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The sizes of maximal \((v,k,k-2,k-1)\) optical orthogonal codes

Abstract

An optical orthogonal code (OOC) is a family of binary sequences having good auto- and cross-correlation properties. Let \(\Phi (v,k,\lambda _{a},\lambda _{c})\) denote the largest possible size among all \((v,k,\lambda _{a},\lambda _{c})\)-OOCs. A \((v,k,\lambda _{a},\lambda _{c})\)-OOC with \(\Phi (v,k,\lambda _{a},\lambda _{c})\) codewords is said to be maximal. In this paper, we research into maximal \((v,k,k-2,k-1)\)-OOCs and determine the exact value of \(\Phi (v,k,k-2,k-1)\). This generalizes the result on the special case of \(k=4\) by Huang and Chang in 2012. Distributions of differences with maximum multiplicity are analyzed by several classes to deal with the general case for all possible v and k.

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Acknowledgements

The authors would like to thank Prof. Yanxun Chang for many valuable suggestions and comments. They also wish to thank Prof. Marco Buratti and two anonymous referees for carefully reading the manuscript and suggesting several corrections and improvements.

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Correspondence to Junling Zhou.

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Supported by NSFC Grants 11571034 and 11971053.

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Appendix

Appendix

Proof of Lemma 2.1

Denote \(A=\{x_{0},x_{1},\ldots ,x_{m-1}\}\) where \(x_{i}=i\tau \) for \(0\le i\le m-1\). List the differences of \(\Delta A\) in the following \(m\times m\) matrix M:

$$\begin{aligned} M=\begin{pmatrix} *&{}\quad \tau &{}\quad 2\tau &{}\quad \cdots &{}\quad (m-2)\tau &{}\quad (m-1)\tau \\ -\tau &{}\quad *&{}\quad \tau &{}\quad \cdots &{}\quad (m-3)\tau &{}\quad (m-2)\tau \\ -2\tau &{}\quad -\tau &{}\quad *&{}\quad \cdots &{}\quad (m-4)\tau &{}\quad (m-3)\tau \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ -(m-2)\tau &{}\quad -(m-3)\tau &{}\quad -(m-4)\tau &{}\quad \cdots &{}\quad *&{}\quad \tau \\ -(m-1)\tau &{}\quad -(m-2)\tau &{}\quad -(m-3)\tau &{}\quad \cdots &{}\quad -\tau &{}\quad *\end{pmatrix}, \end{aligned}$$

where the (ij)-entry denotes the difference \(x_{j}-x_{i}\) (\(i\ne j\)).

Clearly, when \({{\text {ord}}}(\tau )=m\) (so A forms a subgroup of \({\mathbb {Z}}_v\)), we have \(-i\tau \equiv (m-i)\tau ~(\bmod ~v)\) and thus \(\Delta A=[(i\tau )^{m}:1\le i\le m-1].\) When \({{\text {ord}}}(\tau )=m+1\), we have \(-i\tau \equiv (m+1-i)\tau ~(\bmod ~v)\). Hence it is not difficult to have that \( \Delta A=[(i\tau )^{m-1}:1\le i\le m].\)

Let \({{\text {ord}}}(\tau )=d\ge m+2\). It is obvious that \(\tau \not \equiv -i\tau \)\((\bmod ~v)\) for all \(1\le i\le m-1\) and thus \(\lambda _{\Delta X}(\pm \tau )=m-1\). Besides, for \(2\le i\le m-1\), we have \(i\tau \equiv -(d-i)\tau \)\((\bmod ~v)\) occurring in M at most \((m-i)+(i-2)=m-2\) times. Hence, \(\lambda _{\Delta A}(\pm \tau )=m-1\) and \(\lambda _{\Delta A}(\sigma )\le m-2\) for \(\sigma \ne \pm \tau \). This completes the proof. \(\square \)

Proof of Lemma 2.2

Obviously we have

$$\begin{aligned} \Delta Y=(\bigcup _{0\le i\le \frac{h}{d}-1}\Delta Y_{i})\bigcup (\bigcup _{\begin{array}{c} 0\le i,j\le \frac{h}{d}-1\\ i\ne j \end{array}}\Delta (Y_{i},Y_{j}))=D\cup D'. \end{aligned}$$

Since \(Y_{i}=y_{i}+\left\langle \tau \right\rangle \) for \(0\le i\le \frac{h}{d}-1\), we have \(\Delta Y_{i}=\Delta (\left\langle \tau \right\rangle )=[(i\tau )^{d}:1\le i\le d-1]\) by Lemma 2.1. Hence, \(D=\bigcup _{0\le i\le \frac{h}{d}-1}\Delta Y_{i}=[(i\tau )^{h}:1\le i\le d-1]\), proving part (ii). It follows that every nonzero element of \(\left\langle \tau \right\rangle \) is a difference of multiplicity h in D and thus also in \(\Delta Y\) as \(|Y|=h\). Consequently, \(D\cap D'=\emptyset \), proving part (i).

Next we show that, for any \( 0\le i,j\le \frac{h}{d}-1\) with \(i\ne j\),

$$\begin{aligned} \lambda _{\Delta (Y_{i},Y_{j})}(\delta )=d\ \mathrm{for\ every\ } \delta \in \Delta (Y_{i},Y_{j}). \end{aligned}$$
(6)

Because \(\delta \in \Delta (Y_{i},Y_{j}),\) assume that \(\delta =(y_{j}+a\tau )-(y_{i}+b\tau )=y_j-y_i+\alpha \) where \(\alpha =a\tau -b\tau \in \Delta (\left\langle \tau \right\rangle )\cup \{0\}\). Applying Lemma 2.1 yields \(\lambda _{\Delta (\left\langle \tau \right\rangle )}(\alpha )=d\) if \(\alpha \ne 0\). Clearly \(0=i\tau -i\tau \) for all \(0\le i\le d-1\). Thus (6) holds.

Let \(0\le i\le \frac{h}{d}-1\) and i be fixed. Then we have

$$\begin{aligned} \Delta (Y_{i},Y_{j})\cap \Delta (Y_{i},Y_{l})=\emptyset ,\ 0\le j,l\le \frac{h}{d}-1,i\ne j\ne l \ne i. \end{aligned}$$
(7)

Otherwise, we have \(y_{j}-y_{i}+\alpha = y _{l}-y_{i}+\beta \) for some \(\alpha ,\beta \in \left\langle \tau \right\rangle =\left\langle {v\over d}\right\rangle \) so that \(y_{j}=y_{l}+\beta -\alpha \), which contradicts with that \(y_j\ne y_l\) and \(y_{j},y_{l}\in {\mathbb {Z}}_{\frac{v}{d}}\).

Let \(\delta \in D'\) be fixed. By (7), given \(0\le i\le {h\over d}-1\), if \(\delta \in \bigcup _{i\ne j}\Delta (Y_i,Y_j)\), then there exists at most one j with \(0\le j\le {h\over d}-1\) such that \(\delta \in \Delta (Y_{i},Y_{j})\). Hence, we have \(|I|\le \frac{h}{d}\), i.e., \(|I|=\frac{h}{d}-a\) for some integer a with \(0\le a\le {h\over d}-1\), where

$$\begin{aligned} I=\{(i,j):\delta \in \Delta (Y_{{i}},Y_{j}),0\le i,j\le \frac{h}{d}-1,i\ne j\}. \end{aligned}$$

Then combining with (6) and using part (i), we have that \(\lambda _{\Delta Y}(\delta )=\lambda _{D'}(\delta )=|I|\cdot d=h-ad\), proving (iii). \(\square \)

Proof of Lemma 2.3

  1. (i)

    Let \({{\text {O}}}(X)\in \Lambda (k-1).\) Suppose that \(\lambda _{\Delta X}(\tau )=\lambda (\Delta X)=k-1\). Then clearly we have \(\tau \ne 0\) and \(|X\cap (X+\tau )|=k-1\). Obviously \(|X\cap (X+(v-\tau ))|=k-1\) if and only if \(|X\cap (X+\tau )|=k-1\). Thus there is \(\tau \) with \(1\le \tau \le \lfloor {v\over 2}\rfloor \) such that \(|X\cap (X+\tau )|=k-1\). This proves that \(\Lambda (k-1)\) is contained in the right-hand side set. Conversely, let \(X\in \Omega (v,k)\) satisfy \(|X\cap (X+\tau )|=k-1\) for some \(1\le \tau \le \lfloor {v\over 2}\rfloor \). Then \(\lambda _{\Delta X}(\tau )= k-1\). We need to show that \(\lambda (\Delta X)=k-1\) (meaning X belongs to a full orbit) and then \({{\text {O}}}(X)\in \Lambda (k-1).\) Obviously \(\lambda (\Delta X)\ge \lambda _{\Delta X}(\tau )= k-1\). If \(\lambda (\Delta X)=k\), then there exists \(\sigma \in {\mathbb {Z}}_{v}^*\) such that \(X=X+\sigma \). Denote \({{\text {ord}}}(\sigma )=d\). By Corollary 1.4, we have \(X=X_0+\left\langle \sigma \right\rangle \) for some \(X_0\in \Omega (\frac{v}{d},\frac{k}{d}).\) Note that \(d\ge 2\) as \(\sigma \ne 0\). By Lemma 2.2, \(\lambda _{\Delta X}(\tau )=k\) or \(\lambda _{\Delta X}(\tau )\le k-d\le k-2\), contradicting to \(\lambda _{\Delta X}(\tau )=k-1\). This proves (i).

  2. (ii)

    This follows immediately by noting that we have shown any \(X\in \Gamma (k-1)\) belongs to a full orbit. \(\square \)

Proof of Lemma 3.1

Write X as

$$\begin{aligned} X=Y\cup Z=(Y'+\left\langle \tau \right\rangle )\cup \{x,x+\tau ,\ldots ,x+(m-1)\tau \}, \end{aligned}$$

where \(Y=Y+\tau \) can be written as \(Y=Y'+\left\langle \tau \right\rangle \) for some \(Y'=\{y_{0},y_{1},\ldots ,y_{\frac{k-m}{d}-1}\}\in \Omega ({v\over d},{k-m\over d})\) by applying Corollary 1.4. For convenience, denote \(y_{\frac{k-m}{d}}=x\), \(Y_{\frac{k-m}{d}}=Z\), and \(Y_{i}=\{y_{i}\}+\left\langle \tau \right\rangle \) for \(0\le i\le \frac{k-m}{d}-1\). Then \(X=\bigcup _{i=0}^{\frac{k-m}{d}}Y_{i}\) and \(\Delta X=D\cup D'\), where

$$\begin{aligned} D=\bigcup _{0\le i\le \frac{k-m}{d}}\Delta Y_{i},\ D'=\bigcup _{\begin{array}{c} 0\le i,j\le \frac{k-m}{d}\\ i\ne j \end{array}}\Delta (Y_{i},Y_{j}). \end{aligned}$$

By Lemma 2.2, we have \(D=[(i\tau )^{k-m}:1\le i\le d-1]\cup \Delta Z\). So it is clear that all differences of D belong to \(\left\langle \tau \right\rangle \), giving that if \(\delta \in \Delta X\) and \(\delta \notin \left\langle \tau \right\rangle \) then \(\delta \in D'\).

Let \(\delta \in \Delta (Y_{i},Y_{j})\) where \(0\le i,j\le \frac{k-m}{d}\) and \(i\ne j\). Then we have

$$\begin{aligned} \lambda _{\Delta (Y_{i},Y_{j})}(\delta )=\left\{ \begin{array}{ll} d, &{}0\le i,j\le \frac{k-m}{d}-1,i\ne j, \\ m, &{}\frac{k-m}{d}\in \{i,j\},i\ne j.\end{array}\right. \end{aligned}$$
(8)

Equation (8) holds for \(0\le i,j\le \frac{k-m}{d}-1\) and \(i\ne j\) by similar analysis to the proof of (6) in Lemma 2.2. Now we prove the case that \(i=\frac{k-m}{d}\) or \(j=\frac{k-m}{d}\) with \(i\ne j\). W.l.o.g. let \(i=\frac{k-m}{d}\) and \(i\ne j\). Since \(\alpha \in \Delta (Y_{\frac{k-m}{d}},Y_{j})\), we may write \(\alpha =y_{j}-y_{\frac{k-m}{d}}+\beta \) for some \(\beta \in \Delta (Z_{0},\left\langle \tau \right\rangle )\) with \(Z_{0}=\{0,\tau ,\ldots ,(m-1)\tau \}.\) It is not difficult to learn that \(\Delta (Z_{0},\left\langle \tau \right\rangle ) =[(i\tau )^m:0\le i\le d-1]\) by noting that \(i\tau =(l+i)\tau -l\tau \) for all \(0\le l\le m-1\). Hence \(\lambda _{\Delta (Y_{\frac{k-m}{d}},Y_{j})}(\alpha )=\lambda _{\Delta (Z_{0},\left\langle \tau \right\rangle )}(\beta )=m\).

Let \(0\le i\le \frac{k-m}{d}\) be fixed. By the same argument for (7) in the proof of Lemma 2.2, we have

$$\begin{aligned}&\Delta (Y_{i},Y_{j})\cap \Delta (Y_{i},Y_{l})=\emptyset ,\ \Delta (Y_{j},Y_{i})\cap \Delta (Y_{l},Y_{i})=\emptyset ,\nonumber \\ {}&\quad 0\le j,l\le \frac{k-m}{d}-1,i\ne j\ne l \ne i. \end{aligned}$$
(9)

Let \(\delta \in D'\) be fixed. By (9), given \(0\le i\le {k-m\over d}\), if \(\delta \in \bigcup _{j\notin \{i, {k-m\over d}\}}\Delta (Y_i,Y_j)\), then there exists at most one j with \(0\le j\le {k-m\over d}-1\) such that \(\delta \in \Delta (Y_{i},Y_{j})\). Similarly, there exists at most one i with \(0\le i\le {k-m\over d}-1\) such that \(\delta \in \Delta (Y_{i},Y_{k-m\over d})\). Hence, we have \(|I|\le \frac{k-m}{d}+1\) and \(|I'|\le 1,\) where

$$\begin{aligned}&I=\{(i,j):\delta \in \Delta (Y_{{i}},Y_{j}),0\le i\le \frac{k-m}{d},0\le j\le \frac{k-m}{d}-1,i\ne j\},\\&I'=\{i:\delta \in \Delta (Y_{i},Y_{\frac{k-m}{d}}),0\le i\le \frac{k-m}{d}-1\}. \end{aligned}$$

Combining with (8) yields that

$$\begin{aligned} \lambda _{D}(\delta )=\left\{ \begin{array}{ll} (|I|-1)d+(|I'|+1)m, &{}\mathrm{if\ }(\frac{k-m}{d},j)\in I\ \mathrm{for\ some\ } j, \\ |I|d+|I{'}|m, &{}\mathrm{otherwise}.\end{array}\right. \end{aligned}$$

Now we prove the conclusion by contradiction. Note that if \(m=1\) then \(Z=\{x\}\) is independent of \(\tau \). So w.l.o.g. we let \(G_Y=\left\langle \tau \right\rangle \) when \(m=1\). Assume that \(\lambda _{\Delta X}(\sigma )=k-1\) and \(\sigma \notin \left\langle \tau \right\rangle \). Then \(\sigma \in D'\) by previous arguments. We consider the following cases.

  1. (a)

    \((\frac{k-m}{d},j)\in I\) and \(|I'|=0\). Then \(\lambda _{\Delta X}(\sigma )=(|I|-1)d+m\). Because \(|I|\le \frac{k-m}{d}+1\) and \(\lambda _{\Delta X}(\sigma )=k-1\), we must have \(k-1=\lambda _{\Delta X}(\sigma )=k-ad\) for some positive integer a, which contradicts to \(d\ge m+2\ge 3\).

  2. (b)

    \((\frac{k-m}{d},j)\in I\) and \(|I'|=1\). In such a case we have \(\lambda _{\Delta X}(\sigma )=(|I|-1)d+2m=k-ad+m\) for some nonnegative integer a, contradicting to \(\lambda _{\Delta X}(\sigma )=k-1\) because \(d\ge m+2\).

  3. (c)

    \((\frac{k-m}{d},j)\not \in I\) and \(|I'|=1\). Then \(\lambda _{\Delta X}(\sigma )=|I|d+m\) and \(|I|\le \frac{k-m}{d}\). Similarly to (a), we have a contradiction \(k-1=k-ad\) (\(a\ge 1\)).

  4. (d)

    \((\frac{k-m}{d},j)\not \in I\) and \(|I'|=0\). Then \(\lambda _{\Delta X}(\sigma )\le {k-m\over d}\cdot d= k-m.\) Since \(\lambda _{\Delta X}(\sigma )= k-1,\) we have \(m=1\). Thus \(X=Y\cup \{x\}\) and \(Y=Y+\tau \). It is easy to show \(\Delta X=\Delta Y\cup \Delta (\{x\},Y)\cup \Delta (Y,\{x\})\). By Lemma 2.2, \(\lambda _{\Delta Y}(\sigma )=k-1-ad\) for some nonnegative integer a. Hence \(\lambda _{\Delta Y}(\sigma )=k-1\) or \(\lambda _{\Delta Y}(\sigma )\le k-4\) as \(d\ge 3\). Clearly \(\lambda (\Delta (Y,\{x\}))=\lambda (\Delta (\{x\},Y)) =1\). From \(\lambda _{\Delta X}(\sigma )= k-1,\) we must have \(\lambda _{\Delta Y}(\sigma )= k-1\), yielding \(\sigma \in G_Y=\left\langle \tau \right\rangle \), a contradiction.

This completes the proof. \(\square \)

Proof of Lemma 3.2

The existence of \(\hat{X}\) and x is obvious if \(X\in \Gamma _{1,1}\) because \(d=k+1\). So we let \(X\in \Gamma _{2,2}\) be of Type II with respect to \((Y,Z,\tau ,d,m,z)\) where \(d=m+1\). Then \(X=Y\cup Z\) where \(Y=Y+\tau \), \(Z=\{z,z+\tau ,\ldots ,z+(m-1)\tau \}\), and \({{\text {ord}}}(\tau )=d=m+1\). Clearly \(Z=(z+\left\langle \tau \right\rangle )\setminus \{z+m\tau \}\). By applying Corollary 1.4, we have \(Y=Y_0+\left\langle \tau \right\rangle \) for some \(Y_0=\{y_{0},y_{1},\ldots ,y_{\frac{k-m}{d}-1}\}\in \Omega ({v\over d},{k-m\over d})\). Then letting \(\hat{X}=(Y_0\cup \{z\})+\left\langle \tau \right\rangle \) and \(x=z+m\tau \) yields \(X=\hat{X}\setminus \{x\}\). Obviously \(\left\langle \tau \right\rangle \le G_{\hat{X}}\) and thus \(\hat{X}\) generates a short orbit. This proves the existence of \(\hat{X}\) and x.

In the sequel we suppose that \(X=(Y_0+\left\langle \tau \right\rangle )\cup Z=\hat{X}\setminus \{x\}\) as above and suppose \(\tau \) is a generator of \(G_{\hat{X}}\). Obviously

$$\begin{aligned} \Delta X=\Delta \hat{X}\setminus (\Delta (X,\{x\})\cup \Delta (\{x\},X)). \end{aligned}$$

For convenience, denote \(y_{\frac{k-m}{d}}=z\) and \(Y_{i}=\{y_{i}\}+\left\langle \tau \right\rangle \) for \(0\le i\le \frac{k-m}{d}\). Then \(\hat{X}=\bigcup _{i=0}^{\frac{k-m}{d}}Y_{i}\) and

$$\begin{aligned} \Delta \hat{X}= D\cup D',\ \mathrm{where}\ D=\bigcup _{{ 0\le i\le \frac{k-m}{d}}}\Delta Y_{i},\ D'=\bigcup _{\begin{array}{c} 0\le i,j\le \frac{k-m}{d}\\ i\ne j \end{array}}\Delta (Y_{i},Y_{j}). \end{aligned}$$

By Lemma 2.2, for any \(\delta \in {\mathbb {Z}}_v^*\), we have \(\lambda _{\Delta \hat{X}}(\delta )=k+1\) or \(\lambda _{\Delta \hat{X}}(\delta )\le k+1-d\le k-1\) because \(d\ge 2\). Obviously, \(\lambda (\Delta (X,\{x\}))=\lambda (\Delta (\{x\},X)) =1\). Now suppose that \(\lambda _{\Delta X}(\sigma )=k-1\) for some \(\sigma \in {\mathbb {Z}}_v^*\). Then (i) \(\lambda _{\Delta \hat{X}}(\sigma )=k+1\) and \(\lambda _{\Delta (\{x\},X)}(\sigma )=\lambda _{\Delta (X,\{x\})}(\sigma )=1\), or (ii) \(\lambda _{\Delta \hat{X}}(\sigma )=k-1\), \(d=2\), and \(\lambda _{\Delta (\{x\},X)}(\sigma )=\lambda _{\Delta (X,\{x\})}(\sigma )=0\). Now we show the latter case (ii) is impossible. Otherwise, we have \(d=2\), \(m=1\) and hence \(\hat{X}\) can be written as \(\hat{X}=X_0+\left\langle {v\over 2}\right\rangle \) where \(X_0=\{x_0,x_1,\ldots ,x_{k-1\over 2}\}\in \Omega ({v\over 2},{k+1\over 2})\) and \(x_0\equiv x\) (mod \({v\over 2}\)). From \(\lambda _{\Delta \hat{X}}(\sigma )=k-1\), it follows that \(\lambda _{\Delta X_0}(\sigma _0)={k-1\over 2}\), where \(\sigma _0\equiv \sigma \) (mod \({v\over 2}\)) and \(\sigma _0\in {\mathbb {Z}}_{v\over 2}\). Consider the expression of \(\sigma _0\) as the differences \(x_{k_i}-x_{j_i}\) where \(1\le i\le {k-1\over 2}\) and \(0\le j_i,k_i\le {k-1\over 2}\). It is easy to show that \(x_0\in \{x_{ j_i}:1\le i\le {k-1\over 2}\}\cup \{x_{ k_i}:1\le i\le {k-1\over 2}\}\), meaning \(\lambda _{\Delta (X,\{x\})}(\sigma )=1\) or \(\lambda _{\Delta (\{x\},X)}(\sigma )=1\), a contradiction. To sum up, if \(\lambda _{\Delta X}(\sigma )=k-1\) then \(\lambda _{\Delta \hat{X}}(\sigma )=k+1\). Conversely, if \(\lambda _{\Delta \hat{X}}(\sigma )=k+1\), then \(\sigma \) can be expressed as differences of \(\hat{X}\) in exactly \(k+1\) times, meaning \(\lambda _{\Delta (\{x\},X)}(\sigma )=\lambda _{\Delta (X,\{x\})}(\sigma )=1\) and hence \(\lambda _{\Delta X}(\sigma )=k-1\). As a result, \(\lambda _{\Delta X}(\sigma )=k-1\) if and only if \(\lambda _{\Delta \hat{X}}(\sigma )=k+1\), or equivalently, \(\sigma \in G_{\hat{X}}\). \(\square \)

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Fang, Z., Zhou, J. The sizes of maximal \((v,k,k-2,k-1)\) optical orthogonal codes. Des. Codes Cryptogr. (2020) doi:10.1007/s10623-020-00714-1

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Keywords

  • List of differences
  • Maximal
  • Optical orthogonal code
  • Orbit
  • Stabilizer

Mathematics Subject Classification

  • 05B30
  • 94B65
  • 94C30