On a Hyperplane Arrangement Problem and Tighter Analysis of an Error-Tolerant Pooling Design

Transcription

1 On a Hypeplane Aangement Poblem and Tighte Analysis of an Eo-Toleant Pooling Design Hung Q Ngo August 19, 2006 Abstact In this pape, we fomulate and investigate the following poblem: given integes d, k and whee k > 1, d 2, and a pime powe, aange d hypeplanes on F k to maximize the numbe of -dimensional subspaces of F k each of which belongs to at least one of the hypeplanes The poblem is motivated by the need to give tighte bounds fo an eo-toleant pooling design based on finite vecto spaces 1 Intoduction Designing good eo-toleant pooling design is a cental poblem in the aea of non-adaptive goup testing [9, which has many pactical applications including DNA libay sceening [8, 10, 21, multiple access contol [5 7, 17, 26, and eo coecting/detecting supeimposed codes [11 15, to name a few To date, thee ae elatively few papes addessing the poblem of designing and analyzing eotoleant pooling designs [1,3,4,16,19,20,22,24 In [22, Ngo and Du intoduced a non-adaptive pooling design based on finite vecto spaces, which was late found to be highly eo-toleant by D yachkov et a [10 The analysis of the design in [10 was not vey tight In this pape, we give a tighte analysis of the design This is done via fomulating a new and vey inteesting hypeplane aangement poblem on finite fields To fomally descibe ou poblem, we fist need a few definitions A 01-matix M is said to be d-disjunct if and only if no column is contained in the union of d othes (Hee, columns ae viewed as chaacteistic vectos of sets of ows) A d-disjunct matix coesponds pecisely to a pooling design which can identify at most d negative items Fo the design to toleate a few eos in outcomes, it is not sufficient fo a column to just not be coveed by d othes A d z -disjunct matix is a matix whee, given any d + 1 columns C 0, C 1,, C d, the set C 0 \ C 1 C d has at least z elements It is easy to show that a d z -disjunct matix can detect z 1 eos and coect (z 1)/2 eos The constuction in [22 is as follows Let be a pime powe and m, k, be integes such that m > k > 1 Let M (m, k, ) be the 01-matix whose ows ae indexed by -dimensional subspaces of F m and whose columns ae indexed by k-dimensional subspaces of F m M (m, k, ) has a 1 in ow R and column C if and only if R is a subspace of C It is easy to see that M (m, k, d) is d-disjunct (the containment method by Macula [18) Late, D yachkov et al [10 ealized that we do not have to take = d fo M (m, k, ) to be d-disjunct ( could be a lot smalle than d, even = 1 woks sometimes) Moeove, the constuction can, in geneal, toleate a lot of eos Specifically, thei main esult was that, Compute Science and Engineeing Depatment, SUNY at Buffalo, Amhest, NY 14260, USA hungngo@csebuffaloedu This wok is patially suppoted by NSF CAREER Awad CCF

2 fo any d + 1 k-dimensional subspaces C 0, C 1,, C d of F m, the numbe of -dimensional subspaces each of which belongs to C 0 but not othe C i is at least [ [ [ k d + (d 1), and that the bound is tight fo d + 1 Hee, fo any non-negative integes m, n, [ n m denotes the Gaussian coefficient, to be defined in the next section The numbe of columns of M (m, k, ) is [ m k, exponentially lage than +1 Hence, it is desiable to devise tight bounds fo the case when + 1 < d [ m k 1 In this pape, we patially addess this poblem In the pocess, we fomulate a new to the best of the autho s knowledge hypeplane aangement poblem on finite fields The est of this pape is oganized as follows Section 2 motivates the hypeplane aangement poblem and pesents peliminay esults on the poblem Section 3 gives tighte bounds fo the oiginal goup testing poblem using esults fom Section 2 Section 4 concludes the pape with additional emaks and a conjectue 2 An extemal hypeplane aangement poblem on finite fields 21 Motivation and notations Hencefoth, we shall use (a; ) n (o (a) n fo shot) to denote the -shifted factoial: (a) n = (a; ) n := (1 a)(1 a) (1 a n 1 ) The [ -analogue of a natual numbe n is denoted by [n, and the Gaussian coefficient is denoted by n m They ae defined as follows [0 := 0 [n := n 1, n 1 { 0 when n < m := othewise [ n m () n () n m () m = (1 n )(1 n m+1 ) (1 m )(1 ) We shall dop the subscipt and wite [n and [ n m when thee is no potential confusion as to what is Ou notations ae standad in the -seies liteatue [2 Fo any vecto space X, let X denote the set of all -dimensional subspaces of X, and dim(x) the dimension of X Then, it is well known (see, eg [25) that [ dim(x) X = Fo any vecto spaces X and Y, X Y = X Y (1) because any vecto space which is a subspace of X and a subspace of Y is also a subspace of the vecto space X Y Note that, in geneal X Y is not a vecto space, and The matix M (m, k, ) is d z -disjunct fo X Y span(x Y ) z = min{ C 0 \ C 1 C d : C 0, C 1, C d ae d + 1 diffeent k-dimensional subspaces of F m } 2

3 Thus, we want to find k-dimensional subspaces C 0, C 1,, C d of F m that minimizes the uantity Fo any i {1,, d}, let H i = C i C 0, then C 0 \ C 1 C d C 0 \ C 1 C d = C 0 \ (C 1 C 0 ) (C d C 0 ) = C 0 \ (C 1 C 0 ) (C d C 0 ) = C 0 \ H 1 H d Fo C 0 \ C 1 C d to be minimized, we can assume that all H i ae hypeplanes of C 0 The numbe of hypeplanes of C 0 is [ k k 1 = [k Thus, when d [k we can also assume that the Hi ae diffeent hypeplanes of C 0 ; because, given d hypeplanes H 1,, H d, we can take the span of each of them with a vecto v / C 0 to econstuct the C i Fo the goup testing poblem, we will addess the case when d > [k in a late section In this section, we only conside the case when d [k Because C 0 \ C 1 C d is minimized when H 1 H d is maximized, the above discussion motivates the following poblem Poblem 1 (Ou Hypeplane Aangement Poblem) Given a k-dimensional vecto space C ove F, and an intege d such that 1 d [k, find d hypeplanes H 1,, H d of C that maximizes the following uantity H 1 H d At least, find good uppe bounds fo the uantity The esult in [10 can be estated as follows Theoem 21 (D yachkov et al) Given integes, and d [k Let H 1,, H d be d diffeent hypeplanes of a k-dimensional vecto space ove F Then, [ [ H 1 H d d (d 1) (2) The bound is tight when d Initial obsevations By inclusion-exclusion, we have H 1 H d = = = d ( 1) i 1 i=1 d ( 1) i 1 i=1 d ( 1) i 1 i=1 T {1,,d} T =i T {1,,d} T =i T {1,,d} T =i t T t T H t H t [ ( dim t T H t As we will see late, it is not easy to detemine the dimension of the intesection of a given numbe of abitay hypeplanes That is why inclusion-exclusion does not help us diectly solve the poblem Next, fo any two vecto spaces X and Y, dim(x) + dim(y ) = dim(span(x Y )) + dim(x Y ) ) (3) 3

4 In paticula, if X is a hypeplane and Y is a pope subspace of a k-dimensional vecto space, then eithe Y X o dim(y ) = dim(x Y ) + 1 To see this, suppose Y has dimension l If Y X, then dim(span(x Y )) = k, which implies dim(y ) = k + dim(x Y ) dim(x) = dim(x Y ) + 1 In wods, a hypeplane eithe contains Y o cut into Y at one dimension lowe than that of Y This obsevation leads to the following simple yet impotant lemma Lemma 22 Let H 1,, H x be some x hypeplanes of an l-dimensional vecto space ove F whose intesection is I = H 1 H x Let H be any hypeplane not containing I, and set Y i = H H i, i {1,, x} Then, fo any subset S {1,, x}, we have ( ) ( ) dim H i = dim Y i + 1 (4) i S Poof Because H does not contain I, H does not contain i S H i fo any S Thus, ( ) ( dim H i = dim H ) ( ) ( ) H i + 1 = dim (H i H) + 1 = dim Y i + 1 i S i S i S i S i S The Y i actually ae hypeplanes of H What this lemma tells us is that, the inte-elationship (in tems of dimensions of intesections) between the hypeplanes H 1,, H x is the same as the inte-elationship between the hypeplanes Y 1,, Y x of H The hypeplanes Y 1,, Y x fom a down-scaled pictue of H 1,, H x inside H Conside an i-dimensional subspace X of a k-dimensional vecto space S ove F Let l be an intege whee i l k Then, the numbe of l-dimensional subspaces of S containing X is [ k i l i In paticula, when i = the numbe of hypeplanes that contains X is [ [ k () 2 = = + 1 (5) () 1 Lastly, the following identity is the -analog of the Pascal s tiangle identity fo binomial coefficients [25: [ [ [ n n 1 n 1 = + n m (6) m m m 1 23 The cases of 4 and 5 hypeplanes Using the basic obsevations in the pevious section, when thee ae a constant numbe of hypeplanes it is possible to enumeate all possible classes of aangements (with espect to ou objective function) In this section, we will compute the objective function fo all aangements of 4 and 5 hypeplanes These aangements will seve as the base case to pove geneic bounds in the next section We will be woking on an l-dimensional vecto space S ove F, namely S is isomophic to F l Fo any set of (at least two) hypeplanes H, let x(h) be the maximum numbe of hypeplanes in H whose intesection has dimension l 2 Note that 2 x(h) + 1 Also define g(h) = H We fist conside the 4-hypeplane case H H 4

5 Lemma 23 Let H = {H 1, H 2, H 3, H 4 } be a set of 4 hypeplanes of F l (i) If x(h) = 4, then This case can only hold when 3 g(h) = g (4) 1 := 4 [ l 1 3 [ l 2 (7) (ii) If x(h) = 3, then g(h) = g (4) 2 := 4 [ l 1 5 [ l [ l 3 (8) (iii) If x(h) = 2, then thee ae two cases: [ [ [ [ g(h) = g (4) l 1 l 2 l 3 l 4 3 := (9) [ [ [ g(h) = g (4) l 1 l 2 l 3 4 := (10) Moeove, g (4) 1 g (4) 2 g (4) 3 g (4) 4 Poof Cases (i) and (ii) follow staightfowadly fom the inclusion-exclusion fomula (3) and Lemma 22 Suppose x(h) = 2, then W = H 1 H 2 H 3 has dimension l 3 If H 4 does not contain W then H 1 H 2 H 3 H 4 has dimension l 4, and the fomula fo g (4) 3 follows fom (3) again (Note that, when l 3 all fomulas follow tivially) Thus, the last case is when W H 4 Let V i = H 4 H i, fo i = 1, 2, 3 Because x(h) = 2, the V i ae thee diffeent hypeplanes of H 4 Moeove, V 1 V 2 V 3 = W, and dim(v i ) = l 2 fo i = 1, 2, 3 We can compute g(h) as follows, noting Lemma 22, g(h) = H 1 H 2 H 3 + H 4 \ H 1 H 2 H 3 = H 1 H 2 H 3 + H 4 \ V 1 V 2 V 3 = H 1 H 2 H 3 + H 4 V 1 V 2 V 3 ( [ [ [ ) [ l 1 l 2 l 3 l 1 = = g (4) 4 ( 3 [ [ ) l 2 l 3 2 Lemma 24 Let H = {H 1, H 2, H 3, H 4, H 5 } be a set of 5 hypeplanes of F l (i) If x(h) = 5, then This case can only hold when 4 (ii) If x(h) = 4, then g(h) = g (5) 1 := 5 g(h) = g (5) 2 := 5 This case can only hold when 3 [ l 1 [ l [ l 2 [ l (11) [ l 3 (12) 5

6 (iii) If x(h) = 3, then thee ae thee cases: [ [ [ g(h) = g (5) l 1 l 2 l 3 3 := , (13) [ [ [ [ g(h) = g (5) l 1 l 2 l 3 l 4 4 := , (14) [ [ [ g(h) = g (5) l 1 l 2 l 3 5 := (15) The last case (of g (5) 5 ) can only hold when 3) (iv) If x(h) = 2, then thee ae fou cases: [ [ [ [ g(h) = g (5) l 1 l 2 l 3 l 4 6 := , (16) [ [ [ [ [ g(h) = g (5) l 1 l 2 l 3 l 4 l 5 7 := , (17) [ [ [ [ g(h) = g (5) l 1 l 2 l 3 l 4 8 := , (18) [ [ [ g(h) = g (5) l 1 l 2 l 3 9 := (19) The last case (of g (5) 9 can only hold when 3) Moeove, g (5) 1 g (5) 2 g (5) 3 g (5) 4 g (5) 5 ; and g(5) 4 g (5) 6 g (5) 7 g (5) 8 g (5) 9 Also, g(5) 5 g (5) 6 when 4 Poof Cases (i) and (ii) follow staightfowadly fom the inclusion-exclusion fomula (3) and Lemma 22 Suppose x(h) = 3, and assume V = H 1 H 2 H 3 has dimension l 2 Let V i = H 4 H i, fo i = 1, 2, 3, and U = H 1 H 2 H 3 H 4 Since x(h) = 3, H 4 does not contain V and thus dim(u) = l 3 by Lemma (22) We conside thee cases as follows Case 1: H 5 contains some V i fo i = 1, 2, 3 Note that H 5 cannot contain two diffeent V i because the span of two diffeent V i is exactly H 4 Without loss of geneality, assume V 1 H 5 In this case H 1, H 2, H 3, H 4 intesect H 5 at 3 diffeent hypeplanes (of H 5 ), because H 1 and H 4 intesect H 5 at the same hypeplane V 1 Lemma 22 and the inclusion-exclusion fomula (3) gives H 1 H 5 = H 1 H 2 H 3 + H 4 \ H 4 H 1 H 4 H 2 H 4 H 3 + = H 5 \ H 5 H 1 H 5 H 4 ( [ [ ) ([ [ [ ) l 1 l 2 l 1 l 2 l = g (5) 3 Case 2: H 5 contains U but does not contain any W i fo i = 1, 2, 3 In this case, H 1, H 2, H 3, H 4 intesect H 5 at 4 diffeent hypeplanes all of which contains U It follows that H 1 H 5 = H 1 H 2 H 3 + H 4 \ H 4 H 1 H 4 H 2 H 4 H 3 + = H 5 \ H 5 H 1 H 5 H 4 ( [ [ ) ([ l 1 l 2 l ([ [ [ ) l 1 l 2 l = g (5) 5 [ l [ l 3 ) + 6

7 Case 3: H 5 does not contain U This is the situation of Lemma 22 We have H 1 H 5 = H 1 H 4 + H 5 \ H 5 H 1 H 5 H 4 ( [ [ [ ) l 1 l 2 l 3 = ([ [ [ [ ) l 1 l 2 l 3 l = g (5) 4 If x(h) = 2, then W = H 1 H 2 H 3 has dimension l 3 The fomula fo g (5) 6 comes fom the case when H 4 and H 5 both contain W ; g (5) 7 is obtained when W H 4 but W H 5 o vice vesa; g (5) 8 is obtained when W is neithe a subspace of H 4 o H 5 and H 5 does not contain the intesection U = H 1 H 2 H 3 H 4, and g (5) 9 is obtained when W is neithe a subspace of H 4 no H 5, yet H 5 does contain U The computation is simila to the pevious case 24 Tighte bounds and the packing aangement We fist conside the simplest case when = 1 The total numbe of lines (ie 1-dimensional subspaces) of a k-dimensional vecto space S ove F is [ k = [k = 1 1 Let V be any k 2-dimensional subspace of S, and H 1,, H +1 be the set of all hypeplanes containing V Then, the inclusion-exclusion fomula (3) gives [ [ H 1 H +1 = ( + 1) 1 1 The following theoem follows immediately = ( + 1) k k = 1 Theoem 25 When = 1 and d + 1, the maximum value of H 1 H d is exactly [k, the total numbe of lines in S One way to obtain this maximum is to have + 1 of the hypeplanes contain a ()-dimensional subspace of S Fo the est of this section, we can assume 2 We fist give a paticula aangement called the packing aangement which poves to be optimal in cetain cases Definition 26 (Packing Aangement) Suppose 1 + < d Let S be the k-dimensional vecto space that the hypeplanes belong to Let V be any ()-dimensional subspace of S, and W be any ()-dimensional subspace of V The packing aangement of d hypeplanes is an aangement in which + 1 hypeplanes, say H 1,, H +1, all contain V and the est of the hypeplanes contain W We could define the packing aangement fo lage values of d Howeve, fo the puposes of this pape d is sufficient The following lemma tells us the cost of this aangement Lemma 27 Conside 1 + < d , and let H 1,, H d be in the packing configuation Then, [ H 1 H d = d ( d( + 1) ( ) ) [ [ + (d 1) (20) 7

8 Poof Without loss of geneality, assume H 1,, H +1 intesect at a ()-dimensional subspace V and the est of the hypeplanes contain a ()-dimensional subspace W V Conside any H i whee +1 < i d Let V j = H i H j, fo j {1,, i 1} Note that all V j contain W ; and, due to Lemma 22 it is easy to see that V 1,, V +1 ae diffeent hypeplanes of H i Moeove, the total numbe of hypeplanes in H i that contain W is exactly 1 + Hence, Conseuently, {V 1,, V +1 } = {V 1,, V i 1 } H i \ H 1 H i 1 = H i \ H 1 H i H i 1 H i = H i \ V 1 V i 1 = H i \ V 1 V +1 = H i V 1 V +1 [ [ = ( + 1) + [ Finally, d +1 d H i = H i + H i \ H 1 H i 1 i=1 i=1 i=+2 [ [ ([ [ [ ) = ( + 1) + (d 1) ( + 1) + [ = d ( d( + 1) ( ) ) [ [ + (d 1) Theoem 28 Suppose d [k Conside any d hypeplanes H 1,, H d of a k-dimensional vecto space S ove F Let x be a maximal numbe of hypeplanes intesecting in a ()-dimensional subspace V C Then, [ [ [ ([ [ ) H 1 H d d + d + x(x d 1) Poof Without loss of geneality, assume H 1,, H x intesect at V of dimension (), and no othe H i contains V We invoke Lemmas 22 and fomula (3) again Since H i does not contain V, it is easy to see that, fo 1 j x, the vecto spaces V j = H i H j ae all distinct with dimension one less than H j Also, the intesection of the V j has dimension one less than V It follows that [ H i \ H 1 H x = H i \ V 1 V x = x [ + (x 1) [ Conseuently, d x d H i = H i + H i \ H 1 H i 1 i=1 i=1 i=x+1 x d H i + H i \ H 1 H x i=1 i=x+1 [ [ ([ [ [ ) = x (x 1) + (d x) x + (x 1) [ [ [ ([ [ ) = d + d + x(x d 1) (21) 8

9 We get Theoem 21 fo fee Coollay 29 (Same as Theoem 21) Suppose 2 d +1 Then, fo any d hypeplanes H 1,, H d of a k-dimensional vecto space C ove F we have [ [ H 1 H d d (d 1) (22) Moeove, thee exists an aangement of hypeplanes achieving the ight hand side Poof Without loss of geneality, suppose H 1,, H x intesect at some ()-dimensional subspace V, and no othe H i contains V Note that 2 x d Thus x(x d 1) d The tiangle identity (6) gives [ [ [ = k Relation (21) implies [ [ [ H 1 H d d + d d [ [ = d (d 1) ([ [ ) The ineuality is tight because euality can be obtained by choosing d hypeplanes H 1,, H d all of which contain a ()-dimensional subspace V Theoem 210 Suppose d + 2 and k > 2 Then, fo any d hypeplanes H 1,, H d of a k-dimensional vecto space S ove F we have [ [ [ H 1 H d d (2d 3) + (d 2) (23) Moeove, when d = + 2 the packing aangement achieves the bound Poof Without loss of geneality, suppose H 1,, H x intesect at some ()-dimensional subspace V, and no othe H i contains V Note that, in this case 2 x + 1, and thus x(x d 1) 2(2 d 1) = 2(d 1) Relation (21) implies [ [ [ ([ [ ) H 1 H d d + d 2(d 1) [ = d (2d 3) [ + (d 2) [ When d = + 2, we only have to veify that the ight hand side of (23) is the same as that of (20), which is mechanical Theoem 211 Suppose d + 3 and k > 2 Then, fo any d hypeplanes H 1,, H d of a k-dimensional vecto space S ove F we have [ [ [ H 1 H d d (3d 7) + (2d 6) (24) Moeove, when d = + 3 the packing aangement achieves the bound 9

10 Poof Let x = x({h 1,, H d }) Without loss of geneality, suppose H 1,, H x intesect at some ()-dimensional subspace V, and no othe H i contains V Conside two cases as follows Case 1: x 3 Note that, x + 1 d 2, and thus x(x d 1) 3(3 d 1) = 3(d 2) Relation (21) implies [ [ [ ([ [ ) H 1 H d d + d 3(d 2) [ = d (3d 7) [ + (2d 6) Case 2: x = 2 Applying Lemma 23 with l = k we get [ H 1 H 2 H 3 H 4 max{g (4) 3, g(4) 4 } = g(4) 3 = 4 6 [ [ + 4 [ [ k 4 Conside any H i with i > 4 Fo j = 1, 2, 3, 4, let V j = H i H j Then, the V j ae fou diffeent hypeplanes of H i, because x = 2 Applying Lemma 23 with l = we get [ [ [ V 1 V 2 V 3 V 4 min{g (4) k 4 1, g(4) 2, g(4) 3, g(4) 4 } = g(4) 4 = Hence, [ [ H i \ H 1 H 2 H 3 H 4 = H i V 1 V 2 V 3 V [ 3 Putting them all togethe, we get d d H i = H 1 H 2 H 3 H 4 + H i \ H 1 H 2 H 3 H 4 i=1 i=5 ( [ [ [ [ ) k ([ [ [ [ ) k 4 (d 4) [ [ [ [ k 4 = d (4d 10) + (6d 20) (3d 12) [ k 4 It is easy to see that, when k > 2 the last expession is at most the ight hand side of (24) Lastly, when d = + 3 the fact that the packing aangement achieves the bound (24) is staightfowad Theoem 212 Suppose d + 4 and k > 2 Then, fo any d hypeplanes H 1,, H d of a k-dimensional vecto space S ove F we have [ H 1 H d d (4d 13) [ + (3d 12) Moeove, when d = + 4 the packing aangement achieves the bound [ (25) Poof The poof is simila to the pevious theoem with thee cases to conside: x 4, x = 3, and x = 2 This time we make use of Lemma 24 and its vaious elations 10

11 3 Tighte analysis of M (m, k, ) The esults of the pevious section help us analyze the M (m, k, ) constuction Fistly, we show that M (m, k, 1) is not a good design when d + 1 The esult is a diect coollay of Theoem 25 Coollay 31 When d + 1, then M (m, k, 1) is not d-disjunct Poof Let C 0 be a k-dimensional subspace of F m Let H 1,, H +1 be hypeplanes of C 0 chosen accoding to Theoem 25 Let v be any vecto in F m not belonging to C 0 Fo i = 1,, d, let C i = span{h i, v} Choose abitaily k-dimensional subspaces C +2,, C d Then, it is easy to see that C 0 \ C 1 C d = Secondly, the numbe of columns of M (m, k, ) is [ m k, which is exponentially lage than [k, the numbe of hypeplanes in a k-dimensional vecto space The following theoem shows a limit of the pooling design Theoem 32 If d [k, then M (m, k, ) is not d-disjunct Poof Conside any k-dimensional subspace C 0 of F m Let H 1,, H [k be the set of all hypeplanes of C 0 Let v be any vecto in F m \ C 0 Fo any i = 1,, [k, define C i = span{v, H i } Fo i = [k + 1,, d, choose k-dimensional subspaces C i abitaily as long as they have not been chosen befoe Then, C 0 \ C 1 C d = 0, namely M (m, k, ) is not d-disjunct This is because any - dimensional subspace of C 0 is also a subspace of some H i, i = 1, [k; thus, it is also an -dimensional subspace of C i Hencefoth, we only need to conside the case when 2 and + 2 d [k The following coollaies follow fom Theoems 210, 211, and 212, espectively Coollay 33 When d + 2 and m > k > 2, M (m, k, ) is d z -disjunct, whee [ [ [ [ k z = d + (2d 3) (d 2) Moeove, the constuction is exactly d z -disjunct when d = + 2 Coollay 34 When d + 3 and m > k > 2, M (m, k, ) is d z -disjunct, whee [ [ [ [ k z = d + (3d 7) (2d 6) Moeove, the constuction is exactly d z -disjunct when d = + 3 Coollay 35 When d + 4 and m > k > 2, M (m, k, ) is d z -disjunct, whee [ [ [ [ k z = d + (4d 13) (3d 12) Moeove, the constuction is exactly d z -disjunct when d = + 4 Remak 36 Note that, to apply all the thee coollaies above, we only need to find the ange of d which makes z > 0 It tuns out that this ange is uite lage, and the task is mechanical We omit this step hee Also, it is staightfowad to check that the bounds in Theoems 210, 211, and 212 ae bette than that of Theoem 21 11

12 4 Discussions It is vey natual to ask the convese of ou hypeplane aangement poblem, leading to the following: Poblem 2 (Second Hypeplane Aangement Poblem) Given a k-dimensional vecto space C ove F, and an intege d such that 1 d [k, find d hypeplanes H 1,, H d of C that minimizes the following uantity H 1 H d Histoically, thee have been uite a lot of studies on hypeplane aangements The extemal poblems such as the poblem of dividing a space into as many egions as possible given a fixed numbe of hypeplanes ae mostly on infinite vecto spaces Aangement poblems and esults on finite fields mostly ae about algebaic and stuctual infomation (Möbius functions, Poincaé polynomials, ) o topological stuctues The eade is efeed to [23 fo a good teatment of such poblems Ou two hypeplane aangement poblems ae new, to be best of the autho s knowledge It is possible to show that the packing aangement is the best fo d = + 5 ( 3) and so on, but the cuent method becomes too tedious to be useful We conjectue that the packing aangement is best fo Poblem 1 when 1 + < d We also leave open Poblem 2 at this point Refeences [1 M AIGNER, Seaching with lies, J Combin Theoy Se A, 74 (1996), pp [2 G E ANDREWS, The theoy of patitions, Cambidge Univesity Pess, Cambidge, 1998 Repint of the 1976 oiginal [3 D J BALDING, W J BRUNO, E KNILL, AND D C TORNEY, A compaative suvey of non-adaptive pooling designs, in Genetic mapping and DNA seuencing (Minneapolis, MN, 1994), Spinge, New Yok, 1996, pp [4 D J BALDING AND D C TORNEY, Optimal pooling designs with eo detection, J Combin Theoy Se A, 74 (1996), pp [5 A E F CLEMENTI, A MONTI, AND R SILVESTRI, Selective families, supeimposed codes, and boadcasting on unknown adio netwoks (extended abstact), in Poceedings of the Twelfth Annual ACM-SIAM Symposium on Discete Algoithms (Washington, DC, 2001), Philadelphia, PA, 2001, SIAM, pp [6 A DE BONIS AND U VACCARO, Efficient constuctions of genealized supeimposed codes with applications to goup testing and conflict esolution in multiple access channels, in Algoithms ESA 2002, vol 2461 of Lectue Notes in Comput Sci, Spinge, Belin, 2002, pp [7, Constuctions of genealized supeimposed codes with applications to goup testing and conflict esolution in multiple access channels, Theoet Comput Sci, 306 (2003), pp [8 D-Z DU, F HWANG, W WU, AND T ZNATI, New constuction fo tansvesal design, Jounal of Computational Biology, 13 (2006), pp [9 D-Z DU AND F K HWANG, Combinatoial goup testing and its applications, Wold Scientific Publishing Co Inc, Rive Edge, NJ, 1993 [10 A D YACHKOV, F HWANG, A MACULA, P VILENKIN, AND C W WENG, A constuction of pooling designs with some happy supises, Jounal of Computational Biology, 12 (2005), pp [11 A G D YACHKOV, A J MACULA, AND V V RYKOV, New applications and esults of supeimposed code theoy aising fom the potentialities of molecula biology, in Numbes, infomation and complexity (Bielefeld, 1998), Kluwe Acad Publ, Boston, MA, 2000, pp [12 A G D YACHKOV, A J MACULA, JR, AND V V RYKOV, New constuctions of supeimposed codes, IEEE Tans Infom Theoy, 46 (2000), pp [13 A G D YACHKOV AND V V RYKOV, A suvey of supeimposed code theoy, Poblems Contol Infom Theoy/Poblemy Upavlen Teo Infom, 12 (1983), pp

13 [14 W H KAUTZ AND R C SINGLETON, Nonandom binay supeimposed codes, IEEE Tans Inf Theoy, 10 (1964), pp [15 H K KIM AND V LEBEDEV, On optimal supeimposed codes, J Combin Des, 12 (2004), pp [16 E KNILL, W J BRUNO, AND D C TORNEY, Non-adaptive goup testing in the pesence of eos, Discete Appl Math, 88 (1998), pp [17 X MA AND L PING, Coded modulation using supeimposed binay codes, IEEE Tans Infom Theoy, 50 (2004), pp [18 A J MACULA, A simple constuction of d-disjunct matices with cetain constant weights, Discete Math, 162 (1996), pp [19, Eo-coecting nonadaptive goup testing with d e -disjunct matices, Discete Appl Math, 80 (1997), pp [20 S MUTHUKRISHNAN, On optimal stategies fo seaching in pesence of eos, in Poceedings of the Fifth Annual ACM-SIAM Symposium on Discete Algoithms (Alington, VA, 1994), New Yok, 1994, ACM, pp [21 H Q NGO AND D-Z DU, A suvey on combinatoial goup testing algoithms with applications to DNA libay sceening, in Discete mathematical poblems with medical applications (New Bunswick, NJ, 1999), vol 55 of DIMACS Se Discete Math Theoet Comput Sci, Ame Math Soc, Povidence, RI, 2000, pp [22, New constuctions of non-adaptive and eo-toleance pooling designs, Discete Math, 243 (2002), pp [23 P ORLIK AND H TERAO, Aangements of hypeplanes, vol 300 of Gundlehen de Mathematischen Wissenschaften [Fundamental Pinciples of Mathematical Sciences, Spinge-Velag, Belin, 1992 [24 J K PERCUS, O E PERCUS, W J BRUNO, AND D C TORNEY, Asymptotics of pooling design pefomance, J Appl Pobab, 36 (1999), pp [25 J H VAN LINT AND R M WILSON, A couse in combinatoics, Cambidge Univesity Pess, Cambidge, second ed, 2001 [26 J K WOLF, Bon again goup testing: multiaccess communications, IEEE Tansaction on Infomation Theoy, IT-31 (1985), pp