Difference between revisions of "Open Problems:95"

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|title=Non-Adaptive Group Testing
 
 
|source=wola19
 
|source=wola19
 
|who=Oliver Gebhard
 
|who=Oliver Gebhard
 
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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?