Two constructions of balanced Boolean functions with optimal algebraic immunity, high nonlinearity and good behavior against fast algebraic attacks

Abstract

In this paper, two constructions of Boolean functions with optimal algebraic immunity are proposed. They generalize previous ones respectively given by Rizomiliotis (IEEE Trans Inf Theory 56:4014–4024, 2010) and Zeng et al. (IEEE Trans Inf Theory 57:6310–6320, 2011) and some new functions with desired properties are obtained. The functions constructed in this paper can be balanced and have optimal algebraic degree. Further, a new lower bound on the nonlinearity of the proposed functions is established, and as a special case, it gives a new lower bound on the nonlinearity of the Carlet-Feng functions, which is slightly better than the best previously known ones. For $n\le 19$, the numerical results reveal that among the constructed functions in this paper, there always exist some functions with nonlinearity higher than or equal to that of the Carlet-Feng functions. These functions are also checked to have good behavior against fast algebraic attacks at least for small numbers of input variables.

Appendix

1.1 The proofs of Lemmas 5–8

Proof of Lemma 5

When $x\in [0,\frac{1}{4}]$, let $\varphi (x)=\sin \pi x-\pi x-(8\sqrt{2}-4\pi )x^2$. Then $\frac{d\varphi }{dx}=\pi \cos \pi x-\pi -2(8\sqrt{2}-4\pi )x$, and $\frac{d^2\varphi }{dx^2}=-\pi ^2\sin \pi x-2(8\sqrt{2}-4\pi )$. We have $\frac{d^2\varphi }{dx^2}|_{x=0}=-2(8\sqrt{2}-4\pi )>0$ and $\frac{d^2\varphi }{dx^2}|_{x=\frac{1}{4}}=-\frac{\sqrt{2}}{2}\pi ^2-2(8\sqrt{2}-4\pi )<0$. Thus there exists exactly one point $x_0 \,(0<x_0<\frac{1}{4})$ such that $\frac{d^2\varphi }{dx^2}|_{x=x_0}=0$, since $\frac{d^2\varphi }{dx^2}$ is strictly decreasing for $x\in [0,\frac{1}{4}]$.

Then $\frac{d^2\varphi }{dx^2}>0$ when $x\in [0,x_0)$ and $\frac{d^2\varphi }{dx^2}<0$ when $x\in (x_0,\frac{1}{4}]$. Consequently, $\frac{d\varphi }{dx}$ is strictly increasing for $x\in [0,x_0]$ and is strictly decreasing for $x\in [x_0,\frac{1}{4}]$. Therefore, we have $\frac{d\varphi }{dx}>0$ for $x\in (0,x_0]$ since $\frac{d\varphi }{dx}|_{x=0}=0$. On the other hand, $\frac{d\varphi }{dx}|_{x=\frac{1}{4}}=(\frac{\sqrt{2}}{2}+1)\pi -4\sqrt{2}<0$. Thus, there exists exactly one point $x_1\,(x_0<x_1<\frac{1}{4})$ such that $\frac{d\varphi }{dx}|_{x=x_1}=0$. Furthermore, $\frac{d\varphi }{dx}>0$ for $x\in (0,x_1)$ and $\frac{d\varphi }{dx}<0$ for $x\in (x_1,\frac{1}{4}]$.

Thus, $f(x)$ is strictly increasing for $x\in [0,x_1]$ and is strictly decreasing for $x\in [x_1,\frac{1}{4}]$. Since $f(0)=f(\frac{1}{4})=0$, we have $f(x)\ge 0$ with $x\in [0,\frac{1}{4}]$.

The proof of the case $x\in [\frac{1}{4},\frac{1}{2}]$ is the same as that of the case $x\in [0,\frac{1}{4}]$, and we omit its proof here.$\square$

Proof of Lemma 6

If $a> 0$, then

$$\begin{aligned} \frac{1}{a(i+1)+b}\le \frac{1}{ax+b}\quad \mathrm{for\ }x\in [i,i+1], \end{aligned}$$

i.e.,

$$\begin{aligned} \frac{1}{a(i+1)+b}\le \int _{i}^{i+1}\frac{1}{ax+b}\,\mathrm d x. \end{aligned}$$

Thus, we have

$$\begin{aligned} \sum _{i=L_1}^{L_2}\frac{1}{ai+b}&= \sum _{i=L_1-1}^{L_2-1}\frac{1}{a(i+1)+b}\le \sum _{i=L_1-1}^{L_2-1}\int _{i}^{i+1}\frac{1}{ax+b}\,\mathrm d x\\&= \int _{L_1-1}^{L_2}\frac{1}{ax+b}\,\mathrm d x=\frac{1}{a}\ln \left| \,\frac{aL_2+b}{a(L_1-1)+b}\,\right| . \end{aligned}$$

For the case $a<0$, the inequality in this lemma can be similarly proved.$\square$

Proof of Lemma 7

Note that

$$\begin{aligned}&1-\frac{\sqrt{2}}{4}\pi +(-8+4\sqrt{2}+\frac{3 }{\sqrt{2}}\pi )x-2(-8+4\sqrt{2}+\sqrt{2}\pi )x^2\\&\quad =2(-8+4\sqrt{2}+\sqrt{2}\pi )(x-N_1)(N_2-x), \end{aligned}$$

by Lemma 5

$$\begin{aligned} \frac{1}{\sin \pi x}&\le \frac{1}{2(-8+4\sqrt{2}+\sqrt{2}\pi )(x-N_1)(N_2-x)}\nonumber \\&= \frac{1}{N_3}\left( \frac{1}{x-N_1}+ \frac{1}{-x+N_2}\right) \end{aligned}$$

(16)

for $x\in [\frac{1}{4},\frac{1}{2}]$, where $N_1$, $N_2$ and $N_3$ are defined as in Lemma 7, respectively. By Lemma 5 and the inequality (16), we have

$$\begin{aligned} \frac{1}{\cos \frac{j\pi }{2^n-1}}=\frac{1}{\sin \left( \frac{\pi }{2}-\frac{j\pi }{2^n-1}\right) }\le \frac{1}{N_3}\left( \frac{1}{-\frac{j}{2^n-1}+\frac{1}{2}-N_1}+ \frac{1}{\frac{j}{2^n-1}-\frac{1}{2}+N_2}\right) \end{aligned}$$

for $1\le j\le 2^{n-2}-1$. Thus, by Lemma 6, the sum in this lemma satisfies

$$\begin{aligned} \sum _{j=1}^{2^{n-2}-1} \dfrac{1}{\cos \frac{j\pi }{2^n-1}}&\le \frac{1}{N_3}\left( \sum _{j=1}^{2^{n-2}-1}\frac{1}{-\frac{j}{2^n-1}+\frac{1}{2}-N_1} +\sum _{j=1}^{2^{n-2}-1}\frac{1}{\frac{j}{2^n-1}-\frac{1}{2}+N_2}\right) \\&\le \frac{2^n-1}{N_3}\left( \ln \left| \,\frac{-\frac{1}{2^n-1}+\frac{1}{2}-N_1}{-\frac{2^{n-2}}{2^n-1}+\frac{1}{2}-N_1}\,\right| +\ln \left| \,\frac{\frac{2^{n-2}-1}{2^n-1}-\frac{1}{2}+N_2}{-\frac{1}{2}+N_2}\,\right| \right) . \end{aligned}$$

Note that the real number

$$\begin{aligned} \ln \frac{\frac{1}{2}-N_1}{\frac{1}{4}-N_1}+\ln \frac{N_2-\frac{1}{4}}{N_2-\frac{1}{2}} \end{aligned}$$

is the limit of the sequence

$$\begin{aligned} \ln \left| \,\frac{-\frac{1}{2^n-1}+\frac{1}{2}-N_1}{-\frac{2^{n-2}}{2^n-1}+\frac{1}{2}-N_1}\,\right| +\ln \left| \,\frac{\frac{2^{n-2}-1}{2^n-1}-\frac{1}{2}+N_2}{-\frac{1}{2}+N_2}\,\right| . \end{aligned}$$

Since the sequences

$$\begin{aligned} \ln \left( -\frac{1}{2^n-1}+\frac{1}{2}-N_1\right) -\ln \left( -\frac{2^{n-2}}{2^n-1}+\frac{1}{2}-N_1\right) \text{ and } \ln \left( \frac{2^{n-2}-1}{2^n-1}-\frac{1}{2}+N_2\right) \end{aligned}$$

are both increasing with respect to $n$, then

$$\begin{aligned} \sum _{j=1}^{2^{n-2}-1} \dfrac{1}{\cos \frac{j\pi }{2^n-1}} \le \frac{2^n-1}{N_3}\left( \ln \frac{\frac{1}{2}-N_1}{\frac{1}{4}-N_1}+\ln \frac{N_2-\frac{1}{4}}{N_2-\frac{1}{2}}\right) . \end{aligned}$$

This finishes the proof. $\square$

Proof of Lemma 8

By Lemma 5, we have

$$\begin{aligned} \frac{1}{\sin \pi x}\le \frac{1}{\pi }\left( \frac{1}{x}+\frac{-N_4}{ N_4x-\pi }\right) , \end{aligned}$$

(17)

where $N_4$ is defined as in Lemma 8. Since $0\le \frac{j+\frac{1}{2}}{2^n-1}\le \frac{1}{4}$ for $0\le j\le 2^{n-2}-1$, by Lemma 6 and the inequality (17), we have

$$\begin{aligned} \sum _{j=0}^{2^{n-2}-1} \dfrac{1}{\sin \frac{(j+\frac{1}{2})\pi }{2^n-1}}&\le \frac{1}{\pi }\left( \sum _{j=0}^{2^{n-2}-1}\frac{2^n-1}{j+ \frac{1}{2}}+\sum _{j=0}^{2^{n-2}-1}\frac{-N_4(2^n-1)}{ N_4j+\frac{1}{2}N_4-\pi (2^n-1)}\right) \\&= \frac{2^n-1}{\pi }\left( 2+\sum _{j=1}^{2^{n-2}-1}\frac{1}{j+\frac{1}{2}}+\sum _{j=0}^{2^{n-2}-1}\frac{1}{ -j-\frac{1}{2}+\frac{\pi (2^n-1)}{N_4}}\right) \\&\le \frac{2^n-1}{\pi }\left( 2+(n-1)\ln 2+\ln \frac{\pi (2^n-1)-\frac{1}{2}N_4}{-2^{n-2}N_4+\pi (2^n-1)-\frac{1}{2}N_4}\right) . \end{aligned}$$

Since the sequence

$$\begin{aligned} \ln \frac{\pi (2^n-1)-\frac{1}{2}N_4}{-2^{n-2}N_4+\pi (2^n-1)-\frac{1}{2}N_4} \end{aligned}$$

is decreasing with respect to $n$, for $n\ge 5$,

$$\begin{aligned} \ln \frac{\pi (2^n-1)-\frac{1}{2}N_4}{-2^{n-2}N_4+\pi (2^n-1)-\frac{1}{2}N_4}\le \ln \left( \frac{31\pi -\frac{1}{2}N_4}{31\pi -\frac{17}{2}N_4}\right) . \end{aligned}$$

Therefore,

$$\begin{aligned} \sum _{j=0}^{2^{n-2}-1} \dfrac{1}{\sin \frac{(j+\frac{1}{2})\pi }{2^n-1}}\le \frac{2^n-1}{\pi }\left( 2+(n-1)\ln 2+\ln \left( \frac{31\pi -\frac{1}{2}N_4}{31\pi -\frac{17}{2}N_4}\right) \right) , \end{aligned}$$

this completes the proof.$\square$