Information Technology Reference
In-Depth Information
Proof.
Observe that for every
x
S
∈
W
the function
σ
x
S
(
x
T
)isapermutation
on
U
. Similarly, for every
x
T
∈
U
the function
σ
x
T
(
x
S
)isapermutationon
W
.
Hence, by Lemma 7,
f
is bent with respect to
U
⊕
U
and bent with respect to
W
⊕
W
. Using Lemma 7, the partial dual of
f
with respect to
W
⊕
W
can be
written as
f
W⊕W
(
x
T
,x
S
,y
T
,y
S
)=
σ
z
S
(
x
T
)
S
=
σ
−
1
·
y
T
+
x
S
·
z
S
+
g
(
x
T
,z
S
)
,
x
T
(
y
S
)
.
Now we first find
z
S
by solving the system of
k
equations implied by
σ
x
T
(
z
S
)=
y
S
.
Then
z
S
can be used to find
σ
z
S
(
x
T
). The solution is given in the theorem.
Starting from Theorem 4, the preceding theorem together with Theorem 3 can
be used to construct further bent-negabent functions of the form (1). If
m
=2,
it is easy to check that we do not obtain any new bent-negabent functions. But
for larger
m
the function
f
and a partial dual of
f
generally have a different
structure. However, explicit expressions for the partial dual of
f
can look rather
cumbersome, so we illustrate the application of Theorem 8 by an example.
Example 9.
Take
m
=3and
k
=2inTheorem4.Then
f
reads
f
(
x
1
,x
2
,x
3
,y
1
,y
2
,y
3
)=
σ
(
x
1
,x
2
,x
3
)
·
(
y
1
,y
2
,y
3
)+
g
(
x
1
,x
2
,x
3
)
,
where
σ
(
x
1
,x
2
,x
3
)=(
x
1
,x
1
+
ψ
(
x
2
)
,φ
(
x
2
)+
x
3
)
g
(
x
1
,x
2
,x
3
)=
h
(
x
2
)
.
,sothat
W
=
V
, and apply Theorem 8. Then
f
W⊕W
is
the usual dual of
f
and given by
Now set
S
=
{
0
,
1
,
2
}
f
(
x
1
,x
2
,x
3
,y
1
,y
2
,y
3
)=
σ
(
y
1
,y
2
,y
3
)
(
x
1
,x
2
,x
3
)+
g
(
y
1
,y
2
,y
3
)
,
·
where
σ
(
y
1
,y
2
,y
3
)=(
y
1
,ψ
−
1
(
y
1
+
y
2
)
,y
3
+
φ
(
ψ
−
1
(
y
1
+
y
2
)))
g
(
y
1
,y
2
,y
3
)=
h
(
ψ
−
1
(
y
1
+
y
2
))
.
f
is by Theorem 3 negabent.
The function
5 A Bound on the Degree
It is well known that, if
n>
1, bent Boolean functions in 2
n
variables have
a maximum algebraic degree of
n
[1]. If
n
∈{
2
,
3
}
,themaximumdegreeofa
