WHEN IS INDIVIDUAL TESTING OPTIMAL FOR
NONADAPTIVE GROUP TESTING?∗
S. H. HUANG† AND F. K. HWANG†
Abstract. The combinatorial group testing problem is, assuming the existence of up to d defectives among n items, to identify the defectives by as few tests as possible. In this paper, we study the problem for what values of n, given d, individual testing is optimal in nonadaptive group testing. Let N(d) denote the largest n for fixed d for which individual testing is optimal. We will show that N(d) = (d + 1)2 under a prevalent constraint in practical nonadaptive algorithms and prove that N(d) = (d + 1)2for d = 1, 2, 3, 4 without any constraint.
Key words. nonadaptive group testing, disjunct matrix, union-free matrix AMS subject classifications. 05D05, 05B20
PII. S0895480199359247
1. Introduction. In combinatorial group testing, a prototype problem called
the ( ¯d, n) problem is to assume that there are up to d defectives among n given items,
and the problem is to separate the good items from the defective ones by group tests. A (group) test is administered on an arbitrary subset S of the items with two possible outcomes; a negative outcome means S contains no defectives and a positive outcome means S contains at least one defective, not knowing exactly how many or which ones. A group testing algorithm is optimal if it minimizes the worst-case number of tests required.
A group testing algorithm is sequential if the tests can be done sequentially and the outcomes of previous tests are known at the time to determine the current test. A group testing algorithm is nonadaptive if all tests must be specified at once. A nonadaptive algorithm can be represented by a 0-1 matrix where columns are items, rows are tests, and a 1-entry in cell (i, j) means item j is contained in test i. Note that a column can be viewed as a subset whose elements are indices of the rows incident to the column. Thus we can talk about the union of columns. Group testing has applications to blood testing, electrical and chemical testing, coding, multiaccess channel conflict resolution, etc. Recently, nonadaptive group testing has been shown to play a crucial role in the clone library screening problem.
Kautzand Singleton [8] introduced the notions of “ ¯d-separable” and “d-disjunct”
of 0-1 matrix Mt×n; the former requires that no two unions of up to d columns are
identical, while the latter requires that no union of d columns contains a column not in the union. They showed that both properties guarantee Mt×nto be a nonadaptive
( ¯d, n) algorithm, while the d-disjunct property has an extra feature of simplifying the
process of identifying defectives. These two properties were also called r-union-free and r-cover-free [3, 4] in extremal set theory.
A trivial d-disjunct algorithm not using the idea of group testing would test the
n items individually, which requires n tests. Thus it is of interest to know for what
values of n, given d, individual testing is optimal.
∗Received by the editors July 6, 1999; accepted for publication (in revised form) June 7, 2001;
published electronically October 23, 2001.
http://www.siam.org/journals/sidma/14-4/35924.html
†Department of Applied Mathematics, National Chiao-Tung University, Hsinchu 30050, Taiwan,
Republic of China (u8622522@math.nctu.edu.tw, fhwang@math.nctu.edu.tw). This research was partially supported by Republic of China National Science Council grant NSC 90-2115-M-009-007.
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A similar question as posed in the title has been asked on sequential group testing for exactly d defectives. (Thus the state of the last item can be deduced without testing.) Hu, Hwang, and Wang [5] conjectured that individual testing is optimal if and only if n ≤ 3d. They proved the “necessary” part, but the sufficient condition is proved only for n ≤ 2.5d, improving an earlier sufficient condition n ≤ 2d of Hwang [7]. Du and Hwang [1] further improved the sufficient condition to n ≤ 2.625d, but the conjecture remains open.
Back to the nonadaptive case, let N(d) denote the largest n for fixed d for which individual testing is optimal. Bassalygo (see [2]) first gave a lower bound.
Lemma 1.1. N(d) ≥d+22 .
Erd¨os, Frankl, and F¨uredi [3] conjectured that
lim N(d)/d2= 1 (weaker version),
N(d) ≤ (d + 1)2 (stronger version)
and stated without giving details that they can prove the stronger version for d ≤ 3. In this paper, we will prove N(d) ≤ (d + 1)2for d + 1 a prime power. Thus
d + 2
2
≤ N(d) ≤ (d + 1)2.
This establishes N(d) = O(d2), as opposed to N(d) = O(d) in the sequential case. We
will also show that under a prevalent constraint in practical nonadaptive algorithms,
N(d) = (d + 1)2. Finally, we prove N(d) = (d + 1)2 for d=1, 2, 3, 4 without any
constraint.
2. A necessary condition. Let t(d, n) denote the minimum number of tests a
nonadaptive algorithm requires, given d and n. We first make an observation. Lemma 2.1. The existence of a d-disjunct matrix Mt×n with t < n implies
N(d) ≤ t.
Proof. t(d, n) is clearly nondecreasing in n. Hence, the existence of a d-disjunct Mt×n with n > t implies t + 1 items can be done in t tests, or, equivalently, N(d) ≤
t.
Let λcc denote the inner product of two columns c and c. Define ¯λ = maxλcc over all pairs of columns. A 0-1 matrix is called a weight-w matrix if each column is a w-set. The following lemma is well known [10].
Lemma 2.2. A weight-w matrix is (w/¯λ − 1)-disjunct.
Corollary 2.3. If ¯λ = 1, then a weight-w matrix is (w − 1)-disjunct. The main result in this section is the following theorem.
Theorem 2.4. N(d) ≤ (d + 1)2 for d + 1, a prime power.
Proof. By Lemma 2.1 and Corollary 2.3, it suffices to construct a weight-(d + 1)
matrix M(d+1)2×nwith ¯λ = 1 and (d + 1)2< n.
It is well known [9] that if d + 1 is a prime power, then there exists a set of
d mutually orthogonal latin squares (MOLS). Each such square will generate d + 1
columns (as (d + 1)-subsets of {0, 1, 2, . . . , (d + 1)2− 1}) of M, one for each set of cells
(i, j) having the same entry k, 0 ≤ k ≤ d, and cell (i, j) is translated to the number
i(d + 1) + j. Clearly, columns generated from the same square have λcc = 0. Due to orthogonality, columns generated from different latin squares satisfy λcc = 1. Thus the d MOLS generate a total of (d + 1)d columns of M with ¯λ = 1. Finally, consider the (d + 1) × (d + 1) matrix S, where the entry in cell (i, j) is just (i, j). Clearly, the columns (rows) of S have λcc = 0, and each row-column pair has λrc = 1.
Furthermore, the set of cells having the same entry from a latin square must be a
transversal; i.e., they have distinct row indices and distinct column indices. Let s be
a transversal, r a row, and c a column of S. Then λsr= λsc= 1. Thus we can add the
2(d + 1) rows and columns of S to be columns of M and preserve the ¯λ = 1 property. The total number of columns in M is now
(d + 1)d + 2(d + 1) = (d + 1)(d + 2).
Note that the base set of the columns is the set {0, 1, . . . , (d + 1)2− 1}. By treating
the base set as the set of tests, then we have constructed a (d + 1)2× (d + 1)(d + 2)
matrix with ¯λ = 1.
3. A necessary and sufficient condition under ¯λ = 1. Constructing
ef-ficient d-disjunct matrices is a difficult task, with the simplest and most prevalent method being given by Corollary 2.3, i.e., constructing matrices with ¯λ = 1. In this section, we study the problem given in the title of this paper under the constraint ¯λ = 1.
Let Mt×n be a 0-1 matrix. Let G(M) denote the graph with the rows of M as
vertices and an edge between two vertices if and only if the inner product of the two corresponding rows (viewed as subsets of {1, 2, . . . , n}) is not zero. Let dG(v) denote
the degree of v in G.
Lemma 3.1. Suppose M is weight-(d + 1) and ¯λ = 1. Then G(M) consists of n
edge-disjoint Kd+1 (complete graph of order d + 1).
Proof. Each column generates a Kd+1. The ¯λ = 1 property forces the Kd+1’s to
be edge disjoint.
Recently, W. T. Huang and Hwang observed (private communication) that a previous result of Weideman and Raghavarao [12] for the ¯d-separable matrix can be
extended to the following lemma.
Lemma 3.2. Any d-disjunct matrix with ¯λ = 1 can be reduced to one where no
column weight exceeds d + 1 with the d-disjunct property preserved.
Let n(d, t) denote the largest n such that there exists a d-disjunct Mt×n. A
column in M is called isolated if there exists a row incident to this column only. It is easily observed in the following lemma.
Lemma 3.3. Let Mt×n be a d-disjunct matrix containing an isolated column.
Then
n ≤ 1 + n(d, t − 1).
Proof. Deleting the isolated column and its incident row does not affect the d-disjunct property.
Note that to determine N(d), we need to find an Mt×n satisfying t < n and
minimizing n. A matrix with no nonisolated column always satisfies t ≥ n, and hence cannot be such a candidate, and consequently is of no interest. Therefore, we assume from now on that we consider only matrices with an isolated column.
Dyachkov and Rykov [2] proved the following lemma.
Lemma 3.4. Let M be a d-disjunct matrix. Then a column with weight at most
d must be isolated.
Lemma 3.5. Let M be a d-disjunct matrix with ¯λ = 1 and no column weight less
than d + 1. Then n ≤ t(t−1−r)(d+1)d , where t − 1 ≡ r (mod d).
Proof. By Lemmas 3.2 and 3.4, we may obtain a constant weight-w matrix M∗
with w = d + 1 from M. Consider G(M∗). Each vertex has a maximum possible
degree of t − 1. By Lemma 3.1, d | dG(v). Hence, dG(v) ≤ (t − 1) − r. Thus the total
number of edges in G satisfies
n × d + 1 2 = ΣdG2(v) ≤t[(t − 1) − r]2 , or n ≤ (t(t − 1 − r) (d + 1)d .
We are now ready to prove the main result of this section.
Theorem 3.6. Under ¯λ = 1, N(d) = (d + 1)2 for d + 1, a prime power.
Proof. By Theorem 2.4, it suffices to prove N(d) ≥ (d + 1)2, which is done by
proving the nonexistence of a d-disjunct matrix M[(d+1)2−1]×(d+1)2 with ¯λ = 1.
Let Mt×nbe a d-disjunct matrix with ¯λ = 1 and t = (d + 1)2− 1. By Lemma 3.2,
we may assume that every column of M has weight at most d + 1.
Case (i). No column of M has weight less than d + 1. Since t − 1 = (d + 1)2− 2 =
(d + 1)d + d − 1, t − 1 − r = (d + 1)d. By Lemma 3.5,
n ≤t(d + 1)dd(d + 1) = t.
Case (ii). There exists a column of M with weight at most d. Let C be the set of
columns with weight at most d. By Lemma 3.4, each column c ∈ C is isolated; i.e., there exists a row r(c) incident only to c. Let M be obtained from M by deleting c
and {r(c) : c ∈ C}. Then M is a (t− | C |) × (n− | C |) weight-(d + 1) matrix with
¯λ = 1, since t− | C | −1 = d(d + 1) + d − 1− | C |, t − 1− | C | −r ≤ d(d + 1). By Lemma 3.5, (n− | C |) ≤ (t− | C |)(d + 1)d d(d + 1) ,
which again leads to n ≤ t.
4. N(d) for small d. Bassalygo (see [2]) proved the following lemma.
Lemma 4.1. Let M be a d-disjunct matrix and c a column of M with weight w.
Then t(d, n) ≥ w + t(d − 1, n − 1).
Spencer [11] proved the following lemma Lemma 4.2. n(1, t) = ( t
t
2).
Theorem 4.3. N(1) = 4.
Proof. By Theorem 2.4, N(1) ≤ 4. Since n(1, 3) = 3, N(1) ≥ 4. Hence, N(1) = 4.
Lemma 4.4. n(2, 7) = 7.
Proof. Let M7×nbe a 2-disjunct matrix. If ¯λ = 1, then Lemma 4.4 follows from
Theorem 3.6. Therefore, we may assume the existence of two columns, c and c, with
λcc > 1. Then c is either isolated or has weight at least 4. (i) c is isolated. By Lemmas 1.1 and 3.3,
n ≤ 1 + n(2, 6) = 1 + 6 = 7.
(ii) c has weight at least 4. Deleting c and its incident rows, the reduced matrix
M can have at most three rows and is 1-disjunct. By Lemma 4.2, M can have at
most three columns. Then n ≤ 4. Theorem 4.5. N(2) = 9.
Proof. Since d + 1 = 3 is a prime power, by Theorem 2.4, N(2) ≤ 9. Therefore,
it suffices to prove N(2) ≥ 9, or n(2, 8) = 8.
Let M8×n be 2-disjunct. By Theorem 3.6, we may assume the existence of two
columns, c and c, with λcc > 1.
(i) c is isolated. Then by Lemmas 3.3 and 4.4
n ≤ 1 + n(2, 7) = 8.
(ii) c has weight at least 4. Then by Lemmas 4.1 and 4.2
t(2, 8) ≥ 4 + t(1, 7), or n(2, 8) ≤ 7.
Corollary 4.6. N(3) ≥ 13.
Proof. By Lemmas 3.4 and 4.1, t ≥ (d + 1) + t(d − 1, n − 1) for a d-disjunct
matrix Mt×n (with at least one nonisolated column), since n(2, 8) = 8, t(2, 12) ≥ 9.
Therefore, t ≥ 4 + t(2, n − 1) ≥ 4 + 9 = 13 for n ≥ 13. It implies that t(3, n) = n for
n ≤ 13, i.e., N(3) ≥ 13.
Let I(c) be a collection of different b-subsets, b ≥ 2, of {1, 2, 3, . . . , t} denoting the intersection property of column c with other columns. For example, I(c) =
{(1, 2), (1, 3, 4)} means that there exists at least one column c1intersecting c at rows
1 and 2, and there exists at least one column c2 intersecting c at rows 1, 3, 4.
Let M1 be a t × n1 weight-4 2-disjunct matrix with no isolated column. Then
¯λ ≤ 2. We will do some deletions on M1to reduce weight 4 to weight 3 such that the
2-disjunct property is still preserved in (M1)i, i ≥ 0, which is the reduced matrix after
the ith deletion. Note that deleting a 1-entry of c will affect other I(c) in general.
Therefore, after each deletion, we need to reconsider the I(c) of the reduced matrix, where c has weight 4.
Consider I(cj), j ≤ n1, of (M1)i, i ≥ 0, and I(cj) = ∅, where cj has weight 4. The
deletion rule is as follows:
(1) If I(cj) ⊆ {(xj, yj), (xj, zj), (xj, vj); xj = yj = zj = vj ∈ {1, 2, 3, . . . , t}} in
(M1)i, then delete the 1-entry in row xj of cj; hence the reduced column has weight 3
and has an inner product at most one with any other column of (M1)i. (M1)iremains
2-disjunct by Lemma 3.2.
(2) If I(cj) = {(xj, yj), (xj, zj), (yj, zj); xj = yj = zj ∈ {1, 2, 3, . . . , t}} in (M1)i,
then do nothing at this moment until the last step.
Finally, we will get a reduced matrix (M1)f with no case (1) after f deletions. If
case (2) does not occur, we are done. If case (2) occurs, then there exists a t × n 1
submatrix M
1contained in (M1)f which has the following four properties:
(1) M
1has at least four columns and each column of M1 has weight 4.
(2) ∀cj ∈ C(M1), I(cj) has the form {(xj, yj), (xj, zj), (yj, zj); xj = yj = zj ∈
{1, 2, 3, . . . , t}}.
(3) cj∈ M1 and |cicj| = 2 =⇒ ci∈ M1.
(4) M
1 doesn’t contain a submatrix with fewer columns having properties (1),
(2), and (3). Let M
1 be a q × n1 submatrix of M1, where the q rows of M1are the collection
of rows {xi, yi, zi : ci∈ C(M), I(ci) = {(xi, yi), (xi, zi), (yi, zi)}}. Then we have the
following lemma.
Lemma 4.7. n 1≥ q.
Proof. Suppose c ∈ M
1 and I(c) = {(x, y), (x, z), (y, z)}. Then each row x, y, z
must have at least three 1-entries in M
1. For example, x appears once in c, once in a
column intersecting c at (x, y), and once in a column intersecting c at (x, z). Suppose
c has a fourth 1-entry v also in M
1. Then v ∈ {x, y, z} for some column c with
I(c) = {(x, y), (x, z), (y, z)}. Therefore, row v has at least four 1-entries. Let k
be the number of columns with four 1-entries in M. Then counting the number of
1-entries by column and by row separately, we have 3n
1+ k ≥ 3q + k, or n1≥ q.
Lemma 4.8. Suppose c = {x, y, z, v} and I(c) = {(x, y), (x, z), (y, z)} ∀c ∈
C(M
1). Then c
{x, y, z} = ∅ implies v /∈ c in a 2-disjunct matrix.
Proof. Without loss of generality, assume that c contains x. Suppose to the
contrary that c also contains v. By the definition of I(c), there exists a column c
containing (y, z). Then cccontains c, contradicting the 2-disjunct.
Lemma 4.9. n(2, 9) = 12.
Proof. The proof uses the Steiner triple system with v = 9 and b = 12 [9].
Lemma 4.10. n(2, 10) ≤ 13.
Proof. Let M10×nbe a 2-disjunct matrix. If M has an isolated column, then by
Lemmas 3.3 and 4.9, n ≤ 1+ n(2, 9) = 1+12 = 13. If ¯λ = 1 and there exists no isolated column in M, then r ≡ 10 − 1 ≡ 1 (mod 2). By Lemma 3.5, n ≤ 10(10−1−1)3·2 = 13.
Therefore, we may assume M has no isolated column and ¯λ > 1. Let column c have maximum weight of M. Then c has weight at least 4.
(i) c has weight greater than 4. Then by Lemmas 4.1 and 4.2,
t(2, 14) ≥ 5 + t(1, 13) = 11, or n(2, 10) ≤ 13.
(ii) c has weight 4. Then M is 2-disjunct with ¯λ = 2, and each column has weight 3 or 4. Let M = M1M2, where M1 is a 10 × n1 weight-4 matrix, and M2 is a
10 × (n − n1) weight-3 matrix.
We first make an observation. Because M is 2-disjunct with ¯λ = 2 and has maximum weight 4, it forces both M1, M2to be 2-disjunct with M1having ¯λ = 2, M2
having ¯λ = 1, and ∀c1∈ M1, c2∈ M2, λc1c2≤ 1.
Next we want to claim that M1with ¯λ = 2 can be reduced to a matrix with ¯λ = 1
by deleting some 1-entries and remains 2-disjunct.
By the method we used previously, we get matrix (M1)f, M1, M1from (M1)f. If
M
1, hence M1, does not exist, then we are done. Otherwise, by relabeling the rows,
we may assume that c1= {1, 2, 3, 4}. The completion of I(c1) = {(1, 2), (1, 3), (2, 3)}
implies the following submatrices must be in M
1 (see Figure 1):
Note that the completion of I(c2), I(c3), I(c4) requires at least one more row. In
fact, if q = 4, then (b) shows the only way to complete I(c2), I(c3), and I(c4). This
case can be taken care of by deleting the circled 1’s. It is easily verified that any two columns intersect at most once after the deletion. Hence, ¯λ = 1 in M
1. In other
words, M can be reduced to a matrix which remains 2-disjunct with ¯λ = 1 and has column weight 3 or 4. Hence, by Lemma 3.5, t − 1 − r = 8, n ≤ 10×8
3×2 = 13.
If q ≥ 5, then there are other ways to complete I(c2), I(c3), and I(c4). By
Lemma 4.7, n
1≥ q ≥ 5. Let c5 be a new column in M1. Since M1 is minimal, I(c5)
must intersect {(1, 2), (1, 3), (2, 3)}. Without loss of generality, assume (1, 2) ∈ I(c5).
Since c5 intersects all c1, c2, c3, and c4, by Lemma 4.8, c5 cannot intersect the
fourth 1-entries of these columns. Therefore, the other two 1-entries of c5 must take
up new rows, say, rows 8 and 9 (see Figure 2(a)).
C1 C2 C3 C4 C1 C2 C3 C4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 (a) (b)
Fig. 1. Forced submatrices in M
1. c1 c2 c3 c4 c5 c1 c2 c3 c4 c5 c6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 (a) (b)
Fig. 2. Larger forced submatrices.
Inspecting the relation between c5and the other ci, we notice its first two 1-entries
are symmetric, and so are its last two. Therefore, we may assume (1, 8) ∈ I(c5). To
complete I(c5), there must exist c6 containing rows 1 and 8. Since c6 intersects
all other columns except c4, among the existing rows, it can contain only row 7.
Therefore, the fourth 1-entry of c6 must be a new row, say, row 10. However, we
still need at least one more row to complete I(c2), I(c3), and I(c4). Hence, the total
number of rows exceed 10. Lemma 4.11. n(3, 13) = 13.
Proof. Let M13×n be a 3-disjunct matrix. If ¯λ = 1, then Lemma 4.11 follows
from Theorem 3.6. Therefore, we may assume the existence of two columns c and c
with λcc > 1. Then c is either isolated or has weight at least 5. (i) c is isolated. By Lemma 3.3 and Corollary 4.6,
n ≤ 1 + n(3, 12) = 1 + 12 = 13.
(ii) c has weight at least 5. Deleting c and its incident rows, the reduced matrix
M can have at most 8 rows and is 2-disjunct. By n(2, 8) = 8, then n ≤ 9.
Lemma 4.12. n(3, 14) = 14.
Proof. Let M14×n be a 3-disjunct matrix. If ¯λ = 1, then Lemma 4.12 follows
from Theorem 3.6. Therefore, we may assume the existence of two columns c and c
with λcc > 1. Then c either is isolated or has weight at least 5.
(i) c is isolated. By Lemmas 3.3 and 4.11,
n ≤ 1 + n(3, 13) = 1 + 13 = 14.
(ii) c has weight at least 5. Deleting c and the incident rows, the reduced matrix
M can have at most 9 rows and is 2-disjunct. By Lemma 4.9,
n ≤ 1 + n(2, 9) = 13.
Theorem 4.13. N(3) = 16.
Proof. Since d + 1 = 4 is a prime power, by Theorem 2.4, N(3) ≤ 16. Therefore,
it suffices to prove N(3) ≥ 16 or n(3, 15) = 15. Let M15×n be a 3-disjunct matrix.
By Theorem 3.6, we may assume the existence of two columns c and c with λ cc > 1. (i) c is isolated. Then by Lemmas 3.3 and 4.12
n ≤ 1 + n(3, 14) = 15.
(ii) c has weight at least 5. Then by Lemmas 4.1 and 4.10
n ≤ 1 + n(2, 10) ≤ 1 + 13 = 14.
With similar but slightly more complicated arguments, we can also prove N(4) = 25 [6].
5. Conclusion. In this paper, we studied the problem of when individual testing
is optimal for nonadaptive group testing. We showed that N(d) ≤ (d + 1)2 is a
necessary condition by constructing the d-disjunct matrix M[(d+1)2−1]×(d+1)2, where d + 1 is a prime power. Besides, we showed that N(d) ≥ (d + 1)2 is a sufficient
condition under ¯λ = 1. Hence, under ¯λ = 1 and d + 1 a prime power, N(d) = (d + 1)2
is a necessary and sufficient condition of the problem we studied. We also prove that
N(d) = (d + 1)2 for d = 1, 2, 3, 4 without any constraint, giving further support to the
conjecture N(d) = (d + 1)2 for d + 1, a prime power. However, for d + 1, not a prime
power, we still know little about N(d).
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