Claim: In the class of S1,2,2 -free planar graphs, there are no augmenting graphs with
a black vertex of high degree.
When finding augmenting graphs, we consider only those which are minimal. The
definition of minimal augmenting graph tells us that for every subset A  W ,
A  N B ( A) .
Hall’s Theorem states that a bipartite graph H with bipartitions B and W has a perfect
matching iff N ( A)  A for all subsets A of W.
Therefore, by definition, every minimal augmenting graph must contain a perfect
matching. (Augmenting graphs always have an odd number of vertices so there will be
one vertex which is not matched).
Note: We consider only connected graphs.
If we start with a black vertex v1 of unbounded degree, let Y denote the set of white
neighbors of v1 and let Z denote the set of k black vertices matched with the vertices of
Y. Therefore, the underlying structure of every minimal augmenting graph of this type
will look like:
Y
…
Z
We break up the proof into a few parts:
1. Suppose there exist pairs of vertices of Z which share a neighborhood in Y of
cardinality >2.
This graph is no longer planar.
2. Suppose there exist pairs of vertices of Z which share a neighborhood in Y of
cardinality=2.
a. Suppose both of these vertices have degree>2. Not planar.
b. Suppose both of these vertices have degree=2. Delete one of them. Also,
delete one of the two neighbors in Y. We now consider 3 cases:
i. If the remaining neighbor of Y (call it “a”) is not adjacent to any
other vertices, then we obtain an induced P2 emanating from v1 by
deleting only those two vertices.
ii. If the remaining neighbor a  Y is adjacent to a finite number of
additional vertices of Z, then delete them to obtain an induced P2.
iii. If the remaining neighbor a  Y is adjacent to an infinite number
of additional vertices in Z, then we want to show that there exist
infinite vertices of Z with degree=2. Note: We will refer back to
this procedure as (*). The neighbors of a must have a minimum
degree of 2 since by definition of perfect matching each is matched
to another vertex of Y as well as to a. If there exist neighbors of a
which also have high degree, then there must also exist neighbors
of a with degree ≤2. Otherwise we would have an infinite number
of vertices of high degree, and by the pigeonhole property, their
neighborhoods would intersect in sets larger then cardinality 2 and
form a nonplanar graph.(See note below) Therefore, in the case
where a has an infinite number of neighbors in Z, delete that
remaining neighbor a and you will obtain an infinite number of
vertices in Z with degree 1 in Y. Since a had an infinite number
of neighbors in Z, no other vertex of Y may have an infinite
number of neighbors in Z. Now look at the neighbors of each
vertex in the infinite neighborhood of a. None of these vertices
may have an infinite neighborhood in Z, otherwise the graph
would no longer be planar. Therefore, delete the finite number of
vertices in their neighborhoods and you will obtain an induced P2.
Note: even though we are talking about an infinite number of
vertices, the number of vertices in Y and Z must be the same, or
differ only by one since there is a perfect matching in the graph,
this is why the pigeonhole principle may be applied here.
c. Suppose one of these vertices has degree>2 and one has degree=2.
We fulfill the minimal and planar requirements. However, we show again
how to obtain either an induced P2 by deleting finitely many vertices or an
induced S1,2,2 . Delete the vertex of deg >2 as well as one of its neighbors,
call this a, in the shared neighborhood of Y. Again we consider 3 cases:
i. If a has no additional neighbors in Z, this will give you an induced
P2.
ii. If a has finitely many additional neighbors in Z, then delete them
to obtain an induced P2.
iii. If a has infinitely many additional neighbors in Z, then follow the
procedure in (*) (i.e.-delete a to obtain infinitely many P2’s
emanating from v1.).
3. Suppose there exist pairs of vertices of Z which share a neighborhood in Y of
cardinality=1.
Call the pair (u,v) and call the vertex to which they are both adjacent a,
where a  Y . Consider the following cases:
a. deg(u)=1 and deg(v)=2. If a has a finite number of additional neighbors
in Z, then delete v and a’s finite number of neighbors and we obtain an
induced P2 by deleting finitely many vertices. If a has an infinite number
of additional neighbors in Z, then follow (*).
b. deg(u)=2 and deg(v)=2. Follow the same procedure as above. Also,
delete the neighbor of u in Y which is not a.
c. deg(u)=2 and deg(v)=3. Follow the same procedure as above. Also,
delete the two neighbors of u which are not a.
d. deg(u)=3 and deg(v)=3. Follow the same procedure.
e. deg(u)>3 or deg(v)>3 (i.e.-have unbounded degree). Then there exist
infinite pairs of vertices which either follow the form of 2b, or which will
be of degree 1 and hence an induced graph can easily be found.
4. Suppose there do not exist any pair of vertices of Z which share
neighborhoods in Y.
The only possibility is the original underlying graph:
…
It is obvious that this graph contains an induced S1,2,2 .