# Review of the Lineal Complexity Calculation through Binomial Decomposition-Based Algorithms

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## Abstract

**:**

## 1. Introduction

## 2. Shift Registers and the Concept of Linear Complexity

- L binary stages, which are interconnected and numbered $(0,1,2,\dots ,L-1)$ from left to right. Each stage stores a unique bit.
- The L-degree feedback or connection polynomial$$p(x)={x}^{L}+{c}_{1}{x}^{L-1}+{c}_{2}{x}^{L-2}+\dots +{c}_{L-1}x+{c}_{L}$$
- A non-zero initial state (stage contents) at the initial instant.

**Definition**

**1.**

#### An LFSR-Based Sequence Generator

- (a)
- A PN-sequences $\{{a}_{n}\}$ generated by an L-stage LFSR and a shifted version of such a sequence, notated $\{{b}_{n}\}$. Both sequences are related by the expression $\{{b}_{n}\}=\{{a}_{n+p}\}$, p being an integer. Thus, $\{{b}_{n}\}$ is nothing but the PN-sequence $\{{a}_{n}\}$ circularly rotated p positions with $(p=0,1,2\dots ,{2}^{L}-2)$.
- (b)
- A simple decimation rule defined as:$$\left(\right)$$

- All the generalized self-shrunken sequences are balanced apart from the identically 1 sequence [25] (Theorem 1).
- By construction, the family of generalized self-shrunken sequences consists of ${2}^{L}-1$ sequences of ${2}^{L-1}$ bits each of them. Thus, the length of any generalized sequence will be ${2}^{L-1}$ or divisors. At any rate, the length of these sequences will always be a power of 2.
- The family of generalized sequences plus the identically null sequence has structure of Abelian group where the group operation is the bit-wise sum mod 2. the neutral element is the identically null sequence and every sequence is its own inverse element [25] (Theorem 2).
- The sequence produced by the self-shrinking generator is a member of this family for $p={2}^{L-1}$, see [22].

**Example**

**1.**

## 3. Binomial Sequences

#### 3.1. Introduction to Binomial Sequences

**Definition**

**2.**

- Given the binomial sequence $\left\{\left(\right),\genfrac{}{}{0pt}{}{n}{k}\right\}$ with $k={2}^{m}+i$ where m is a non-negative integer and the index i takes values in the interval $0\le i<{2}^{m}$, then we have that [12] (Proposition 3):
- (a)
- The binomial sequence $\{\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{n}{k}$ has length $l={2}^{m+1}$.
- (b)
- The formation rule of this binomial sequence is:$${\left\{\left(\right),\genfrac{}{}{0pt}{}{n}{{2}^{m}+i}\right\}}_{}0\le n{2}^{m+1}$$

- The linear complexity of the binomial sequence $\left\{\left(\right),\genfrac{}{}{0pt}{}{n}{{2}^{m}+i}\right\}$ with m and i defined as above is $LC={2}^{m}+i+1$, see [12] (Theorem 13).
- Every binary sequence ${\{{s}_{n}\}}_{n\ge 0}$ whose length is a power of 2 can be written as linear combination of binomial sequences [12] (Theorem 2). This combination is called the Binomial Decomposition of ${\{{s}_{n}\}}_{n\ge 0}$. Such a decomposition allows us to analyze fundamental properties of the sequence, e.g., length and linear complexity.
- Given a sequence ${\{{s}_{n}\}}_{n\ge 0}$ with binomial decomposition $\{{s}_{n}\}={\sum}_{i=1}^{r}\left\{\left(\right),\genfrac{}{}{0pt}{}{n}{{k}_{i}}\right\}$, where $0\le {k}_{1}<{k}_{2}<\cdots <{k}_{r}$ are integer indices, then its linear complexity is given by $LC={k}_{r}+1$, see [12] (Corollary 14).
- Given a sequence ${\{{s}_{n}\}}_{n\ge 0}$ with binomial decomposition $\{{s}_{n}\}={\sum}_{i=1}^{r}\left\{\left(\right),\genfrac{}{}{0pt}{}{n}{{k}_{i}}\right\}$, where $0\le {k}_{1}<{k}_{2}<\cdots <{k}_{r}$ are integer indices, then its length l is that of the binomial sequence $\left\{\left(\right),\genfrac{}{}{0pt}{}{n}{{k}_{r}}\right\}$, i.e., the length of the binomial sequence of maximum index in its binomial decomposition, see [32] (Theorem 1).

#### 3.2. Binomial Decomposition of GSS-Sequences

## 4. Different Algorithms to Compute the Linear Complexity of a Sequence

- For the sake of readability, In the sequel the binomial coefficient $\left(\right)$ just denotes the k-$th$ binomial sequence.
- The term ${\left(\right)}_{\genfrac{}{}{0pt}{}{n}{k}}$ represents the sub-sequence of $\left(\right)$ between the i-$th$ and j-$th$ bits.
- The term ${\left(\right)}_{\genfrac{}{}{0pt}{}{n}{k}}$ stands for the sub-sequence corresponding to the j first bits of $\left(\right)$.

#### 4.1. Berlekamp-Massey Algorithm

#### 4.2. Binomial Decomposition Algorithm or BD-Algorithm

- According to Item 3 (in Section 3.1), the sequence $seq$ of length $l={2}^{m}$ can be decomposed into r binomial sequences of the form:$$seq=\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{n}{{k}_{1}}.$$
- According to Item 4 (in Section 3.1), the lineal complexity of $seq$ is that of the binomial sequence of maximum index $\left(\right)$ in its binomial decomposition. Since the indices of the binomial sequences are written in increasing order, then $LC$ is computed by means of the following equation:$$LC={k}_{r}+1.$$

Algorithm 1 The BD-algorithm. |

Require:$seq$: the sequence to be analyzed |

$binom=[\varnothing ]$, ${k}_{r}=0$ |

for $i=0;ilength(seq);i$++ do |

if $se{q}_{i}==1$ then |

$seq+=\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{n}{i}$ |

$binom.add(i)$ |

${k}_{r}=i$ |

end if |

end for |

return $binom$ and $LC={k}_{r}+1$: binomial decomposition and $LC$ of $seq$. |

#### Improvement of the BD-Algorithm

#### 4.3. Half-Interval Search Algorithm

#### 4.3.1. Symmetry of the Binomial Sequences

**Theorem**

**1.**

- If k the index of the binomial sequence is $k<\frac{l}{2}$, then the two sub-sequences in Equation (2) are equal.
- If k the index of the binomial sequence is $k\ge \frac{l}{2}$, then the two sub-sequences in Equation (2) are written as:$${\left(\right)}_{\genfrac{}{}{0pt}{}{n}{k}}l$$

**Proof.**

- Since $k<\frac{l}{2}$, then k can be written as $k={2}^{j}+i$, where j and i are non-negative integers such that $j<m-1$ and $0\le i<{2}^{j}$. According to Item 1(a) in Section 3.1, the binomial sequence $\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{n}{k}$ has length $\tilde{l}={2}^{j+1}$ where the maximum length is ${\tilde{l}}_{max}={2}^{m-1}$ when $j=m-2$ and the minimum length ${\tilde{l}}_{min}={2}^{0}$ when $j=0$. At any rate, $\tilde{l}$ is a power of 2 as well as $\tilde{l}<{2}^{m}$ and, therefore, the first and second sub-sequences in Equation (2) are equal.
- Since $k\ge \frac{l}{2}={2}^{m-1}$, then k can be written as $k={2}^{m-1}+i$ with $0\le i<{2}^{m-1}$. According to Item 1(a) in Section 3.1, the binomial sequence $\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{n}{k}$ has length $\tilde{l}=l={2}^{m}$. Moreover, according to Item 1(b) in Section 3.1$${\left(\right)}_{\genfrac{}{}{0pt}{}{n}{k}}\frac{l}{2},l-1={\left(\right)}_{\genfrac{}{}{0pt}{}{n}{i}}\frac{l}{2}$$Thus, the sub-sequence ${\left(\right)}_{\genfrac{}{}{0pt}{}{n}{k}}$ satisfies the Equation (3) as well as the $\frac{l}{2}$ first terms are zeros.

**Proposition**

**1.**

**Proof.**

Algorithm 2 Classification of the Binomial sequences. |

Given the sub-sequence ${\left(\right)}_{\genfrac{}{}{0pt}{}{n}{k}}$: |

if $k<\frac{l}{2}$ then |

${\left(\right)}_{\genfrac{}{}{0pt}{}{n}{k}}l)$ |

else |

${\left(\right)}_{\genfrac{}{}{0pt}{}{n}{k}}l$ |

end if |

- ${M}_{0}$ and ${M}_{1}$ are $((i-1)\times \frac{l}{2})$ sub-matrices that, according to Theorem 1, satisfy the equality ${M}_{0}={M}_{1}$.
- ${M}_{2}$ is the $((r-i+1)\times \frac{l}{2})$ identically null sub-matrix.
- ${M}_{3}$ is the $((r-i+1)\times \frac{l}{2})$ sub-matrix representing the decomposition of a new sequence of length $\frac{l}{2}$ coming from the bit-wise sum of the two halves of $seq$. Therefore, from ${M}_{3}$ the matrix representation can be extended recursively.

**Example**

**2.**

#### 4.3.2. Description of the Half-Interval Search Algorithm

Algorithm 3 The half-interval search Algorithm |

Require:$seq$: sequence to be analyzed |

$k=0$ |

while $length(seq)>1$ do |

$l=length(seq)$ |

sum = $se{q}_{0,\frac{l}{2}-1}$ + $se{q}_{\frac{l}{2},l-1}$ |

if $sum\ne {0}_{\frac{l}{2}}$ then |

$seq=sum$ |

$k+=\frac{l}{2}$ |

else |

$seq=se{q}_{0,\frac{l}{2}-1}$ |

end if |

end while |

return k: maximum index k and $LC$ of $seq$. |

**Example**

**3.**

- Step 1: $\begin{array}{cc}\begin{array}{}\\ \\ sum=\end{array}& \begin{array}{ccc}& 0001& 1101\\ +& 1000& 1011\\ & 1001& 0110\end{array}\end{array}$

- Step 2: $\begin{array}{cc}\begin{array}{}\\ \\ sum=\end{array}& \begin{array}{ccc}& 10& 01\\ +& 01& 10\\ & 11& 11\end{array}\end{array}$

- Step 3: $\begin{array}{cc}\begin{array}{}\\ \\ sum=\end{array}& \begin{array}{ccc}& 1& 1\\ +& 1& 1\\ & 0& 0\end{array}\end{array}$

- Step 4: $\begin{array}{cc}\begin{array}{}\\ \\ sum=\end{array}& \begin{array}{cc}& 1\\ +& 1\\ & 0\end{array}\end{array}$

- Output: the maximum binomial sequence $\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{n}{12}$.

#### 4.4. Matrix Binomial Decomposition or m-BD Algorithm

**Example**

**4.**

_{16}will havec

_{3}= c

_{4}= c

_{6}= c

_{8}= c

_{9}= c

_{10}= c

_{12}= 1 while the remaining components equal zero. The coefficients c

_{i}= 1 correspond to the binomial sequences $\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{n}{i}$ that appear in the binomial decomposition of seq

_{16}.

**Remark**

**1.**

#### 4.4.1. Description of the m-BD Algorithm

Algorithm 4 The m-BD Algorithm |

Require:$seq=[{s}_{0},{s}_{1},\dots ,{s}_{{2}^{m}-1}]$ and the binomial matrix ${H}_{m}=({\tilde{h}}_{0},{\tilde{h}}_{1},\dots ,{\tilde{h}}_{{2}^{m}-1})$ |

$i={2}^{m}-1$ |

${i}_{max}=0$ |

while $i>0$ do |

${c}_{i}=[{s}_{0},{s}_{1},\dots ,{s}_{{2}^{m}-1}]\xb7{\tilde{h}}_{i}$ |

if ${c}_{i}==0$ then |

$i--$ |

else |

${i}_{max}=i$ |

$i=0$ |

end if |

end while |

return $LC={i}_{max}+1$: Linear complexity of $seq$. |

#### 4.4.2. Sequences with Maximum $LC$:

**Theorem**

**2.**

**Proof.**(⇒)

**Corollary**

**1.**

**Corollary**

**2.**

#### 4.4.3. Sequences with Quasi-Maximum $LC$

**Theorem**

**3.**

- The sequence $\{{s}_{n}\}$ has an even number of ones.
- It satisfies the equality:$$\sum _{i=0}^{l/2-1}{s}_{2\xb7i}=1.$$

**Proof.**(⇒)

- $\{{s}_{n}\}$ must have an even number of ones, otherwise by Theorem 2 the sequence would have maximum linear complexity.
- Quasi-maximum linear complexity implies that ${c}_{{2}^{m}-2}=1$, but ${c}_{{2}^{m}-2}$ is the product mod 2 of the sequence $[{s}_{0},{s}_{1},\dots ,{s}_{{2}^{m}-1}]$ multiplied by the column ${\tilde{h}}_{{2}^{m}-2}$ in the binomial matrix (the $1010\dots 10$ column), thus$${c}_{{2}^{m}-2}=\sum _{i=0}^{l/2-1}{s}_{2\xb7i}.$$Hence, ${c}_{{2}^{m}-2}=1$ when the number of terms $({s}_{2\xb7i})$ (terms with even indices) equal to 1 is an odd number.

- (⇐)

- If the sequence $\{{s}_{n}\}$ has an even number of ones, then ${c}_{{2}^{m}-1}=0$.
- If $\{{s}_{n}\}$ satisfies the equality$$\sum _{i=0}^{l/2-1}{s}_{2\xb7i}=1,$$

## 5. Algorithm Comparison

#### 5.1. Algorithm Analysis

#### 5.2. Experimental Results

#### 5.3. Different Use-Cases

## 6. Conclusions

## Author Contributions

## Funding

## Conflicts of Interest

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**Figure 3.**Comparison between the algorithms in the $LC$ calculation for all the possible GSS-sequences of a given length (Half-Interval scale).

**Figure 5.**Comparison between the algorithms in the $LC$ calculation for all the possible GSS-sequences of a given length (m-BD scale).

p-Rotation | $\{{\mathit{b}}_{\mathit{n}}\}$ Sequences | GSS-Sequences |
---|---|---|

0 | $\mathbf{1}$ 0 $\mathbf{1}$ $\mathbf{1}$ $\mathbf{1}$ 0 0 | 1111 |

1 | $\mathbf{0}$ 1 $\mathbf{1}$ $\mathbf{1}$ $\mathbf{0}$ 0 1 | 0110 |

2 | $\mathbf{1}$ 1 $\mathbf{1}$ $\mathbf{0}$ $\mathbf{0}$ 1 0 | 1100 |

3 | $\mathbf{1}$ 1 $\mathbf{0}$ $\mathbf{0}$ $\mathbf{1}$ 0 1 | 1001 |

4 | $\mathbf{1}$ 0 $\mathbf{0}$ $\mathbf{1}$ $\mathbf{0}$ 1 1 | 1010 |

5 | $\mathbf{0}$ 0 $\mathbf{1}$ $\mathbf{0}$ $\mathbf{1}$ 1 1 | 0101 |

6 | $\mathbf{0}$ 1 $\mathbf{0}$ $\mathbf{1}$ $\mathbf{1}$ 1 0 | 0011 |

PN-sequence | $\mathbf{1}$ 0 $\mathbf{1}$ $\mathbf{1}$ $\mathbf{1}$ 0 0 |

Binom. Coeff. | Binomial Sequences $\{\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{\mathit{n}}{\mathit{k}}\}$ | Length | Linear Complexity |
---|---|---|---|

$\left(\right)$ | $\left(\right)$ | ${l}_{0}=1$ | $L{C}_{0}=1$ |

$\left(\right)$ | $\left(\right)$ | ${l}_{1}=2$ | $L{C}_{1}=2$ |

$\left(\right)$ | $\left(\right)$ | ${l}_{2}=4$ | $L{C}_{2}=3$ |

$\left(\right)$ | $\left(\right)$ | ${l}_{3}=4$ | $L{C}_{3}=4$ |

$\left(\right)$ | $\left(\right)$ | ${l}_{4}=8$ | $L{C}_{4}=5$ |

$\left(\right)$ | $\left(\right)$ | ${l}_{5}=8$ | $L{C}_{5}=6$ |

$\left(\right)$ | $\left(\right)$ | ${l}_{6}=8$ | $L{C}_{6}=7$ |

$\left(\right)$ | $\left(\right)$ | ${l}_{7}=8$ | $L{C}_{7}=8$ |

Step | Op. | Seq. | Bit Position | |||
---|---|---|---|---|---|---|

0 | 4 | 8 | 12 | |||

1 | $seq$ | 0 0 0 1 | 1 1 0 1 | 1 0 0 0 | 1 0 1 1 | |

+ | $\left(\right)$ | 0 0 0 1 | 0 0 0 1 | 0 0 0 1 | 0 0 0 1 | |

2 | = | $seq$ | 0 0 0 0 | 1 1 0 0 | 1 0 0 1 | 1 0 1 0 |

+ | $\left(\right)$ | 0 0 0 0 | 1 1 1 1 | 0 0 0 0 | 1 1 1 1 | |

3 | = | $seq$ | 0 0 0 0 | 0 0 1 1 | 1 0 0 1 | 0 1 0 1 |

+ | $\left(\right)$ | 0 0 0 0 | 0 0 1 1 | 0 0 0 0 | 0 0 1 1 | |

4 | = | $seq$ | 0 0 0 0 | 0 0 0 0 | 1 0 0 1 | 0 1 1 0 |

+ | $\left(\right)$ | 0 0 0 0 | 0 0 0 0 | 1 1 1 1 | 1 1 1 1 | |

5 | = | $seq$ | 0 0 0 0 | 0 0 0 0 | 0 1 1 0 | 1 0 0 1 |

+ | $\left(\right)$ | 0 0 0 0 | 0 0 0 0 | 0 1 0 1 | 0 1 0 1 | |

6 | = | $seq$ | 0 0 0 0 | 0 0 0 0 | 0 0 1 1 | 1 1 0 0 |

+ | $\left(\right)$ | 0 0 0 0 | 0 0 0 0 | 0 0 1 1 | 0 0 1 1 | |

7 | = | $seq$ | 0 0 0 0 | 0 0 0 0 | 0 0 0 0 | 1 1 1 1 |

+ | $\left(\right)$ | 0 0 0 0 | 0 0 0 0 | 0 0 0 0 | 1 1 1 1 | |

end | = | $seq$ | 0 0 0 0 | 0 0 0 0 | 0 0 0 0 | 0 0 0 0 |

$seq=\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{n}{3}+\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{n}{6}$ | ||||||

$LC={k}_{r}+1=12+1=13$ |

seq | 0 0 0 1 | 1 1 0 1 | 1 0 0 0 | 1 0 1 1 |
---|---|---|---|---|

${\left(\right)}_{\genfrac{}{}{0pt}{}{n}{3}}l)$ | 0 0 0 1 | 0 0 0 1 | 0 0 0 1 | 0 0 0 1 |

${\left(\right)}_{\genfrac{}{}{0pt}{}{n}{4}}l)$ | 0 0 0 0 | 1 1 1 1 | 0 0 0 0 | 1 1 1 1 |

${\left(\right)}_{\genfrac{}{}{0pt}{}{n}{6}}l)$ | 0 0 0 0 | 0 0 1 1 | 0 0 0 0 | 0 0 1 1 |

${\left(\right)}_{\genfrac{}{}{0pt}{}{n}{8}}l$ | 0 0 0 0 | 0 0 0 0 | 1 1 1 1 | 1 1 1 1 |

${\left(\right)}_{\genfrac{}{}{0pt}{}{n}{9}}l$ | 0 0 0 0 | 0 0 0 0 | 0 1 0 1 | 0 1 0 1 |

${\left(\right)}_{\genfrac{}{}{0pt}{}{n}{10}}l$ | 0 0 0 0 | 0 0 0 0 | 0 0 1 1 | 0 0 1 1 |

${\left(\right)}_{\genfrac{}{}{0pt}{}{n}{12}}l$ | 0 0 0 0 | 0 0 0 0 | 0 0 0 0 | 1 1 1 1 |

$se{q}_{16}=\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{n}{3}+\left(\right)open="("\; close=")">\genfrac{}{}{0pt}{}{n}{6}$ |

Algorithms | Length Required | Complexity | Seq. Restrictions |
---|---|---|---|

Berlekamp-Massey algorithm | $2\ast l$ | O(${l}^{2}$) | None |

BD-algorithm | $l-logl$ | O($r\xb7l$) | Length power of 2 |

Half-interval search algorithm | $l-logl$ | O(l) | Length power of 2 |

m-BD algorithm | l | O(${l}^{2}$) - $\Omega $(l) | Length power of 2 |

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Martin-Navarro, J.L.; Fúster-Sabater, A.
Review of the Lineal Complexity Calculation through Binomial Decomposition-Based Algorithms. *Mathematics* **2021**, *9*, 478.
https://doi.org/10.3390/math9050478

**AMA Style**

Martin-Navarro JL, Fúster-Sabater A.
Review of the Lineal Complexity Calculation through Binomial Decomposition-Based Algorithms. *Mathematics*. 2021; 9(5):478.
https://doi.org/10.3390/math9050478

**Chicago/Turabian Style**

Martin-Navarro, Jose Luis, and Amparo Fúster-Sabater.
2021. "Review of the Lineal Complexity Calculation through Binomial Decomposition-Based Algorithms" *Mathematics* 9, no. 5: 478.
https://doi.org/10.3390/math9050478