User:Ahughes6/Example/Testing

d-separable

Definition: A $t$ x $n$ matrix $M$ is $d$-separable if and only if $\forall S_{1}\neq S_{2}\subseteq [n]$ where $|S_{1}|,|S_{2}|\leq d$ such that $\bigcup _{j\in S_{1}}M_{j}\neq \bigcup _{i\in S_{2}}M_{i}$

Decoding algorithm

First we will describe another way to look at the problem of group testing and how to decode it from a different notation. We can give a new interpretation of how group testing works as follows:

Group Testing: Given input $M$ and $\mathbf {r}$ such that $\mathbf {r} =M\mathbf {x}$ output $\mathbf {x}$

• Take $M_{j}$ to be the $j^{th}$ column of $M$
• Define $S_{M_{j}}\subseteq [t]$ so that $M_{j}(i)=1$ if and only if $i\in S_{M_{j}}$
• This gives that $S_{\mathbf {r} }=\bigcup _{j\in [n],\mathbf {x} _{j}=1}S_{M_{j}}$

This formalizes the relation between $\mathbf {x}$ and the columns of $M$ and $\mathbf {r}$ in a way more suitable to the thinking of $d$-separable and $d$-disjunct matrices. The algorithm to decode a $d$-separable matrix is as follows:

Given a $t$ x $n$ matrix $M$ such that $M$ is $d$-separable:

1. For each $T\subseteq [n]$ such that $|T|\leq d$ check if $S_{\mathbf {r} }=\bigcup _{j\in T}S_{M_{j}}$

This algorithm runs in time $n^{{\mathcal {O}}(d)}$.

d-disjunct

In literature disjunct matrices are also called super-imposed codes and d-cover-free families.

Definition: A $t$ x $n$ matrix $M$ is d-disjunct if $\forall S\subseteq [n]$ such that $|S|\leq d$, $\forall j\notin S$ $\exists i$ such that $M_{i,j}=1$ but $\forall k\in S,M_{i,k}=0$. Denoting $M_{a}$ is the $a^{th}$ column of $M$ and $S_{M_{a}}\subseteq [t]$ where $M_{a}(b)=1$ if and only if $b\in S_{M_{a}}$ gives that $M$ is $d$-disjunct if and only if $S_{M_{j}}\subsetneq \cup _{k\in S}S_{M_{k}}$

Claim: $M$ is $d$-disjunct implies $M$ is $d$-separable

Proof: (by contradiction) Let $M$ be a $t$ x $n$ $d$-disjunct matrix. Assume for contradiction that $M$ is not $d$-separable. Then there exists $T_{1},T_{2}\in [n]$ and $T_{1}\neq T_{2}$ with $|T_{1}|,|T_{2}|\leq d$ such that $\bigcup _{i\in T_{1}}M_{i}=\cup _{i\in T_{2}}S_{M_{i}}$. This implies that $\exists j\in T_{2}\setminus T_{1}$ such that $S_{M_{j}}\subseteq \bigcup _{k\in T_{1}}T_{M_{k}}$. This contradicts the fact that $M$ is $d$-disjunct. Therefore $M$ is $d$-separable. $\Box$

Decoding Algorithm

The algorithm for $d$-separable matrices was still a polynomial in $n$. The following will give a nicer algorithm for $d$-disjunct matrices which will be a $d$ multiple instead of raised to the power of $d$ given our bounds for $t$. The algorithm is as follows in the proof of the following lemma:

Lemma 1: There exists an ${\mathcal {O}}(nt)$ time decoding for any $d$-disjunct $t$ x $n$ matrix.

• Observation 1: For any matrix $M$ and given $M\mathbf {x} =\mathbf {r}$ if $\mathbf {r} _{i}=1$ it implies $\exists j$ such that $M_{i,j}=1$ and $\mathbf {x} _{j}=1$ where $1\leq i\leq t$ and $1\leq j\leq n$. The opposite is also true. If $\mathbf {r} _{i}=0$ it implies $\forall j$ if $M_{i,j}=1$ then $\mathbf {x} _{j}=0$. This is the case because $\mathbf {r}$ is generated by taking all of the logical or of the $\mathbf {x} _{j}$'s where $M_{i,j}=1$.
• Observation 2: For any $d$-disjunct matrix and every set $T=\{j|\mathbf {x} _{j}=1\}$ where $|T|\leq d$ and for each $j\notin T$ where $1\leq j\leq n$ there exists some $i$ where $1\leq i\leq t$ such that $M_{i,j}=1$ but $M_{i,l}=0{\text{ }}\forall l\in T$. Thus, if $\mathbf {r} _{i}=0$ then $\mathbf {x} _{j}=0$.

Proof of Lemma 1: Given as input $\mathbf {r} \in \{0,1\}^{t},M$ use the following algorithm:

1. For each $j\in [n]$ set $\mathbf {x} _{j}=1$
2. For $i=1\ldots t$, if $\mathbf {r} _{i}=0$ then for all $j\in [n]$, if $M_{i,j}=1$ set $\mathbf {x} _{j}=0$

By Observation 1 we get that any position where $mathbf{r}_{i}=0$ the appropriate $\mathbf {x} _{j}$'s will be set to 0 by step 2 of the algorithm. By Observation 2 we have that there is at least one $i$ such that if $\mathbf {x} _{j}$ is supposed to be 1 then $M_{i,j}=1$ and, if $\mathbf {x} _{j}$ is supposed to be 1, it can only be the case that $\mathbf {r} _{i}=1$ as well. Therefore step 2 will never assign $\mathbf {x} _{j}$ the value 0 leaving it as a 1 and solving for $\mathbf {x}$. This takes time ${\mathcal {O}}(nt)$ overall. $\Box$

Upper Bounds for Non-Adaptive Group Testing

The results for these upper bounds rely mostly on the properties of $d$-disjunct matrices. Not only are the upper bounds nice, but from Lemma 1 we know that there is also a nice decoding algorithm for these bounds. First the following lemma will be proved since it is relied upon for both constructions:

Lemma 2: Given $1\leq d\leq n$ let $M$ be a $t$ x $n$ matrix and:

1. $\forall j\in [n]{\text{, }}|S_{M_{j}}|\geq w_{min}$
2. $\forall i\neq j\in [n]{\text{, }}|S_{M_{i}}\cap S_{M_{j}}|\leq a_{max}$

for some integers $a_{max}\leq w_{min}\leq t$ then $M$ is $\geq d'\lfloor {\frac {w_{min}-1}{a_{max}}}\rfloor$-disjunct.

Note: these conditions are stronger than simply having a subset of size $d$ but rather applies to any pair of columns in a matrix. Therefore no matter what column $i$ that is chosen in the matrix, that column will contain at least $w_{min}$ 1's and the total number of shared 1's by any two columns is $a_{max}$.

Proof of Lemma 2: Fix an arbitrary $S\subseteq [n],|S|\leq d,j\notin S$ and a matrix $M$. There exists a match between $i\in S{\text{ and }}j\notin S$ if column $i$ has a 1 in the same row position as in column $j$. Then the total number of matches is $\leq a_{max}\cdot d\leq a_{max}\cdot ({\frac {w_{min}-1}{a_{max}}})=w_{min}-1<{\text{ }}w_{min}$, i.e. a column $j$ has a fewer number of matches than the number of ones in it. Therefore there must be a row with all 0s in $S$ but a 1 in $j$. $\Box$

We will now generate constructions for the bounds.

Randomized Construction

This first construction will use a probabilistic argument to show the property wanted, in particular the Chernoff bound. Using this randomized construction gives that $t(d,n)\leq {\mathcal {O}}(d^{2}\log {n})$. The following lemma will give the result needed.

Theorem 1: There exists a random $d$-disjunct matrix with ${\mathcal {O}}(d^{2}\log {n})$ rows.

Proof of Theorem 1: Begin by building a random $t$ x $n$ matrix $M$ with $t=cd^{2}\log {n}$ (where $c$ will be picked later). It will be shown that $M$ is $\Omega (d)$-disjunct. First note that $M_{i,j}\in \{0,1\}$ and let $M_{i,j}=1$ independently with probability ${\frac {1}{d}}$ for $i\in [t]$ and $j\in [n]$. Now fix $j\in [n]$. Denote the $j^{th}$ column of $M$ as $T_{j}\subseteq [t]$. Then the expectancy is $\mathbb {E} [|T_{j}|]={\frac {t}{d}}$. Using the Chernoff bound, with $\mu ={\frac {1}{2}}$, gives $\mathrm {Pr} [|T_{j}|<{\frac {t}{2d}}]\leq e^{\frac {-t}{12d}}=e^{\frac {-cd\log {n}}{12}}\leq n^{-2d}[$if $c\geq 24]$. Taking the union bound over all columns gives $\mathrm {Pr} [\exists j$, $|T_{j}|<{\frac {t}{2d}}]\leq n\cdot n^{-2d}\leq n^{-d}$. This gives $\mathrm {Pr} [\forall j$, $|T_{j}|\geq {\frac {t}{2d}}]\geq 1-n^{-d}$. Therefore $w_{min}\geq {\frac {t}{2d}}$ with probability $\geq 1-n^{-d}$.

Now suppose $j\neq k\in [n]$ and $i\in [t]$ then $\mathrm {Pr} [M_{i,j}=M_{i,k}=1]={\frac {1}{d^{2}}}$. So $\mathbb {E} [|T_{j}\cap T_{k}|]={\frac {t}{d^{2}}}$. Using the Chernoff bound on this gives $\mathrm {Pr} [|T_{j}\cap T_{k}|<{\frac {2t}{d^{2}}}]\leq e^{\frac {-t}{3d^{2}}}=e^{-2\log {n}}\leq n^{-4}[$if $c\geq 12]$. By the union bound over $(j,k)$ pairs $\mathrm {Pr} [\exists (j,k)$ such that $|T_{j}\cap T_{k}|<{\frac {2t}{d^{2}}}]\leq n^{2}\cdot n^{-4}=n^{-2}$. This gives that $a_{max}\leq {\frac {2t}{d^{2}}}$ and $w_{min}\geq {\frac {t}{2d}}$ with probability $\geq 1-n^{-d}-n^{-2}\geq 1-{\frac {1}{n}}$. Note that by changing $c$ the probability $1-{\frac {1}{n}}$ can be made to be $1-{\frac {1}{poly(n)}}$. Thus $d'=\lfloor {\frac {{\frac {t}{2d}}-1}{\frac {2t}{d^{2}}}}\rfloor \approx {\frac {d}{4}}$. By setting $d$ to be $4d$, the above argument shows that $M$ is $d$-disjunct.

Note that in this proof $t=d^{2}\log {n}$ thus giving the upper bound of $t(d,n)\leq {\mathcal {O}}(d^{2}\log {n})$. $\Box$

Strongly Explicit Construction

It is possible to prove a bound of $t(d,n)\leq {\mathcal {O}}(d^{2}\log ^{2}{n})$ using a strongly explicit code. Although this bound is worse by a $\log {n}$ factor it is preferable because this produces a strongly explicit construction instead of a randomized one.

Theorem 2: There exists a strongly explicit $d$-disjunct matrix with ${\mathcal {O}}(d^{2}\log ^{2}{n})$ rows.

This proof will use the properties of concatenated codes along with the properties of disjunct matrices to construct a code that will satisfy the bound we are after.

Proof of Theorem 2: Let $C\subseteq \{0,1\}^{t},|C|=n$ such that $C=\{\mathbf {c} _{1},\ldots ,\mathbf {c} _{n}\}$. Denote $M_{C}$ as the matrix with its $i^{th}$ column being $\mathbf {c} _{i}$. If $C^{*}$ can be found such that

1. $\forall i\in C^{*}{\text{, }}|\mathbf {c} _{i}|\geq w_{min}$
2. $\forall \mathbf {c} ^{1}\neq \mathbf {c} ^{2}\in C^{*}{\text{, }}|\{i|\mathbf {c} _{i}^{1}=\mathbf {c} _{i}^{2}=1\}|\leq a_{max}$,

then $M_{C^{*}}$ is $\lfloor {\frac {w_{min}-1}{a_{max}}}\rfloor$-disjunct. To complete the proof another concept must be introduced. This concept uses code concatenation to obtain the result we want.

Kautz-Singleton '64

Let $C^{*}=C_{out}\circ C_{in}$. Let $C_{out}$ be a $[q,k]_{q}$-Reed-Solomon code. Let $C_{in}=[q]\rightarrow \{0,1\}^{q}$ such that for $i\in [q]$, $c_{in}(i)=(0,\ldots ,0,1,0,\ldots ,0)$ where the 1 is in the $i^{th}$ position. Then $n=q^{k}$, $t=q^{2}$, and $w_{min}=q$.

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Example: Let $k=1,q=3,C_{out}=\{(0,0,0),(1,1,1),(2,2,2)\}$. Below, $M_{C}$ denotes the matrix of codewords for $C_{out}$ and $M_{C^{*}}$ denotes the matrix of codewords for $C^{*}=C_{out}\circ C_{in}$, where each column is a codeword. The overall image shows the transition from the outer code to the concatenated code.

File:Https://wiki.cse.buffalo.edu/cse545/files/26/example 0.png

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Divide the rows of $M_{C^{*}}$ into sets of size $q$ and number them as $(i,j)\in [q]{\text{ x }}[q]$ where $i$ indexes the set of rows and $j$ indexes the row in the set. If $M_{(i,j),k_{1}}=M_{(i,j),k_{2}}=1$ then note that $\mathbf {c} _{k_{1}}(i)=\mathbf {c} _{k_{2}}(i)=j$ where $\mathbf {c} _{k_{1}},\mathbf {c} _{k_{2}}\in C_{out}$. So that means $|M_{k_{1}}\cap M_{k_{2}}|=q-\Delta (\mathbf {c} _{k_{1}},\mathbf {c} _{k_{2}})$. Since $\Delta (\mathbf {c} _{k_{1}},\mathbf {c} _{k_{2}})\geq q-k+1$ it gives that $|M_{k_{1}}\cap M_{k_{2}}|\leq k-1$ so let $a_{max}=k-1$. Since $t=q^{2}$, the entries in each column of $M_{C^{*}}$ can be looked at as $q$ sets of $q$ entries where only one of the entries is nonzero (by definition of $C_{in}$) which gives a total of $q$ nonzero entries in each column. Therefore $w_{min}=q$ and $d=_{def}\lfloor {\frac {w_{min}-1}{a_{max}}}\rfloor$ (so $M_{C^{*}}$ is $d$-disjunct).

Now pick $q$ and $k$ such that $\lfloor {\frac {q-1}{k-1}}\rfloor =d$ (so $\lfloor {\frac {q}{k}}\rfloor \approx d$). Since $q^{k}=n$ we have $k={\frac {\log {n}}{\log {q}}}\leq \log {n}$. Since $q\approx kd$ and $t=q^{2}$ it gives that $t=q^{2}\approx (kd)^{2}\leq (d\log {n})^{2}$. $\Box$

Thus we have a strongly explicit construction for a code that can be used to form a group testing matrix and so $t(d,n)\leq (d\log {n})^{2}$.

For non-adaptive testing we have shown that $\Omega (d\log {n})\leq t(d,n)$ and we have that (i) $t(d,n)\leq {\mathcal {O}}(d^{2}\log ^{2}{n})$ (strongly explicit) and (ii) $t(d,n)\leq {\mathcal {O}}(d^{2}\log {n})$ (randomized). As of recent work by Porat and Rothscheld they presented a explicit method construction (i.e. deterministic time but not strongly explicit) for $t(d,n)\leq {\mathcal {O}}(d^{2}\log {n})$, however it is not shown here. There is also a lower bound for disjunct matrices of $t(d,n)\geq \Omega ({\frac {d^{2}}{\log {d}}}\log {n})$ which is not shown here either.

See Also

• Group Testing
• Code Concatenation

References

1. Atri Rudra's course on Error Correcting Codes: Combinatorics, Algorithms, and Applications (Spring 2010), Lectures 28, 29.