Upper bounds on the minimum distance for turbo codes using CPP interleavers

Abstract

Analysis of error correction performance for error correcting codes is very important when using such codes in digital communication systems. At medium-to-high signal-to-noise ratios, the distance spectrum of the error correcting code represents a good indicator for the error correction performance of the code. It is desired that the minimum distance of the code is as large as possible and that the corresponding multiplicity (i.e. the number of codewords having the weight equal to the minimum distance) is as small as possible. If we know an upper bound of the minimum distance of the code, then we have a good indication about the capabilities and the limitations of the code. One of the classes of the error correcting codes with the best performance is that of turbo codes. For such codes, establishing upper bounds on the minimum distance is challenging because it depends on the interleaver component of the turbo code. In this paper we consider turbo codes with component convolutional codes as in the Long Term Evolution standard. The interleaver lengths are of the form $16 \varPsi$ or $48 \varPsi$, with $\varPsi$ a product of different prime numbers greater than three. The first achievement in the paper is that for these interleaver lengths, we show that cubic permutation polynomials (CPP), with some constraints on the coefficients, when $3 \not \mid (p_i-1)$ for a prime $p_i > 3$, always have a true inverse CPP. The most accurate upper bounds on the minimum distance for turbo codes are achieved by identifying bit information sequences leading to a certain weight of the corresponding turbo-codeword. In this paper we have indentified such bit information sequences by means of the full range dual impulse method to estimate the weight of the turbo-codewords. For the previously mentioned turbo codes and CPP interleavers, we show that the minimum distance is upper bounded by the values of 38, 36, and 28, for three different classes of coefficients. Previously, it was shown that for the same interleaver lengths and for quadratic PP (QPP) interleavers, the upper bound of the minimum distance is equal to 38. Several examples show that $d_{min}$-optimal CPP interleavers are better than $d_{min}$-optimal QPP interleavers because the multiplicities corresponding to the minimum distances for CPPs are about a half of those for QPPs. A theoretical explanation in terms of nonlinearity degrees for this result is given for all considered interleaver lengths and for the class of CPPs for which the upper bound is equal to 38.

Appendices

Appendix 1

Proof of Lemma  2:

Table 13 Determining coefficients $\rho _2$ and $\rho _1$ of the inverse CPP $\rho (x)$ depending on the coefficients $f_3$, $f_2$ and $f_1$ when $L = 16\varPsi$ ($k{_\varPsi } = 2\varPsi \ (\mathrm {mod}\ 8)$)

Table 14 Determining coefficients $\rho _2$ and $\rho _1$ of the inverse CPP $\rho (x)$ depending on the coefficients $f_3$, $f_2$ and $f_1$ when $f_3 = 2\varPsi$ and $L=48\varPsi$ ($k_{\varPsi } = (2\varPsi ) \ (\mathrm {mod}\ 24)$)

$\rho (x)$ is an inverse CPP of $\pi (x)$ if

$$\begin{aligned} \pi (\rho (x)) = x \ (\mathrm {mod}\ L), \forall x \in \{0, 1, \dots , L-1 \}. \end{aligned}$$

(76)

Taking into account Lemma 1, after some algebraic manipulations, Eq. (76) is equivalent to

$$\begin{aligned}&(f_1 \rho _1 - 1) \cdot x + (f_1 \rho _2 + f_2 \rho _1^2) \cdot x^2 + (f_1 \rho _3 + 2 f_2 \rho _2 \rho _1 + f_3 \rho _1^3)\nonumber \\&\quad \cdot x^3 + (3 f_3 \rho _1^2 \rho _2 + 2 f_2 \rho _3 \rho _1 + f_2 \rho _2^2) \cdot x^4 \nonumber \\&\quad + (3 f_3 \rho _1^2 \rho _3 + 3 f_3 \rho _1 \rho _2^2) \cdot x^5\nonumber \\&\quad + (f_3 \rho _2^3 + f_2 \rho _3^2) \cdot x^6 + 3 f_3 \rho _1 \rho _3^2 \cdot x^7 = 0 \ (\mathrm {mod}\ L), \nonumber \\&\quad \forall x \in \{0, 1, \dots , L-1 \}. \end{aligned}$$

(77)

Because $\pi (x)$ and $\rho (x)$ are true CPPs, from Lemma 1 it results that $\rho _3 = f_3 = k_3 \cdot 2 \varPsi$, with $k_3 \in \{ 1, 2, 3 \}$. Because $p_i$ is odd $\forall i \in \{ 1, 2, \dots , N_p \}$, $\varPsi$ from (6) is also odd. Then, we can have $\varPsi = 1 \ (\mathrm {mod}\ 8)$, $\varPsi = 3 \ (\mathrm {mod}\ 8)$, $\varPsi = 5 \ (\mathrm {mod}\ 8)$ or $\varPsi = 7 \ (\mathrm {mod}\ 8)$. Then, $2 \varPsi =2 \ (\mathrm {mod}\ 8)$ or $2 \varPsi =6 \ (\mathrm {mod}\ 8)$. Because every $p_i$ is odd and $3 \not \mid p_i$, we can have $\varPsi = 1 \ (\mathrm {mod}\ 24)$, $\varPsi = 5 \ (\mathrm {mod}\ 24)$, $\varPsi = 7 \ (\mathrm {mod}\ 24)$, $\varPsi = 11 \ (\mathrm {mod}\ 24)$, $\varPsi = 13 \ (\mathrm {mod}\ 24)$, $\varPsi = 17 \ (\mathrm {mod}\ 24)$, $\varPsi = 19 \ (\mathrm {mod}\ 24)$, or $\varPsi = 23 \ (\mathrm {mod}\ 24)$. Then $2\varPsi =2 \ (\mathrm {mod}\ 24)$, $2\varPsi = 10 \ (\mathrm {mod}\ 24)$, $2\varPsi =14 \ (\mathrm {mod}\ 24)$ or $2\varPsi =22 \ (\mathrm {mod}\ 24)$.

Table 15 Determining coefficients $\rho _2$ and $\rho _1$ of the inverse CPP $\rho (x)$ depending on the coefficients $f_3$, $f_2$ and $f_1$ when $f_3 = 2\varPsi$ and $L=48\varPsi$ ($k_{\varPsi } = (2\varPsi ) \ (\mathrm {mod}\ 24)$)

Thus, for $L = 16 \varPsi$ and $\rho _2 = f_2 = k_2 \cdot 2 \varPsi$, with $k_2 \in \{ 0, 1, 2, 3 \}$, (77) is equivalent to

$$\begin{aligned}&(f_1 \rho _1 - 1) \cdot x + 2 k_2 \varPsi \cdot ( f_1 + \rho _1^2)\nonumber \\&\quad \cdot x^2 + 2\varPsi \cdot (k_3 f_1 + 4 k_2^2 \varPsi \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 4 k_2 \varPsi ^2 \cdot (3 k_3 \rho _1^2 + 2 k_3 \rho _1 + 2 k_2^2 \varPsi )\nonumber \\&\quad \cdot x^4 + 4 k_3 \varPsi ^2 \cdot (2 k_2^2 \varPsi \rho _1 + 3 k_3 \rho _1^2)\nonumber \\&\quad \cdot x^5 + 8 k_2 k_3^2 \varPsi ^3 \cdot x^6 \nonumber \\&\quad + 8 k_3^3 \varPsi ^3 \rho _1 \cdot x^7 = 0 \ (\mathrm {mod}\ 16 \varPsi ),\nonumber \\&\quad \forall x \in \{0, 1, \dots , 16 \varPsi -1 \}. \end{aligned}$$

(78)

For $L = 16 \varPsi$, $f_2 = k_2 \cdot 2 \varPsi$ and $\rho _2 = ((k_2 +2) \ (\mathrm {mod}\ 4)) \cdot 2 \varPsi$, with $k_2 \in \{ 0, 1, 2, 3 \}$, (77) is equivalent to

$$\begin{aligned}&(f_1 \rho _1 - 1) \cdot x + 2 \varPsi \cdot (2 f_1 + f_1 k_2 + k_2 \rho _1^2)\nonumber \\&\quad \cdot x^2 + 2\varPsi \cdot (k_3 f_1 + 4 k_2^2 \varPsi \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 4 \varPsi ^2 \cdot (3 k_2 k_3 \rho _1^2 + 2 k_2 k_3 \rho _1 \nonumber \\&\quad + 2 k_2^3 \varPsi + 2 k_3 \rho _1^2) \cdot x^4 + 4 k_3 \varPsi ^2 \cdot (2 k_2^2 \varPsi \rho _1 + 3 k_3 \rho _1^2) \cdot x^5 \nonumber \\&\quad + 8 k_2 k_3^2 \varPsi ^3 \cdot x^6 + 8 k_3^3 \varPsi ^3 \rho _1 \cdot x^7 = 0 \ (\mathrm {mod}\ 16 \varPsi ),\nonumber \\&\quad \forall x \in \{0, 1, \dots , 16 \varPsi -1 \}. \end{aligned}$$

(79)

For $L = 48 \varPsi$, $\rho _2 = f_2 = k_2 \cdot 6 \varPsi$, with $k_2 \in \{ 0, 1, 2, 3 \}$, (77) is equivalent to

$$\begin{aligned}&(f_1 \rho _1 - 1) \cdot x + 6 k_2 \varPsi \cdot ( f_1 + \rho _1^2) \cdot x^2 + 2\varPsi \nonumber \\&\quad \cdot (k_3 f_1 + 12 k_2^2 \varPsi \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 12 k_2 \varPsi ^2 \cdot (3 k_3 \rho _1^2 + 2 k_3 \rho _1 + 2 k_2^2 \varPsi ) \nonumber \\&\quad \cdot x^4 + 12 k_3 \varPsi ^2 \cdot (2 k_2^2 \varPsi \rho _1 + k_3 \rho _1^2)\nonumber \\&\quad \cdot x^5 + 24 k_2 k_3^2 \varPsi ^3 \cdot x^6 \nonumber \\&\quad + 24 k_3^3 \varPsi ^3 \rho _1 \cdot x^7 + 16 k_3^4 \varPsi ^4 \cdot x^9 = 0 \ (\mathrm {mod}\ 48 \varPsi ),\nonumber \\&\quad \forall x \in \{0, 1, \dots , 48 \varPsi -1 \}. \end{aligned}$$

(80)

For $L = 48 \varPsi$, $f_2 = k_2 \cdot 6 \varPsi$ and $\rho _2 = ((k_2 + 2) \ (\mathrm {mod}\ 4)) \cdot 6 \varPsi$, with $k_2 \in \{ 0, 1, 2, 3 \}$, (77) is equivalent to

$$\begin{aligned}&(f_1 \rho _1 - 1) \cdot x + 6 \varPsi \cdot ( k_2 f_1 + k_2 \rho _1^2 + 2 f_1) \cdot x^2 + 2\varPsi \nonumber \\&\quad \cdot (k_3 f_1 + 12 k_2^2 \varPsi \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 12 \varPsi ^2 \cdot (3 k_2 k_3 \rho _1^2 + 2 k_2 k_3 \rho _1 + 2 k_2^3 \varPsi + 2 k_3 \varPsi ^2 \rho _1^2)\nonumber \\&\quad \cdot x^4 + 12 k_3 \varPsi ^2 \cdot (2 k_2^2 \varPsi \rho _1 + k_3 \rho _1^2) \cdot x^5 \nonumber \\&\quad + 24 k_2 k_3^2 \varPsi ^3 \cdot x^6 + 24 k_3^3 \varPsi ^3 \rho _1 \nonumber \\&\quad \cdot x^7 + 16 k_3^4 \varPsi ^4 \cdot x^9 = 0 \ (\mathrm {mod}\ 48 \varPsi ), \nonumber \\&\quad \forall x \in \{0, 1, \dots , 48 \varPsi -1 \}. \end{aligned}$$

(81)

Because $(2 \varPsi ) \mid L$, from (78), (79), (80), or (81), we have

$$\begin{aligned} (f_1 \rho _1 - 1) \cdot x = 0 \ (\mathrm {mod}\ 2 \varPsi ), \forall x \in \{0, 1, \dots , L-1 \}.\qquad \qquad \end{aligned}$$

(82)

Equation (82) is equivalent to

$$\begin{aligned}&f_1 \rho _1 = 1 \ (\mathrm {mod}\ 2 \varPsi ) \Leftrightarrow f_1 \rho _1 = 2 \varPsi \cdot k + 1 \ (\mathrm {mod}\ L),\nonumber \\&\quad \text {with} ~k \in \{0, 1, 2, \dots , 7\} ~\text {when} ~L=16 \varPsi ,\nonumber \\&\quad ~\text {and} ~k \in \{0, 1, 2, \dots , 23\} ~\text {when} ~L=48 \varPsi . \end{aligned}$$

(83)

According to Theorem 57 from [34], we note that congruence $f_1 \rho _1 = 2 \varPsi \cdot k + 1 \ (\mathrm {mod}\ L)$ has only one solution $\rho _1$ modulo L when $L=16 \varPsi$ or when $L = 48 \varPsi$ and $f_1 = 1 \ (\mathrm {mod}\ 3)$ or $f_1 = 2 \ (\mathrm {mod}\ 3)$, because $\gcd (f_1, L) = 1$. When $L = 48 \varPsi$, with $\varPsi = 1 \ (\mathrm {mod}\ 3)$, $f_1 = 0 \ (\mathrm {mod}\ 3)$, and $k \in \{1, 4, 7, \dots , 22 \}$, or when $L = 48 \varPsi$, with $\varPsi = 2 \ (\mathrm {mod}\ 3)$, $f_1 = 0 \ (\mathrm {mod}\ 3)$, and $k \in \{2, 5, 8, \dots , 23 \}$, congruence $f_1 \rho _1 = 2 \varPsi \cdot k + 1 \ (\mathrm {mod}\ L)$ has three solutions modulo L because $\gcd (f_1, L) = 3$ and $3 \mid (2 \varPsi \cdot k + 1)$, but we will show that only the solution that fulfills condition $\rho _1 = 0 \ (\mathrm {mod}\ 3)$ is valid and it is unique.

Table 16 Determining coefficients $\rho _2$ and $\rho _1$ of the inverse CPP $\rho (x)$ depending on the coefficients $f_3$, $f_2$ and $f_1$ when $f_3 = 4\varPsi$ and $L=48\varPsi$ ($k_{\varPsi } = (2\varPsi ) \ (\mathrm {mod}\ 24)$)

With (83), (78) is fulfilled if and only if

$$\begin{aligned}&k \cdot x + k_2 \cdot ( f_1 + \rho _1^2) \cdot x^2 \nonumber \\&\quad + (k_3 f_1 + 4 k_2^2 \varPsi \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 2 k_2 \varPsi \cdot (3 k_3 \rho _1^2 + 2 k_3 \rho _1 + 2 k_2^2 \varPsi )\nonumber \\&\quad \cdot x^4 + 2 k_3 \varPsi \cdot (2 k_2^2 \varPsi \rho _1 + 3 k_3 \rho _1^2)\nonumber \\&\quad \cdot x^5 + 4 k_2 k_3^2 \varPsi ^2 \cdot x^6 \nonumber \\&\quad + 4 k_3^3 \varPsi ^2 \rho _1 \cdot x^7 = 0 \ (\mathrm {mod}\ 8),\nonumber \\&\quad \forall x \in \{0, 1, \dots , 7 \}. \end{aligned}$$

(84)

When $2 \varPsi =2 \ (\mathrm {mod}\ 8)$, (84) is equivalent to

$$\begin{aligned}&k \cdot x + k_2 \cdot ( f_1 + \rho _1^2) \cdot x^2 \nonumber \\&\quad + (k_3 f_1 + 4 k_2^2 \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 2 k_2 \cdot (3 k_3 \rho _1^2 + 2 k_3 \rho _1 + 2 k_2^2) \nonumber \\&\quad \cdot x^4 + 2 k_3 \cdot (2 k_2^2 \rho _1 + 3 k_3 \rho _1^2) \nonumber \\&\quad \cdot x^5 + 4 k_2 k_3^2 \cdot x^6 \nonumber \\&\quad + 4 k_3^3 \rho _1 \cdot x^7 = 0 \ (\mathrm {mod}\ 8), \forall x \in \{0, 1, \dots , 7 \}. \end{aligned}$$

(85)

When $2 \varPsi =6 \ (\mathrm {mod}\ 8)$, (84) is equivalent to

$$\begin{aligned}&k \cdot x + k_2 \cdot ( f_1 + \rho _1^2) \cdot x^2\nonumber \\&\quad + (k_3 f_1 + 4 k_2^2 \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 2 k_2 \cdot ( k_3 \rho _1^2 + 2 k_3 \rho _1 + 2 k_2^2) \cdot x^4 + 2 k_3 \nonumber \\&\quad \cdot (2 k_2^2 \rho _1 + k_3 \rho _1^2) \cdot x^5 + 4 k_2 k_3^2 \cdot x^6 \nonumber \\&\quad + 4 k_3^3 \rho _1 \cdot x^7 = 0 \ (\mathrm {mod}\ 8), \forall x \in \{0, 1, \dots , 7 \}. \end{aligned}$$

(86)

With (83), (79) is fulfilled if and only if

$$\begin{aligned}&k \cdot x + ( 2 f_1 + f_1 k_2 + k_2 \rho _1^2) \cdot x^2 \nonumber \\&\quad + (k_3 f_1 + 4 k_2^2 \varPsi \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 2 \varPsi \cdot (3 k_2 k_3 \rho _1^2 + 2 k_2 k_3 \rho _1 + 2 k_2^3 \varPsi + 2 k_3 \rho _1^2) \cdot x^4 + 2 k_3 \varPsi \nonumber \\&\quad \cdot (2 k_2^2 \varPsi \rho _1 + 3 k_3 \rho _1^2) \cdot x^5 \nonumber \\&\quad + 4 k_2 k_3^2 \varPsi ^2 \cdot x^6 + 4 k_3^3 \varPsi ^2 \rho _1 \cdot x^7 = 0 \ (\mathrm {mod}\ 8),\nonumber \\&\quad \forall x \in \{0, 1, \dots , 7 \}. \end{aligned}$$

(87)

When $2 \varPsi =2 \ (\mathrm {mod}\ 8)$, (87) is equivalent to

$$\begin{aligned}&k \cdot x + 2 \cdot (2 f_1 + f_1 k_2 + k_2 \rho _1^2) \cdot x^2 \nonumber \\&\quad + (k_3 f_1 + 4 k_2^2 \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 2 \cdot (3 k_2 k_3 \rho _1^2 + 2 k_2 k_3 \rho _1 + 2 k_2^3 + 2 k_3 \rho _1^2)\nonumber \\&\quad \cdot x^4 + 2 k_3 \cdot (2 k_2^2 \rho _1 + 3 k_3 \rho _1^2) \cdot x^5 \nonumber \\&\quad + 4 k_2 k_3^2 \cdot x^6 + 4 k_3^3 \rho _1 \cdot x^7 = 0 \ (\mathrm {mod}\ 8),\nonumber \\&\quad \forall x \in \{0, 1, \dots , 7 \} \end{aligned}$$

(88)

and when $2 \varPsi =6 \ (\mathrm {mod}\ 8)$, (87) is equivalent to

$$\begin{aligned}&k \cdot x + 2 \cdot (2 f_1 + f_1 k_2 + k_2 \rho _1^2) \cdot x^2\nonumber \\&\quad + (k_3 f_1 + 4 k_2^2 \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 2 \cdot (k_2 k_3 \rho _1^2 + 2 k_2 k_3 \rho _1 + 2 k_2^3 + 2 k_3 \rho _1^2) \nonumber \\&\quad \cdot x^4 + 2 k_3 \cdot (2 k_2^2 \rho _1 + k_3 \rho _1^2) \cdot x^5 \nonumber \\&\quad + 4 k_2 k_3^2 \cdot x^6 + 4 k_3^3 \rho _1 \cdot x^7 = 0 \ (\mathrm {mod}\ 8), \nonumber \\&\quad \forall x \in \{0, 1, \dots , 7 \}. \end{aligned}$$

(89)

Solutions $(k, \rho _1)$ of Eqs. (85), (86), (88), and (89) for each value of $f_3 \in \{ 2\varPsi , 4\varPsi , 6\varPsi \}$, $f_2 \in \{ 0, 2\varPsi , 4\varPsi , 6\varPsi \}$, and $f_1 \ (\mathrm {mod}\ 8) \in \{ 1, 3, 5, 7 \}$, can be found using specific software programs. We have used symbolic calculus in Matlab for this goal. These solutions are unique for each value of $f_1 \ (\mathrm {mod}\ 8)$ and they are given in Table 13, unified for the cases when $2 \varPsi = 2 \ (\mathrm {mod}\ 8)$ and $2 \varPsi = 6 \ (\mathrm {mod}\ 8)$.

Table 17 Determining coefficients $\rho _2$ and $\rho _1$ of the inverse CPP $\rho (x)$ depending on the coefficients $f_3$, $f_2$ and $f_1$ when $f_3 = 4\varPsi$ and $L=48\varPsi$ ($k_{\varPsi } = (2\varPsi ) \ (\mathrm {mod}\ 24)$)

With (83), (80) is fulfilled if and only if

$$\begin{aligned}&k \cdot x + 3 k_2 \cdot ( f_1 + \rho _1^2) \cdot x^2 \nonumber \\&\quad + (k_3 f_1 + 12 k_2^2 \varPsi \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 6 k_2 \varPsi \cdot (3 k_3 \rho _1^2 + 2 k_3 \rho _1 + 2 k_2^2 \varPsi ) \cdot x^4 + 6 k_3 \varPsi \nonumber \\&\quad \cdot (2 k_2^2 \varPsi \rho _1 + k_3 \rho _1^2) \cdot x^5 + 12 k_2 k_3^2 \varPsi ^2 \cdot x^6 \nonumber \\&\quad + 12 k_3^3 \varPsi ^2 \rho _1 \cdot x^7 + 8 k_3^4 \varPsi ^3 \cdot x^9 = 0 \ (\mathrm {mod}\ 24),\nonumber \\&\quad \forall x \in \{0, 1, \dots , 23 \}. \end{aligned}$$

(90)

When $2 \varPsi =2 \ (\mathrm {mod}\ 24)$, (90) is equivalent to

$$\begin{aligned}&k \cdot x + 3 k_2 \cdot ( f_1 + \rho _1^2) \cdot x^2 \nonumber \\&\quad + (k_3 f_1 + 12 k_2^2 \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 6 k_2 \cdot (3 k_3 \rho _1^2 + 2 k_3 \rho _1 + 2 k_2^2) \nonumber \\&\quad \cdot x^4 + 6 k_3 \cdot (2 k_2^2 \rho _1 + k_3 \rho _1^2) \cdot x^5 \nonumber \\&\quad + 12 k_2 k_3^2 \cdot x^6 \nonumber \\&\quad + 12 k_3^3 \rho _1 \cdot x^7 + 8 k_3^4 \cdot x^9 = 0 \ (\mathrm {mod}\ 24), \nonumber \\&\quad \forall x \in \{0, 1, \dots , 23 \}. \end{aligned}$$

(91)

When $2 \varPsi =10 \ (\mathrm {mod}\ 24)$, (90) is equivalent to

$$\begin{aligned}&k \cdot x + 3 k_2 \cdot ( f_1 + \rho _1^2) \cdot x^2 \nonumber \\&\quad + (k_3 f_1 + 12 k_2^2 \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 6 k_2 \cdot (3 k_3 \rho _1^2 + 2 k_3 \rho _1 + 2 k_2^2) \nonumber \\&\quad \cdot x^4 + 6 k_3 \cdot (2 k_2^2 \rho _1 + k_3 \rho _1^2) \cdot x^5 \nonumber \\&\quad + 12 k_2 k_3^2 \cdot x^6 \nonumber \\&\quad + 12 k_3^3 \rho _1 \cdot x^7 + 16 k_3^4 \cdot x^9 = 0 \ (\mathrm {mod}\ 24),\nonumber \\&\quad \forall x \in \{0, 1, \dots , 23 \}. \end{aligned}$$

(92)

When $2 \varPsi =14 \ (\mathrm {mod}\ 24)$, (90) is equivalent to

$$\begin{aligned}&k \cdot x + 3 k_2 \cdot ( f_1 + \rho _1^2) \cdot x^2 \nonumber \\&\quad + (k_3 f_1 + 12 k_2^2 \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 6 k_2 \cdot (k_3 \rho _1^2 + 2 k_3 \rho _1 + 2 k_2^2) \nonumber \\&\quad \cdot x^4 + 6 k_3 \cdot (2 k_2^2 \rho _1 + 3 k_3 \rho _1^2) \nonumber \\&\quad \cdot x^5 + 12 k_2 k_3^2 \cdot x^6 \nonumber \\&\quad + 12 k_3^3 \rho _1 \cdot x^7 + 8 k_3^4 \cdot x^9 = 0 \ (\mathrm {mod}\ 24), \nonumber \\&\quad \forall x \in \{0, 1, \dots , 23 \}. \end{aligned}$$

(93)

When $2 \varPsi =22 \ (\mathrm {mod}\ 24)$, (90) is equivalent to

$$\begin{aligned}&k \cdot x + 3 k_2 \cdot ( f_1 + \rho _1^2) \cdot x^2\nonumber \\&\quad + (k_3 f_1 + 12 k_2^2 \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 6 k_2 \cdot (k_3 \rho _1^2 + 2 k_3 \rho _1 + 2 k_2^2) \nonumber \\&\quad \cdot x^4 + 6 k_3 \cdot (2 k_2^2 \rho _1 + 3 k_3 \rho _1^2) \nonumber \\&\quad \cdot x^5 + 12 k_2 k_3^2 \cdot x^6 \nonumber \\&\quad + 12 k_3^3 \rho _1 \cdot x^7 + 16 k_3^4 \cdot x^9 = 0 \ (\mathrm {mod}\ 24), \nonumber \\&\quad \forall x \in \{0, 1, \dots , 23 \}. \end{aligned}$$

(94)

Table 18 Determining coefficients $\rho _2$ and $\rho _1$ of the inverse CPP $\rho (x)$ depending on the coefficients $f_3$, $f_2$ and $f_1$ when $f_3 = 6\varPsi$ and $L=48\varPsi$ ($k_{\varPsi } = (2\varPsi ) \ (\mathrm {mod}\ 24)$)

With (83), (81) is fulfilled if and only if

$$\begin{aligned}&k \cdot x + 3 \cdot ( 2 f_1 + 3 f_1 k_2 + \rho _1^2) \cdot x^2 \nonumber \\&\quad + (k_3 f_1 + 12 k_2^2 \varPsi \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 6 \varPsi \cdot (3 k_2 k_3 \rho _1^2 \nonumber \\&\quad + 2 k_2 k_3 \rho _1 + 2 k_2^3 \varPsi + 2 k_3 \rho _1^2) \cdot x^4 + 6 k_3 \varPsi \nonumber \\&\quad \cdot (2 k_2^2 \varPsi \rho _1 + k_3 \rho _1^2) \cdot x^5 \nonumber \\&\quad + 12 k_2 k_3^2 \varPsi ^2 \cdot x^6 + 12 k_3^3 \varPsi ^2 \rho _1 \nonumber \\&\quad \cdot x^7 + 8 k_3^4 \varPsi ^3 \cdot x^9 = 0 \ (\mathrm {mod}\ 24), \forall x \in \{0, 1, \dots , 23 \}.\nonumber \\ \end{aligned}$$

(95)

When $2 \varPsi =2 \ (\mathrm {mod}\ 24)$, (95) is equivalent to

$$\begin{aligned}&k \cdot x + 3 \cdot ( 2 f_1 + 3 f_1 k_2 + \rho _1^2) \cdot x^2 \nonumber \\&\quad + (k_3 f_1 + 12 k_2^2 \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 6 \cdot (3 k_2 k_3 \rho _1^2 + 2 k_2 k_3 \rho _1 + 2 k_2^3 + 2 k_3 \rho _1^2) \nonumber \\&\quad \cdot x^4 + 6 k_3 \cdot (2 k_2^2 \rho _1 + k_3 \rho _1^2) \nonumber \\&\quad \cdot x^5 + 12 k_2 k_3^2 \cdot x^6 \nonumber \\&\quad + 12 k_3^3 \rho _1 \cdot x^7 + 8 k_3^4 \cdot x^9 = 0 \ (\mathrm {mod}\ 24), \nonumber \\&\quad \forall x \in \{0, 1, \dots , 23 \}. \end{aligned}$$

(96)

When $2 \varPsi =10 \ (\mathrm {mod}\ 24)$, (95) is equivalent to

$$\begin{aligned}&k \cdot x + 3 \cdot ( 2 f_1 + 3 f_1 k_2 + \rho _1^2) \cdot x^2 \nonumber \\&\quad + (k_3 f_1 + 12 k_2^2 \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 6 \cdot (3 k_2 k_3 \rho _1^2 + 2 k_2 k_3 \rho _1 + 2 k_2^3 + 2 k_3 \rho _1^2) \nonumber \\&\quad \cdot x^4 + 6 k_3 \cdot (2 k_2^2 \rho _1 + k_3 \rho _1^2) \cdot x^5 \nonumber \\&\quad + 12 k_2 k_3^2 \cdot x^6 + 12 k_3^3 \rho _1 \cdot x^7 + 16 k_3^4 \cdot x^9 = 0 \ (\mathrm {mod}\ 24), \nonumber \\&\quad \forall x \in \{0, 1, \dots , 23 \}. \end{aligned}$$

(97)

When $2 \varPsi =14 \ (\mathrm {mod}\ 24)$, (95) is equivalent to

$$\begin{aligned}&k \cdot x + 3 \cdot ( 2 f_1 + 3 f_1 k_2 + \rho _1^2) \cdot x^2\nonumber \\&\quad + (k_3 f_1 + 12 k_2^2 \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 6 \cdot (k_2 k_3 \rho _1^2 + 2 k_2 k_3 \rho _1 + 2 k_2^3 + 2 k_3 \rho _1^2)\nonumber \\&\quad \cdot x^4 + 6 k_3 \cdot (2 k_2^2 \rho _1 + 3 k_3 \rho _1^2) \cdot x^5 \nonumber \\&\quad + 12 k_2 k_3^2 \cdot x^6 + 12 k_3^3 \rho _1 \cdot x^7 + 8 k_3^4 \cdot x^9 = 0 \ (\mathrm {mod}\ 24), \nonumber \\&\quad \forall x \in \{0, 1, \dots , 23 \}. \end{aligned}$$

(98)

When $2 \varPsi =22 \ (\mathrm {mod}\ 24)$, (95) is equivalent to

$$\begin{aligned}&k \cdot x + 3 \cdot ( 2 f_1 + 3 f_1 k_2 + \rho _1^2) \cdot x^2\nonumber \\&\quad + (k_3 f_1 + 12 k_2^2 \rho _1 + k_3 \rho _1^3) \cdot x^3 \nonumber \\&\quad + 6 \cdot (k_2 k_3 \rho _1^2 + 2 k_2 k_3 \rho _1 + 2 k_2^3 + 2 k_3 \rho _1^2)\nonumber \\&\quad \cdot x^4 + 6 k_3 \cdot (2 k_2^2 \rho _1 + 3 k_3 \rho _1^2) \cdot x^5 \nonumber \\&\quad + 12 k_2 k_3^2 \cdot x^6 + 12 k_3^3 \rho _1 \cdot x^7 \nonumber \\&\quad + 16 k_3^4 \cdot x^9 = 0 \ (\mathrm {mod}\ 24), \forall x \in \{0, 1, \dots , 23 \}. \end{aligned}$$

(99)

Solutions $(k, \rho _1)$ of equations (91)–(94) and (96)–(99) for each value of $f_3 \in \{ 2\varPsi , 4\varPsi , 6\varPsi \}$, $f_2 \in \{ 0, 6\varPsi , 12\varPsi ,$ $18\varPsi \}$, and $f_1 \ (\mathrm {mod}\ 24) \in \{ 1, 3, \dots , 23 \}$, found by software means, are given in Tables 14, 15, 16, 17 and 18. When the congruence equation $f_1 \rho _1 = 2 \varPsi \cdot k +1 \ (\mathrm {mod}\ 48 \varPsi )$ has three solutions in variable $\rho _1 \ (\mathrm {mod}\ 48 \varPsi )$, the valid solution (i.e. that fulfills one of the equations (91)–(94)) is only $\rho _1 = f_1 \ (\mathrm {mod}\ 24)$. We note that, because of condition $(f_1 + f_3)\ne 0 \ (\mathrm {mod}\ 3)$ for $L = 48 \varPsi$, for a certain value of $f_3$, $f_1 \ (\mathrm {mod}\ 24)$ can take only 8 different values from the 12 ones from the set $\{ 1, 3, \dots , 23 \}$. For example, if $f_3 = 2 \varPsi$ and $k_{ \varPsi } = (2 \varPsi ) \ (\mathrm {mod}\ 24) = 2$, then $f_3 = 2 \ (\mathrm {mod}\ 3)$ and $f_1 \ (\mathrm {mod}\ 24) \in \{ 3, 5, 9, 11, 15, 17, 21, 23 \}$. $\square$

Appendix 2

According to the results from [33], the nonlinearity degree of CPP interleavers of even lengths is equal to

$$\begin{aligned} \zeta _{CPP} = L / ( \gcd \big (\gcd (3 f_3, L), \gcd (2 f_2, L)) + N_{k_0,QNP_1} \big ), \end{aligned}$$

(100)

where $N_{k_0,QNP_1}$ is the number of the common solutions $k_0$ of congruence equations

$$\begin{aligned} \left\{ \begin{array}{l} 3 f_3 k_0 = L/2 ~ \ (\mathrm {mod}\ L) \\ (2 f_2 + L/2) k_0 = L/2 ~ \ (\mathrm {mod}\ L). \end{array} \right. \end{aligned}$$

(101)

For coefficients given in Table 7, we have

$$\begin{aligned} \gcd ( 3 f_3 , L) = \gcd ( (4-k_L) \cdot k_L \cdot k_3 \cdot 2 \varPsi , k_L \cdot 16 \varPsi ) = k_L \cdot 2 \varPsi . \end{aligned}$$

(102)

Because $\gcd ( 3 f_3 , L) \mid (L/2)$, the solutions of the first equation from (101) are of the form

$$\begin{aligned}&k_{0, \text {eq1}}(i) = k_{0,f_3} + L \cdot i / \gcd ( 3 f_3 , L) = k_{0,f3} + 8 \cdot i, ~\text {with } \nonumber \\&\quad i=0, 1, \dots , \gcd (3 f_3, L) - 1, \end{aligned}$$

(103)

where $k_{0,f3}$ is equal to

$$\begin{aligned}&k_{0,f_3} = \bigg ( \frac{3 f_3}{\gcd (3 f_3, L)} \bigg )^{-1} \nonumber \\&\quad \cdot \frac{L}{2 \cdot \gcd (3 f_3, L)} \ (\mathrm {mod}\ ( L/\gcd (3 f_3, L) ) ) \nonumber \\&\quad = ( (4-k_L) \cdot k_3 )^{-1} \cdot 4 \ (\mathrm {mod}\ 8) \nonumber \\&\quad = \left\{ \begin{array}{l} 1 \cdot 4 \ (\mathrm {mod}\ 8) = 4, \text { for}~ k_L = 3 ~\text {and}~ k_3 = 1, ~\text {or}~ k_L = 1 ~\text {and} ~k_3 = 3, \\ 3 \cdot 4 \ (\mathrm {mod}\ 8) = 4, \text { for}~ k_L = 1 ~\text {and}~ k_3 = 1, ~\text {or}~ k_L = 3 ~\text {and} ~k_3 = 3. \end{array} \right. \nonumber \\ \end{aligned}$$

(104)

Thus, the solutions of the first equation from (101) are

$$\begin{aligned} k_{0, \text {eq1}}(i) = 4 + 8 \cdot i, \text { with } i=0, 1, \dots , k_L \cdot 2 \varPsi - 1. \end{aligned}$$

(105)

Similarly, we have

$$\begin{aligned}&\gcd ( 2 f_2 + L/2 , L) = \gcd ( k_L \cdot 4 \varPsi \cdot (k_2 + 2), k_L \cdot 16 \varPsi )\nonumber \\&\quad = \left\{ \begin{array}{l} k_L \cdot 4 \varPsi , \text { for } k_2 \in \{ 1, 3 \}, \\ k_L \cdot 8 \varPsi , \text { for } k_2 = 0, \\ k_L \cdot 16 \varPsi , \text { for } k_2 = 2. \end{array} \right. \end{aligned}$$

(106)

Because $\gcd ( 2 f_2 + L/2 , L) \mid (L/2)$ only for $k_2 \in \{ 0, 1, 3 \}$, the second equation from (101) has solutions only for these three values of $k_2$. These solutions are of the form

$$\begin{aligned}&k_{0, \text {eq2}}(i) = k_{0,f_2} + L \cdot i / \gcd ( 2 f_2 + L/2 , L), \nonumber \\&\quad \text { with } i=0, 1, \dots , \gcd (2 f_2 + L/2 , L) - 1, \end{aligned}$$

(107)

where $k_{0,f_2}$ is equal to

$$\begin{aligned}&k_{0,f_2} = \bigg ( \frac{2 f_2 + L/2}{\gcd (2 f_2 + L/2, L)} \bigg )^{-1} \nonumber \\&\quad \cdot \frac{L}{2 \cdot \gcd (2 f_2 + L/2, L)} \ (\mathrm {mod}\ ( L/\gcd (2 f_2 + L/2, L) ) ) \nonumber \\&\quad = \left\{ \begin{array}{l} (k_2 + 2)^{-1} \cdot 2 \ (\mathrm {mod}\ 4) = (k_2 + 2) \cdot 2 \ (\mathrm {mod}\ 4) = 2, \text { for } k_2 \in \{ 1, 3 \}, \\ 1 \cdot 1 \ (\mathrm {mod}\ 2) = 1, \text { for } k_2 = 0. \end{array} \right. \nonumber \\ \end{aligned}$$

(108)

Thus, the solutions of the second equation from (101) are

$$\begin{aligned} k_{0, \text {eq2}}(i) = \left\{ \begin{array}{l} 2 + 4 \cdot i, \text { with } i=0, 1, \dots , k_L \cdot 4 \varPsi - 1, \text { for } k_2 \in \{ 1, 3 \}, \\ 1 + 2 \cdot i, \text { with } i=0, 1, \dots , k_L \cdot 8 \varPsi - 1, \text { for} k_2 = 0. \end{array} \right. \end{aligned}$$

(109)

Because

$$\begin{aligned} k_{0, \text {eq1}}(i) \ (\mathrm {mod}\ 8) = 4, \forall i = 0, 1, \dots , k_L \cdot 2 \varPsi - 1, \end{aligned}$$

(110)

and

$$\begin{aligned} \left\{ \begin{array}{l} k_{0, \text {eq2}}(i) \ (\mathrm {mod}\ 8) \in \{ 2, 6 \}, \forall i = 0, 1, \dots , k_L \cdot 4 \varPsi - 1, \text { for } k_2 \in \{ 1, 3 \}, \\ k_{0, \text {eq2}}(i) \ (\mathrm {mod}\ 8) \in \{ 1, 3, 5, 7 \}, \forall i = 0, 1, \dots , k_L \cdot 8 \varPsi - 1, \text { for } k_2 = 0, \end{array} \right. \end{aligned}$$

(111)

it results that the two equations from (101) have no common solutions, i.e. $N_{k_0,QNP_1} = 0$. Because for $k_3 \in \{ 1, 3 \}$

$$\begin{aligned}&\gcd \big ( \gcd (3 f_3, L), \gcd (2 f_2, L) \big ) \nonumber \\&\quad = \left\{ \begin{array}{l} \gcd \big ( k_L \cdot 2 \varPsi , k_L \cdot 4 \varPsi \big ) = k_L \cdot 2 \varPsi , \text { for } k_2 \in \{ 1, 3 \}, \\ \gcd \big ( k_L \cdot 2 \varPsi , k_L \cdot 16 \varPsi \big ) = k_L \cdot 2 \varPsi , \text { for } k_2 = 0, \\ \gcd \big ( k_L \cdot 2 \varPsi , k_L \cdot 8 \varPsi \big ) = k_L \cdot 2 \varPsi , \text { for } k_2 = 2, \end{array} \right. \nonumber \\ \end{aligned}$$

(112)

from (100) we have

$$\begin{aligned} \zeta _{CPP} = \frac{L}{\gcd \big ( \gcd (3 f_3, L), \gcd (2 f_2, L) \big ) } = \frac{ k_L \cdot 16 \varPsi }{ k_L \cdot 2 \varPsi } = 8. \end{aligned}$$

(113)