Linear covering codes and error-correcting codes for limited-magnitude errors

Abstract

The concepts of a linear covering code and a covering set for the limited-magnitude-error channel are introduced. A number of covering-set constructions, as well as some bounds, are given. In particular, optimal constructions are given for some cases involving small-magnitude errors. A problem of Stein is partially solved for these cases. Optimal packing sets and the corresponding error-correcting codes are also considered for some small-magnitude errors.

Appendix

1.1 Proof of Theorem 5

In the setting of Theorem 5 we have $q$ odd. Let $\varsigma _r(q)$ denote the number of cyclotomic cosets of odd size. In [8], an expression for $\varsigma _1(q)$ was given, in a slightly different notation. Here we will use a similar method to show that $\varsigma _r(q)=\vartheta _r(q_0)$ for all $r$.

Let

$$\begin{aligned} \mathbb {Z}_q^* =\bigl \{a\mid \gcd (a,q)=1, a\in [1,q-1]\bigr \}, \end{aligned}$$

and for $d|q$, let

$$\begin{aligned} \mathbb {Z}_{q,d}=\left\{ a\frac{q}{d} \, \Big |\, a\in \mathbb {Z}_d^*\right\} . \end{aligned}$$

In particular, $\mathbb {Z}_{q,1} =\left\{ 0\right\}$. The size of $\mathbb {Z}_{q,d}$ is $\varphi (d)$, where $\varphi (\cdot )$ is Euler’s totient function. We have the following disjoint-union decomposition [8, Lemma 4]:

$$\begin{aligned} \mathbb {Z}_q= \bigcup _{d\mid q} \mathbb {Z}_{q,d}. \end{aligned}$$

(22)

Lemma 4

Let $d$ be a divisor of $q$. For any $b\in \mathbb {Z}_{q,d}$, the least positive $\ell$ such that $2^\ell b\equiv b\;(\mathrm{mod}\; q)$ is exactly $\ell _d$.

Proof

Let $b=a\cdot q/d$. Since $\gcd (a,q)=1$ by definition, we have the following equivalences:

$$\begin{aligned} 2^\ell \equiv 1 \quad (\mathrm{mod}\; d)\, \Leftrightarrow \, 2^\ell \frac{q}{d}\equiv \frac{q}{d} \quad (\mathrm{mod}\; q) \,\Leftrightarrow \, 2^\ell a\frac{q}{d}\equiv a\frac{q}{d} \quad (\mathrm{mod}\; q). \end{aligned}$$

$\square$

We observe that Lemma 4 implies that $\mathbb {Z}_{q,d}$ is the disjoint union of $\varphi (d)/\ell _d$ cosets of size $\ell _d$.

In [8, Lemma 5] the following result was given.

Lemma 5

• (i) For odd $d=p_1^{e_1}p_2^{e_2}\ldots p_s^{e_s}$, with $p_i$ distinct odd primes, we have

  $$\begin{aligned} \ell _d={{\mathrm{lcm}}}\left( \ell _{p_1^{e_1}}, \ell _{p_2^{e_2}}, \ldots , \ell _{p_s^{e_s}}\right) . \end{aligned}$$

  In particular, $v_2(\ell _d)=i$ if and only if $\max _{1\le j\le s}v_2(\ell _{p_j^{e_j}})=i$.

• (ii) For any odd prime $p$, suppose $2^{\ell _p}=1+p^{u_p}s_p$ with $p\not \mid s_p$. Then

  $$\begin{aligned} \ell _{p^k}= {\left\{ \begin{array}{ll} \ell _p &{}\quad \! \text {if } k\le u_p,\\ p^{k-u_p} \ell _p &{}\quad \! \text {if } k>u_p. \end{array}\right. } \end{aligned}$$

From Lemma 5 we get the following more general result:

Lemma 6

For any odd $q$, and $d_1, d_2, \ldots , d_r$ divisors of $q$ we have

$$\begin{aligned} {{\mathrm{lcm}}}(\ell _{d_1}, \ell _{d_2}, \ldots ,\ell _{d_r}) = \ell _{ {{\mathrm{lcm}}}(d_1,d_2,\ldots ,d_r) }. \end{aligned}$$

Proof

Let $p_j$, $j=1,2,\ldots ,s$, be the set of all primes dividing $q$, and let $d_i= \prod _{j=1}^s p_j^{e_{i,j} }$. Then

$$\begin{aligned} {{\mathrm{lcm}}}(d_1,d_2,\ldots ,d_r)= \prod _{j=1}^s p_j^{\max _{1\le i \le r}e_{i,j}}. \end{aligned}$$

By Lemma 5 we get

$$\begin{aligned} {{\mathrm{lcm}}}_{1\le i\le r} (\ell _{d_i})&= {{\mathrm{lcm}}}_{1\le i\le r} {{\mathrm{lcm}}}_{1\le j \le s} (\ell _{p_j^{e_{i,j}}}) = {{\mathrm{lcm}}}_{1\le j \le s} {{\mathrm{lcm}}}_{1\le i\le r} (\ell _{p_j^{e_{i,j}}}) \\&= {{\mathrm{lcm}}}_{1\le j \le s} (\ell _{ p_j^{\max _{1\le i \le r}e_{i,j}}}) = \ell _{ {{\mathrm{lcm}}}(d_1,d_2,\ldots ,d_r) }. \end{aligned}$$

$\square$

Now, consider a cyclotomic coset in $\mathbb {Z}_q^r$, generated by $(a_1,a_2,\ldots a_r)$. Suppose $a_i\in \mathbb {Z}_{q,d_i}$. Then the size of the coset is ${{\mathrm{lcm}}}(\ell _{d_1},\ell _{d_2},\ldots , \ell _{d_r})=\ell _{{{\mathrm{lcm}}}(d_1,d_2,\ldots ,d_r)}$, and the number of such cosets is

$$\begin{aligned} \frac{\varphi (d_1)\varphi (d_2)\ldots \varphi (d_r)}{\ell _{{{\mathrm{lcm}}}(d_1,d_2,\ldots ,d_r)}}. \end{aligned}$$

We get cosets of odd order if and only if $d_i|q_0$ for all $i$. Hence we get

$$\begin{aligned} \varsigma _r(q)= \sum _{d_1\mid q_0} \sum _{d_2\mid q_0}\cdots \sum _{d_r\mid q_0} \frac{\varphi (d_1)\varphi (d_2)\ldots \varphi (d_r)}{\ell _{{{\mathrm{lcm}}}(d_1,d_2,\ldots ,d_r)}} =\sum _{d\mid q_0} \frac{\Phi _r(d)}{\ell _{d}}, \end{aligned}$$

(23)

where

$$\begin{aligned} \Phi _r(d)= \sum _{ {{\mathrm{lcm}}}(d_1,d_2,\ldots ,d_r)=d } \varphi (d_1)\varphi (d_2)\ldots \varphi (d_r). \end{aligned}$$

It follows that

$$\begin{aligned} \sum _{c| d} \Phi _r(c)&= \sum _{d_1| d,d_2| d,\ldots ,d_r| d} \varphi (d_1)\varphi (d_2)\ldots \varphi (d_r)\\&= \left( \sum _{d_1| d}\varphi (d_1) \right) \left( \sum _{d_2| d}\varphi (d_2) \right) \ldots \left( \sum _{d_r| d}\varphi (d_r) \right) = \left( \sum _{a | d}\varphi (a) \right) ^r = d^r. \end{aligned}$$

Using Möbius inversion, we get

$$\begin{aligned} \Phi _r(d)= \sum _{c| d}\mu \left( \frac{d}{c} \right) c^r. \end{aligned}$$

(24)

Substituting this expression in (23) we get

$$\begin{aligned} \varsigma _r(q)=\vartheta _r(q_0). \end{aligned}$$

This completes the proof of Theorem 5.

For $r=1$, the expression in (23) was given in [8, Theorem 2], in a slightly different notation. In the same theorem we determined $\theta _{2,0,1}(q)$. For $r>1$, the result is new.

Example 8

If $q_0= p^a$, where $p$ is a prime and $a\ge 1$, the divisors of $q_0$ are $p^b$, $b\in [0,a]$. If $d=p^b>1$, then $\mu (d/c)\ne 0$ for $c\mid d$ exactly for $c=p^a$ and $c=p^{a-1}$. Since $\mu (1)=1$ and $\mu (p)=-1$, we get

$$\begin{aligned} \vartheta _r(p^a)=1+ \sum _{b=1}^a \frac{p^{br}-p^{(b-1)r}}{\ell _{p^b}}. \end{aligned}$$

In particular, if $v_2\left( 2^{\ell _p}-1 \right) =1$, that is $u_p=0$ (which is the situation in most cases), Lemma 5 implies that $\ell _{p^b}=p^{b-1}\ell _p$ for all $b\ge 1$. Hence, in this case we get

$$\begin{aligned} \vartheta _1(p^a)=1+ \frac{1}{\ell _p} \sum _{b=1}^a \left( p-1 \right) = 1+\frac{a(p-1)}{\ell _p}. \end{aligned}$$

and for $r>1$ we get

$$\begin{aligned} \vartheta _r(p^a)=1+ \frac{1}{\ell _p} \sum _{b=1}^a \left( p^{b(r-1)+1}-p^{(b-1)(r-1)} \right) = 1+\frac{(p^r-1)(p^{a(r-1)}-1)}{(p^{r-1}-1)\ell _p}. \end{aligned}$$

In particular,

$$\begin{aligned} \vartheta _1(p^r)=1+\frac{r(p-1)}{\ell _p} \quad \text { and }\quad \vartheta _r(p)=1+\frac{p^r-1}{\ell _p}, \end{aligned}$$

and so $\vartheta _1(p^r)<\vartheta _r(p)$. This is a special case of Lemma 10 below.

1.2 Proof of Theorem 8

The expressions for $\omega _{2,0,r}(q)$ in Theorem 5 and $\theta _{2,0,r}(q)$ in (12), and their proofs, are closely related. In the same way we will get closely related expressions and proofs for $\omega _{2,2,r}(q)$ and $\theta _{2,2,r}(q)$. For $\theta _{2,2,1}(q)$, the expression in Theorem 8 was given in [9], in a slightly different notation. For general $r$ we get a proof that generalizes its proof.

Consider the coset generated by a non-zero $\mathbf{a}=(a_1,a_2,\ldots , a_r)\in \mathbb {Z}_q^r$. We first remark that if $-\mathbf{a}\in \sigma (\mathbf{a})$, that is, $\sigma (-\mathbf{a})=\sigma (\mathbf{a})$, then $\left| \sigma (\mathbf{a})\right|$ is even: if $u>0$ is minimal such that $-\mathbf{a}=2^u\mathbf{a}\;(\mathrm{mod}\; q)$, then $\left| \sigma (\mathbf{a})\right| =2u$.

The coset $\sigma (\mathbf{a})$ has odd size if and only if $\sigma (a_i)$, $1\le i \le r$, all have odd size. In this case $\sigma (\mathbf{a})$ and $\sigma (-\mathbf{a})$ are disjoint, and the $(\left| \sigma (\mathbf{a})\right| +1)/2$ elements $2^{2i}\mathbf{a}$, $i\in [0,(\left| \sigma (\mathbf{a})\right| -1)/2]$ will cover $\sigma (\mathbf{a})\cup \sigma (-\mathbf{a})$, and the union cannot be covered by fewer elements. We select these elements in a covering set. Hence we get a contribution $(\left| \sigma (\mathbf{a})\right| +1)/4$ to $\omega _{2,2,r}(q)$ from the coset $\sigma (\mathbf{a})$ and $(\left| \sigma (\mathbf{a})\right| +1)/4=(\left| \sigma (-\mathbf{a})\right| +1)/4$ from the coset $\sigma (-\mathbf{a})$. The number of such cosets is $\vartheta _r(q_0)$ as was shown in the proof of Theorem 5.

If $\left| \sigma (\mathbf{a})\right|$ is even, but $\sigma (-\mathbf{a})\ne \sigma (\mathbf{a})$ (that is, the two sets are disjoint), then we select the $\left| \sigma (\mathbf{a})\right| /2$ elements $2^{2i}\mathbf{a}$, $i\in [0,\left| \sigma (\mathbf{a})\right| /2-1]$ to cover $\sigma (\mathbf{a})\cup \sigma (-\mathbf{a})$. The contribution to $\omega _{2,2,r}(q)$ from the cosets $\sigma (\mathbf{a})$ and $\sigma (-\mathbf{a})$ is therefore $\left| \sigma (\mathbf{a})\right| /4+\left| \sigma (-\mathbf{a})\right| /4$.

Now, consider the situation when $\sigma (-\mathbf{a})=\sigma (\mathbf{a})$. As before, let $u>0$ be the minimal integer such that $-\mathbf{a}=2^u\mathbf{a}\;(\mathrm{mod}\; q)$. If $u$ is even, then the $u/2=\left| \sigma (\mathbf{a})\right| /4$ elements $2^{2i}\mathbf{a}$, $i\in [0,u/2-1]$ cover $\sigma (\mathbf{a})$. Finally, if $u$ is odd, then the $(u+1)/2=(\left| \sigma (\mathbf{a})\right| +2)/4$ elements $2^{2i}\mathbf{a}$, $i\in [0,(u-1)/2]$ cover $\sigma (\mathbf{a})$. We see that $u$ is odd if and only if $\sigma (a_i)$ is singly even for all $i$. In the proof of [9, Theorem 6], it was shown that this occurs exactly when $a_i\in \mathbb {Z}_{q,d_i}$ for some $d_i|q_1$. A proof similar to the proof in Appendix 10 shows that the number of such cosets is $\vartheta _r(q_1)$. Summing over all the cosets, we get the expression in Theorem 8.

1.3 A result for $\vartheta _r(q)$

A simple, but useful relation is the following.

Lemma 7

If $d_1|d_2$, then $\ell _{d_1} \le \ell _{d_2}$.

Proof

By definition, $d_2| 2^{\ell _{d_2}}-1$ and so $d_1| 2^{\ell _{d_2}}-1$, which implies that $\ell _{d_1} \le \ell _{d_2}$ (and in fact $\ell _{d_1}|\ell _{d_2}$). $\square$

We recall that $\Phi _r(d)$ was defined by (24). This is a multiplicative function, as the following lemma shows.

Lemma 8

If $\gcd (d_1,d_2)=1$, then $\Phi _r(d_1d_2)= \Phi _r(d_1) \Phi _r(d_2)$.

Proof

If $c|d_1d_2$, then $c=c_1c_2$, where $c_1|d_1$ and $c_2|d_2$. Hence

$$\begin{aligned} \Phi _r(d_1d_2)= \sum _{c| d_1 d_2}\mu \left( \frac{d_1d_2}{c} \right) c^r= \left( \sum _{c\mid d_1}\mu \left( \frac{d_1}{c} \right) c^r \right) \left( \sum _{c\mid d_2}\mu \left( \frac{d_2}{c} \right) c^r \right) = \Phi _r(d_1) \Phi _r(d_2). \end{aligned}$$

$\square$

For a prime $p$, define $\Delta _r(p^\beta )$ by

$$\begin{aligned} \Delta _r(p^\beta )={\left\{ \begin{array}{ll} 1 &{}\quad \! \beta =0, \\ p^{r\beta }-p^{r(\beta -1)} &{}\quad \! \text {otherwise.} \end{array}\right. } \end{aligned}$$

For convenience, we let $\Delta (p^\beta )=\Delta _1(p^\beta )$.

Lemma 9

If $p$ is a prime and $\beta \ge 1$, then

$$\begin{aligned} \Phi _r(p^\beta )=\Delta _r(p^\beta )=\sum _{j=0}^{r-1} \Delta (p^{\beta -j}). \end{aligned}$$

Proof

We have

$$\begin{aligned} \Phi _r(p^\beta )=p^{r\beta }-p^{r(\beta -1)}=\Delta _r(p^\beta )=\sum _{j=0}^{r-1}\left( p^{r\beta -j}-p^{r\beta -j-1} \right) =\sum _{j=0}^{r-1} \Delta (p^{\beta -j}). \end{aligned}$$

$\square$

We now give a main lemma on $\vartheta _r(q)$.

Lemma 10

For all $r\ge 2$ and odd $q$ we have $\vartheta _1(q^r)\le \vartheta _r(q)$.

Proof

By definition, $\vartheta _r(q)=\sum _{d|q}\frac{1}{\ell _d} \Phi _r(d)$. Let $q= \prod _{i=1}^{s}p_i^{\alpha _i}$ be the prime factorization of $q$. The divisors of $q^r$ are all numbers of the form $\prod _{i=1}^{s}p_i^{\beta _i}$ where $0\le \beta _i\le r\alpha _i$. Using Lemmas 8 and 9, we get

$$\begin{aligned} \Phi _r\left( \prod _{i=1}^{s}p_i^{\beta _i} \right)&=\prod _{i=1}^{s} \Phi _r\left( p_i^{\beta _i}\right) =\prod _{\begin{array}{c} 1\le i \le s\\ \beta _i>0 \end{array}} \sum _{j_i=0}^{r-1} \Delta \left( p_i^{r\beta _i-j_i}\right) . \end{aligned}$$

Hence

$$\begin{aligned} \vartheta _r(q) = \sum _{\beta _1=0}^{\alpha _1} \sum _{\beta _2=0}^{\alpha _2} \cdots \sum _{\beta _s=0}^{\alpha _s} \frac{1}{\ell _{ \prod _{1\le i \le s}p_i^{\beta _i}}} \prod _{\begin{array}{c} 1\le i \le s\\ \beta _i>0 \end{array}} \sum _{j_i=0}^{r-1} \Delta \left( p_i^{r\beta _i-j_i}\right) . \end{aligned}$$

Similarly, we get

$$\begin{aligned} \vartheta _1(q^r)= \sum _{\beta _1=0}^{r\alpha _1} \sum _{\beta _2=0}^{r\alpha _2} \cdots \sum _{\beta _s=0}^{r\alpha _s} \frac{1}{ \ell _{ \prod _{1\le i \le s} p_i^{\beta _i} } } \prod _{i=1}^s \Delta \left( p_i^{\beta _i}\right) . \end{aligned}$$

In order to compare the two expressions we note all the summands are non-negative, and of the form $\prod _{i=1}^s \Delta (p_i^{\gamma _i})$. One can verify that the coefficient of $\prod _{i=1}^s \Delta (p_i^{\gamma _i})$ in $\vartheta _r(q)$ is

$$\begin{aligned} C_r=\ell ^{-1}_{ \prod _{1\le i \le s}p_i^{\left\lceil \gamma _i/r \right\rceil }}, \end{aligned}$$

whereas its coefficient in $\vartheta _1(q^r)$ is

$$\begin{aligned} C_1=\ell ^{-1}_{ \prod _{1\le i \le s}p_i^{\gamma _i}}. \end{aligned}$$

Since $\left\lceil \gamma _i/r \right\rceil \le \gamma _i$ we have

$$\begin{aligned} \left. \prod _{1\le i \le s}p_i^{\left\lceil \gamma _i/r \right\rceil } \right| \prod _{1\le i \le s}p_i^{\gamma _i}, \end{aligned}$$

and so $C_r\ge C_1$ by Lemma 7. Hence $\vartheta _r(q)\ge \vartheta _1(q^r)$. $\square$

We illustrate the proof by a simple example.

Example 9

Let $q=p^2\pi$ where $p$ and $\pi$ are distinct odd primes, and let $r=2$. Then

$$\begin{aligned} \vartheta _2(q)&= 1 + \frac{\Delta _2(p)}{\ell _p} + \frac{\Delta _2(p^2)}{\ell _{p^2}} + \frac{\Delta _2(\pi )}{\ell _{\pi }} + \frac{\Delta _2(p)\Delta _2(\pi )}{\ell _{p\pi }} + \frac{\Delta _2(p^2)\Delta _2(\pi )}{\ell _{p^2\pi }} \\&= 1 + \frac{\Delta (p^2)+\Delta (p)}{\ell _p} + \frac{\Delta (p^4)+\Delta (p^3)}{\ell _{p^2}} \\&\quad \ + \frac{\Delta (\pi ^2)+\Delta (\pi )}{\ell _\pi } + \frac{ \left( \Delta (p^2)+\Delta (p) \right) \left( \Delta (\pi ^2)+\Delta (\pi ) \right) }{\ell _{p\pi }} \\&\quad \ + \frac{ \left( \Delta (p^4)+\Delta (p^3) \right) \left( \Delta (\pi ^2)+\Delta (\pi ) \right) }{\ell _{p^2\pi }}, \end{aligned}$$

and

$$\begin{aligned} \vartheta _1(q^2)&= 1 + \frac{\Delta (p)}{\ell _p} + \frac{\Delta (p^2)}{\ell _{p^2}} + \frac{\Delta (p^3)}{\ell _{p^3}} + \frac{\Delta (p^4)}{\ell _{p^4}} \\&\quad \ + \frac{\Delta (\pi )}{\ell _\pi } + \frac{\Delta (p)\Delta (\pi )}{\ell _{p\pi }} + \frac{\Delta (p^2)\Delta (\pi )}{\ell _{p^2\pi }} + \frac{\Delta (p^3)\Delta (\pi )}{\ell _{p^3\pi }} + \frac{\Delta (p^4)\Delta (\pi )}{\ell _{p^4\pi }} \\&\quad \ + \frac{\Delta (\pi ^2)}{\ell _{\pi ^2}} + \frac{\Delta (p)\Delta (\pi ^2)}{\ell _{p\pi ^2}} + \frac{\Delta (p^2)\Delta (\pi ^2)}{\ell _{p^2\pi ^2}} + \frac{\Delta (p^3)\Delta (\pi ^2)}{\ell _{p^3\pi ^2}} + \frac{\Delta (p^4)\Delta (\pi ^2)}{\ell _{p^4\pi ^2}}. \end{aligned}$$

For example, for $\Delta (p^3)\Delta (\pi )$, the coefficients are $1/\ell _{p^3\pi }$ and $1/\ell _{p^2\pi }$ respectively, and $\ell _{p^2\pi }\le \ell _{p^3\pi }$.