Low-Weight Superimposed Codes and Their Applications

Abstract

A (k, n)-superimposed code is a well known and widely used combinatorial structure that can be represented by a $t\times n$ binary matrix such that for any k columns of the matrix and for any column $\mathbf{c}$ chosen among these k columns, there exists a row in correspondence of which column $\mathbf{c}$ has an entry equal to 1 and the remaining $k-1$ columns have entries equal to 0. Due to the many situations in which superimposed codes find applications, there is an abundant literature that studies the problem of constructing (k, n)-superimposed codes with a small number t of rows. Motivated by applications to conflict-free communication in multiple-access networks, group testing, and data security, we study the problem of constructing superimposed codes that have the additional constraints that the number of 1’s in each column of the matrix is constant, and equal to an input parameter w. Our results improve on the known literature in the area. We also extend our findings to other important combinatorial structures, like selectors.

Appendix A: Proof of Fact 1

We first state and prove the following technical estimate.

Fact 2

Given integers a, b, c with $c \le a \le b$, it holds that

$$\begin{aligned} \displaystyle \frac{a}{b} \cdot \frac{a-c}{b-c} \le \left( \frac{a-\frac{c}{2}}{b-\frac{c}{2}} \right) ^2 \end{aligned}$$

(17)

Proof

Since $c \le a \le b$, we have $a(a-c) \le b(b-c)$ and then $a(a-c) c^2 \le b(b-c) c^2$. Adding the quantity $4ab(a-c)(b-c)$ to both members of the above inequality, we have $a(a-c) c^2 + 4ab(a-c)(b-c)\le b(b-c) c^2 + 4ab(a-c)(b-c)$, that immediately gives $a(a-c)(2b-c)^2 \le b(b-c)(2a-c)^2$ proving the fact.    $\square$

We can now prove Fact 1, that is,

Given integers a, b, c with $c \le a \le b$, it holds that:

${\displaystyle \left( {\begin{array}{c}a\\ c\end{array}}\right) }\times {\displaystyle \left( {\begin{array}{c}b\\ c\end{array}}\right) ^{-1}} \le \left( \displaystyle \frac{a- \frac{c-1}{2}}{b-\frac{c-1}{2}} \right) ^{c}$

Proof

$$\begin{aligned} {\displaystyle \left( {\begin{array}{c}a\\ c\end{array}}\right) }\times {\displaystyle \left( {\begin{array}{c}b\\ c\end{array}}\right) ^{-1}}= & {} \frac{a!}{(a-c)!} \cdot \frac{(b-c)!}{b!} \nonumber \\= & {} \frac{a}{b} \cdot \frac{a-1}{b-1} \cdot \ldots \cdot \frac{a-(c-2)}{b-(c-2)} \cdot \frac{a-(c-1)}{b-(c-1)}. \end{aligned}$$

(18)

To upper bound expression (18), we consider two cases according to the situation in which c is even or odd.

• Let c be even. The c factors in the product $\frac{a}{b} \cdot \frac{a-1}{b-1} \cdot \ldots \cdot \frac{a-(c-2)}{b-(c-2)} \cdot \frac{a-(c-1)}{b-(c-1)}$ can be paired two by two as follows:

  $$\begin{aligned} \frac{a-i}{b-i} \cdot \frac{a-(c-1-i)}{b-(c-1-i)} \qquad \qquad \text{ for } i=0, \ldots , \lceil \frac{c-1}{2} \rceil -1 \end{aligned}$$

  (19)

  Considering that $\frac{a-(c-1-i)}{b-(c-1-i)} = \frac{a-i-(c-1-2i)}{b-i-(c-1-2i)},$ we can apply Fact 2 to each pair in (19) having

  $$\begin{aligned} \frac{a-i}{b-i} \cdot \frac{a-(c-1-i)}{b-(c-1-i)}= & {} \frac{a-i}{b-i} \cdot \frac{a-i-(c-1-2i)}{b-i-(c-1-2i)} \\\le & {} \left( \frac{a-i-\frac{c-1-2i}{2}}{b-i-\frac{c-1-2i}{2}} \right) ^2 = \left( \frac{a-\frac{c-1}{2}}{b-\frac{c-1}{2}} \right) ^2 \end{aligned}$$

  for $i=0, \ldots , \lceil \frac{c-1}{2} \rceil -1$ and Fact 1 follows in this case.

• Let c be odd. The product $\frac{a}{b} \cdot \frac{a-1}{b-1} \cdot \ldots \cdot \frac{a-(c-2)}{b-(c-2)} \cdot \frac{a-(c-1)}{b-(c-1)}$ has an odd number of factors and, except for the factor $\displaystyle \frac{a-\frac{c-1}{2}}{b-\frac{c-1}{2}}$ in the middle, the remaining $c-1$ factors can be paired two by two as in (19). Reasoning as in the c even case, we can prove Fact 1 also in the odd case.    $\Box$