Bent and Semi-bent Functions via Linear Translators

Abstract

The paper is dealing with two important subclasses of plateaued functions: bent and semi-bent functions. In the first part of the paper, we construct mainly bent and semi-bent functions in the Maiorana-McFarland class using Boolean functions having linear structures (linear translators) systematically. Although most of these results are rather direct applications of some recent results, using linear structures (linear translators) allows us to have certain flexibilities to control extra properties of these plateaued functions. In the second part of the paper, using the results of the first part and exploiting these flexibilities, we modify many secondary constructions. Therefore, we obtain new secondary constructions of bent and semi-bent functions not belonging to the Maiorana-McFarland class. Instead of using bent (semi-bent) functions as ingredients, our secondary constructions use only Boolean (vectorial Boolean) functions with linear structures (linear translators) which are very easy to choose. Moreover, all of them are very explicit and we also determine the duals of the bent functions in our constructions. We show how these linear structures should be chosen in order to satisfy the corresponding conditions coming from using derivatives and quadratic/cubic functions in our secondary constructions.

A Appendix

The following proposition is related to Proposition 4 in Sect. 4.

Proposition 8

Let H be defined by Eq. (4.1), $\gamma$ and $\delta$ be defined as in Proposition 4. Then the dual of H is $\tilde{H}(x,y)=Tr_{1}^{m}\left( y \phi ^{-1}(x) \right) + h(\phi ^{-1}(x))$ where $\phi ^{-1}={\pi _2}^{-1} \circ \rho ^{-1} \circ {\pi _1}^{-1}$ and $\rho ^{-1}(x)$ is given as follows.

• (i) If $\gamma$ is a 0-linear structure of f, $\delta$ is a 0-linear structure of f and g, then

  $$\begin{aligned} \rho ^{-1}(x)=x+\gamma f(x)+\delta \left[ g(x)(1+f(x)) + g(x+\gamma )f(x)\right] . \end{aligned}$$

• (ii) If $\gamma$ is a 0-linear structure of f, $\delta$ is a 1-linear structure of f and $\delta +\gamma$ is a 0-linear structure of g, then

  $$\begin{aligned} \rho ^{-1}(x)&=x+\gamma \left[ g(x)+f(x)\big (1+g(x)+g(x+\gamma )\big )\right] +\delta \left[ g(x)(1+f(x))\right. \\&\quad \left. +\,g(x+\gamma )f(x)\right] . \end{aligned}$$

• (iii) If $\delta$ is a 0-linear structure of g, $\gamma$ is a 0-linear structure of f and g, then

  $$\begin{aligned} \rho ^{-1}(x)=x+\gamma \left[ f(x)(1+g(x)) + f(x+\delta )g(x)\right] +\delta g(x). \end{aligned}$$

• (vi) If $\delta$ is a 0-linear structure of g, $\gamma$ is a 1-linear structure of g and $\delta +\gamma$ is a 0-linear structure of f, then

  $$\begin{aligned} \rho ^{-1}(x)&=x+\gamma \left[ f(x)(1+g(x)) + f(x+\delta )g(x)\right] +\delta \left[ f(x)(1+g(x))\right. \\&\quad \left. +\,\big (1+f(x+\delta )\big )g(x)\right] . \end{aligned}$$

• (v) If $\delta$ is a 1-linear structure of f or $\delta$ is a 0-linear structure of g, then

  $$\begin{aligned} \rho ^{-1}(x)=x+\gamma \left[ f(x)(1+g(x+\delta )) + \big (1+f(x)\big )g(x)\right] +\delta f(x). \end{aligned}$$

• (vi) If $\gamma$ is a 1-linear structure of g, $\delta$ is a 1-linear structure of f and g, then

  $$\begin{aligned} \rho ^{-1}(x)=x+\gamma g(x)+\delta \left[ f(x)(1+g(x)) + f(x+\gamma )g(x)\right] . \end{aligned}$$

Proof

We give only the proof for the case (i). Assume that $\gamma$ is a 0-linear structure of f, $\delta$ is a 0-linear structure of f and g, then we claim that

$$\begin{aligned} \rho ^{-1}(x) = {\left\{ \begin{array}{ll} x &{} \quad \text {if}\; f(x)=0\;\; \text {and}\;\; g(x)=0\\ x+\delta &{} \quad \text {if}\;\ f(x)=0 \;\;\text {and}\;\; g(x)=1 \\ x+\gamma &{} \quad \text {if}\; f(x)=1\;\; \text {and}\;\; g(x+\gamma )=0 \\ x+\gamma +\delta &{} \quad \text {if}\; f(x)=1\;\;\text {and}\;\; g(x+\gamma )=1 \end{array}\right. } \end{aligned}$$

(A.1)

Let $\rho (y)=a$. Then,

$$\begin{aligned} y+\gamma f(y) + \delta g(y)= & {} a \end{aligned}$$

(A.2)

Taking f of both sides gives $f(y+\gamma f(y) + \delta g(y))=f(a)$. Since $\gamma$ and $\delta$ are 0-linear structures of f, we get

$$\begin{aligned} f(y)= & {} f(a). \end{aligned}$$

(A.3)

Note that, $\big (f(a),g(a)\big ) \in \left\{ (0,0),(0,1),(1,0),(1,1)\right\}$. These four cases correspond to the cases in Eq. (A.1). We prove only the first case in Eq. (A.1) and the proofs of other cases are similar. Hence, we assume that $\left( f(a),g(a)\right) =(0,0)$. Then, by Eq. (A.3), $f(y)=0$ and by Eq. (A.2), $y+\delta g(y)=a$. Taking g of both sides and using that $\delta$ is a 0-linear structure of g, we obtain that $g(y+\delta g(y))=g(y)=g(a)$. As $g(a)=0$ by our assumption, we get $g(y)=0$ and putting $f(y)=g(y)=0$ in Eq. (A.2) we conclude that $y=a$.

Finally, the Eq. (A.1) can be written in the form

$$\begin{aligned} \rho ^{-1}(x)=x+\gamma f(x)+\delta \left[ g(x)(1+f(x)) + g(x+\gamma )f(x)\right] . \end{aligned}$$

The following five propositions are related to Proposition 7 in Sect. 6.

Proposition 9

Let f and g be functions from ${\mathbb F}_{2^{m}}$ to ${\mathbb F}_{2^{}}$. For $i\in \{1,2,3\}$ set $\phi _i(y):=y+\gamma _i f(y)+\delta _i g(y)$ where

• (i) $\gamma _1$, $\gamma _2$, $\gamma _3$ are elements of ${\mathbb F}_{2^{m}}^\star$ which are 0-linear structures of f;

• (ii) $\delta _1$, $\delta _2$ and $\delta _3$ are elements of ${\mathbb F}_{2^{m}}^\star$ which are 1-linear structures of f;

• (iii) $\gamma _1+\delta _1$, $\gamma _2+\delta _2$, $\gamma _3+\delta _3$ are 0-linear structures of g;

• (vi) $\gamma _1 +\gamma _2$ and $\gamma _1 +\gamma _3$ are 0-linear structures of g.

Then the function h defined on ${\mathbb F}_{2^{m}}\times {\mathbb F}_{2^{m}}$ by

$$\begin{aligned} h(x,y)&=Tr_{1}^{m} \Big ( x \phi _1(y)\Big )Tr_{1}^{m} \Big ( x \phi _2(y)\Big )+Tr_{1}^{m} \Big ( x \phi _1(y)\Big )Tr_{1}^{m} \Big ( x \phi _3(y)\Big ) \\&\quad +Tr_{1}^{m} \Big ( x \phi _2(y)\Big )Tr_{1}^{m} \Big ( x \phi _3(y)\Big ) \end{aligned}$$

is bent and the dual of h is given by

$$\begin{aligned} \tilde{h}(x,y)&= Tr_{1}^{m} \Big ( y \phi _1^{-1}(x)\Big )Tr_{1}^{m} \Big ( y \phi _2^{-1}(x)\Big )+Tr_{1}^{m} \Big ( y \phi _1^{-1}(x)\Big )Tr_{1}^{m} \Big ( y \phi _3^{-1}(x)\Big )\\&\quad +Tr_{1}^{m} \Big ( y \phi _2^{-1}(x)\Big )Tr_{1}^{m} \Big ( y \phi _3^{-1}(x)\Big ) \end{aligned}$$

where

$$\begin{aligned} \phi _i^{-1}(x)&= x+\gamma \left[ g(x)+f(x)\big (1+g(x)+g(x+\gamma )\big )\right] \\&\quad + \delta \left[ g(x)(1+f(x)) + g(x+\gamma )f(x)\right] . \end{aligned}$$

Proposition 10

Let f and g be functions from ${\mathbb F}_{2^{m}}$ to ${\mathbb F}_{2^{}}$. For $i\in \{1,2,3\}$ set $\phi _i(y):=y+\gamma _i f(y)+\delta _i g(y)$ where

• (i) $\gamma _1$, $\gamma _2$, $\gamma _3$ are elements of ${\mathbb F}_{2^{m}}^\star$ which are 0-linear structures of f and g;

• (ii) $\delta _1$, $\delta _2$ and $\delta _3$ are elements of ${\mathbb F}_{2^{m}}^\star$ which are 0-linear structures of g;

• (iii) $\delta _1 +\delta _2$ and $\delta _1 +\delta _3$ are 0-linear structures of f.

Then the function h defined on ${\mathbb F}_{2^{m}}\times {\mathbb F}_{2^{m}}$ by

$$\begin{aligned} h(x,y)&=Tr_{1}^{m} \Big ( x \phi _1(y)\Big )Tr_{1}^{m} \Big ( x \phi _2(y)\Big )+Tr_{1}^{m} \Big ( x \phi _1(y)\Big )Tr_{1}^{m} \Big ( x \phi _3(y)\Big ) \\&\quad +Tr_{1}^{m} \Big ( x \phi _2(y)\Big )Tr_{1}^{m} \Big ( x \phi _3(y)\Big ) \end{aligned}$$

is bent and the dual of h is given by

$$\begin{aligned} \tilde{h}(x,y)&= Tr_{1}^{m} \Big ( y \phi _1^{-1}(x)\Big )Tr_{1}^{m} \Big ( y \phi _2^{-1}(x)\Big )+Tr_{1}^{m} \Big ( y \phi _1^{-1}(x)\Big )Tr_{1}^{m} \Big ( y \phi _3^{-1}(x)\Big ) \\&\quad +Tr_{1}^{m} \Big ( y \phi _2^{-1}(x)\Big )Tr_{1}^{m} \Big ( y \phi _3^{-1}(x)\Big ) \end{aligned}$$

where $\phi _i^{-1}(x)= x+\gamma \left[ f(x)(1+g(x)) + f(x+\delta )g(x)\right] +\delta g(x)$.

Proposition 11

Let f and g be functions from ${\mathbb F}_{2^{m}}$ to ${\mathbb F}_{2^{}}$. For $i\in \{1,2,3\}$ set $\phi _i(y):=y+\gamma _i f(y)+\delta _i g(y)$ where

• (i) $\gamma _1$, $\gamma _2$, $\gamma _3$ are elements of ${\mathbb F}_{2^{m}}^\star$ which are 1-linear structures of g;

• (ii) $\delta _1$, $\delta _2$ and $\delta _3$ are elements of ${\mathbb F}_{2^{m}}^\star$ which are 0-linear structures of g;

• (iii) $\gamma _1+\delta _1$, $\gamma _2+\delta _2$, $\gamma _3+\delta _3$ are 0-linear structures of f;

• (vi) $\delta _1 +\delta _2$ and $\delta _1 +\delta _3$ are 0-linear structures of f.

Then the function h defined on ${\mathbb F}_{2^{m}}\times {\mathbb F}_{2^{m}}$ by

$$\begin{aligned} h(x,y)&=Tr_{1}^{m} \Big ( x \phi _1(y)\Big )Tr_{1}^{m} \Big ( x \phi _2(y)\Big )+Tr_{1}^{m} \Big ( x \phi _1(y)\Big )Tr_{1}^{m} \Big ( x \phi _3(y)\Big ) \\&\quad +Tr_{1}^{m} \Big ( x \phi _2(y)\Big )Tr_{1}^{m} \Big ( x \phi _3(y)\Big ) \end{aligned}$$

is bent and the dual of h is given by

$$\begin{aligned} \tilde{h}(x,y)&= Tr_{1}^{m} \Big ( y \phi _1^{-1}(x)\Big )Tr_{1}^{m} \Big ( y \phi _2^{-1}(x)\Big )+Tr_{1}^{m} \Big ( y \phi _1^{-1}(x)\Big )Tr_{1}^{m} \Big ( y \phi _3^{-1}(x)\Big )\\&\quad +Tr_{1}^{m} \Big ( y \phi _2^{-1}(x)\Big )Tr_{1}^{m} \Big ( y \phi _3^{-1}(x)\Big ) \end{aligned}$$

where

$$\begin{aligned} \phi _i^{-1}(x)&=x+\gamma \left[ f(x)(1+g(x)) + f(x+\delta )g(x)\right] \\&\quad +\delta \left[ f(x)(1+g(x)) + \big (1+f(x+\delta )\big )g(x)\right] . \end{aligned}$$

Proposition 12

Let f and g be functions from ${\mathbb F}_{2^{m}}$ to ${\mathbb F}_{2^{}}$. For $i\in \{1,2,3\}$ set $\phi _i(y):=y+\gamma _i f(y)+\delta _i g(y)$ where

• (i) $\gamma _1$, $\gamma _2$, $\gamma _3$ are elements of ${\mathbb F}_{2^{m}}^\star$ which are 1-linear structures of f and g;

• (ii) $\delta _1$, $\delta _2$ and $\delta _3$ are elements of ${\mathbb F}_{2^{m}}^\star$ which are 1-linear structures of f;

• (iii) $\delta _1 +\delta _2$ and $\delta _1 +\delta _3$ are 0-linear structures of g.

Then the function h defined on ${\mathbb F}_{2^{m}}\times {\mathbb F}_{2^{m}}$ by

$$\begin{aligned} h(x,y)&=Tr_{1}^{m} \Big ( x \phi _1(y)\Big )Tr_{1}^{m} \Big ( x \phi _2(y)\Big )+Tr_{1}^{m} \Big ( x \phi _1(y)\Big )Tr_{1}^{m} \Big ( x \phi _3(y)\Big ) \\&\quad +Tr_{1}^{m} \Big ( x \phi _2(y)\Big )Tr_{1}^{m} \Big ( x \phi _3(y)\Big ) \end{aligned}$$

is bent and the dual of h is given by

$$\begin{aligned} \tilde{h}(x,y)&= Tr_{1}^{m} \Big ( y \phi _1^{-1}(x)\Big )Tr_{1}^{m} \Big ( y \phi _2^{-1}(x)\Big )+Tr_{1}^{m} \Big ( y \phi _1^{-1}(x)\Big )Tr_{1}^{m} \Big ( y \phi _3^{-1}(x)\Big )\\&\quad +Tr_{1}^{m} \Big ( y \phi _2^{-1}(x)\Big )Tr_{1}^{m} \Big ( y \phi _3^{-1}(x)\Big ) \end{aligned}$$

where $\phi _i^{-1}(x)=x+\gamma \left[ f(x)(1+g(x+\delta )) + \big (1+f(x)\big )g(x)\right] +\delta f(x)$.

Proposition 13

Let f and g be functions from ${\mathbb F}_{2^{m}}$ to ${\mathbb F}_{2^{}}$. For $i\in \{1,2,3\}$ set $\phi _i(y):=y+\gamma _i f(y)+\delta _i g(y)$ where

• (i) $\gamma _1$, $\gamma _2$, $\gamma _3$ are elements of ${\mathbb F}_{2^{m}}^\star$ which are 1-linear structures of g;

• (ii) $\delta _1$, $\delta _2$ and $\delta _3$ are elements of ${\mathbb F}_{2^{m}}^\star$ which are 1-linear structures of f and g;

• (iii) $\gamma _1 +\gamma _2$ and $\gamma _1 +\gamma _3$ are 0-linear structures of f.

Then the function h defined on ${\mathbb F}_{2^{m}}\times {\mathbb F}_{2^{m}}$ by

$$\begin{aligned} h(x,y)&=Tr_{1}^{m} \Big ( x \phi _1(y)\Big )Tr_{1}^{m} \Big ( x \phi _2(y)\Big )+Tr_{1}^{m} \Big ( x \phi _1(y)\Big )Tr_{1}^{m} \Big ( x \phi _3(y)\Big ) \\&\quad +Tr_{1}^{m} \Big ( x \phi _2(y)\Big )Tr_{1}^{m} \Big ( x \phi _3(y)\Big ) \end{aligned}$$

is bent and the dual of h is given by

$$\begin{aligned} \tilde{h}(x,y)&= Tr_{1}^{m} \Big ( y \phi _1^{-1}(x)\Big )Tr_{1}^{m} \Big ( y \phi _2^{-1}(x)\Big )+Tr_{1}^{m} \Big ( y \phi _1^{-1}(x)\Big )Tr_{1}^{m} \Big ( y \phi _3^{-1}(x)\Big )\\&\quad +Tr_{1}^{m} \Big ( y \phi _2^{-1}(x)\Big )Tr_{1}^{m} \Big ( y \phi _3^{-1}(x)\Big ) \end{aligned}$$

where $\phi _i^{-1}(x)=x+\gamma g(x)+\delta \left[ f(x)(1+g(x)) + f(x+\gamma )g(x)\right]$.