This paper is availiable online at http://www.math.nthu.edu.tw/tjm/
GROUP TESTING IN BIPARTITE GRAPHS¤
Su-Tzu Juan and Gerard J. Changy
Abstract. This paper investigates the group testing problem in graphs as
follows. Given a graph G = (V ; E), determine the minimum number t(G) such that t(G) tests are sufficient to identify an unknown edge e with each test specifies a subsetX µ V and answers whether the unknown edge e is in G[X] or not. Damaschke proved that dlog2e(G)e · t(G) · dlog2e(G)e + 1 for any graph G; where e(G) is the number of edges of G. While there are infinitely many complete graphs that attain the upper bound, it was conjectured by Chang and Hwang that the lower bound is attained by all bipartite graphs. This paper verifies the conjecture for bipartite graphs G with e(G) · 24 or 2k¡ 1< e(G) · 2k¡ 1+ 2k¡ 3+ 2k¡ 6+ 19 ¢2k¡ 72 ¡ 1 for k ¸ 5.
1. INTRODUCTION
The idea of group testing originated from the blood testing in 1942 by Dorfman, who published the first paper [8] on this topic. While traditional group testing literature employs probabilistic models, Li [12] was the first to study combinatorial group testing as follows. Consider a population V of n items consisting of an unknown subset D µ V of d defectives. The problem is to identify the set D by a sequence of group tests. Each test is on a subset X of V with two possible outcomes: a pure outcome indicates thatX \ D = ;, and a contaminated outcome indicates that X \ D 6= ;. The goal is to minimize the number M(d; n) of tests under the worst scenario. A best algorithm under this goal is called a minmax algorithm. For a good reference, see the book by Du and Hwang [9].
Received January 15, 2000, revised March 15, 2000. Communicated by F. K. Hwang.
2001 Mathematics Subject Classification: 05C99.
Key words and phrases: Group testing, algorithm, complete graph, bipartite graph, induced subgraph. ¤Supported in part by the National Science Council under grant NSC89-2115-M009-007.
yE-mail: gjchang@math.nctu.edu.tw.
As the sample space of the problem consists of¡nd¢ samples, we have the fol-lowing information-theoretic lower bound
M(d; n) ¸ dlog2 µ n d ¶ e;
where dxe (bxc) denotes the smallest (largest) integer not less (greater) than x. Using a bisection method, it is easy to get
M(1; n) = dlog2ne:
On the other hand, it is hard to determineM(d; n) for d ¸ 2. Even for the case of d = 2, we only know that
dlog2 µ n 2 ¶ e · M(2; n) · dlog2 µ n 2 ¶ e + 1:
Toward the study ofM(2; n), Chang and Hwang [4, 5] considered the problem of identifying two defectives in two disjoint setsA and B, each containing exactly one defective. At first, it seems that one cannot do better than working on the two disjoint sets separately. Surprisingly, a small example with jAj = 3 and jBj = 5 shows that 4 = dlog2(3 ¢5)e tests is enough rather than identifying the defectives inA and B separately, which takes dlog23e + dlog25e = 5 tests. In general, they [5] proved that the minmax number to identify the only defective inA and the only defective inB is
dlog2(mn)e;
where m = jAj and n = jBj. By associating each item to a vertex, Spencer [4] observed that the sample space of this problem can be represented by a bipartite graph where each edge represents a sample inA £ B. (Throughout this paper we presume that the reader is familiar with the basic-theoretic notations. See [3, 13] if necessary.) Chang and Hwang [4] conjectured that a bipartite graph with 2k (k ¸ 1) edges always has an induced subgraph with 2k¡ 1 edges, or equivalently, t(G) = dlog2e(G)e for any bipartite graph G. While the conjecture remains open,
it has stimulated forthcoming research casting group testing on graphs.
Aigner [1] proposed the following problem: Given a graphG = (V; E), deter-mine the minimum numbert(G) such that t(G) tests are sufficient in the worst case to identify an unknown edgee when each test specify a subset X µ V and answers whether the unknown edgee is in G[X] or not, where G[X] is the subgraph of G induced by the vertex set X. It is then clear that t(G) = 0 if G has exactly one edge, and otherwise
t(G) = 1 + min
The information-theoretic lower bound for this parameter is dlog2e(G)e · t(G);
where e(G) denotes the number of edges in G. Chang and Hwang’s result [5] becomes that
t(Km;n) = dlog2e(Km;n)e = dlog2(mn)e
for complete bipartite graphsKm;n, and their conjecture is
Conjecture 1 [4]. For any bipartite graphG; we have t(G) = dlog2e(G)e. From the result in [6], it follows thatt(Kn) · dlog2e(Kn)e + 1, and there are
infinitely many complete graphs attaining the upper bound. Alth¨ofer and Triesch [2] showed thatt(G) · dlog2e(G)e+1 for bipartite graphs, and t(G) · dlog2e(G)e+3 for arbitrary graphs. Damaschke [7] proved thatt(G) · dlog2e(G)e+1 for arbitrary graphs. In fact, he proved a more general result thatt(G) = dlog2e(G)e for a graph G with 2k¡ 1< e(G) · 2k¡ 1+171
64 ¢2
k¡ 1
2 whenk ¸ 13 and e(G) 2 [1; 14][[17; 25][
[33; 45] [ [65; 83] [ [129; 155] [[257; 295] [ [513; 568] [[1025; 1105] [[2049; 2165]. The attempt of this paper is to determine the largest number f(k) such that t(G) = dlog2e(G)e for any bipartite graph G with 2k¡ 1 < e(G) · f(k). Note that
Conjecture 1 says f(k) = 2kfor k ¸ 0. In this paper, we verify the conjecture for
k · 4, and show that f(k) ¸ 2k¡ 1+ 2k¡ 3+ 2k¡ 6+ 19 ¢2k¡ 7
2 ¡ 1 for k ¸ 5.
2. GRAPHSGWITHt(G) = dlog2e(G)e
It is of our interest to study which graphs G satisfy t(G) = dlog2e(G)e. The first well-known result of this kind is
Theorem 2 [5]. For any complete bipartite graphKm;n; we have
t(Km;n) = dlog2e(Km;n)e = dlog2(mn)e:
It is not hard to see that acyclic graphs also have this property. Theorem 3. For any acyclic graphG; we have t(G) = dlog2e(G)e.
Proof. Removing successively vertices of degree one, we can get induced sub-graphs ofG whose numbers of edges range from 1 to e(G). This together with the information-theoretic lower bound gives the theorem.
Theorem 4 [7]. For any graphG with 2k¡ 1< e(G) · 2k¡ 1+171 64 ¢2
k¡ 1
2 and
k ¸ 13; and e(G) 2 [1; 14] [ [17; 25] [ [33; 45] [ [65; 83] [ [129; 155] [ [257; 295] [ [513; 568] [ [1025; 1105] [ [2049; 2165]; we have t(G) = dlog2e(G)e.
In the remaining part of this paper, we employ Damaschke’s techniques towards Conjecture 1. For a graphG, denote by ±(G) the minimum degree of a vertex in G.
Lemma 5. IfG is a bipartite graph with ±(G) ¸ n; then e(G) ¸ n2.
Proof. The lemma follows from the fact that any part of the vertex set ofG has at leastn vertices and any vertex is of degree at least n.
Lemma 6. Ifn2¡ 1 · b < (n + 1)2¡ 1 and a = b ¡ n + 1; then any bipartite graphG with e(G) ¸ a has an induced subgraph H with a · e(H) · b.
Proof. Choose an induced subgraphH of G with as few vertices as possible such thata · e(H). Assume e(H) ¸ b + 1. By the choice of H, for any vertex x of degree ±(H) in V (H), we have e(H ¡ x) · a ¡ 1 < b + 1 · e(H), which implies that
±(H) = degH(x) = e(H ) ¡ e(H ¡ x) ¸ b ¡ a + 2 = n + 1:
Assume that ±(H) = n + i, where i ¸ 1. Then, according to Lemma 5, e(H) ¸ (n + i)2. Therefore,
e(H ¡ x) = e(H) ¡ ±(x) ¸ (n + i)2¡ (n + i) ¸ n2+ n
= (n + 1)2¡ 1 ¡ n > b ¡ n = a ¡ 1;
a contradiction. Hencea · e(H) · b as desired.
Lemma 7. Suppose verticesx and y are in the same part of a bipartite graph G. If degG(x)+ degG(y) ¸ 2m; then G has an induced subgraph with exactly 2m edges.
Proof. SupposeH is the subgraph of G induced by C [ fx; yg, where C is the set of all neighbors of x and y. As Pv2CdegH(v) = degG(x) + degG(y) ¸ 2m anddegH(v) is 1 or 2 for any vertex v in C, there is a subset D µ C such that
P
v2DdegH(v) = 2m. Hence D [ fx; yg induces a subgraph with exactly 2m
edges.
Proof. The theorem is clearly true for k · 1. Now consider a bipartite graph G of 2k vertices for 2 · k · 4. It is sufficient to prove that G has an induced subgraph with2k¡ 1 edges. By Lemma 7, we may assume
(¤) degG(x) + degG(y) < 2k¡ 1
for any two verticesx and y in the same part of G.
This in turn implies that for2 · k · 3 every vertex of G has degree at most two, which allows the existence of an induced subgraph of 2k¡ 1 edges. So now
consider the case ofk = 4.
According to (¤), any part of G has at most one vertex of degree at least 4. Furthermore, either there is some part in which there are some vertices whose degree sum is8, or else the degree sequence of each part is (4; 3; 3; 3; 3) or (3; 3; 3; 3; 3; 1). For the former case, those vertices of degree sum 8 together with their neighbors induce a subgraph of 8 edges. For the later case, choose a vertex x in part A with exactly 3 neighbors y1; y2; y3 in B. Then, choose a vertex z in B ¡ fy1; y2; y3g
with exactly 3 neighbors w1; w2; w3 in A ¡ fxg. At least one of w1; w2; w3, say
w1, is of degree3. Then G ¡ fx; z; w1g is an induced subgraph of G with exactly
8 edges.
Theorem 9.f (k + 1) ¸ 2f(k) + 1 ¡ bpf(k) + 1c.
Proof. Supposen2¡ 1 · f(k) < (n +1)2¡ 1, i.e., n = bpf (k) + 1c. We only
need to show that for any bipartite graphG with 2f (k) +1¡ n edges, t(G) · k+1. Choosingb = f (k) and applying Lemma 6, we infer that G has an induced subgraph H with
f (k) + 1 ¡ n · e(H) · f(k): And hence
f (k) + 1 ¡ n · e(G ¡ E(H)) · f(k): Therefore, t(H) · k and t(G ¡ E(H)) · k, which imply
t(G) · 1 + maxft(H); t(G ¡ E(H))g · k + 1: This completes the proof of the theorem.
To estimate a good lower bound for f (k) by using the above theorem, we consider the sequencefbk : k ¸ 4g defined by b4= 16 and
bk = 2bk¡ 1+ 1 ¡ bpbk¡ 1+ 1c for k ¸ 5. It is clear that f(k) ¸ bk for k ¸ 4. Note that b5= 2 ¢16 + 1 ¡ bp16 + 1c = 29:
Lemma 10. For anyk ¸ 5; we have bk+ 1 · 15 ¢2k¡ 4.
Proof. First,b5+1 = 29+1 = 15¢25¡ 4. Supposek ¸ 5 and bk+ 1 · 15 ¢2k¡ 4
holds. Then
bk+1+ 1 = 2bk+ 1 ¡ b
p
bk+ 1c + 1 · 2bk+ 2 · 2(15 ¢2k¡ 4) = 15 ¢2(k+1)¡ 4
and so the lemma follows from induction.
Theorem 11. Fork ¸ 5; we have bk¸ 2k¡ 1+ 2k¡ 3+ 2k¡ 6+ 19 ¢2
k¡ 7 2 ¡ 1.
Proof. The theorem is true for k = 5 as b5= 29 = 25¡ 1+ 25¡ 3+ 25¡ 6+ 19 ¢
25¡ 72 ¡ 1. Suppose k ¸ 6 and the theorem is true for k ¡ 1. Then
bk= 2bk¡ 1+ 1 ¡ bpbk¡ 1+ 1c (by the definition of bk) ¸ 2bk¡ 1+ 1 ¡ p15 ¢2 k¡ 5 2 (by Lemma 10) ¸ 2(2k¡ 2+ 2k¡ 4+ 2k¡ 7+ 19 ¢2k¡ 82 ¡ 1) + 1 ¡ p15 ¢2 k¡ 5 2
(by the induction hypothesis)
= 2k¡ 1+ 2k¡ 3+ 2k¡ 6+ (19p2 ¡ 2p15)2k¡ 72 ¡ 1
¸ 2k¡ 1+ 2k¡ 3+ 2k¡ 6+ 19 ¢2k¡ 72 ¡ 1 (since 19p2 ¡ 2p15 > 19):
The theorem then follows
Corollary 12. If G is a bipartite graph with 2k¡ 1< e(G) · 2k¡ 1+ 2k¡ 3+ 2k¡ 6+ 19 ¢2k¡ 72 ¡ 1 and k ¸ 5; then t(G) = dlog2e(G)e.
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Su-Tzu Juan and Gerard J. Chang Department of Applied Mathematics National Chiao Tung University Hsinchu, Taiwan 300, R.O.C. E-mail: gjchang@math.ntu.edu.tw
