CS 388C: COMBINATORICS AND GRAPH THEORY
Lecture 17
Scribes: Pravesh Kothari, Aibo Tian
March 22
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Three Solutions to a Problem on the Test
Problem. Prove that every bipartite graph G with ` edges has a matching of size at least
`
, where ∆(G) is the maximum degree of vertices in G.
∆(G)
We will provide 3 different solutions to this problem.
Solution 1: From the Kőnig-Egerváry Theorem.
Theorem 1 (Kőnig-Egerváry). The size of a largest matching in a bipartite graph is equal
to the size of the smallest vertex cover.
To use the above theorem to complete a solution to the problem, we must show that the size
`
of the smallest vertex cover in the graph is at least ∆(G)
. A single vertex can cover at most
l
vertices. This completes the
∆(G) edges. Thus to cover all l edges, we need at least ∆(G)
proof.
Solution 2: Using a Result From Homework 3 .
Lemma 1. Any bipartite graph with maximum degree d is a union of d matchings.
Thus, G must be a union of ∆(G) matchings. Using the Pigeon Hole Principle, there exists
`
a matching that has at least ∆(G)
edges. This completes the proof.
Solution 3: Using Edge Coloring.
Definition 1 (Proper Edge Coloring). Let G(V, E) be a graph. A proper k−edge coloring
of G is a function c : E → [k] such that for every two edges e1 = {u1 , v1 } and e2 = {u2 , v2 },
e1 ∩ e2 6= ∅ ⇒ c(e1 ) 6= c(e2 ).
Definition 2 (Edge Chromatic Number). The edge chromatic number of a graph G(V, E)
is the the least positive integer k such that there exists a proper k-edge coloring of G.
Observation. For a graph G(V, E) and a proper k-edge coloring of G given by c : E → [k],
let Ei = {e ∈ E | c(e) = i}. Then for each i ∈ [k], Ei forms a matching of G.
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Proof. Easy. Observe that no two edges in a color class share a vertex.
Thus, to complete the proof, one must show that G can be edge-colored by at most ∆(G)
colors. Lemma 1 above shows that any bipartite graph with maximum degree d is a union
of d matchings. Let M1 , M2 , ..., Md be these matchings for d = ∆(G). Then we have that
E = M1 ∪ M2 ∪ ... ∪ Md from lemma 1. We can assume without loss of generality, that the
matchings M1 , . . . , Md are disjoint, that is they form a partition of the edges. (Make Mi a
partition of E by assigning an edge e ∈ E to the matching Mi with the least index among
those that contain e, if needed.) Let this partition thus created be M10 , M20 , ..., Md0 . Note that
Mi0 is a matching for each i ∈ [d]. Now let c : E → [d] be a coloring of edges of G such that
c(e) = i ⇔ e ∈ Mi0 . Since each Mi0 is a matching, no two edges in Mi0 share a vertex and
thus c is a proper d-edge coloring of G. By the lemma G can be colored by d = ∆(G) colors.
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More on Edge Colorings
Theorem 2 (Vizing). Any graph G of max degree ∆ can be edge-colored with at most ∆ + 1
colors.
Observation. A complete graph G can be edge-colored by ∆ + 1 colors, if G has odd number
of vertices. A complete graph G can be edge-colored by ∆ colors, if G has even number of
vertices.
Proof for this observation is left as an exercise. We now prove an interesting result in
Matching theory.
Definition 3 (Matched and Free Vertices). A vertex v ∈ V for a graph G(V, E) is said to
be matched in a matching M if there is an edge (v, x) for some x ∈ V such that {v, x} ∈ M .
Otherwise v is said to be free.
Definition 4 (M-augmenting Path). Given a matching M in a graph G(V, E), a path
u1 , u2 , ..., uk in G is called M-augmenting, if {ui , ui+1 } ∈ M ⇒ i is even and k is even.
In other words, each edge on the path alternates between an edge from M and an edge from
E \ M and the end points are free.
Theorem 3 (Berge). A matching M in a graph G(V, E) is a maximum cardinality matching
if and only if there is no M -augmenting path in the graph.
Proof. Suppose M is a maximum cardinality matching. We show that G(V, E) has no M augmenting path in it. Suppose to the contrary, there exists a path P = u1 , u2 , ..., uk of
length k which is M -augmenting. We show that there is a matching M 0 with cardinality
|M | + 1. This yields a contradiction and gives us one direction of the required result. Notice
that since k is even there are odd number of edges on the path. Also since u1 and uk
are free, the number of edges from M are exactly 1 less than the number of edges from
E \ M . Consider the set M 0 = M \ P ∪ P \ M . Observe that |M 0 | = |M | + 1. Also observe
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that no edges except the ones in P are modified. And the edges from P selected in M 0 do
not intersect in any vertex. Thus M 0 is a matching of higher cardinality than M . This is
contradiction.
Now suppose M is a matching in G(V, E) such that there is no M -augmenting path in G. We
claim that M is a maximum cardinality matching. Suppose not. Then there is a matching
M 0 which is of larger cardinality than M . Consider the set of edges formed by T = M ∪ M 0 .
Then each vertex in V has degree at most 2 in T . Thus T consists of paths and cycles. We
obtain an M -augmenting path from T to complete the proof. First note that there must be
at least one path containing more edges from M 0 than from M in T since if there isn’t, then
each cycle and path in T has equal number of edges from M and M 0 (they must alternate as
each vertex has degree at most 1 in any matching). But |M 0 | > |M |. Thus let P be such an
odd length path in T . Notice that since P is odd length and has more edges from M 0 than
from M , the end points of P must be free in M . This yields that P is an M -augmenting
path in T and thus an M -augmenting path in G. This completes the proof.
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