Information Technology Reference
In-Depth Information
Thus, if
N
is a permutation, then
Tr
(
H
(
x
)+
H
(
x
+
α
)) = 1 for all
x
,
i.e.
,
α
is
a 1-linear structure for
Tr
(
H
(
x
)). Conversely, assume that
Tr
(
H
(
x
)+
H
(
x
+
α
)) = 1
for all
x
∈
F
2
n
.
(6)
Let
y, z
∈
F
2
n
be such that
N
(
y
)=
N
(
z
). If
Tr
(
H
(
y
)+
H
(
z
)) = 0 then
N
(
y
)+
N
(
z
)=
L
(
y
+
z
)=0
,
and hence
y
+
z
. Further, (6) forces
y
=
z
. To complete the proof,
observe that
Tr
(
H
(
y
)+
H
(
z
)) = 1 is impossible, since it implies
∈{
0
,α
}
N
(
y
)+
N
(
z
)=
L
(
y
+
z
)+
γ
=0
,
which contradicts the assumption that
γ
is not in the image set of
L
.
Lemmas 1, 2 in combination with Lemma 4 imply the following classes of PP.
Theorem 5.
Let
L
∈
F
2
n
[
X
]
be a linearized polynomial, defining a 2- to -1 map-
ping with kernel
{
0
,α
}
.Furtherlet
H
∈
F
2
n
[
X
]
,
β
∈
F
2
n
and
γ
∈
F
2
n
be not in
the image set of
L
.
(a)
The polynomial
L
(
X
)+
γTr
H
(
X
2
+
αX
)+
βX
is PP if and only if
Tr
(
βα
)=1
.
(b)
The polynomial
L
(
X
)+
γTr
(
H
(
X
)+
H
(
X
+
α
)+
βX
)
is PP if and only if
Tr
(
βα
)=1
.
Remark 1.
To apply Theorem 5 we need to have a linearized 2- to -1 polynomial
with known kernel and image set. An example of such a polynomial is
X
2
k
+
α
2
k
−
1
X
where 1
∈
F
2
n
.Thekernelofits
≤
k
≤
n
−
1withgcd(
k, n
)=1and
α
associated mapping is
{
0
,α
}
and the image set is
H
α
−
2
k
(0). Moreover, any linear
2- to -1 mapping with kernel
(or image set
H
α
−
2
k
(0)) can be obtained as
a left (or right) composition of this mapping with an appropriate bijective linear
mapping.
{
0
,α
}
The next result is a direct consequence of Lemmas 3 and 4.
Theorem 6.
Let
L
∈
F
2
n
[
X
]
be a linearized polynomial defining a 2- to -1 map-
ping with kernel
{
0
,α
}
.Let
β, γ
∈
F
2
n
and
γ
do not belong to the image set of
2
n
L
.If
3
≤
s
≤
−
2
is of binary weight
≥
2
, then the polynomial
L
(
X
)+
γTr
(
δX
s
+
βX
)
is PP if and only if
s
=2
i
+2
j
,
(
δα
2
j
)
2
n
−
i
+(
δα
2
i
)
2
n
−
j
=0
and
Tr
(
δα
2
i
+2
j
+
βα
)=1
.
