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.

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 (i, j)-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

• (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).

• (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.

• (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 $d\ge m+2\ge 3$.

• (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 $d\ge m+2$.

• (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 $k-1=k-ad$ ($a\ge 1$).

• (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, $\sigma \in G_{\hat{X}}$. $\square$