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Ma/CS 6b
Class 5: Graph Connectivity
By Adam Sheffer
Communications Network
We are given a set of routers and wish to
connect pairs of them to obtain a
connected communications network.
 The network should be reliable – a few
malfunctioning routers should not disable
the entire network.

◦ What condition should we
require from the network?
◦ That after removing any 𝑘
routers, the network
remains connected.
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𝑘-connected Graphs
A graph 𝐺 = (𝑉, 𝐸) is said to be
𝑘-connected if 𝑉 > 𝑘 and we cannot
disconnect 𝐺 by removing 𝑘 − 1 vertices
from 𝑉.
 Is the graph in the figure

◦ 1-connected? Yes.
◦ 2-connected? Yes.
◦ 3-connected? No!
Connectivity

Which graphs are 1-connected?
◦ These are the connected graphs ( 𝑉 > 1).
The connectivity of a graph 𝐺 is the
maximum 𝑘 such that 𝐺 is 𝑘-connected.
 The graph in the figure has a
connectivity of 2.
 What is the connectivity of the
complete graph 𝐾𝑛 ? 𝑛 − 1.

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Hypercube

1
0
A hypercube is a generalization of
the cube to any dimension.
◦ In 𝑑-dimensions, we consider the vertices of
the hypercube as the points with coordinates
of 0 and 1.
◦ Two vertices are adjacent if they differ in a
single coordinate.
(1,1)
(0,1)
(0,0)
(1,0)
Hypercube Properties


1
0
The points of a 𝑑-dimensional hypercube
correspond to the 𝑑-dimensional points with
0-1 coordinates.
◦ Thus, it has 2𝑑 vertices.
Two vertices are adjacent if they have 𝑑 − 1
common coordinates.
◦ Thus, there are 2𝑑−1 𝑑 edges.
(1,1)
(0,1)
(0,0)
(1,0)
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Hypercube graphs

The hypercube graph 𝑄𝑑 corresponds to
the 𝑑-dimensional hypercube.
◦ A vertex for each vertex of the hypercube.
◦ An edge for each edge of the hypercube.
◦ The graph 𝑄𝑑 has 2𝑑 vertices and 2𝑑−1 𝑑
edges. It is 𝑑-regular.
4d-hypercube
Hypercube graph Properties

Problem. Is 𝑄𝑑 a bipartite graph?

Answer. Yes!
◦ On one side we place every vertex with an
even number of 1-coordinates. On the other,
the vertices with an odd number of
1-coordintes.
4d-hypercube
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More Properties

Problem. What is the connectivity of 𝑄𝑑 ?

Answer.
◦ We can disconnect 𝑄𝑑 by removing the 𝑑
neighbors of any vertex of 𝑄𝑑 . So the
connectivity is at most 𝑑.
◦ We prove by induction that it is exactly 𝑑.
𝑄4
Proof by Induction

We prove by induction that the
connectivity of 𝑄𝑑 is 𝑑.
◦ Induction basis. Easy to check for 𝑑 = 1,2.
◦ Induction step. We can consider 𝑄𝑑 as two
copies 𝑄, 𝑄′ of 𝑄𝑑−1 with a perfect matching
between their vertices.
𝑄
𝑄′
𝑄𝑑
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Completing the Proof
𝑆 – a minimum disconnecting set of 𝑄𝑑 .
 To complete the proof, we claim that
𝑆 = 𝑑.

◦ To disconnect 𝑄 from 𝑄′ , the size of 𝑆 must
be at least 2𝑑−1 ≥ 𝑑.
◦ By the induction hypothesis, disconnecting 𝑄
(or 𝑄′ ) requires removing 𝑑 − 1 vertices of 𝑄.
To disconnect 𝑄 in 𝑄𝑑 , we must remove at
least one additional vertex from 𝑄′ .
𝑘-edge Connected Graphs
A graph 𝐺 = (𝑉, 𝐸) is said to be 𝑘-edgeconnected if 𝑉 > 1 and we cannot
disconnect 𝐺 by removing at most 𝑘 − 1
edges from 𝐸.
 Is the graph in the figure

◦ 1-edge-connected? Yes.
◦ 2-edge-connected? Yes.
◦ 3-edge-connected? Yes.
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𝑘-edge-connectivity

Which graphs are 1-edge-connected?
◦ These are the connected graphs ( 𝑉 > 1).
The edge-connectivity of a graph 𝐺 is the
maximum 𝑘 such that 𝐺 is
𝑘-edge-connected.
 The graph in the figure has an
edge-connectivity of 3.
 What is the edge-connectivity of
the complete graph 𝐾𝑛 ? 𝑛 − 1.

Edge Connectivity and Minimum
Degree

What is the relation between the
minimum degree and the
edge-connectivity of a graph?
◦ Can they be equal? Yes
◦ Can the edge connectivity be smaller? Yes
◦ Can the minimum degree be smaller?
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Edge Connectivity and Minimum
Degree (cont.)
Claim. In any graph, the edge connectivity
is at most the minimum degree.
 Proof.

◦ Consider a vertex 𝑣 of minimum degree, and
denote this degree as 𝑑.
◦ By removing the 𝑑 edges that are adjacent to
𝑣, we disconnect the graph.
Connectivity and Edge-Connectivity

What is the relation between the
connectivity and the edge-connectivity of
a graph?
◦ Can they be equal? Yes
◦ Can the connectivity be smaller? Yes
◦ Can the edge-connectivity be smaller?
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Connectivity and Edge-Connectivity
Claim. In any graph 𝐺 = 𝑉, 𝐸 , the
connectivity is at most the
edge-connectivity.
 Proof.

◦ Let 𝐹 ⊂ 𝐸 be a minimum set of edges whose
removal disconnects 𝐺.
◦ We need to find a set of at most 𝐹 vertices
that disconnects 𝐺.
◦ We divide the analysis into two cases:
 Some 𝑣 ∈ 𝑉 is not adjacent to any edge of 𝐹.
 Every vertex of 𝑉 is adjacent to an edge of 𝐹.
Analysis of Case 1

Assume that a vertex 𝑣 ∈ 𝑉 is not
adjacent to any edge of 𝐹.
◦ The removal of 𝐹 disconnects 𝐺.
◦ Let 𝐶 be the connected component of 𝑣 in
𝐺 − 𝐹.
◦ Removing the vertices 𝑉 ′ in 𝐶 that are
adjacent to an edge of 𝐹 disconnects 𝑣 from
the vertices of 𝐺 − 𝐶.
◦ Since 𝐹 is minimal, each
𝐹
edge of 𝐹 has at most
𝑣
one endpoint in 𝐶, so
𝑉′ ≤ 𝐹 .
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Analysis of Case 2

Assume that every vertex of 𝑉 is adjacent to an
edge of 𝐹.
◦ If 𝐺 is not complete, there exists 𝑣 ∈ 𝑉 such
that deg 𝑣 < 𝑉 − 1. Removing the set 𝑉 ′ of
neighbors of 𝑣 disconnects 𝐺.
◦ 𝐶 – the connected component of 𝑣 in 𝐺 − 𝐹.
◦ Every vertex of 𝑉 ′ is either in 𝐶 or connected
to 𝑣 by an edge of 𝐹.
◦ Since no edge of 𝐹 is between
two vertices of 𝐶, we have
𝐹
′
𝑉 ≤ |𝐹|.
𝑣
Analysis of Case 2 (cont.)

Assume that every vertex of 𝑉 is adjacent
to an edge of 𝐹.
◦ If 𝐺 is a complete graph, the connectivity is
𝑉 − 1 and edge-connectivity is 𝑉 − 1.
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Connectivity in 3-regular Graphs

Claim. If 𝐺 = 𝑉, 𝐸 is a connected
3-regular graph, then the connectivity
and the edge-connectivity of 𝐺 are equal.
Proof

𝑘𝑣 – (vertex) connectivity.
𝑘𝑒 – edge connectivity.

We already know that 𝑘𝑣 ≤ 𝑘𝑒 .

◦ Thus, it suffices to find a set of 𝑘𝑣 edges that
disconnects 𝐺.
◦ 𝑆 – a minimum set of 𝑘𝑣 vertices whose
removal disconnects 𝐺 into 𝐶1 and 𝐶2 (which
might not be connected
themselves).
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Proof (cont.)

Every vertex of 𝑣 ∈ 𝑆 is connected to
both 𝐶1 and 𝐶2 , since 𝑆 is a minimum
disconnecting set.
◦ Due to 3-regularity, either 𝑣 is adjacent to a
single vertex of 𝐶1 , or to a single vertex of 𝐶2 .
◦ That is, by removing a single edge, we can
disconnect 𝑣 from either 𝐶1 or from 𝐶2 .
◦ By removing such an edge from each 𝑣 ∈ 𝑆,
we obtain a disconnecting set of 𝑘𝑣 edges.
Recap
We proved that in every graph:
Minimum
Edge
Connectivity ≤
≤
degree
connectivity


This implies that small minimum degree
implies small connectivity.
◦ Does large minimum degree imply large
connectivity.
◦ No!
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Highly Connected Subgraphs

Recall. The average degree of a graph
𝐺 = 𝑉, 𝐸 is
1
2𝐸
deg 𝐺 =
෍ deg 𝑣 =
.
|𝑉|
𝑉
𝑣∈𝑉

Theorem (Mader `72). Let 𝑘 be a positive
integer and let 𝐺 = 𝑉, 𝐸 be a graph
such that deg 𝐺 ≥ 4𝑘. Then there exists
a 𝑘 + 1 -connected subgraph 𝐻 ⊂ 𝐺
with minimum degree > deg(𝐺)/2.
Proof
Set 𝑑 = deg 𝐺 ≥ 4𝑘.
 Let 𝐻 = (𝑉 ′ , 𝐸 ′ ) be a subgraph of 𝐺 such
𝑑
that 𝑉 ′ ≥ 2𝑘 and 𝐸 ′ >
𝑉′ − 𝑘 .

2
◦ Among the subgraphs that satisfy the above,
we take as 𝐻 one that minimizes 𝑉 ′ .

Such subgraphs exist since 𝐺 is one:
◦ Since 𝑑 ≥ 4𝑘, there exists a vertex of degree
at least 4𝑘, and thus 𝑉 ≥ 4𝑘 + 1.
2𝐸 𝑉
𝑑
𝑑
𝐸 =
⋅
= 𝑉 >
𝑉 −𝑘 .
𝑉
2
2
2
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Proof (2)
Set 𝑑 = deg 𝐺 ≥ 4𝑘.
 𝐻 = (𝑉 ′ , 𝐸 ′ ) – subgraph of 𝐺 such that
𝑑
𝑉 ′ ≥ 2𝑘 and 𝐸 ′ >
𝑉′ − 𝑘 .

2

′
If 𝑉 = 2𝑘, we get the contradiction
𝑑
𝑑
′
′
𝐸 >
𝑉 − 𝑘 = 𝑘 ≥ 2𝑘 2
2
2
′
2𝑘
𝑉
>
=
≥ 𝐸′ .
2
2
Proof (3)

𝐻 = (𝑉 ′ , 𝐸 ′ ) – the minimal subgraph of 𝐺
such that
𝑉 ′ ≥ 2𝑘 and 𝐸 ′ >
𝑑
2
𝑉′ − 𝑘 .
◦ We actually have 𝑉 ′ ≥ 2𝑘 + 1.
◦ Removing a vertex of degree ≤ 𝑑/2 decreases
𝑑
𝑑
𝑑
𝑉 ′ − 𝑘 by and 𝐸 ′ by at most .
2
2
′
◦ That is, both 𝑉 ≥ 2𝑘 and
𝑑
𝐸′ >
𝑉 ′ − 𝑘 remains valid,
2
contradicting the minimality of 𝐻.
2
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Proof (4)

𝐻 = (𝑉 ′ , 𝐸 ′ ) – the minimal subgraph of 𝐺
𝑑
such that 𝑉 ′ ≥ 2𝑘 and 𝐸 ′ > 2 𝑉 ′ − 𝑘 .
◦ We proved that every vertex of 𝑉 ′ is of degree
larger than 𝑑/2 in 𝐻.
◦ We need to show that 𝐻 is (𝑘 + 1)-connected.
◦ Assume for contradiction that the removal of a
set 𝑆 of 𝑘 vertices disconnects 𝐻.
◦ We can thus partition the vertices of 𝑉 ′ ∖ 𝑆 into
the subsets 𝑉1 , 𝑉2 with no edges between them.
Proof (5)

Assume for contradiction that the removal of a
set 𝑆 of 𝑘 vertices whose removal disconnects 𝐻.
◦ Partition the vertices of 𝑉 ′ ∖ 𝑆 into subsets
𝑉1 , 𝑉2 with no edges between them.
◦ 𝐻1 – the subgraph induced by 𝑉1 ∪ 𝑆.
◦ Consider a vertex 𝑣 ∈ 𝑉1 and notice that all the
edges that are adjacent to it remain in 𝐻1 .
◦ Since deg 𝑣 >
𝑑
2
≥ 2𝑘, we have 𝑉1 ∪ 𝑆 > 2𝑘.
𝑆
𝑉1
𝑉2
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Illustration
𝑉1
𝑉2
𝑆
𝐻1
𝐻2
𝑆
𝑆
Proof (6)



The subgraph 𝐻1 has more than 2𝑘 vertices.
In order not to contradict the minimality of 𝐻,
the number of edges of 𝐻1 must be
𝑑
≤
𝑉 ∪𝑆 −𝑘 .
2 1
Similarly, the number of edges of 𝐻2 must be
𝑑
≤
𝑉 ∪𝑆 −𝑘 .
2 2
𝑑
𝑑
𝐸′ ≤
𝑉1 ∪ 𝑆 − 𝑘 +
𝑉 ∪𝑆 −𝑘
2
2 2
𝑑
=
𝑉′ − 𝑘
2
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The End
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