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Proof. Let us fix r, n, p, i, A, and B as in the lemma. By Lemma 4 , if a coordinate
is irrelevant for A , then it is also irrelevant for B and vice versa.
In the case n = r ,wehave A = B and this family must be a singleton, so
that the lemma is trivially true. From now on, we assume that n>r and hence
the notion of r -cross-intersecting families is meaningful for n
1 coordinates.
Let q =( p 1 ,...,p i− 1 ,p i +1 ,...,p n ). For any l
[ p i ], let
A l =
{
x
A : x i = l
}
,
B l =
,
and let A l and B l stand for the families obtained from A l and B l , respectively,
by dropping their i th coordinates. By definition, we have A l ,B l
{
y
B : y i = l
}
S q ,and
|
A
|
=
l |
= l |
[ p i ] ,
the families A l and B m are r -cross-intersecting, since the vectors in A l differ from
the vectors in B m in the i th coordinate, and therefore the r indices where they
agree must be elsewhere.
Let Z denote the maximum product
A l |
and
|
B
|
B l |
. Furthermore, for any two distinct elements l, m
A |·|
B |
|
of an r -cross-intersecting
pair A ,B
= m .
Adding an irrelevant i th coordinate to the maximal r -cross-intersecting pair
A ,B
S q .Wehave
|
A l |·|
B m |≤
Z for all l, m
[ p i ] with l
S q , we obtain a pair A ,B
S p with
|
A |·|
B |
= p i Z . Thus, using
the maximality of A and B ,wehave
|
A
|·|
B
|≥
p i Z .Let l 0 be chosen so as to
maximize
|
A l 0 |·|
B l 0 |
, and let c =
|
A l 0 |·|
B l 0 |
/Z .
Assume first that c
1. Then we have
p i Z
Z = p i Z.
≤|
A
|·|
B
|
=
[ p i ] |
A l |·|
B m |≤
l,m
l,m
[ p i ]
Hence, we must have equality everywhere. This yields that c = 1 and that A l
and B m form a maximal r -cross-intersecting pair for all l, m
[ p i ], l
= m . This
also implies that
|
A l |
=
|
A m |
for l, m
[ p i ], from where the statement of the
lemma follows, provided that p i =2.
If p i
3, then all families A l must be equal to one another, since one member
in a maximal r -cross-intersecting family determines the other (by Lemma 4 ). This
contradicts our assumption that the i th coordinate was relevant for A .
Thus, we may assume that c> 1.
For m
[ p i ], m
= l 0 ,wehave
|
A l 0 |·|
B m |≤
Z =
|
A l 0 |·|
B l 0 |
/c .Thus,
|
B m |≤|
B l 0 |
/c,
(1)
= m∈ [ p i ] |
which yields that
|
B
|
B m |≤
(1 + ( p i
1) /c )
|
B l 0 |
. By symmetry, we
also have
|
A m |≤|
A l 0 |
/c
(2)
for m
= l 0 and
|
A
|≤
(1 + ( p i
1) /c )
|
A l 0 |
. Combining these inequalities, we
obtain
p i Z
1) /c ) 2
1) /c ) 2 cZ.
≤|
A
|·|
B
|≤
(1 + ( p n
|
A l 0 |·|
B l 0 |
=(1+( p i
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