Difference between revisions of "Open Problems:95"

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In (non-adaptive) quantitative group testing, one has a population of $n$ individuals, among which $k=n^c$ (for some constant $c\in(0,1)$) are sick. The goal is to identity the $k$ sick individuals by performing $m$ non-adaptive tests. In each test, one specifies a subset $S\subseteq [n]$ and learns whether $S$ contains one or more sick individuals.
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In (non-adaptive) quantitative group testing, one has a population of $n$ individuals, among which $k=n^c$ (for some constant $c\in(0,1)$) are sick. The goal is to identify the $k$ sick individuals by performing $m$ non-adaptive tests. In each test, one specifies a subset $S\subseteq [n]$ and learns whether $S$ contains one or more sick individuals.
  
 
By a counting argument, one gets a lower bound of $m = \Omega\bigl( \frac{k}{\log k}\log \frac{n}{k} \bigr)$ tests; however, the best known upper bound is $m = O( k\log \frac{n}{k} )$.  
 
By a counting argument, one gets a lower bound of $m = \Omega\bigl( \frac{k}{\log k}\log \frac{n}{k} \bigr)$ tests; however, the best known upper bound is $m = O( k\log \frac{n}{k} )$.  
  
 
'''Question:''' Can one get rid of the $\log k$ factor in the lower bound; or, conversely, improve the upper bound to match it?
 
'''Question:''' Can one get rid of the $\log k$ factor in the lower bound; or, conversely, improve the upper bound to match it?

Latest revision as of 20:40, 26 August 2019

Suggested by Oliver Gebhard
Source WOLA 2019
Short link https://sublinear.info/95

In (non-adaptive) quantitative group testing, one has a population of $n$ individuals, among which $k=n^c$ (for some constant $c\in(0,1)$) are sick. The goal is to identify the $k$ sick individuals by performing $m$ non-adaptive tests. In each test, one specifies a subset $S\subseteq [n]$ and learns whether $S$ contains one or more sick individuals.

By a counting argument, one gets a lower bound of $m = \Omega\bigl( \frac{k}{\log k}\log \frac{n}{k} \bigr)$ tests; however, the best known upper bound is $m = O( k\log \frac{n}{k} )$.

Question: Can one get rid of the $\log k$ factor in the lower bound; or, conversely, improve the upper bound to match it?