Sarah Bleiler
June 28, 2005
Claim: A planar, S1,2,2 -free minimal augmenting strip with underlying path of odd
size where V (W )  {v1, v3 ,..., v2 n1} and V ( B)  {v2 , v4 ,..., v2 n} must have the following
qualities:
1. v2 must be duplicated
2. No black or white vertex may have more than one twin.
3. Any other black vertex, say v2i , may be duplicated as long as v2i1 or v2i1 is
the corresponding white compensator (WC).
4. v2n may not be duplicated
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Consider underlying paths of odd size. WOLOG, let the vertices of the underlying paths
be represented by V (W )  {v1, v3 ,..., v2 n1} and V ( B)  {v2 , v4 ,..., v2 n} .
These paths are not augmenting since W  B . Therefore, we must duplicate a black
vertex to satisfy the requirement that W  B  1. We consider 2 cases:
A. Duplicate the first black vertex which is a neighbor to the white endpoint (i.e. v2 ).
This graph is planar, S1,2,2 -free, and minimal augmenting.
B. Duplicate any other black vertex.
This is not minimal augmenting since the neighborhood of the white vertex which
is an endpoint contains one black vertex.
Now we work with the graph represented in A.
We consider 3 cases:
(i)
(ii)
Duplicate v2 again
If the WC (white compensator) is the duplicate of v1 or v3 , then the graph
is no longer planar. If the WC is the duplicate any other vertex of
{v5 , v7 ,..., v2 n1} then the remaining graph is not minimal augmenting since
we will always have a subset A  W where A  N B ( A) . Therefore, v2
cannot have more than one twin.
Duplicate the next black vertex (i.e. v4 )
Let the WC be the duplicate v3 or v5 , then we have a valid augmenting
graph. However, if the WC is the duplicate of any other vertex then again
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Sarah Bleiler
June 28, 2005
the remaining graph is not minimal augmenting since we will always have
a subset A  W where A  N B ( A) .
(iii)
Duplicate any other black vertex except the endpoint (i.e. v2i with
i  1, 2, n ).
Let the WC be the duplicate of v2i1 or v2i1 , then we have a valid
augmenting graph.
However, if the WC is the duplicate of any other vertex then again the
remaining graph is not minimal augmenting since we will always have a
subset A  W where A  N B ( A) .
(iv)
Duplicate the black endpoint (i.e. v2n )
Let the WC be the duplicate of v2 n1 , then we create a redundant set U and
the graph augmentation can be reduced to that of H-U.
Let the WC be the duplicate of any other vertex, then not minimal.
Therefore, the black endpoint cannot be duplicated.
At this point we could have duplicated any one of the black vertices (say v2i from
i  2...n 1). The WC for this duplicated black vertex had to be either the twin of v2i1 or
v2i1 . We now consider the duplication of another black vertex in 3 more cases:
(i)
(ii)
(iii)
Duplicate v2i again
We automatically obtain a graph which is not planar.
Duplicate either v2i 2 or v2i 2 (i.e.-the black “neighbors” of v2i ).
Let the WC be the duplicate of the white vertex remaining who has no
twin. Then this graph is valid.
Let the WC be the duplicate of the white vertex who already has a twin.
Then the graph is no longer planar.
Let the WC be the duplicate of any other white vertex in the graph, then
the remaining graph is not minimal augmenting since we will always have
a subset A  W where A  N B ( A) .
Duplicate any other black vertex which is not a black “neighbor” of v2i .
Let the WC be the duplicate of v2i1 or v2i1 . Then this graph is valid.
Let the WC be the duplicate of any other white vertex in the graph, then
the remaining graph is not minimal augmenting since we will always have
a subset A  W where A  N B ( A) .
Therefore, we have shown that in order to have a minimal augmenting graph, the WC
must always be the duplicate of v2i1 or v2i1 , where v2i is the duplicated black vertex and
where no white vertex may be duplicated more than once.
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