A new strongly competitive group testing algorithm with small sequentiality

Abstract

In many fault detection problems, we want to identify all defective items from a sample set of items using the minimum number of tests. Group testing is for the scenario where each test is performed on a subset of items, and tells whether the subset contains at least one defective item or not. In practice, the number of defective items in the sample set is usually unknown. In this paper, we investigate new algorithms for the group testing problem with unknown number of defective items. We consider the scenario where the performance of a group testing algorithm is measured by two criteria: the primary criterion is the number of tests performed, which measures the total cost spent; and the secondary criterion is the number of stages the algorithm works in, which is referred to as the sequentiality of the algorithm in this paper and measures the minimum amount of time required by using the algorithm to identify all the defective items. We present a new algorithm Recursive Binary Splitting (RBS) for the above group testing problem with unknown number of defective items, and prove an upper bound on the number of tests required by RBS. The computational results show that RBS exhibits very good practical performance, measured in terms of both the above two criteria.

Appendices

Appendix 1: Calculating the sequentiality of Algorithm DP

From the description of Algorithm DP (following Lemma 2.1) in Sect. 2, for an input set $S$, if $|S|=1$ or $S$ is tested to be pure, then clearly $Seq(\text{ DP }, S)=1$. If $|S|>1$ and $S$ is tested to be contaminated, then Procedure DIG is applied to $S$, to identify and remove one defective item and an unspecified number of good items from $S$. We use $S'$ to denote the resulting set obtained from $S$ after the removal. Then, $S'$ is bisected into two subsets $S_1$ and $S_2$ (if $|S'|>1$), and the two subsets $S_1$ and $S_2$ are processed independently in the same way as set $S$ (i.e. procedures DP($S_1$) and DP($S_2$) are recursively called). Since the callings of DP($S_1$) and DP($S_2$) are independent with each other, they can be performed in parallel. Thus, we have the following recursive formula, for calculating the sequentiality of Algorithm DP on a contaminated input set $S$ with $|S|>1$

$$\begin{aligned} Seq(\text{ DP }, S)= \left\{ \begin{array}{ll} 1 + T(\text{ DIG }, S),&{} \text{ if } \; S'=\emptyset ; \\ 1 + T(\text{ DIG }, S)+1=2 + T(\text{ DIG }, S),&{} \text{ if } \; |S'|=1; \\ 1 + T(\text{ DIG }, S)+\max \big \{Seq(\text{ DP }, S_1), \,Seq(\text{ DP }, S_2)\big \},&{} \text{ if } \; |S'|>1. \end{array} \right. \end{aligned}$$

(5)

For any input set $S, Seq(\text{ DP }, S)$ can be calculated by using the recursive formula in Equ. (5), together with the formulae given above for the base cases where $|S|=1$ or $S$ is pure.

Appendix 2: Calculating the sequentiality of Algorithm RBS

The sequentiality of Algorithm RBS can be analyzed in a similar way as Algorithm DP. For an input set $S$ of items and a status indicator $c\in \{1,*\}$ on $S$, we use $Seq(\text{ RBS }, S, c)$ to denote the sequentiality of Algorithm RBS on input $(S,c)$. Trivially, if $S=\emptyset$, we let $Seq(\text{ RBS }, S, c)=0$ for any $c\in \{1,*\}$. If $|S|=1$, then clearly $Seq(\text{ RBS }, S, 1)=0$ and $Seq(\text{ RBS }, S, *)=1$. For inputs $|S|>1$ and $c=*$, from Algorithm RBS we have

$$\begin{aligned} Seq(\text{ RBS }, S, *)= \left\{ \begin{array}{ll} 1,&{} \text{ if } \text{ the } \text{ test } \text{ result } \text{ of } S\, \text{(in } \text{ Step } \text{2) } \text{ is } \text{ pure }; \\ 1 +Seq(\text{ RBS }, S, 1),&{} \text{ otherwise. } \end{array} \right. \end{aligned}$$

(6)

For inputs $|S|>1$ and $c=1$, from Algorithm RBS, $S$ is partitioned into two subsets $S_1$ and $S_2$. We have the following three possibilities on the test results of $S_1$ and $S_2$ in Step 4:

Case 1. The test result is pure for $S_1$, and is contaminated for $S_2$. In this case, according to Step 4(a), Procedure DIG is applied to $S_2$, to identify and remove one defective item and an unspecified number of good items from $S_2$. We use $S_2'$ to denote the resulting subset obtained from $S_2$ after the removal, then RBS$(S_2',*)$ is recursively called after Procedure DIG.

Case 2. The test result is contaminated for $S_1$, and is pure for $S_2$. Similarly, according to Step 4(b), in this case Procedure DIG is applied to $S_1$, to identify and remove one defective item and an unspecified number of good items from $S_1$. We use $S_1'$ to denote the resulting subset obtained from $S_1$ after the removal, then RBS$(S_1',*)$ is recursively called after Procedure DIG.

Case 3. The test results are contaminated for both $S_1$ and $S_2$. Then, according to Step 4(c), RBS$(S_1,1)$ and RBS$(S_2,1)$ are recursively called. The callings of RBS$(S_1,1)$ and RBS$(S_2,1)$ are independent with each other, and so they can be performed in parallel.

From the above three cases, and since the two tests on $S_1$ and $S_2$ in Step 4 can be performed in parallel, we have the following recursive formula for calculating the sequentiality of Algorithm RBS with inputs $|S|>1$ and $c=1$

$$\begin{aligned} Seq(\text{ RBS }, S, 1)= \left\{ \begin{array}{ll} 1+ T(\text{ DIG }, S_2)+Seq(\text{ RBS }, S_2', *),&{} \text{ if }\; S_1\; \hbox {is pure and}\; S_2\; \hbox {is} \\ &{} \text{ contaminated; } \\ 1+ T(\text{ DIG }, S_1)+Seq(\text{ RBS }, S_1', *),&{} \text{ if }\; S_1\; \hbox {is contaminated} \\ &{} \text{ and }\; S_2\; \hbox {is pure;} \\ 1+ \max \left\{ \begin{array}{l} Seq(\text{ RBS },\; S_1, 1) \\ Seq(\text{ RBS }, \;S_2, 1) \end{array} \right\} ,&{} \begin{array}{l} \text{ if } \text{ both } \;S_1\; \hbox {and}\; S_2\; \hbox {are} \\ \text{ contaminated. } \end{array} \end{array} \right. \end{aligned}$$

(7)

By combining the above, for any inputs $S$ and $c\in \{1,*\}, Seq(\text{ RBS }, S, c)$ can be calculated by using the recursive formulae in Eqs. (6) and (7), together with the formulae for the base cases $|S|\le 1$ given above.