs with tail in and arcs with head in .
- shortest path length
supplies
demands
Slide37
May 2014
Connectivity
37
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
.
Transportation problem was introduced by Kantorovich (1939), solved by Hitchcock (1941) and many others.
■Slide38
Cuts and Connectivity
May 2014
Connectivity
38It 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.
Slide39
May 2014
Connectivity
39
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
.
■
Slide40
May 2014
Connectivity
40
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
.
Slide41
May 2014
Connectivity
41
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
w.l.o.g that
is
disconnected.
has
vertices, hence by induction
, hence
has
vertices in
.
Slide42
May 2014
Connectivity
42
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.
Slide43
Edge Connectivity
May 2014
Connectivity
43Perhaps 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
.
Slide44
May 2014
Connectivity
44
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 has no path from
to .
The converse is false. The 3 edges of
are disconnecting set, but not an edge cut. Still, there
is:
Proposition
. Every
minimal
disconnecting set is an edge cut.
Let
and
have more than one component.
Slide45
May 2014
Connectivity
45
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
.
Slide46
May 2014
Connectivity
46
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
be the vertex set
.
Slide47
May 2014
Connectivity
47
Since and belong to different components of , is a separating set.
The vertices of can be associated with distinct edges connecting to , hence
. ■
Slide48
-connected Graphs
May 2014
Connectivity48A communication network is fault-tolerant if there are alternative paths between vertices.The more vertex disjoint paths (except ends) the better.
Lemma. A graph 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
.
Slide49
May 2014
Connectivity
49
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.
Slide50
May 2014
Connectivity
50
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 these 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.
Slide51
May 2014
Connectivity
51
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 induction there are
vertex-disjoint paths
and
.
Slide52
May 2014
Connectivity
52
Since is connected, there is a -path , 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.
■
Slide53
May 2014
Connectivity
53
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.
(homework)
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
.
Slide54
May 2014
Connectivity
54
Theorem. (Menger 1927, Göring’s proof 2000)Let be non adjacent. Then
. Proof. Let
Let us show that
.
Let
be a
largest set
of internally disjoint
.
Each
path of
must meet at least one vertex of
any smallest
-vertex-cut, as otherwise
would have been connected.
Hence, the
-vertex-cut
must have at least
distinct vertices, yielding
.
Slide55
May 2014
Connectivity
55
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
t
he interim vertices of the
-paths define an
-vertex-cut
and
follows immediately.
Slide56
May 2014
Connectivity
56
Set . Because is a subgraph of there is
By induction there is .
Slide57
May 2014
Connectivity
57
We may assume that .Otherwise,
and 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
, there is
, yielding
.
Slide58
May 2014
Connectivity
58
Let
the set of vertices reachable from
in
. There is
.
Since
,
cannot be an
-vertex-cut of
(
).
Therefore, there must exist an
-path in
, that includes
, where w.l.o.g
and
.
Slide59
May 2014
Connectivity
59
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
.
Slide60
May 2014
Connectivity
60
The would then be an -path in 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
-paths
in
, and each
must lie on one and only of them.
Slide61
May 2014
Connectivity
61
Assume w.l.o.g that
, and . Likewise, there are
internally disjoint -paths
in
, obtained by contracting
to
, such that
, and
.
Slide62
May 2014
Connectivity
62
It follows that there are internally disjoint -paths in ,
, and .
Consequently
.
■
Slide63
May 2014
Connectivity
63
Example. Show that
for every , where is the number of connected components. Proof. By induction on . If
has only isolated vertices, so
and an equality holds.
Let
. Since the removal of an edge can turn a connected component into two, there is
(1)
.
Assume by induction that
(2)
.
Substitution of (2) in (1) yields
Slide64
May 2014
Connectivity
64
.