Matching

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Lecture 11
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Matching
A set of edges which do not share a vertex is a
matching.
Application Wireless Networks may consist of
nodes with single radios, no interference
constraint
A set of links can be scheduled only if the set
is a matching
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Input Queued Switches
The links scheduled must be a matching
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Maximum Matching
Matching of largest cardinality
A node is free if none of its incident edges is
in a matching.
Augmenting path is a path of alternating sequence
of matched and unmatched edges with free end
nodes.
Size of matching increases if e2 is replaced by
e1, e3
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A ? B (A B) ? (B A)
Suppose M is a matching and P is an augmenting
path.
Then M ? P is also a matching
Also, M ? P M 1
A matching is maximum if and only if there is no
augmenting path.
If there is no augmenting path, then the matching
is not maximum.
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Let there be a matching M of greater cardinality
than M.
Consider the graph M ? M
A vertex can have one incident edge from M, or
one incident edge from M, or one incident edge
from M and another from M
Clearly, a vertex can have at most 2 incident
edges
So the new graph consists of disjoint paths and
cycles.
The cycles must be even cycles with alternating
edges from M and M
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Paths will also have edges altering between M and
M
Since M has a cardinality greater than M, there
must be at least one path which starts and ends
in M.
This is an augmenting path for M.
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Maximum Matching Algorithm
Start with an empty set.
1) Search for an augmenting path.
If there is an augmenting path, then update the
matching, else terminate.
Go to (1)
How does one find an augmenting path?
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Bipartite Graph
Augmenting path can be found by Hungarian tree
method
Breadth first search like method
Consider one of the two partitions S.
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If we happen to reach a free vertex in T, then
we have found an augmenting path.
Suppose all the vertices in T are matched, then
we follow the matched edges back to S, and
subsequently all free edges from the newly
encountered ones in S.
If we reach a free vertex in T we are done,
otherwise again follow back the matched edges
into S.
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Dont include free edges from S into already
reached vertices in T. So vertices in T are never
repeated
We would never reach the same vertices in S
anyway.
Vertices in S are either matched or free.
We take only matched edges to reach S from T.
A matched edge can not reach a free vertex. A
matched edge can not reach an already reached
matched vertex, because it has been reached using
a different matched edge.
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Hungarian tree search finishes in O(E)
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Is it possible that there is an augmenting path,
but the Hungarian tree does not yield one?
An augmenting path has odd length.
Any odd length path must have one end in S and
another end in T.
An augmenting path must have one free vertex in S
and another in T.
Consider an augmenting path from the free vertex
in S.
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The first edge is from a free vertex in S. The
second edge is a matched edge from T. The third
edge is a free edge from S. The fourth is a
matched edge from T.
We are looking at all such paths in the Hungarian
method.
If there is an augmenting path, we would find one!
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Complexity Analysis
Once an augmenting path is found, one new vertex
is matched.
Need to look for augmenting paths at most V
times.
Looking for an augmenting path consume O(E).
So, overall O(VE).
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Nonbipartite Graph
Basic algorithm remains same.
Need to find augmenting paths in different ways.
We can not partition the vertex set into two
sets, and start with free vertices in just one
set.
As a result need to consider all free vertices.
You may clearly reach some free vertex in your
search because an augmenting path ends in a free
vertex.
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But reaching a free vertex does not mean that we
have reached an augmenting path.
So we can not rule out reaching free vertices (as
an augmenting path starts and ends in a free
vertex). At the same time when we reach a free
vertex, we do not know whether we have an
augmenting path or not!
Odd cycles are the problems.
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So for nonbipartite graphs, we start from a
single free vertex at a time.
Also, whenever odd cycles are detected they are
shrunk into one vertex, called blossoms.
After discovering augmenting paths, the blossoms
are expanded.
Need an O(V4) algorithm.
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Weighted Matching
Now, every edge has a weight.
Weight of a matching is the sum of the weights of
its edges.
Need to find a maximum weighted matching. More
general case of maximum matching (maximum
matching is maximum weighted matching with edge
weight 1)
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Is it sufficiently general to consider a graph
with nonnegative weights?
Is it sufficiently general to consider complete
graphs? (A complete graph is one where there is
an edge between every pair of vertices).
For bipartite graphs, is it sufficient to
consider equal size partitions only.
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Bipartite Graphs
Will consider bipartite graphs. Partitions are X
and Y and X Y We assume that there is
always an edge between vertex x in partition X,
and vertex y in partition Y. Every edge has a
nonnegative weight.
A matching where every vertex is matched is a
perfect matching. Such a bipartite graph has a
perfect matching.
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Vertex Labeling
We assign a nonnegative real number with each
vertex. This nonnegative real number is the label
of the vertex. Label of vertex x is l(x).
A label is feasible if for every edge (x,
y), l(x) l(y) ? w(x, y)
An example of a feasible label is l(y) 0 for
all vertices in Y and l(x) maxyw(x, y)
l(x) l(y) maxyw(x, y) ? w(x, y)
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Equality Subgraph
Equality subgraph consists of all vertices. An
edge (x, y) is included in the equality subgraph
if l(x) l(y) w(x, y).
If the equality subgraph Gl has a perfect
matching M?, then M? is the maximum weighted
matching in the original graph G.
Weight of M? ?e?M w(e)
?u, u, is matched l(u) (since M? is a
matching and Gl is an equality subgraph)
?u, u, is in V l(u) (since M? is a
perfect matching)
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Consider any other matching M in the graph G.
Weight of M ?e?M w(e)
??u, u, is matched l(u) (since M is
matching)
??u, u, is in V l(u) (since labels are
nonnegative)
Weight of M?
So, if we can adjust the labels so that the
equality subgraph has a perfect matching, then we
are done. Why?
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If a perfect matching exists, then it is the
maximum matching.
So, existence of a perfect matching can be tested
by computing the maximum matching in the equality
subgraph.
Supposing, the equality subgraph had all the
edges in the original graph. Then it is a
complete bipartite graph. Hence, it would have a
perfect matching.
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So, if the equality subgraph does not have a
perfect matching, we should adjust the labels so
that the number of edges in the equality subgraph
increases.
However, any adjustment should be such that the
labels are feasible.
Adjustment should aim at satisfying equality
constraints in the edges.
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Maximum Weighted Matching Algorithm
Start with the following labels
l(y) 0 for all vertices in Y and l(x)
maxyw(x, y)
2. Construct the equality subgraph.
Find a maximum matching in the equality subgraph.
If this is a perfect matching, then stop.
Otherwise, readjust the labels and go to step (2).
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Readjustment of Labels
Construct the Hungarian tree w.r.t. the current
maximum matching in the equality subgraph.
Start from the free vertices in X. Free vertices
exist since the matching is not perfect.
The hungarian tree will not yield an augmenting
path in the equality subgraph.
Let S be the set of vertices in X in the tree,
and T be the set of vertices in Y.
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Observe that there are no edges between S and Y
T in the equality subgraph (else hungarian tree
has not been grown properly).
At the same time, there are edges between S and Y
T in the original graph.
Reason Current matching is not maximum in G, as
G has a perfect matching, and this is not a
perfect matching. So the hungarian tree must
have an augmenting path in G. But we know
there is no augmenting path in the current
hungarian tree. This means that the current
hungarian tree can grow further in G, and this is
possible only if there are edges between S and
Y-T.
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Decrease the labels in S uniformly, and increase
those in T uniformly so that the labels are valid.
Edges between S and Y-T in G will now start
entering the equality subgraph.
Stop when the first edge enters.
A new vertex in Y is added to the Hungarian
tree. If this vertex is free we have an
augmenting path, otherwise the Hungarian tree
grows.
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If this vertex is free, stop the relabelling and
go to step (2).
Otherwise, relabel once again.
Hungarian tree can not grow indefinitely. So we
always improve the matching before going to step
(2).
Now, we improve the matching till we find a
perfect matching in the equality subgraph.