Combinatorial Group Testing for DNA Library Screening

Transcription

1 0-0 Combinatorial Group Testing for DNA Library Screening Ying Miao University of Tsukuba

2 0-1 Contents 1. Introduction 2. An Example 3. General Case 4. Consecutive Positives Case 5. Bayesian Network Pool Result Decoder Algorithm 6. Concluding Remarks

3 Introduction Group Testing R. Dorfman, The detection of defective members of a large population, Ann. Math. Stat. Vol. 14, pp , 1943 Test a large number of blood samples to determine the presence of a rare disease efficiently.

4 0-3 Methods Commonsense Method Test all items one by one. Advanced Method (group testing) Combine several items into groups (or pools) and then test them. Assumption: The tests always give the correct answer, unless otherwise stated. A negative result ensures that none of the items are positive. A positive result reveals that at least one of the items is positive. Further tests are required to reveal which particular items are in fact positive.

5 0-4 Objectives Minimizing the number of tests Limiting number of pools Limiting pool sizes Tolerating a few errors etc. These objectives are often contradicting, thus testing strategies are applicant dependent.

6 0-5 Probabilistic Group Testing Some probability p of having a defective is fixed, and a probabilistic model is used to describe the probability distribution of positive items. Goal: Minimize the expected number of tests to identify the defective set. Combinatorial Group Testing It is often assumed that the number of defectives is d for some fixed positive integer d, and a deterministic model is used. Goal: Minimize the number of tests under a worst-case scenario.

7 0-6 Adaptive Algorithm Tests are conducted one by one. The outcomes of previous tests are assumed known at the time of determining the current test. Non-Adaptive Algorithm No such information is available in determining a test. Multi-stage algorithm: Tests are divided into several stages where the stages are considered adaptive but all tests in the same stage are treated as non-adaptive. Error Tolerant group testing algorithm can detect or correct some errors in test outcomes.

8 0-7 DNA Library Screening (pooling designs, mapping genomes.) A DNA sequence is a sequence with four alphabets A, C, G, T, called nucleotides. A DNA sequence is stored in a library by cutting into shorter segments called clones. Basic problem of DNA library screening Determine, in an efficient fashion, whether a particular DNA segment, called a probe, is contained in a clone. A clone is positive for a probe if it contains the probe, and negative otherwise.

9 0-8 DNA library screening introduces several new constraints to group testing: The population of clones is usually large, so collecting a pool of clones and preparing it for probing is not easy. DNA screening is error prone, hence error-tolerant capability is desirable. Most large libraries intend to automate this process by employing robots, and preprogramming the pools, where non-adaptive algorithms are preferred. In practice, s-stage algorithms with small s (e.g. preferable. 2) are often

10 0-9 Other Applications 1. Superimposed Codes: Codes for satellite communication. 2. Frameproof Codes: Codes which avoid coalitions of users forging the signature of a user not in the coalition.

11 An Example A non-adaptive group testing algorithm can be modeled as a pooling design. Let C = {c 1,..., c n } be a set of clones (or items, samples), and let G = {G 1,..., G m } be a set of m subsets of X called pools (or groups). We refer to (C, G) as an (n, m)-nagta. An (n, m)-nagta (C, G) can be represented by an m n incidence matrix H = (h ij ), where the columns correspond to n clones, and the rows to m pools: { 1, if cj G h ij = i, 0, otherwise.

12 0-11 We want to identify a positive subset U C efficiently (minimize the number m of tests for a given number n of clones). This can be done by using a test function f : 2 C {0, 1}: f(y ) = { 1, if Y U, 0, if Y U =, for any Y C. The result (or, outcome) vector of (C, G), given U, is R(U) = (f(g i ) : G i G). We say that (C, G) identifies U if U is determined uniquely from R(U). That is, we require R(U) R(V ) if U V. Under the assumption that U d, we require that H to be dseparable: the unions of d columns of H are all distinct.

13 0-12 A Construction from BIBDs Let (X, B) be a (v, k, 1)-BIBD, and (C, G) the dual incidence structure. We use (C, G) as a (b, v)-nagta, where b = v(v 1) k(k 1). (C, G) satisfies the following properties: (1) Each clone occurs in exactly k tests. (2) Each test contains exactly r = v 1 k 1 clones. (3) Every pair of distinct clones is contained in at most one test. Theorem 2.1 If there exists a (v, k, 1)-BIBD, then there exists a (b, v)- NAGTA with threshold d = k 1. Remark: For fixed k, we have m = O(k n).

14 0-13 Theorem 2.1 says that the positive set U can be identified if U k 1. What happens if U k? Suppose that (C, G) is the (b, v)-nagta of Theorem 2.1, where C = {1,..., b} and G = {G 1,..., G v }. Given the result vector R(U), the following algorithm Identify will identify U if U k 1, and report that U k, otherwise.

15 Algorithm Identify Input: R(U) = (f(g 1 ),..., f(g v )) 1. U = 2. for i 1 to b do H[i] 1 3. for j 1 to v do if f(g j ) = 0 then for all x A j do H[x] 0 4. for i 1 to b do if H[i] = 1 then U U {i} 5. if U k 1 then output U else output the positive subset has size at least k 0-14

16 0-15 Example 2.2 A (9, 3, 1)-BIBD: X = {1, 2, 3, 4, 5, 6, 7, 8, 9}, B = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {1, 4, 7}, {2, 5, 8}, {3, 6, 9}, {1, 5, 9}, {2, 6, 7}, {3, 4, 8}, {1, 6, 8}, {2, 4, 9}, {3, 5, 7}}. The blocks of the dual incidence structure: {1, 4, 7, 10}, {1, 5, 8, 11}, {1, 6, 9, 12}, {2, 4, 9, 11}, {2, 5, 7, 12}, {2, 6, 8, 10}, {3, 4, 8, 12}, {3, 5, 9, 10}, {3, 6, 7, 11}

17 0-16 Carry out Identify with input R(U) = (0, 1, 0, 0, 1, 0, 1, 1, 1) : G G G G The positive set is thus U = {3, 5}.

18 0-17 Strong Group Testing Algorithms (C, G) is a strong NAGTA if for all U C, U d. Identify(R(U)) = U (C, G) is an l-uniform NAGTA if each clone occurs in exactly l tests. The NAGTAs constructed in Theorem 2.1 are strong (thus d-separable) and k-uniform. In fact, they are optimal. Theorem 2.3 Suppose that (C, G) is a proper (i.e. G G, G 2) strong l-uniform (n, m)-nagta with threshold d. Then d l 1. Furthermore, if d = l 1, then n m(m 1) l(l 1), with equality iff (C, G) is the dual of an (m, l, 1)-BIBD.

19 General Case Assumptions: (1) The number of positives is d; (2) The tests always give the correct answer, unless otherwise stated. The incidence matrix H is d-separable if the unions of d columns are all distinct (The corresponding set system is called d-union free). If H is d-separable, then the outcomes uniquely identify the set U of positives. However, the identification may be somewhat involved, because we may have to examine all unions of d columns.

20 0-19 If the union of any ( ) d columns does not cover any column not in this union, then H is a d(= d)-disjunct matrix (The corresponding set system is called d-cover free). When a disjunct matrix H is employed, any clone contained in a pool with negative outcome can be identified as negative, and all other clones as positive. If after the deletion of negative clones, the number of remaining clones is > d, then the false assumption that the number of positives is d can be automatically detected. Remark: This reduces the time complexity from O(n d ) to O(n).

21 0-20 Du and Hwang (00) collected many known results. Ngo and Du (00): Summarized D yachkov and Rykov (82) and D yachkov, Rykov and Rashad (89) s results. Theorem 3.1 Let m(d, n) denote the minimum number of pools needed for the S( d, n) problem, where S( d, n) is the set of all subsets of n items (or columns) with size d. Then, as n and d, d 2 2 log 2 d (1 + o(1)) log 2 n m(d, n) d 2 log 2 e(1 + o(1)) log 2 n.

22 0-21 Constructions for d-disjunct Matrices Deterministic designs Random designs We only describe constructions for deterministic designs. (1) Packing designs (2) Grid designs (3) Macula s construction (4) Ngo and Du s generalized construction

23 0-22 (1) Packing Designs (Kautz and Singleton (64)) In a t-(m, k, 1)-packing, no two blocks have t points in common. Thus, if k > d(t 1) a d-disjunct matrix H can be constructed from a t-(m, k, 1)-packing by simply indexing H s columns by the blocks and H s rows by points. Moreover, if k = d(t 1) + f + 1 (f 0), then H is f-error detecting and f/2 -error correcting.

24 0-23 Problem 3.2 Find the packing number D(m, k, t), the number of blocks in a maximum t-(m, k, 1)-packing. This problem is difficult in general. Theorem 3.3 Let A(n, d, w) denote the size of a maximum constant w-weight binary (n, d) code, then D(m, k, t) = A(m, 2k 2t + 2, k). Let U(m, k, t) = m k m 1 k 1 m t + 1 k t + 1. Schonheim (66) observed that D(m, k, t) U(m, k, t), where equality holds when the design is a t-(m, k, 1)-design.

25 0-24 (2) Grid Designs Robots are sometimes used to assemble the pools. The clones are arranged into rows and columns of a set of r c grids, where each row and column contributes a pool. If only one grid is used, then ambiguity can occur if there are more than one positive clone. To resolve ambiguity, we use more grids s.t. collinear twice, no two clones are Then Hwang (95) showed that the existence of such grids is equivalent to the existence of certain set of mutually orthogonal Latin squares.

26 0-25 This was extended to transversal design by Balding et al. (96). A pooling design is transversal if the pools can be partitioned into parts, each of which is a partition of the clones. Relations of this problem to Coding Theory is also specified in Balding et al. (96). Jimbo and his students (especially, Mutoh) contributed a lot in this research direction.

27 0-26 (3) Macula s Construction (96, 99) Let H(m, d, k) be a (0, 1)-matrix whose rows are indexed by the d- subsets and columns by the k-subsets of {1, 2,..., m} where m/2 k > d 1. H(m, d, k) ij = 1 the i-th d-subset the j-th k-subset. Then H(m, d, k) is d-disjunt with ( m d ) rows and ( m k ) columns.

28 0-27 (4) Ngo and Du s Generalization In packing designs, H was row indexed by all elements and column by selected k-subsets of {1, 2,..., m}, i.e., the rows by all points at rank 1 and columns by selected points at rank k of the Boolean lattice B m. Macula s construction involves taking all points at rank d as rows and rank k as columns. Ngo and Du (99): Take selected points at different ranks of B m.

29 0-28 Let H(m, d, k) be a (0, 1)-matrix whose rows are indexed by the set of all d-matchings (a matching of size d) and columns by the set of all k-matchings on K 2m. H(m, d, k) ij = 1 the i-th d-matching the j-th k-matching. Then H(m, d, k) is d-disjunct. Observe that a k-matching is a k-subset of the set of edges of K 2m. So this construction could be seen as taking from B ( 2m selected points at rank d as rows and selected points at rank k as columns. 2 ) Ngo and Du also showed that H(m, d, k) is d-error detecting and d/2 -error correcting.

30 0-29 Problem 3.4 Besides the Boolean lattice B m, what are other lattices we can use? For example, one candidate is C m,u, the lattice of all m-tuples of Z u, which is a generalization of B m, since B m = C m,2. Ngo ad Du (99): Picking points at levels d and k of the lattice of all subspaces of GF (q) v would also work. Problem 3.5 Can we from the lattice infer some information about the error correcting and detecting capability of the matrix H being constructed?

31 0-30 Error-Tolerant Group Testing A pooling design is e-error detecting (resp. correcting) if it can detect (resp. correct) up to e errors in test outcomes. In other words, if a design is e-error detecting (resp. correcting), then the test outcome vectors form a binary code with min distance at least e + 1 (resp. 2e + 1). For s S( d, n) (the set of all subsets of n clones (or columns) with size at most d), let P (s) denote the union of all columns corresponding to s. Theorem 3.6 Suppose H has the property that s s S( d, n), P (s) and P (s ) have distance k. Then, H is (k 1)-error detecting and (k 1)/2 -error correcting.

32 0-31 Assume the number of errors is f. (C, G) is a solution to the group testing problem with d positives and f-error detection (resp. f/2 -error correction) iff, for any union of d columns, every other column contains at least f + 1 points not in this union. Any t-(m, k, 1)-packing with k d(t 1) + f + 1 is a solution to the group testing problem with at most d positives and f-error detection. Optimal? Allowing for up to f errors, Balding and Torney (96) defined (C, G) to be an optimal solution if it maximizes n (the number of clones) for fixed m (the number of pools) subject to the requirement that whenever the number of errors is f, one can infer from the outcomes either the status of each clone, or that the number of positives is > d.

33 0-32 Let φ(a) denote the set of positive pools when the clones indexed by A are positive and no error occurs. (C, G) is a d-positive, f-error solution, or (d, f)-solution, if from φ(a) we can infer either A or that A > d, even in the presence of up to f errors. This occurs iff d(φ(a), φ(a )) > f A, A 2 {1,2,...,n}, A A, A d, Theorem 3.7 (C, G) σ m d,f (the set of (d, f)-solutions) iff c i \ φ(a) > f A 2 {1,2,...,n}, A d, i {1, 2,..., n} \ A.

34 0-33 Theorem 3.8 If (C, G) σ1,f m, then n = C satisfies n 1 ( ) m, K f m/2 in which K 0 = 1 and, for f even, K f = ( )( ) f/2 m/2 m/2, s=0 s s while for f odd, K f = K f T where T = 2 m/2 /(f + 1). ( )( ) m/2 m/2, (f + 1)/2 (f + 1)/2 Corollary {C 2 {1,2,...,n} : C = m/2 } is optimal in σ m 1,0. 2. If a ( m/2 1)-(m, m/2, 1)-design exists then it is optimal in σ m 1,1.

35 0-34 Theorem 3.10 If (C, G) σ m 2,f, then n ( )( m 2t ) 1 + f 1, t t in which t is the least integer value of t such that m 5t + 2 so that t = (m 2)/5 if f = 0 or 1. f(f 1), t + f Corollary 3.11 If a t -(m, 2t +f 1, 1)-design exists, then it is optimal in σ m 2,f.

36 0-35 In the d = 1 case (Theorem 3.8), For d = 2 (Theorem 3.10), m > log n (1 + o(1)). log 2 m > log n (1 + o(1)). log 5/4 Problem 3.12 Is there any analogue result for σ m d,f with d > 2? In general, we need deeper results and new breakthroughs for errortolerant group testing problems. Problem 3.13 Find bounds similar to those in Theorem 3.1 given the number of clones n, max number of positives d, and max number of errors f.

37 Consecutive Positives Case Bruno et al. (95): Analyzed linear DNA. Various consecutive clones of the DNA have been constructed, which are placed in a linear order corresponding to the order of their appearance in the DNA. Clones are not disjoint, but rather overlap. It may happen that one probe occurs in a relatively large number of clones. We want to know where a particular probe occurs in the linear DNA. Assumption: The probe occurs only once in the DNA. Then positive clones containing this probe are consecutive under the same ordering.

38 0-37 Let C = {c 1, c 2,..., c n } be a set of clones, equipped with the linear order c i c i+1 for 1 i < n. C has the d-consecutive positives property if the set of positive clones is a consecutive set (under the ordering ), and contains at most d clones. Colbourn s Result (99) Theorem 4.1 Non-adaptive group testing for at most d consecutive positives in a linearly ordered n-set of clones can be accomplished with log 2 n + 2d + 1 = O(log 2 n + d) pools. Colbourn (99) noticed all cases d 2 can be reduced to the case d = 2.

39 0-38 Muller and Jimbo s Improvements (04, 06) The incidence matrix H is required to be 2-consecutive positive detectable (2CPD). Let H = [x 0, x 1,..., x n 1 ] be an (0, 1)-matrix. Define y i = x i x i+1, 0 i n 2. Then H is called a 2CPD-matrix iff consist of pairwise distinct vectors. x 0, y 0, x 1, y 1,..., x n 2, y n 2, x n 1 Define y n 1 = x n 1 x 0. Then H is a cyclic 2CPD-matrix iff x 0, y 0, x 1, y 1,..., x n 2, y n 2, x n 1, y n 1 consist of pairwise distinct vectors.

40 0-39 In other words, they considered cyclic sequences with distinct unions. Let P(m) denote the set of all subsets of {0, 1,..., m 1}, and let X = {x 0, x 1,..., x n 1 } P(m). A cyclic sequence of X is a sequence S = (x 0, x 1,..., x n 1 ), x i X, s.t. each x i appears exactly once in S, and the indices of x i S are considered modulo n. S is a cyclic sequence with distinct (consecutive) unions (CSDU) if y i = x i x i+1, 0 i n 1, are pairwise distinct. Remark: In the case d = 2, we can distinguish up to any two consecutive positive clones.

41 0-40 Muller and Jimbo (04) established the existence of m n 2CPDmatrices having a maximum number of columns n = 2 m 1 for any m 3. Then for any n clones, we need only m = log 2 n + 1 pools to distinguish them. Remark: Colbourn (99) needs log 2 n + 3 pools. Muller and Jimbo (04) also investigated 2CPD-matrices having the maximum number of columns where each clone appears the same number of times in the pools. Theorem 4.2 m and 1 w m/2, an optimal non-adaptive group testing strategy for clones having the 2-consecutive positives property with m pools of size ( ( m 1 w 1) and n = m w) clones, where each clone appears in exactly w pools.

42 0-41 Error-Tolerant Group Testing for Consecutive Positives X P(m), let CSDU(X) denote the class of all cyclic sequences with distinct consecutive unions of the element of X. We are interested in the case X = P(m, w), the set of all w-subsets of {0, 1,..., m 1}, where we simply write CSDU(m, w). We may further require any y i = x i x i+1 to be in some prescribed set Y P(m). The sub-class of CSDU(X) with consecutive unions in Y is denoted by CSDU(X Y ). We write CSDU(m, w Y ) for X = P(m, w). We focus on Y = P w (mod 2) (m): The resulting group testing strategy can detect up to one error.

43 0-42 Obviously, any S CSDU(m, w) corresponds to a cyclic 2CPDmatrix with m rows and n = ( m w) columns of weight w. Finding CSDU(X Y ) is very difficult in general, including many unsolved problems such as the middle two levels problem. The middle two levels problem asks whether there is a Hamiltonian cycle in the middle two ranks P(2k + 1, k) and P(2k + 1, k + 1) of the Boolean lattice B 2k+1. Obviously, such Hamiltonian cycles correspond to the elements of CSDU(2k + 1, k k + 1). Shields and Savage (99): The largest value for which a Hamiltonian cycle is known to exist is k = 15.

44 0-43 Muller and Jimbo s Result (06) Theorem 4.3 Suppose m 3 m 4 s.t. CSDU(m w, w w + 2). Then m m w, CSDU(m, w w + 2). It is not difficult to see that the smallest possible m w are given by m w = 2w + 2. Problem 4.4 Is CSDU(2w + 2, w w + 2) for any w 2? This problem seems to be as hard as the middle two levels problem.

45 0-44 We can relax the problem by either increasing m w or by enlarging Y of admissible unions. Problem 4.5 Is CSDU(2w + 2, w P w (mod 2) (2w + 2)) for any w 2? If yes, is there a sequence in each CSDU(2w + 2, w P w (mod 2) (2w + 2)) s.t. at least one consecutive union is of size w + 2? Lemma 4.6 CSDU(m, w P w (mod 2) (m)) for any m m w for the following w and m w : w m w Problem 4.7 Find a general construction principle for infinitely many w.

46 0-45 To obtain group testing strategies with higher error-tolerant ability, we could start with some error-correcting code C of min distance d. Problem 4.8 Find a partition C = X Y s.t. CSDU(X Y ). Any such cyclic sequence would define a group testing strategy which detects up to d 1 errors. Unfortunately, Muller and Jimbo (06) noticed, in general, that this problem seems to be very difficult. Problem 4.9 Derive lower bounds on the length of the longest cyclic sequence with distinct consecutive unions of codewords of C.

47 0-46 Momihara s Result (06) m, Momihara (06) showed the existence of a maximal CSDU(m, 3 3) by applying some sufficient conditions for the existence of a Hamiltonian cycle to the block 0-intersection graph of an optimal (m, 3, 1)- packing.

48 Bayesian Network Pool Result Decoder Algorithm The incidence relation of C and G can also be represented by its Tanner graph E = {{c, G} C G : c G}, which has been studied when investigating low-density parity-check (LDPC) codes. Let X c be a random variable s.t. X c = { 0, if clone c is negative, 1, if clone c is positive. Let Z G be the random variable defined as Z G = c G X c. Then: (1) If G contains only negative clones, then Z G = 0. (2) If G contains at least one positive clone, then Z G = 1.

49 0-48 G. Let S G be a random variable representing the test outcome of pool We should take into account of the error probabilities P (S G = s Z G = z), i.e. P (S G 0 Z G = 0) is the probability of false positive, while P (S G = 0 Z G = 1) is that of false negative. Let X = (X c1,..., X cn ) and S = (S G1,..., S Gm ). Assumption: X c s are independent, each S G depends only on Z G, and S G s are independent. In an experiment, when S is observed, then P (X = x S = s) = P (X = x, S = s)/p (S = s) = KP (X = x, S = s) = KP (X = x)p (S = s X = x) = K c C P (X c = x c ) G G P (S G = s G Z G = z G ) where K = P (S = s) 1 is a constant.

50 0-49 Therefore, P (X c = x S = s) = K P (X d = x d ) P (S G = s G Z G = z G ), x c =x d C G G where x c =x means the sum of all x {0, 1}n with x c = x. Moreover, we assume that P (S G = s G Z G = z G ) are empirically known from previous experiments. We need 2 n 1 = {x {0, 1} n : x c = x} summations to calculate P (X c = x S = s), which is exponential in n. Therefore, an efficient algorithm is needed. Uehara and Jimbo (06) proposed an algorithm based on Bayesian network (graphical model that encodes probabilistic relationships among variables of interest) related to the sum-product algorithm for decoding LDPC codes.

51 0-50 Bayesian Network Pool Result Decoder (BNPD) Algorithm Step 0 For each c G, set P (X c = x). Step 1 (Initialization of Q): For each c G, let Q x (c, G) = P (X c = x), x = 0, 1. Step 2 (Computation of R): For each G c, compute R 0 (G, c) = P (s G 1) + (P (s G 0) P (s G 1)) c G\{c} Q 0 (c, G) and R 1 (G, c) = P (s G 1), where P (s G z) = P (S G = s G Z G = z). Normalize R x as follows: R x (G, c) = R x (G, c), x = 0, 1. R 0 (G, c) + R 1 (G, c)

52 0-51 Step 3 (Computation of Q): For each c G, compute Q x (c, G) = P (X c = x) G (c)\{g} R x (G, c), x = 0, 1, where (c) denotes a set of pools which include c. Normalize Q x as follows: Q x (c, G) = Q x (c, G), x = 0, 1. Q 0 (c, G) + Q 1 (c, G) Step 4 (Iteration): Iterate Step 2 and Step 3 for several times until the values Q x converges.

53 0-52 Step 5 (Computation of marginal probability): For each c G, compute Q x (c) = P (X c = x) G (c) R x (G, c), x = 0, 1. Normalize Q x as follows: Q x (c) = Q x (c), x = 0, 1. Q 0 (c) + Q 1 (c)

54 0-53 Uehara and Jimbo s simulation result (06) shows the positive detecting ability is getting higher when the replication number r becomes larger, and the difference of ability has a big gap between r = 3 and r = 4, so they claimed the desired replication number r is at least 4. MacKay s simulation result (99) for LDPC codes shows the best decoding performance when each bit is checked 3 times (i.e. r = 3). Problem 5.1 How to explain Mackay s and Uehara and Jimbo s simulation results?

55 0-54 If a Tanner graph has no cycle, it is known that each Q x (c) converges to the correct marginal probability P (X c = x S = s) after enough iterations. If it has cycles, we can not say any theoretical result for convergence. However, intuitively, if there are cycles in a Tanner graph, it is desirable that it does not have short cycles.

56 0-55 From a design (X, B), we can construct a Tanner graph by defining B to be the set of vertices representing clones, and X to be that representing pools, and an edge between x X and B B if x B. A packing is -free if non-colinear points P 1, P 2, P 3, if there are two blocks containing P 1, P 2 and P 1, P 3 respectively, then there is no block containing P 2, P 3. Clearly, the Tanner graphs obtained from -free packings are free of 4-cycles and 6-cycles simultaneously, and therefore their girths are 8.

57 0-56 Theorem 5.2 -free (v, 3, 1)-packing (X, B), B v(v + 3)/18. The -free packing number F D(v) is the maximum number of blocks in any -free (v, 3, 1)-packing. Example 5.3 F D(v) = v(v + 3)/18 for 1 v < 10, v 6, and F D(v) = v(v + 3)/18 1 for v = 6. Problem 5.4 Construct infinite series of -free (v, 3, 1)-packings with maximum number of blocks. Problem 5.5 What we can say about -free (v, 4, 1)-packings?

58 Concluding Remarks 1. We described the framework of combinatorial group testing for DNA library screening. 2. We described how to deal the case when there are d positives, and possibly f errors. 3. We described how to deal with consecutive positive clones, and possibly errors. 4. We describe the BNDP algorithm closely related to the decoding of LDPC codes. 5. We mentioned a lot of open problems. Try to solve these problems, and develop more combinatorial methods for DNA library screening.

59 Thank You! 0-58