... the Hadamard difference sets we seek in groups of order 16. All
( 16, 6, 2) difference sets are constructed in this manner; in section 4 we provide the fac-
toring for each of the 27 ( 16, 6, 2) difference ... Each
of the 27 difference sets discovered by Kibler can be rediscovered in this manner. The
results are listed below.
4 All ( 16,6 ,2) Difference sets
We list al...
... n(n +1)/2.
So if S is a subset of the differences, we have:
X
ijk
∈S
X
ijk
≥ n(n +1)/2 where |S| = n (2)
(since X
ijk
=
j+k−1
h=k
X
i1h
the inequalities (2) can be expressed in terms of our ... (I,J) difference triangle set, ∆, as a set of integers
{a
ij
| 1 ≤ i ≤ I,0≤j≤J}such that all the differences a
ij
− a
ik
,1≤i≤I,0≤k<j≤J
are positive and distinct. Let m = m(∆) be the maximum d...
... |A|
15
14
.
Proof. We start with |A + B| ≤ K|A| and |AB| ≤ K|A|. By using Pl¨unnecke’s
inequality (see Ch 6, [14]), we have |A+A| ≤ K
2
|A| and |B +B + B +B| ≤ K
4
|A|. First,
by Cauchy-Schwarz inequality, ... completes the proof.
We note that from the result in [11] and Pl¨unnecke sumset inequality ( see Ch 6, [14]),
we have that if |B| ∼ |A| < p
1/2
then
max(|A + B|, |AB|) |A|
25/2...
... c
2,1
= b, c
2,3
= r and we are done.
Now we may assume C(1, 2) ∩ C(2, 2) = ∅ by (iii). Suppose there exists a ∈ C(1, 2)
and r ∈ C(2, 2) such that a is missing from at least one of C(1, 1) or C(1, ... c
2,3
,c
2,2
,c
2,1
,c
1,1
,c
1,2
.This
coloring will complete a proper list coloring since C(1, 2) ∩ C(2, 2) = ∅. ✷
the electronic journal of combinatorics 9 (20 02), #N8 5
[5] F. G...
... 2-connected, by Condition (3). As G(t)−f
t
(tt
) is 2-connected
unless G(t) is a polygon, Condition (2) implies at least one of H and K is 2-connected.
We must show that one of H and K is an (H ∩ K)-bridge ... an
(H ∩ K)-bridge in G. We claim that G(t) is a bond. The same argument applies to K,
so Condition (2) implies at least one of H and K is an (H ∩ K)-bridge in G.
Let [x, y]=H ∩ K.If...
... also see
that k-COLORABILITY can be transformed into STAR k-COLORABILITY of bipartite
graphs (t = 2). To obtain the case for 3 ≤ t ≤ k simply add in a disjoint K
t
.
8 Questions
We conclude with ... Kostochka, D. R. Woodall, Acyclic colourings of planar graphs
with large girth. J. London Math. Soc. (2) 60 (1999), 344–352.
[7] R. Diestel, Graph Theory, Springer graduate texts (1997).
the ele...
... we need only verify that for each of the 15
pairs (n
, j
)
(0, 1), (0, 2), (1, 0), (1, 3), (1, 4), (2, 1), (2, 2), (2, 4), (3, 0), (3, 3), (4, all),
mod 5 it is true that the right side of ... put x = 1 here, and read off the coefficients of like powers of q, we have (2).
✷
3 Bijective proof
A bijective proof of (2) follows from the theory of 2-cores. The 2-core of a partition π is
obtai...
... W and by V (G) − W are connected.) Note
that ∂ W =
w∈W
∂{w}, where the addition is symmetric difference. In our situation, let
B
i
= ∂ {u, i}, i ∈ {1, 2, 3}, and B
0
= ∂ {u, 1, 2, 3}. Then B
0
, ... not
dual realizable by Lemma 10. By Lemma 13, G is also not dual realizable.
Concerning statement (2), if G is a graph of type G
∗
0
, G
∗
1
or G
∗
2
, then it follows from
Lemma 10 that G is du...