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Ma/CS 6b
Class 4: Matchings in General Graphs
By Adam Sheffer
Reminder: Hall's Marriage Theorem

Theorem. Let 𝐺 = 𝑉1 ∪ 𝑉2 , 𝐸 be a
bipartite graph. There exists a matching
of size 𝑉1 in 𝐺 if and only if for every
𝐴 ⊂ 𝑉1 , we have 𝐴 ≤ 𝑁 𝐴 .

We say that 𝐺 satisfies Hall’s
condition if for every 𝐴 ⊂ 𝑉1 , we
have 𝐴 ≤ 𝑁 𝐴 .
Philip Hall
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𝑘-regular Graphs

A graph 𝐺 = 𝑉, 𝐸 is 𝑘-regular if each
vertex of 𝑉 is of degree 𝑘.

The Petersen graph is a 3-regular graph
with ten vertices:
Example: 𝑘-regular Graphs

For any positive integer 𝑛, what graph do
we know that is 𝑛-regular?
◦ 𝐾𝑛 – the complete graph with 𝑛 vertices
(every two vertices are connected). It is
𝑛 − 1 -regular.
𝐾7
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Example: 2-regular Graphs

For any positive integer 𝑛, what graph do
we know that is 2-regular and has 𝑛
vertices?
◦ 𝐶𝑛 – a cycle of size 𝑛.
𝐶6
1-regular Graphs

Problem. Characterize the graphs that are
1-regular.

Answer. These are the graphs whose
edges form a perfect matching (with no
additional edges).
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2-regular Graphs

Problem. Characterize the graphs that are
2-regular.

Answer. These are the graphs that consist
only of vertex-disjoint cycles.
𝑘-factors

Consider a graph 𝐺 = 𝑉, 𝐸 .
◦ A spanning subgraph of 𝐺 is a subgraph that
contains all of the vertices of 𝑉.
◦ For a positive integer 𝑘, a 𝑘-factor of 𝐺 is a
spanning subgraph of 𝐺 that is 𝑘-regular.
◦ A graph contains a 1-factor iff it contains a
perfect matching.
1-factor
2-factor
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2-factor Theorem

Theorem. Every 2𝑘-regular graph
𝐺 = 𝑉, 𝐺 has a 2-factor (where 𝑘 is a
positive integer).

One of the earliest works on graph theory
– by Petersen in 1891.
Julius Petersen
Recall: Eulerian Cycles
An Eulerian path in a graph 𝐺 is a path
that passes through every edge of 𝐺
exactly once.
 An Eulerian cycle is an Eulerian path that
starts and ends at the same vertex.

a
b
h
c
g
d
f
𝐸 = 𝑎→𝑏→𝑐→𝑑→𝑒→𝑓→𝑔
→𝑕→𝑎→𝑑→𝑕→𝑒→𝑏
→𝑔→𝑐→𝑓→𝑎
e
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Reminder: Graphs Containing
Eulerian Cycles

Theorem. A connected (possibly not
simple) graph 𝐺 = 𝑉, 𝐸 contains an
Eulerian cycle if and only if every vertex of
𝑉 has an even degree.

Proof (of 2-factor Theorem)

We assume that 𝐺 is connected.
◦ (otherwise, we can consider separately each
component).

Since every degree in 𝐺 is 2𝑘, it contains
an Eulerian cycle 𝐶.
◦ We arbitrarily choose a direction for 𝐶.
◦ We split every vertex 𝑣 ∈ 𝑉 into 𝑣 + and 𝑣 − .
◦ An edge in 𝐶 that went from 𝑣 to 𝑢 now goes
from 𝑣 + to 𝑢− .
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Illustration
𝑎
𝑎
𝑑
𝑏
𝑏
𝑒
𝑐
𝑑
𝑒
𝑐
𝑎+ 𝑎−
𝑏+
𝑐
+
𝑏−
𝑐
𝑑+ 𝑑−
𝑒+ 𝑒−
* Unlike our case,
this graph is not
2𝑘-regular.
−
Proof (cont.)

We obtain a graph 𝐺 ′ with 2 𝑉 vertices
and 𝐸 edges.
◦ We ignore the directions in 𝐺 ′ and notice that
it is a bipartite 𝑘-regular graph.
𝑎+ 𝑎−
𝑏+
𝑐+
𝑏−
𝑑+ 𝑑−
𝑒+ 𝑒−
𝑐−
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Bipartite 𝑘-regular Graphs
Claim. Every bipartite 𝑘-regular graph
𝐵 = 𝑉1 ∪ 𝑉2 , 𝐸 contains a perfect
matching.
 Proof.

◦ We have 𝑉1 = 𝑉2 , by double counting the
size of |𝐸|.
◦ We complete the proof by showing that 𝐵
satisfies Hall’s condition.
◦ For any 𝐴 ⊂ 𝑉1 , there are 𝑘 𝐴 edges that
leave 𝐴. Since every vertex of 𝑁 𝐴 is incident
to at most 𝑘 edges from 𝐴, then 𝑁 𝐴 ≥ 𝐴 .
Completing the Proof

We started with a 2𝑘-regular graph 𝐺.
◦ There is an Euler cycle in 𝐺.
◦ Splitting every vertex 𝑣 ∈ 𝑉 into 𝑣 + and 𝑣 − ,
we obtained a bipartite 𝑘-regular graph 𝐵
with the same set of edges.
◦ 𝐵 contains a perfect matching 𝑀.
◦ By merging every pair 𝑣 + and 𝑣 − back to 𝑣, 𝑀
becomes a 2-factor!
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Another Illustration
+
𝑏−
𝑐+
𝑎−
𝑑+
𝑓+
𝑏+
−
𝑎
𝑒+

𝑎
𝑏
𝑒
𝑑−
𝑑
−
𝑐
𝑓−
𝑐
𝑒
𝑓
Merging the vertices of the perfect
matching results in a 2-factor.
Today: Mathematical Facts About
Nicolas Cage
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Matchings in General Graphs
In bipartite graphs, Hall’s condition
characterizes the graphs that contain a
perfect matching.
 We now present a similar condition for
general graphs, which we denote as
Tutte’s condition.

W. T. Tutte
Odd Components

Given a graph 𝐺 = (𝑉, 𝐸), we denote by
𝑞 𝐺 the number of components in 𝐺
that consists of an odd number of
vertices.
𝑞 𝐺 =4
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Tutte’s Condition

We say that a graph 𝐺 = 𝑉, 𝐸 satisfies
Tutte’s condition if for every 𝑆 ⊂ 𝑉 we
have
𝑞 𝐺−𝑆 ≤ 𝑆 .
𝑆 =4
𝑞 𝐺−𝑆 =3
The Easy Direction
Claim. If a graph 𝐺 contains a 1-factor 𝑀,
then it satisfies Tutte’s condition.
 Proof. For any 𝑆 ⊂ 𝑉, consider the
components of 𝐺 − 𝑆.

◦ Every component with an odd number of
vertices must have at least one edge of 𝑀
connected to a vertex of 𝑆.
◦ Thus, 𝑞 𝐺 − 𝑆 ≤ 𝑆 .
𝑆
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The Other Direction

Claim. If a graph 𝐺 = 𝑉, 𝐸 does not
contain a 1-factor, then it does not satisfy
Tutte’s condition.
◦ That is, we need to prove that there exists
𝑆 ⊂ 𝑉 such that 𝑞 𝐺 − 𝑆 > 𝑆 .
◦ Given such a bad set 𝑆, if we remove edges
from 𝐸, then 𝑆 remains a bad set:
 If we split an odd sized component into
several components, at least one of them is
odd-sized.
Illustration

Removing edges can only increase
𝑞(𝐺 − 𝑆).
|𝑆| = 3
𝑞 𝐺−𝑆 =4
|𝑆| = 3
𝑞 𝐺−𝑆 =4
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Rephrasing the Claim

We saw that if a graph 𝐺 contains a bad
set 𝑆, then 𝑆 remains bad after removing
any number of edges.
◦ We say that a graph 𝐺 is edge-maximal
without a 1-factor if it does not contain a 1factor, but adding any one edge to it results in
a 1-factor.
◦ Thus, it suffices to prove:
◦ Claim. If a graph 𝐺 = 𝑉, 𝐸 is edge-maximal
without a 1-factor, then it does not satisfy
Tutte’s condition.
Drowning in a Swimming Pools
 Correlation: 0.666004
* Taken from http://tylervigen.com/
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Odd Number of Vertices

If 𝐺 has an odd number of vertices, it
obviously contains no 1-factor.
◦ In this case, 𝐺 does not satisfy Tutte’s
condition: By taking 𝑆 = ∅ we have
0 = 𝑆 < 𝑞 𝐺 − 𝑆 = 1.

It remains to consider graphs with an
even number of vertices.
Finding a Bad Set
We take 𝑆 ⊂ 𝑉 to be the set of vertices
that are connected to every vertex of 𝐺 (it
is possible that 𝑆 = ∅).
 We partition the analysis into two cases:

◦ Every component of 𝐺 − 𝑆 is a complete
graph (that is, a complete induced subgraph).
◦ There is a component of 𝐺 − 𝑆 that is not a
complete graph.
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The First Case


𝑆 – the set of vertices that are connected to
every vertex of 𝐺.
Consider the case where every connected
component of 𝐺 − 𝑆 is complete.
◦ If 𝑆 ≥ 𝑞 𝐺 − 𝑆 , we get a contradiction to 𝐺
not containing a 1-factor, so 𝑆 is a bad set!
◦ The 1-factor has one edge between every odd
component and a distinct vertex of 𝑆.
𝐺−𝑆
𝑆
Dying in a Helicopter Accident

Correlation: -0.827811
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The Second Case

Assume that a component 𝐶 of 𝐺 − 𝑆 that is
not a complete graph.
◦ There exist two vertices 𝑥, 𝑧 in 𝐶 that
𝑥, 𝑧 ∉ 𝐸 but there exists 𝑦 ∈ 𝑉 ∖ 𝑆 such
that 𝑥, 𝑦 , 𝑦, 𝑧 ∈ 𝐸.
◦ Since 𝑦 ∉ 𝑆, there exists a vertex 𝑤 ∉ 𝑆 such
that 𝑦, 𝑤 ∉ 𝐸.
𝑥
𝑦
𝑆
𝑤
𝑧
Building a New Graph

Recall that adding any edge to 𝐺 results in
a 1-factor contained in it.
◦ 𝑀𝑥𝑧 – 1-factor in 𝐺 after adding 𝑥, 𝑧 .
◦ 𝑀𝑦𝑤 – 1-factor in 𝐺 after adding 𝑦, 𝑤 .
◦ 𝐹 – the set of edges that are either in 𝑀𝑥𝑧 or
in 𝑀𝑦𝑤 (but not in both).
◦ In the graph 𝐺𝐹 = 𝑉, 𝐹 , every vertex is of
degree 0 or 2.
◦ That is, 𝐺𝐹 consists of isolated vertices and
disjoint even length cycles.
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Union of Two Perfect Matchings
𝑧
𝑤
𝑥
𝑧
𝑤
𝑥
𝑦
𝑀𝑥𝑧
𝑦
𝑀𝑦𝑤
𝑧
𝑤
𝑥
𝑦 𝐹
Different Cycles

If 𝑥, 𝑧 and 𝑦, 𝑤 are in different cycles
of 𝐹, we can find a 1-factor in 𝐺:
◦ Take every edge that is in 𝑀𝑥𝑧 ∩ 𝑀𝑦𝑤 .
◦ From the each cycle of 𝐹, we take every other
edge. We choose those such that we avoid
taking 𝑥, 𝑧 and 𝑦, 𝑤 .

This contradicts 𝐺 not containing a 1factor.
𝑧
𝑥
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The Same Cycle

If 𝑥, 𝑧 and 𝑦, 𝑤 are on the same cycle:
◦ If we can take every other edge of the cycle
without taking 𝑥, 𝑧 and 𝑦, 𝑤 , we get a
contradiction as before.
◦ Otherwise, we can use 𝑥, 𝑦 or 𝑦, 𝑧 to
obtain a 1-factor containing neither 𝑥, 𝑧 nor
𝑦, 𝑤 .
𝑦
𝑧
𝑤
𝑥
Concluding the Proof
We proved that either 𝑆 is a bad set, or
we get a contradiction to 𝐺 not
containing a 1-factor.
 That is, we proved that if 𝐺 is edgemaximal without a 1-factor, then there
exists 𝑆 ⊂ 𝑉 such that 𝑆 < 𝑞(𝐺 − 𝑆).

◦ Thus, in every graph 𝐺 without a 1-factor
there exists 𝑆 ⊂ 𝑉 such that 𝑆 < 𝑞(𝐺 − 𝑆).
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Finding a 1-factor in a Graph
As with Hall’s theorem, Tutte’s condition
does not give us an algorithm for finding
a 1-factor in a graph.
 In 6a, we saw an algorithm for finding 1factors in bipartite graphs, using
augmenting paths.

◦ The same algorithm works for general graphs.
◦ We will not repeat the algorithm, but we will
rely on it as a “black box” (also in HW!).
The End
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