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If AI ( h ) <r , then there exists g
= 0 such that either g
An r 1 ( h +1),
and therefore supp ( g )
supp ( h ), or g
An r 1 ( h ), and therefore supp ( g )
supp ( h +1). If supp ( g )
supp ( h ), then we apply Proposition 2 (last sentence
of) to the set F 1 ( z )(where z ranges over F 2 ) and to the function g (resp.
h +1) and we get
|≥ AI ( F ) −r
i =0
n−r +1
i
F 1 ( z )
F 1 ( z )
|
supp ( h )
|≥|
supp ( g )
n− i . The case where supp ( g )
|≥ AI ( F ) −r− 1
i =0
F 1 ( z )
and
|
supp ( h +1)
supp ( h + 1) is similar.
Remark . In [8], the possibility that h could be constant was not taken into
account. The statement of Proposition 5 and the proof of Theorem 1 in this
reference are therefore incomplete. However, the result of [8, Theorem 1] is valid
(it is implied by Theorem 1 of the present paper, reduced to the case m =1).
2.4 Recent Recursive Lower Bounds
Lower bounds on the nonlinearity profile of a Boolean function by
means of the nonlinearity profiles of its derivatives. We denote by D a f
the so-called derivative of any Boolean function f in the direction of a
F 2 :
D a f ( x )= f ( x )+ f ( x + a ) .
The addition is performed mod 2 (i.e. D a f is a Boolean function too). Applying
such discrete derivation several times to a function f leads to the so-called higher
order derivatives D a 1 ···
D a k f ( x )= u∈F 2 f ( x + i =1 u i a i ).
A simple tight lower bound on the r -th order nonlinearity of any function f ,
knowing a lower bound on the ( r
1)-th order nonlinearity of at least one of its
derivatives (in a nonzero direction) is (see [9]):
1
2 max
nl r ( f )
nl r− 1 ( D a f ) .
a∈F 2
When applied repeatedly, the resulting bound
1
2 i
nl r ( f )
max
a 1 ,...,a i ∈F 2
nl r−i ( D a 1 ···
D a i f )
is also tight, but can not lead to bounds equivalent to 2 n− 1 and so is weak with
respect to (1). A potentially stronger lower bound, is valid when a lower bound
on the ( r
1)-th order nonlinearity is known for all the derivatives (in nonzero
directions) of the function.
Proposition 3. [9] Let f be any n -variable function and r a positive integer
smaller than n . We have:
2 2 n
2
a∈F 2
1
2
2 n− 1
nl r ( f )
nl r− 1 ( D a f ) .
This bound also is tight.
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