SCIENCE CHINA Information Sciences, Volume 64 , Issue 5 : 152207(2021) https://doi.org/10.1007/s11432-020-2981-4

## A novel synthesis method for reliable feedback shift registers via Boolean networks

• AcceptedJun 19, 2020
• PublishedMar 25, 2021
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### Acknowledgment

This work was supported by National Natural Science Foundation of China (Grant Nos. 61973078, 61903339), Natural Science Foundation of Jiangsu Province of China (Grant No. BK20170019), Jiangsu Province Six Talent Peaks Project (Grant No. 2015-ZNDW-002), Fundamental Research Funds for the Central Universities (Grant No. 2242019k1G013), and Postgraduate Research Practice Innovation Program of Jiangsu Province (Grant No. KYCX19_0111).

### Supplement

Appendix

In Theorem 3.14, we proved that for any $n$-stage monotonous FSR, the total number of cyclic attractors is no more than $2^{n-2}$, but we did not prove that there exists a cyclic attractor including $\delta_{2^{n}}^{\alpha}$ and $\delta_{2^{n}}^{\frac{\alpha}{2}+2^{n-1}}$ rather than including $\delta_{2^{n}}^{\alpha}$ and $\delta_{2^{n}}^{\frac{\alpha}{2}}$ for any even integer $\alpha\in~[\mathcal~{O}_{2}]$. In the following, we prove that there is definitely no such cycle $\delta_{2^{n}}^{\alpha}\rightarrow~\delta_{2^{n}}^{2(\alpha-2^{n-1})}\rightarrow\cdots\rightarrow~\delta_{2^{n}}^{\frac{\alpha}{2}}\rightarrow~\delta_{2^{n}}^{\alpha}$ but there possibly exists the cycle $\delta_{2^{n}}^{\alpha}\rightarrow~\delta_{2^{n}}^{2(\alpha-2^{n-1})}\rightarrow\cdots\rightarrow~\delta_{2^{n}}^{\frac{\alpha}{2}+2^{n-1}}\rightarrow~\delta_{2^{n}}^{\alpha}$.

Assume that there exists a cyclic attractor with length $\kappa$ in an $n$-stage monotonous FSR: $X_{1}\rightarrow~X_{2}\rightarrow~\cdots\rightarrow~X_{\kappa}$. Then by Definition 3.4, we have that $w(X_{\kappa})\leq~w(X_{\kappa-1})\leq\cdots\leq~w(X_{1})\leq~w(X_{\kappa})$. Therefore, $w(X_{\kappa})=~w(X_{\kappa-1})=\cdots=w(X_{1})=~w(X_{\kappa})$ [2]. If there exists the cycle $\delta_{2^{n}}^{\alpha}\rightarrow~\delta_{2^{n}}^{2(\alpha-2^{n-1})}\rightarrow\cdots\rightarrow~\delta_{2^{n}}^{\frac{\alpha}{2}}\rightarrow~\delta_{2^{n}}^{\alpha}$, then $w(X_{\alpha})=w(X_{\frac{\alpha}{2}})$, where $X_{\alpha}=(a_{1},~a_{2},\ldots,~a_{n})^{\rm~T}\sim~\delta_{2^{n}}^{\alpha},~X_{\frac{\alpha}{2}}=(a'_{1},~a'_{2},\ldots,~a'_{n})^{\rm~T}\sim~\delta_{2^{n}}^{\frac{\alpha}{2}}$, and $\alpha$ satisfies the following condition: $$\alpha=2^{n}-(a_{1}2^{n-1}+a_{2}2^{n-2}+\cdots+a_{n}). \tag{10}$$ Multipling both sides of Eq. (10) by $\frac{1}{2}$ results in \begin{align} \frac{\alpha}{2}=&2^{n-1}-(a_{1}2^{n-2}+a_{2}2^{n-3} +\cdots+a_{n-1}+\frac{a_{n}}{2}) \\ =&2^{n}-(2^{n-1}+a_{1}2^{n-2}+a_{2}2^{n-3} +\cdots+a_{n-1}), \tag{11} \end{align} which implies that $X_{\frac{\alpha}{2}}=(1,~a_{1},\ldots,~a_{n-1})^{\rm~T}$ by Lemma 2.1, and then we have $a_{i}=a'_{i+1}$, $i\in~[1,~n-1]$. It follows from Lemma 3.16 that $\alpha$ is an even integer, and then we have $a_{n}=0$, and $X_{\alpha}=(a_{1},~a_{2},\ldots,~a_{n-1},~0)^{\rm~T}$. Based on the analysis, we have that $w(X_{\alpha})\neq~w(X_{\frac{\alpha}{2}})$ and $w(X_{\alpha})-~w(X_{\frac{\alpha}{2}})=1$, which contradicts with $w(X_{\alpha})=w(X_{\frac{\alpha}{2}})$. Therefore, there does not exist the cycle $\delta_{2^{n}}^{\alpha}\rightarrow~\delta_{2^{n}}^{2(\alpha-2^{n-1})}\rightarrow\cdots\rightarrow~\delta_{2^{n}}^{\frac{\alpha}{2}}\rightarrow~\delta_{2^{n}}^{\alpha}$.

But for another sequence $\delta_{2^{n}}^{\alpha},~\delta_{2^{n}}^{2(\alpha-2^{n-1})},\ldots,~\delta_{2^{n}}^{\frac{\alpha}{2}+2^{n-1}},~\delta_{2^{n}}^{\alpha}$, it is possible to form a cycle. The proof is given as follows. Based on Eq. (8) and Lemma 3.16, the subpaths $\delta_{2^{n}}^{\alpha}\rightarrow~\delta_{2^{n}}^{2(\alpha-2^{n-1})}$ and $\delta_{2^{n}}^{\frac{\alpha}{2}+2^{n-1}}\rightarrow~\delta_{2^{n}}^{\alpha}$ are determined, but the rest path $\delta_{2^{n}}^{2(\alpha-2^{n-1})}\rightarrow~\cdots\rightarrow~\delta_{2^{n}}^{\frac{\alpha}{2}+2^{n-1}}$ with $\delta_{2^{n}}^{2(\alpha-2^{n-1})}\in~\mathcal~{O}_{1}$ is indeterminate, and depends on $b_{i},~i\in~[1,~2^{n-1}]$. Therefore, we only consider these two determined subpaths. That is, if $w(X_{\alpha})=w(X_{2(\alpha-2^{n-1})})$ and $w(X_{\alpha})=w(X_{\frac{\alpha}{2}+2^{n-1}})$, then the cycle is possible; otherwise, it must be impossible.

By doubling both sides of Eq. (10), one has that \begin{equation*} 2\alpha=2^{n+1}-(a_{1}2^{n}+a_{2}2^{n-1}+\cdots+2a_{n}).\end{equation*} Then subtracting both sides of the above equation by $2^{n-1}$ results in $$2(\alpha-2^{n-1})=2^{n}-(a_{1}2^{n}+a_{2}2^{n-1}+\cdots+2a_{n}). \tag{12}$$ Since $\alpha\in~[\mathcal~{S}_{1}]$, which means $a_{1}=0$, Eq. (12) can be rewritten as $$2(\alpha-2^{n-1})=2^{n}-(a_{2}2^{n-1}+\cdots+2a_{n}+0), \tag{13}$$ which implies that $X_{2(\alpha-2^{n-1})}=(a_{2},~a_{3},\ldots,~a_{n},~0)^{\rm~T}$. Therefore, $w(X_{\alpha})=w(X_{2(\alpha-2^{n-1})})$.

On the other hand, adding both sides of Eq. (10) by $2^{n-1}$ results in \begin{align} \frac{\alpha}{2}+2^{n-1} =&2^{n}-(a_{1}2^{n-2}+a_{2}2^{n-3} +\cdots+a_{n-1}) \\ =&2^{n}-(0\cdot2^{n-1}+a_{1}2^{n-2}+a_{2}2^{n-3} +\cdots+a_{n-1}), \tag{14} \end{align} which implies that $X_{\frac{\alpha}{2}+2^{n-1}}=(0,~a_{1},~a_{2},\ldots,~a_{n-1})^{\rm~T}$, and $w(X_{\alpha})=w(X_{\frac{\alpha}{2}+2^{n-1}})$. Therefore, $w(X_{\alpha})=w(X_{\frac{\alpha}{2}+2^{n-1}})=w(X_{2(\alpha-2^{n-1})})$.

Based on the above analysis, we can conclude that for any even integer $\alpha\in~[\mathcal~{O}_{2}]$, the cycle $\delta_{2^{n}}^{\alpha}\rightarrow~\delta_{2^{n}}^{2(\alpha-2^{n-1})}\rightarrow\cdots\rightarrow~\delta_{2^{n}}^{\frac{\alpha}{2}}\rightarrow~\delta_{2^{n}}^{\alpha}$ cannot exist but the cycle $\delta_{2^{n}}^{\alpha}\rightarrow~\delta_{2^{n}}^{2(\alpha-2^{n-1})}\rightarrow\cdots\rightarrow~\delta_{2^{n}}^{\frac{\alpha}{2}+2^{n-1}}\rightarrow~\delta_{2^{n}}^{\alpha}$ may exist. Therefore, for any $n$-stage monotonous FSR, the total number of cyclic attractors is no more than $2^{n-2}$. Additionally, the above analysis also proved that it is impossible to have a cyclic attractor including $(0~1~1~0)^{\rm~T}$ and $(1~0~1~1)^{\rm~T}$ in Remark 3.15.

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• Figure 1

The state transition graph $T(V,~E)$ of system (9), where each $4$-tuple represents a state $(x_{1},~x_{2},~x_{3},~x_{4})^{\rm~T}$.

• Figure 2

The $PT(V,~E&apos;)$ of a $4$-stage monotonous FSR, where each $4$-tuple represents a state $(x_{1},~x_{2},~x_{3},~x_{4})^{\rm~T}$. A node with an arrow in a solid line (or two arraws in dotted lines) represents that it has one successor (or two potential successors).

• Figure 3

The $PT(V,~E&apos;)$ of a $4$-stage monotonous FSR after deleting some edges based on Lemma 3.16, where each $4$-tuple represents a state $(x_{1},~x_{2},~x_{3},~x_{4})^{\rm~T}$, and the dashed line with a “$\times$” means the path has been deleted.

• Figure 4

The pseudo-state transition graph of a $5$-stage monotonous FSR, where each $5$-tuple represents a state $(x_{1},~x_{2},~x_{3},$ $x_{4},~x_{5})^{\rm~T}$.

• Figure 5

The $PT(V,~E&apos;)$ of a $5$-stage monotonous FSR by deleting some edges based on Theorem 3.20, where each $5$-tuple represents a state $(x_{1},~x_{2},~x_{3},~x_{4},~x_{5})^{\rm~T}$, and the dashed line with a “$\times$” means the path has been deleted.

•

Algorithm 1 Find all the states in set $\mathcal~{S}_{2}$ that can reach $\mathcal~{S}_{1}$.

Initial set $\mathcal~{S}_{1}=\{\delta_{2^{n}}^{2^{n-1}+1},~\delta_{2^{n}}^{2^{n-1}+2},\ldots,~\delta_{2^{n}}^{2^{n-1}+2^{n-2}}\}$;

for $i=2^{n-1}+1$ to $2^{n-1}+2^{n-2}$

$x=i$, $F_{i}=\emptyset$;

$F_{i}=F_{i}\cup{\delta_{2^{n}}^{x}}$;

while $(x<~2^{n})\wedge$ ($x$ is even) do

$x=\frac{x}{2}+2^{n-1}$;

$F_{i}\leftarrow~\delta_{2^{n}}^{x}$;

end while

Output $F_{i}$;

end for

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