4. Connectivity
4.1. Connectivity
We recall the definitions connected, disconnected and component.
component.
If G is connected and G − W is disconnected,
disconnected, where W is a set of
vertices, then we say that W separates G, or
or that W is a vertexvertex-cut .
4.1. Connectivity
VertexVertex-cut and vertexvertex-connectivity
EdgeEdge-cut and edgeedge-connectivty
Whitney'
(G)
Whitney's connectivity theorem: (G) (G)
Further theorems for the relation of (G) and (G) for special
graphs
4.2. The Menger Theorem and its consequences
The Whitney'
Whitney's characterization
Application (Reliable network)
4.3. Flows in a graph
The value of a flow and the capacity of a cut
The MaxMax-FlowFlow-MinMin-Cut Theorem
A new proof for the Menger Theorem
4.4. The thoughness of a grpah
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If in G − W two vertices s and t belong to different components then
W separates s from t.
A graph G is k-vertexvertex-connected (k 2) if
either Kk+1
or it has at least k+2 vertices and no set of k – 1 vertices separate
G.
1
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2
Every graph that is not complete has a vertexvertex-cut:
cut: the set of all
vertices distinct from two nonadjacent vertices is a vertexvertex-cut.
cut.
A connected graph is also said to be 1-connected.
connected.
The maximal value of k for which a connected graph G is k-vertexvertexconnected is the vertexvertex-connectivity of G, denoted by (G). If G is
disconnected,
disconnected, we put (G)=0.
The vertexvertex-connectivity – (G) – of a graph G is
the minimum cardinality of a vertexvertex-cut if G is not complete,
complete, and
(G) = n – 1 if G = Kn for some positive integer n.
κ ( G ) = min { W : W ⊂ V ( G ) and W is a vertex - cut }
The two definition is equivalent.
If a graph G is k-vertexvertex-connected then (G) k.
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4
1
The edgeedge-connectivity (G) is defined analogously:
analogously:
ExamplesExamples-1:
A nontrivial graph has vertexvertex-connectivity 0 iff it is disconnected.
disconnected.
A graph G has vertexvertex-connectivity 1 iff G = K2 or G is connected
with cutcut-vertices.
vertices.
An edgeedge-cut X is minimal if no proper subset of X is also an edgeedgecut.
cut.
(G) 2 iff G is nonseparable of order 3 or more.
more.
An edgeedge-cut in a graph G is a set X of edges of G such that G – X is
disconnected.
disconnected.
If X is a minimal edge-cut of a connected graph G, then G – X
contains exactly two components.
A graph G is k-edgeedge-connected (k 2) if it has at least two vertices
and no set of at most k – 1 edges separates G.
The edgeedge-connectivity,
connectivity, (G), of a graph G is the minimum
cardinality of an edgeedge-cut of G if G is nontrivial, and (K1) = 0.
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6
Examples 2:
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(G) = (G) for any graph G?
A graph G is 2-edgeedge-connected iff it is connected, has at least two
vertices and contains no bridge.
(G) = 0 iff G is disconnected or trivial.
(G) = 1
(G) =1 iff G is connected and contains a bridge.
(G) = j
x
Kj
Examples 3:
Kj
(Cn) = (Cn) = 2
(Pj) = (Pj) = 1
(Kn) = (Kn) = n – 1
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It is often easy to determine the connectivity of a given graph. If
1 j n then
(Kj,n) = (Kj,n) = j
7
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8
2
Theorem 4.1. (R. A. Brualdi,
Brualdi, J. Csima,
Csima, 1991): Let G be a connected
graph of order n 3 that is not complete,
complete, For each edgeedge-cut X there
exists a vertexvertex-cut W of G such that |W
|W | | X |.
Case 2: There are vertices u in G1 and v in G2 that are not adjacent in
G.
For each edge e X, we select a vertex for W in the following
way:
Proof.
W.l.o.g.
W.l.o.g. we can suppose that X is a minimal edgeedge-cut of G. Then G –
X is disconnected,
disconnected, which contains exactly two components, say G1
and G2 with order n1 and n2, respectively.
respectively. Then n1 + n2 = n.
n. We
have two cases:
cases:
Otherwise,
Otherwise, select for W the vertex that is incident with e and
belongs to G1.
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9
Case 1: Every vertex of G1 is adjacent to every vertex of G2 in G.
Then | X | = n1n2. Since (n
(n1–1)(n
)(n2–1) 0, it follows that n1n2
n1+n2–1 = n – 1 and so | X | n – 1.
Since G is not complete,
complete, it contains two nonnon-adjacent vertices u
and v. Then W = V(G) – {u,v}
u,v} is a vertexvertex-cut of cardinality n – 2,
and so | W | < | X |.
If u is incident with e, then choose the other vertex (in G2) incident
with e for W.
Now |W |
| X |.
Since u,v V(G – W), and G – W contains no u – v path, so W is a
vertexvertex-cut.
cut.
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10
u
G1 :
G2 :
Theorem 4.2. (H. Whitney, 1932) : If G is a nonnon-trivial graph then
(G) (G)
(G)
Proof.
G:
If we delete all the edges incident with a vertex, the graph
becomes disconnected,
disconnected, so the second inequality holds.
holds.
Let us see the other inequality:
inequality:
If G is a complete graph then (G) = (G) = |G| – 1.
If (G) 1 then (G) = (G).
Suppose G is not complete and (G) = k 2 and
{x1y1, x2y2,…, xkyk} is a set of edges disconnecting G.
v
If G − {x1, x2,…, xk} is disconnected then (G) k.
If G − {x1, x2,…, xk} is connected then each vertex xi has
degree at most k. Deleting the neighbours of x1, we
disconnect G. Hence (G) k.
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3
Theorem 4.2. (H. Whitney, 1932) : If G is a nonnon-trivial graph then
(G) (G)
(G)
Proof.
If we delete all the edges incident with a vertex, the graph
becomes disconnected,
disconnected, so the second inequality holds.
holds.
Let us see the other inequality:
inequality:
If G is a complete graph then (G) = (G) = |G| – 1.
If (G) 1 then (G) = (G).
Suppose G is not complete and (G) = k
edgeedge-cut.
cut.
2. Let | X | = k be an
| X | = (G)
|U |
(G)
Then – by the Theorem 4.1. – there exists a vertexvertex-cut U such
that |U
|U | | X |. Thus
A graph G with (G) = 2,
2, (G) = 3 and (G) = 4.
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14
Let G1 be a component of G – W of smallest order, say n1. Thus
Theorem 4.3.(G. Chartrand, F. Harary,
Harary, 1968 ): Let G be a graph of
order n 2, and let k be an integer that 1 k n – 1. If
Graph Theory 3
n+k −2
deg v ≥
2
n1 ≤
Let v be a vertex of G1. Necessarily, v is adjacent in G only to
vertices of W or to other vertices of G1. Hence
for every v V(G), then G is k-vertexvertex-connected.
connected.
deg v ≤ l + ( n1 − 1 ) ≤ l +
Proof.
Suppose that the theorem is false.
false.
=
Then there is a graph G satisfying the hypothesis of the theorem such
that G is not k-vertexvertex-connected.
connected. Since G is not a complete graph.
Then there exists a vertexvertex-cut W such that |W
|W | = l
n−l
2
n−l
−1
2
n+l−2
n+k −3
≤
2
2
contrary to the hypothesis.
hypothesis.
k – 1.
So, the graph G – W is disconnected of order n – l.
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4
Theorem 4.4.: A nontrivial graph G is k-edge-connected iff there
exists no nonempty proper subset W of V(G) such that the number of
edges joining W and V(G) – W is less than k.
Proof.
A:
B:
Conversely, suppose that G is a k-edgeedge-connected graph.
If there should exist a subset W of V(G) such that j edges, j <
k, join W and V(G) – W, then the deletion of these j edges
produces a disconnected graph, which is again a
contradiction.
Assume that there exists no nonempty proper subset W of
V(G) for which the number of edges joining W and V(G) – W
less than k but that G is not k-edge connected.
The following theorem of Plesnik gives a sufficient condition for
equality to hold between (G) and (G).
Since G is nontrivial, this implies that there exist l edges, 0 <
l < k, such that their deletion from G results in a disconnected
graph H.
Let H1 be a component of H.
The number of edges joining V(H1) and V(G) – W is at most l,
where l < k, and this is a contradiction.
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Theorem 4.5. (Plesnik
(Plesnik,, 1975):
1975): If G is a graph of diameter 2, then
(G) = (G).
Proof.
Let S be an edgeedge-cut with | S | = (G),
(G), and let H1 and H2 be the two
components of G – S, where ni is the order of Hi (i=1,2)
i=1,2) with n1 n2.
Since diam G = 2,
2, every vertex of H2 is adjacent to some vertex of
H1. So
either each vertex of H1 is adjacent to some vertex of H2,
or each vertex of H2 is adjacent to some vertex of H1.
Therefore (G) = | S |
min{
min{n1,n2} = n1.
For u V(H1), let di(u)
(u) denote the number of vertices of Hi (i=1,2)
adjacent to u in G.
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5
Corollary 4.6.: If G is a graph of order n 2 such that
Thus degG u = d1(u) + d2(u), and so
δ ( G ) ≤ degG u = d1 ( u ) + d 2 ( u ) ≤ n1 − 1 + d 2 ( u )
deg u + deg v n – 1,
for each pair of u,v of nonadjacent vertices in G, then (G) = (G)
Proof
≤ λ( G ) − 1 + d2 ( u ) ≤ δ ( G ) − 1 + d2 ( u ).
Therefore,
Therefore, d2(u) 1 for each u V(G); that is, each vertex of H1 is
adjacent to some vertex of H2. Let V(H1) = {u1, u2,…,un1}. Then
λ( G ) = S =
n1
i =1
d 2 ( ui ) =
n1 − 1
i =1
d 2 ( u i ) + d 2 ( u n1 )
If deg u + deg v n – 1, then u and v have at least one common
neighbour.
neighbour.
So, the diameter of the graph is 2 and we can apply the theorem 4.5.
(G)
(G)
n1 – 1 + d2(un ).
1
and so n1 – 1 + d2(un1)
≥ n1 − 1 + d 2 ( u n1 ).
Thus (G) = (G).
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Exercises.
Exercises. (G. Chartrand and L. Lesniak page 74.)
1. Determine the connectivity and edgeedge-connectivity of each
complete k-partite graph.
2. Show that every k-connected graph contains every tree of order
k+1 as a subgraph.
3. Let H = G+K1, where G is k-connected.
connected. Prove that H is (k+1)(k+1)connected.
4. Prove the Corollary 3.19.
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5. Let a,b and c be positive integers with a b c. Prove that there
exists a graph G with (G) = a, (G) = b, and (G) = c.
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A nontrivial graph is connected if between every two distinct
vertices of G there exists at least one path. This fact can be
generalized in many ways,
ways, most of which involve,
involve, either directly or
indirectly,
indirectly, a theorem due to Menger.
In this section we discuss the major ones of these,
these, beginning with
Dirac`s proof (1966)
1966) of Menger`s theorem itself.
itself.
We recall to the definition of separating subset of vertices: A set S of
vertices (or
(or edges) of a graph G is said to separate two vertices u and
v if the removal of the elements of S from G produces a
disconnected graph in which u and v lie in different components.
Certainly then, S is a vertexvertex-cut (edgeedge-cut)
cut) of G.
In the next graph there is a set S = {w1,w2,w3} of vertices that
separate the vertices u and v. No set with fewer than three vertices
separates u and v. As is gauranteed by Menger`s theorem,
theorem, stated
next,
next, there are three internally disjoint u–v paths in G.
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6
4.2. Menger's
Menger's Theorem
w1
A set S of vertices (or
(or edges) of a graph G is said to separate two
vertices u and v if the removal of the elements of S from G produces
a disconnected graph in which u and v lie in different components.
Certainly then, S is a vertexvertex-cut (edgeedge-cut)
cut) of G.
u
w2
v
Two u−v paths are independent if they have only the vertices u and v
in common.
common.
Theorem 4.7. (Menger'
(Menger's theorem,
theorem, 1927):
Let u and v be distinct nonnon-adjacent vertices of a graph G. Then the
minimum number of vertices separating u from v is equal to the
maximum number of independent u−v paths.
w3
We denote by Sk(u,v)
(u,v) the statement that the minimum number of
vertices that separate u and v is k.
25
Proof.
The result is true if u and v lie in different components of G or if u
and v lie in different blocks of G;
So, we may assume that the graphs under consideration are
connected and that u and v lie in the same block.
If Sk(u,v) is true, then the maximum number of internally disjoint u–v
paths in G is at most k.
Thus, if k = 1 then the result is true.
Suppose that the theorem is false. Then there exists a smallest
positive integer t ( 2) such that St(u,v) is true in some graph G but
the maximum number of internally disjoint u–v paths is less than t.
Among all such graphs G of smallest order, let H be one of
minimum size.
We now establish three properties of the graph H.
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Lemma 4.8.: For every two adjacent vertices v1 and v2 of H, where
neither v1 nor v2 is u or v, there exists a set of U of t–1 vertices of H
such that U {vi}, i = 1,2, separates u and v.
Proof.
First, we will show that if e = v1v2 then St(u,v)
(u,v) is false for H – e.
To prove this we claim that St-1(u,v) is true for H – e.
If not then there exists a set W of vertices that separates u and v in
H– e, where |W | t – 2.
Graph Theory 3
A graph illustrating the Menger'
Menger's theorem.
theorem.
Then W separates u and v in both H – v1 and H – v2.
So W
{vi}, i=1,2 separates u and v in H, which is impossible.
Thus, as claimed, St-1(u,v) is true in H – e. So, there exists a set U
which separates u and v in H – e, where |U|= t – 1.
However, then U
Graph Theory 3
{vi}, i = 1,2, separates u and v in H.
28
7
t – 1 vertices
H:
v1
Lemma 4.9. For each vertex w ( u, v ) in H, not both uw and vw are
edges of H.
Proof.
v2
Suppose that it is not true. Then St-1(u,v) is true for H – w.
Then, however, H – w contains t – 1 internally disjoint u – v paths.
paths.
v
u
So, H contains t internally disjoint u – v paths, which is a contradictcontradiction.
St(u,v)
(u,v) is true in H, but the maximum number of internally disjoint u–
v paths is less than t.
NINCS KÉSZ
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Proof.
Suppose that the statement is false.
Define Hu as the subgraph induced by the edges on all u–wi paths in
H that contain only one vertex of W.
Define Hv similarly.
We will show that V(Hu) V(Hv)=W.
Suppose that the above statement is not true. Then both Hu and Hv
have order at least t+2.
Define Hu* to consist of Hu, a new vertex v* together with all edges
v*wi.
Similarly, we define Hv* to consist of Hv, a new vertex u* together
with all edges v*wi.
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Observe that Hu* and Hv* have smaller order than H.
Lemma 4.10.: If W={w1,w2,…,wt} is a set of vertices which separates
u and v in H, then either uwi E(H) for all i (1 i t ) or vwi E(H)
for all i (1 i t ).
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*
*
So, St(u,v)
(u,v) is true in Hu and St(u,v)
(u,v) is true in Hv .
Therefore, there exist t internally disjoint u–v* paths in Hu* and t
internally disjoint u*– v paths in Hv* .
These 2t paths produce t internally disjoint u–v paths in H.
This contradicts the condition that H is among those of graphs for
which the number of internally disjoint paths is less then t.
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8
Thus we may write P: u,u1,u2,…,v where u1,u2
v.
By Lemma 4.8.,
4.8., there exists a set U of t–1 vertices such that both U
{u1} and U {u2} separate u and v.
Since v is not adjacent to u1, therefore every vertex of U
adjacent to u.
{u1} is
Since u is not adjacent to u2, therefore every vertex of U
adjacent to v.
{u2} is
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Theorem 4.12.: If G is a k-connected graph and v,v1,v2,…,v
,…,vk are k+1
distinct vertices of G, then there exist internally disjoint v–vi paths
(1 i k).
Proof.
Construct a new graph H from G by adding a new vertex u together
with the edges uvi, i = 1,2,…,k.
Conversely, suppose that G is a nontrivial graph that is not kconnected but in which every pair of distinct vertices are
connected by at least k internally disjoint paths. Certainly, G is
not complete.
Since G is not kk-connected, (G) < k. Let W be a set of (G)
vertices of G such that G–W is disconnected,
disconnected, and let u and v be
in different components of G–W.
The vertices u and v are necessarily nonadjacent;
nonadjacent; however,
however, by
the hypothesis,
hypothesis, there are at least k internally disjoint u–v paths.
By the Menger'
Menger's theorem,
theorem, u and v cannot be separated by fewer
than k vertices, so a contradiction arises.
arises.
35
34
33
This also produces a contradiction.
Graph Theory 3
Therefore there exists a set U of fewer than k–1 vertices
such that G – uv – U is a disconnected graph.
Therefore, at least one of G – (U {u}) and G – (U {v}) is
> Therefore,
disconnected,
disconnected, implying that (G) < k.
B:
l < k, which is
If uv E(G),
E(G), then the maximum number of internally
disjoint u–v paths in G – uv is l–1 < k–
k–1.
Thus d(u,v)
d(u,v) = 2, which is a contradiction.
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If uv E(G) then, by Theorem 4.7 (G)
contrary to hypothesis.
hypothesis.
By Lemma 4.9.
4.9. d(u,v)
d(u,v) 3.
Let P be a u – v path in H of length d(u,v).
d(u,v).
Theorem 4.11. (Whitney's
(Whitney's characterization): A graph G of order n
2 is k-(vertex)connected (1 k n–1) iff for each pair u,v of distinct
vertices there are at least k internally disjoint u–v paths in G.
Proof.
A:
Let G be a k-connected graph. Assume to the contrary, that
there are two vertices u and v such that the maximum number
of internally disjoint u–v paths in G is l, where l < k.
Now, we can return to prove the statement of the Menger's
Menger's theorem:
Since G is k-connected, H is k-connected.
By Theorem 4.11,
4.11, there exist k internally disjoint u–v paths in H.
The restriction of these paths to G yields the desired internally
disjoint v–vi paths.
One of the interesting properties of 2-connected graphs is that every
two vertices of such graphs lie on a common cycle. There is a
genaralization to k-connected graphs by Dirac:
Dirac:
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36
9
Theorem 4.13.: Let G be a k-connected graph, k 2. Then every k
vertices of G lie on a common cycle of G.
Proof.
Lemma 4.14.: C contains at least l+1 vertices.
Proof.
It is clear that l 2.
Replacing the edge w1w2 on C by the w1–w2 path determined by Q1
and Q2.
By Theorem 4.12.,
4.12., there exist internally disjoint w–wi paths, denoted
by Qi, 1 i l.
Let W be a set of vertices of G. Among all cycles of G, let C be a
cycle containing a maximum number, say l, vertices of W.
Then C can be labeled so that C:w1,w2,…,wl,w1, where wi W for 1
i l.
Suppose the contrary.
For k = 2 the result follows from the Theorem 2.5; hence we assume
that k 3.
We will to show that l = k. Assume to the contrary, that l < k, and let
w be a vertex of W which does not lie on C.
So, we get a cycle containing at least l+1 vertices of W, which is a
contradiction.
Therefore, C contains at least l+1 vertices.
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37
Thus, we may assume that C contains vertices w1,w2,…,wl,wl+1, such
that wi W for 1 i l. and wl+1 W
38
Both the Menger´s theorem and the Whitney‘s characterization have
“edge”-analogue:
Since C contains exactly l vertices of W, there are distinct integers s
and t, 1 s,t
l+1, such that one of the two vs–vt paths, say P,
determined by C contains no interior vertex belonging to W.
Replacing P by the vs–vt path determined by Pi ′, we obtain a cycle of
G containing at least l+1 vertices of W.
Corollary 4.16. : A graph is k-edgeedge-connected (k 2) iff it has at
least two vertices and any two vertices can be joined by k edgeedgedisjoint paths.
Let vi be the first vertex on Pi that belongs to C (possible vi=wi) and
let Pi ′ denote the w–vi subpath of Pi.
Theorem 4.15.(Ford
4.15.(Ford and Fulkerson,
Fulkerson, 1956): Let u and v be distinct
vertices of G. Then the minimal number of edges separating u from v
is equal to the maximal number of edgeedge-disjoint u−v paths.
Since k l+1, we may apply Theorem 4.12 again to conclude that
there exist l+1 internally disjoint w–wi paths, say Pi , 1 i l+1.
Graph Theory 3
This contradiction serves the desired result that l = k.
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40
10
Application of the connectivity.
Corollary 4.17. : A graph is k-edgeedge-connected (k 2) iff it has at
least two vertices and any two vertices can be joined by k edgeedgedisjoint paths.
If we think of a graph as representing a communication network, the
connectivity becomes the smallest number of communication
stations whose breakdown would jeopardise communication in the
system.
With the aid of the earlier theorem it is easy to prove an edge
analogue of the Menger´s theorem.
theorem.
The higher the connectivity and edge connectivity, the more reliable
reliable
the system.
The higher the connectivity and edge connectivity, the more expenexpensive is the system.
So, the tree network – which one can obtained by Kruskal'
Kruskal's
algorithm – is cheap, but it is not very reliable.
This lead to consider the following generalisation of the spanning
spanning
tree problem.
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41
The Reliable Communication Network Problem: Let k be given
positive integer and let G be a weighted graph. Determined a
minimumminimum-weight k-connected spanning subgraph of G.
If k
If k = 1 then this problem reduces to the spanning tree problem.
Graph Theory 3
We shall denote by by f(m,n)
f(m,n) the least number of edges that an mconnected graph on n vertices can have. (It is clear that m < n.)
From the Handshaking Lemma and the Whitney'
Whitney's Connectivity
Theorem it follows immediately that
2 then the problem is unsolved and is known to be difficult.
f ( m,n ) ≥
mn
.
2
We shall prove that equality holds by constructing an m-connected
graph Hm,n on n vertices that has exatly mn/2
mn/2 edges.
If G is a complete graph in which edge is assigned unit weight,
weight,
then the problem has a simple solution.
solution.
42
For a weighted complete graph on n vertices in which each edge is
assigned unit weight,
weight, a minimumminimum-weight m-connected spanning
subgraph is an m-connected graph on n vertices with as few edges as
possible.
possible.
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43
The structure of the construction depends on the parities of n and m.
Graph Theory 3
44
11
0
0
1
0
1
8
1
Case 1. m is even.
even. Let m = 2r. Then H2r,n is constructed as follows.
It has vertices 0,1,…,n0,1,…,n-1 and two vertices i and j are joined if i–r j
i + r (where addition is taken modulo n.)
2
2 7
7
2
7
3
Case 2. m odd,
odd, n even.
even. Let m = 2r + 1. Then H2r+1,n constructed by
first drawing H2r,n and then adding edges joining vertex i to vertex
i+(n/2)
i+(n/2) for 1 i n/2.
5
4
H5,8
H4,8
Theorem 4.17. (Harary
(Harary,, 1962): The graph Hm,n is m-connected.
connected.
Graph Theory 3
4
5
4
5
3 6
Case 3. m odd,
odd, n odd.
odd. Let m = 2r + 1. Then H2r+1,n is constructed
by first drawing H2r,n then adding edges joining vertex 0 to vertices
(n – 1)/2 and (n + 1)/2 and vertex i + (n+1)/2 for 1 i (n – 1)/2.
3 6
6
H5,9
If i = 0 then j = 6,7,1,2.
If i = 1 then j = 7,0,2,3.
45
4.3. Flows and Connectivity
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46
More formally:
formally: if for x V then we denote by
Let G be a (finite) directed graph with vertex set V and edge set E .
Γ + ( x ) = ( y ∈V : xy ∈ E ) ,
are the
the „outgoing“
outgoing“ neighbours,
neighbours,
We will study (static)
static) flows in G from a vertex s (the source)
source) to a
vertex t (the sink).
Γ − ( x ) = ( y ∈V : yx ∈ E ) ,
are the
the „ingoing“
ingoing“ neighbours
A flow f is a nonnon-negative function defined on the edges; the value
f ( xy) is the flow or current in the edge xy . For notational simplicity
we shall write f(x,y)
f(x,y) and a similar convention will be used for other
functions.
functions.
then a flow from s to t satisfies the following condition:
f ( x, y ) =
y∈Γ + ( x )
f ( z, x )
z∈Γ − ( x )
for each x V−
V−{s,t}.
s,t}.
The only condition a flow from s to t has to satisfy the Kirchoff's
current law: the total current flowing into each intermediate vertex
(that is different from s and t) is equal to the total current leaving the
vertex.
Graph Theory 3
47
Graph Theory 3
48
12
Since
What is the maximal flow value from s to t while we have capacity
constraints on edges which restrict the current through the edges?
=
x∈V − {s ,t }
u ∈{s ,t }
f ( x, y ) −
y∈Γ
+
(x)
f ( z,x )
z∈Γ
f ( z ,u ) −
z∈Γ
−
If xy ∈ E , then c(x,y)
c(x,y) 0 is the capacity of the edge.
(x)
We shall assume that the current flowing through the edge
cannot be more than the capacity c(x,y).
c(x,y).
f ( u,z )
y∈Γ
(u )
−
0 =
+
(u )
Given two subsets X,Y of V we write E( X ,Y ) for the set of directed
X−Y edges:
so we find that
f ( s, y ) −
+
y∈Γ ( s )
xy
f ( y,s ) =
−
y∈Γ ( s )
f ( y ,t ) −
−
y∈Γ ( t )
{
}
E( X ,Y ) = xy : x ∈ X , y ∈ Y .
f ( t, y )
+
y∈Γ ( t )
Whenever g : E →
The common value,
value, denoted by v(f),
v(f), is called the value of f or the
amount of flow from s to t.
Graph Theory 3
49
is a function,
function, we put
g ( X ,Y ) =
g( x , y )
e∈ E ( X ,Y )
Graph Theory 3
50
If S is a subset of V containing s but not t then E( S , S ) is called a cut
separating s from t.
5
The capacity of a cut E( S , S ) is denoted by c( S , S ) .
3
Some easy statements:
statements:
4
s
If we delete the edges of a cut then no positivepositive-valued flow from s to
t can be defined on the remainder.
remainder.
2
7
9
6
t
Conversely, if F is a set of edges after whose deletion there is no
flow (v(f)
v(f) = 0) from s to t then F contains a cut.
cut.
A cut with capacity 12
Graph Theory 3
51
Graph Theory 3
52
13
The capacity of a cut is at least as large as the value of any flow:
flow:
min c ( S , S ) ≥ max v ( f )
S is cut
v is flow
Do these values exist?
exist?
Since there are only finitely many cuts,
cuts, there is a cut whose capacity
is minimal.
Let f1,f2,… be a sequence of flows with limnv(fn) = v. Then, by
passing to a subsequence, we may assume that for each xy ∈ E the
the
sequence (fn(x,y))
(x,y)) is convergent, say to f(x,y).
f(x,y).
The function f is a flow with value c, that is a flow with maximal
capacity.
In this way one can show that even if some of the edges have infinite
infinite
capacity, there is a flow with maximal value which can be either
finite or infinite.
The existence of a flow with maximal value is slighly less trivial.
For every flow f we know that
v( f ) ≤
c( x , y )
xy∈E
and so v = sup v(f)
v(f) < .
Graph Theory 3
53
54
We claim that E( S ,S ) is a cut separating s from t with capacity v =
v(f).
v(f).
First, we will prove that S is a cut,
cut, i.e.
i.e. t
Theorem 4.18. (Max
(Max--flow minmin-cut theorem.)
theorem.) The maximal flow
value from s to t is equal to the minimum of the capacities of cuts
sepatating s from t.
Graph Theory 3
S.
Proof.
Graph Theory 3
εi = max { c( xi ,xi+1 )− f ( xi ,xi+1 ), f ( xi+1,xi ) } > 0
55
Graph Theory 3
for every i, 0 i k-1.
Let = min i. Then f can be augmented to a flow f * in the following
way:
if i > f(xi+1,xi) then increase the flow in xi xi+1 by .
if i f(xi+1,xi) then decrease the flow in xi+1xi by .
Define a subset S ⊂V recursively as follows:
Let s S.
If x S and c(x,y)
c(x,y) > f(x,y)
f(x,y) or f(y,x)
f(y,x) > 0 then y S.
Suppose the contrary:
contrary: t belongs to S. Because of the recursive
definition there is a sequence of vertices s = x0, x1, x2, … , xk = t such
that
We have already seen that there is a flow f with maximal value,
value,
say v, and the capacity of every cut is at least v.
So, in order to prove the theorem we have to show that there is a
cut with capacity v.
We will construct the cut in hand.
56
14
The maxmax-flow minmin-cut theorem remains valid if some of the edges
have infinite capacity but the maximal flow value is finite.
So, t S and therefore E( S ,S ) is a cut,
cut, separating s from t.
The proof of the previous theorem also provides a surprisingly
efficient algorithm for finding a flow with maximal value if the
capacity function is integral:
Clearly,
Clearly, f * is a flow,
flow, and its value is v(f *) = v(f)
v(f) + , contradicting
the maximality of f.
We know that v(f)
v(f) is equal to the value of the flow from S to S and
it is defined in the obvious way:
f ( x, y ) −
f ( x, y )
x∈S , y∈S
f0 ( x, y ) = 0
∀xy∈ E
Suppose we have constructed fi, and we find the set S belonging
to fi:
f i is a maximal flow,
If t ∉ S
flow, the algorithm terminates.
terminates.
x∈ S , y∈S
By the definition of S the first sum is exactly
c( x , y ) = c( S , S )
f i can be augmented to a flow fi+1 by increasing
If t ∈ S
the flow along a path from s to t (like in the theorem).
theorem).
x∈ S , y∈ S
and each summand in the second sum is zero.
zero. Hence c( S , S ) = v( f ),
as required.
required.
Since v(fi) is integer, so v(fi-1) < v(fi), and the sequence must end
in at most x,yc(x,y)
c(x,y) steps.
steps.
Graph Theory 3
57
Graph Theory 3
Theorem 4.19. (Integrality
(Integrality Theorem) : If the capacity function is
integral then there is a maximal flow which is also integral.
It is important that the algorithm does not state uniqueness:
uniqueness: it finds
one of the maximal flows (usually there are many)
many) and the theorem
claims that one of maximal flows is integral.
58
H
G
s
t
If instead of one source and one sink we take several of each, the
maxmax-flow minmin-cut theorem remains valid.
valid.
Let us scetch the proof:
proof:
If s1,s2,…,sk are the sources and t1,t2,…,tl are the sinks then E( S ,S ) is
a cut if si S and t j ∈ S for every i,j,
i,j, 1 i k, 1 j l.
Graph Theory 3
59
sources
Graph Theory 3
sinks
60
15
Theorem 4.20. : The maximum of the flow value from a set of
sources to a set of sinks is equal to the minimum of the capacity of
cuts separating the sources from the sinks.
sinks.
v1
What happens if we have capacity restrictions on the vertices, except
the source and the sink?
+
We are given a function c: V−
and every flow f from s to t
V−{s,t}
s,t}
has to satisfy the following inequality:
inequality:
f ( x, y ) =
+
y∈Γ ( x )
−
(2) 6
(1) 6
v3
s
(4) 6
f ( z , x ) ≤ c( x ) for every x ∈V −{ s ,t}
(0) 1
(0)3
(1) 1
(2) 2
(3)5
v8
(1) 5
v5
(0)1
(2) 2
v6 (1) 2
t
(0) 4
(1) 3
(0) 6
(4) 4
z∈Γ ( x )
v2
We can extend the original theorem to this situation as well.
Graph Theory 3
(2) 3
v4
(4) 4
v7
v( f ) = f + ( s ) = f − ( t ) = 6
61
62
Let G be a directed graph and let f be a flow in G . With each
each path P
in the graph we associate a nonnon-negative integer (P) defined by
ε ( P ) = min ε ( a )
a∈A( P )
where
The Labelling Algorithm
Graph Theory 3
Let a = (x,y) be an arc in a directed graph G and, c(a)
c(a) be the capacity
of this edge and let f(a)
f(a) be the flow in the arc. Then we call a
if
f(a)
f(a) < c(a),
c(a),
f-saturated
if
f(a)
f(a) = c(a).
c(a).
(P) is the largest amount by which the flow along P can be increasincreased without violating the condition f(a)
f(a) c(a).
c(a).
f-unsaturated
f(a)
f(a) > 0,
The path P is said to be f-saturated if (P) = 0 an f-unsaturated if
(P) > 0.
if
f-positive
f(a)
f(a) = 0,
if
c( a ) − f ( a ) if a is a forward arc of P
f(a)
if a is reverse arc of P
f-zero
ε( a ) =
How can we decide that a given flow is a maximum flow or not?
An f-incrementing path is an f-unsaturated path from the source s to
the sink t.
Graph Theory 3
63
Graph Theory 3
64
16
The existence of an f-incrementing path P in a network is significant
since it implies that f is not a maximum flow. By sending an
additional flow (P) along P one obtains a new flow f' defined by
s
(1)6
otherwise
for which v(f‘
v(f‘) = v(f)
v(f) + (P) .
v1 (2)3
v4(2)2
t s
(1)1
v5
(s,v1) = 5
(v3,v2) = 5
(3)5
(0)1
(0)3
5
v(3)
2
v1 (2)3
v4(2)2
(3)6
(5)5
f-incrementing path P.
t
(5)5
v5
revised flow based on P.
(v2,v1) = 2
(v3,t) = 4
Theorem 4.21. A flow in a directed graph is a maximum flow iff the
the
graph contains no f-incrementing path.
(2)3
(2)4
(4)4
(0)1
5
v(3)
2
f(a)
(1)1
(5)5
f ′ = f ( a ) − ε ( P ) if a is reverse arc of P
(0)4
(4)4
f ( a ) + ε ( P ) if a is forwarded arc of P
v3
v3
(P) = 2
Graph Theory 3
65
Graph Theory 3
66
The algorithm is ready: Staring with a known flow, for instance the
zero flow, it recursively constructs a sequence of flows of increasing
increasing
value, and terminates with a maximum flow.
v3
How can we find an f-incrementing path?
s
L. R. Ford and D. L. Fulkerson (1957): The Labelling Method.
(1)6
J. Edmond and R. M. Karp (1972): FirstFirst-labelled FirstFirst-scanned
Labelling Method
A tree T in a directed graph is an f-unsaturated tree if
x V(T)
V(T)
for every vertex v of T, the unique (s,v)
s,v) path in T is an f-unsaturunsaturated path.
Graph Theory 3
(0)4
(4)4
67
(1)1
(5)5
(0)1
(2)3
5
v(3)
2
v1 (2)3
v4(2)2
t
(5)5
v5
An f-unsatrated tree.
Graph Theory 3
68
17
The search for an f-incrementing path involves growing an f-unsatunsaturated tree in G .
In the event of breakthrough, the (s,t
(s,t))-path in T is our desired f-inincrementing path.
If T stops growing before reaching t, then f is a maximum flow.
Initially, T consists of just the source s.
The labelling procedure:
At any stage, there are two ways in which the tree may grow:
Assign to the source s the label l(s)
l(s) = .
If there exists an f-satureted arc a in ( S,S ), where S = V(T),
then both a and its head are adjoined to T.
If there exists an f-positive arc a in ( S ,S ), then both a and its
tail are adjoined to T.
The resulting T
either reaches the sink t,
or stops growing before reaching t (breakthrough).
If a is an f-positive arc whose head u is already labelled but whose
tail v is not, then v is labelled l(v)
l(v) = min{
min{l(u),
l(u), f(a)
f(a)}.
In the above cases v is said to be labelled based on u.
69
Graph Theory 3
70
f :=
To scan a labelled vertex u is to label all unlabelled vertices that can
be labelled based on u.
The original Labelling Method is not an efficient procedure, since
since
the number of searching an incrementing path may be equal with the
the
maximum value of the capacity function on E.
The labelling procedure is continued until either the sink t is labelled
(breakthrough) or all labelled vertices have been scanned and no
more vertices can be labelled (implying that f is a maximum flow).
L := {s}
S :=
l(s)
l(s) :=
f := f '
while
vertex u L S
scan u
L := L L(u)
L(u)
find a revised flow
f ' based on P
Y
The Edmonds – Karp improvement uses during the scanning step a
”firstfirst-labelled firstfirst-scanned”
scanned” basis: before scanning a labelled vertex
u scan the vertices that were labelled before u..
Graph Theory 3
71
L: set of labelled vertices.
S: set of scanned vertices.
L(u):
L(u): set of vertices labelled
during scanning of u.
Graph Theory 3
If a is an f-unsaturated arc whose tail u is already labelled but
whose head v is not, then v is labelled l(v)
l(v) = min{
min{l(u),
l(u), c(a)
c(a) – f(a)
f(a)}.
t L
S := S {u
{u}
Graph Theory 3
Stop
Y
L S=
N
72
18
v3
v32
(0)4
(4)4
s
(1)1
(5)5
2
(0)1
5
(2)3 2v(3)
2
(1)6
A flow can be interpreted to a flow in a vertex as well, namely from
the part where all the currents enter it to the part where all the
currents leave it.
t s
(1)1
1 v5
(3)5
(0)1
(0)3
5
v(3)
2
v1 (2)3
v4(2)2
(3)6
(5)5
5 v1 (2)3 1 v4(2)2
(2)4
(4)4
t
(5)5
v5
We can turn each vertex of G into an edge in such a way that that
any current entering (and leaving)
leaving) the vertex will be forced through
the edge.
Replace each vertex x V {s,t}
s,t} by two vertices, say x- and x+, send
each incoming edge to x- and send each outgoing edge to x+.
Finally, add edges from x- to x+ with capacity c(x-,x+) = c(x).
c(x).
Theorem 4.22.: Let G be a directed graph with capacity bounds on
the vertices other then the source s and the sink t. Then the minimum
of the capacity of a vertexvertex-cut is equal to the maximum of the flow
value from s to t.
Graph Theory 3
73
Graph Theory 3
74
Now,
Now, we are ready to give a new proof for the Menger´s theorem.
theorem.
x+
x-
G
x
s
H
Theorem 4.23. (Menger'
(Menger's theorem):
theorem):
ut
u
y
Two s−t paths are independent if they have only the vertices s and t
in common.
common.
z
s
t
u+
y-
(1)
z(2)
y+
Graph Theory 3
z+
Replacing the vertices by edges
75
Let s and t be distinct nonnon-adjacent vertices of a graph G. Then
the minimum number of vertices separating s from t is equal to the
maximum number of independent s−t paths.
Let s and t be distinct vertices of G. Then the minimal number of
edges separating s from t is equal to the maximal number of edgeedgedisjoint s−t paths.
Graph Theory 3
76
19
Proof.
(1) Replace each edge xy of G by two directed edges, xy and yx , and
give each vertex – other than s and t – capacity 1.
By the Theorem 4.18.
4.18. the maximal flow value from s to t is equal
to the minimum of the capacity of a cut separating s from t.
By the Integrality Theorem there is a maximal flow with current 1
or 0 in each edge.
Therefore the maximum flow value from s to t is equal equal to the
maximal number of independent s−t paths.
The minimum of the cut capacity is clearly the minimal number of
vertices separating s from t.
(2) Proving the second statement of the theorem one can proceed as in
(1), except instead of restricting the capacity of the vertices, give
each directed edge capacitx 1.
Graph Theory 3
77
4.4. The toughness of a graph
If G is a noncomplete graph and t is a nonnegative real number such
that t |S| / k(G–
k(G–S) for every vertexvertex-cut S of G, then G is defined ttough. (Here k(G)
k(G) means the number of components of a graph G.)
If G is t-tough and s is a nonnegative real number such that s < t,
t,
then G is also s-tough.
tough.
The maximum real number t for which a graph G is t-tough is called
the toughness of G and denoted by t(G).
Since complete graphs do not contain vertexvertex-cuts,
cuts, this definition
does not apply to such graphs. We define t(Kn) = + for every
positive integer n.
Graph Theory 3
78
Simple statements:
statements:
The toughness of a noncomplete graph is a rational number.
number.
a
t(G) = 0 iff G is disconnected.
disconnected.
If G is noncomplete graph then t(G) = min |S|/k
|S|/k(G–S), where the
minimum is taken over all vertexvertex-cuts S of G.
The toughness of a graph G shows the measure that how tightly the
subgraphs of G are held together.
together.
The smaller the toughness the more vulnerable the graph is:
Graph Theory 3
79
x
s
y
c
w
v
d
S1={u,v,w},
u,v,w},
a 1-tough graph has the property that breaking the graph into k
components requires the removal of at least k vertices.
Breaking a 2-tough graph into k components requires the removal
of at least 2k vertices.
u
b
S2={w},
|S1|/k(
|/k(G–S1) = 3/8,
s
z
S3={u,v}
u,v} are three (of many) vertexvertex-cuts
|S2|/k(
|/k(G–S2) = 1/3,
|S3|/k(
|/k(G–S3) = 2/7
A graph of toughness 2/7.
Graph Theory 3
80
20
Theorem 4.24. (Chvatal
(Chvatal,, 1973): For every noncomplete graph G,
(G)
(G)
.
≤ t(G) ≤
2
(G)
A set S of vertices is independent if every two vertices of S are
independent.
Two nonadjacent vertices in a graph are said to be independent.
The independence number is related to the toughnes in the sense that
among all the vertexvertex-cuts S of a noncomplete graph G, the maximum
value of k(G–S) is (G), so for every vertexvertex-cut S of G, we have that
(G) |S| and k(G–S)
(G).
A parameter that plays an important role in the study of toughness is
the independence number:
number:
The vertex independence number (G) of a graph G is the maximum
cardinality among the independent sets of vertices of G.
According to the simple statement earlier we get
(Cn) = n/2
Graph Theory 3
81
Graph Theory 3
(G)
.
(G)
≥
82
Let S' be a vertexvertex-cut with |S'| = (G). Thus k(G–S') 2, so
S
S′
(G)
≤
≤
.
t(G) = min
k(G − S)
k(G − S)
2
S
k(G − S)
t(G) = min
(Kn) = 1.
Examples:
Examples:
(Kr,s) = max {r,s}
r,s}
Proof.
Theorem 4.25. (Matthews and Summer, 1984): If G is a noncomplete
clawclaw-free graph then
1
(G).
t(G) =
2
Let F be a graph. A graph G is F-free if G contains no induced
subgraph isomorphic to F.
Example:
Example: a K2-free graph is empty.
empty.
A K1,3-free graph is called as a clawclaw-free graph.
Graph Theory 3
83
Graph Theory 3
84
21