Polygonal Designs: Existence and Construction John Hegean Departent of Matheatics, Stanford University, Stanford, CA 9405 Jeff Langford Departent of Matheatics, Drake University, Des Moines, IA 5011 G Bhattacharyya, J Ki and S Y Song Departent of Matheatics, Iowa State University, Aes, Iowa, 50011, U S A Septeber 5, 004 Abstract Polygonal designs for a special class of partially balanced incoplete block designs We resolve the existence proble for polygonal designs with various paraeter sets and present several construction ethods with blocks of sall sizes 1 Introduction and preliinaries Soe results and notation that are used throughout the paper are collected in this section for convenience For positive integers v, b, k, and r with < k < v, 1 r < b, a (doubly regular) incoplete block design with paraeters (v, b, k, r), is a pair (V, B), where V is a set of v eleents, called points or varieties and B is a collection of b k-eleent subsets of V called blocks, satisfying the condition: each point appears in exactly r blocks An incoplete block design is said to be balanced if any two points are contained in exactly λ blocks together It is called partially balanced if every pair of points occurs in a certain nuber of blocks depending on an association relation between the points A special class of partially balanced incoplete block designs called polygonal designs can be defined on a regular polygon The set V fors a v-gon with vertex (point) set V = Z v = {0, 1,,, v 1} This research was supported by NSF grant DMS-05880 1
We define the distance δ(x, y) between points x and y to be the length of the shortest path connecting x and y on V That is, for x, y V v δ(x, y) := in{ x y, v x y }, and thus 0 δ(x, y) Definition 11 Let V = {0, 1,, v 1} be the point set of a regular v-gon, and let < (v 1)/ An incoplete block design (V, B) with paraeters (v, b, k, r) is called a polygonal design with iniu interval, if any two points of V that are at distance + 1 or greater appear together in λ blocks while other pairs do not occur in the blocks at all This design is denoted by PD(v, k, λ; ) We note that a PD(v, k, λ; 0) is a balanced incoplete block design which is also known as a -(v, k, λ) design A nuber of authors have provided solutions to the existence and construction probles of polygonal designs for various cobinations of v, b, k and λ Soe of the relevant references, alost all of which deal priarily with the case = 1, are as follows Hedayat, Rao, and Stufken ([]) introduced polygonal designs for the first tie in 1988 as balanced sapling plans excluding contiguous units pertaining to finite population sapling They showed that v k is a necessary condition for the existence of PD(v, k, λ; 1) They also provided an iterative construction ethod by showing that if a PD(v, k, λ; 1) exists, then a PD(v + α, k, λ ; 1) exists for any positive integer α Stufken, Song, See and Driessel ([7]) showed that if a PD(v, k, λ; ) exists, then b v and v k( + 1) They also showed that a PD(k, k, λ; 1) does not exist for any λ if k 5 In regard to the construction of designs, Colbourn and Ling ([1], []) constructed all PD(v, k, λ; 1) for k = and k = 4 Stufken and Wright ([8]) constructed all possible PD(v, k, λ; 1) with k = 5, 6 and 7, except possibly one, and several designs with block size 9 and 10 In this paper, we study the polygonal designs with v = k( + 1) for an arbitrary We resolve the existence of polygonal designs with v = k( + 1) and k = copletely We present this result in Section In Section, we show that if a PD(k( + 1), k, λ; ) exists, then so does PD((k 1)( + 1), k 1, λ ; ) for soe λ We also show that given a PD(v, k, λ; ) we can construct PD(v + ( + 1)α, k, λ ; ) for any positive integer α with soe λ These results are a generalization of soe of the results provided in [] and [7] In Section 4, we introduce a new construction ethod for PD(v, k, λ; ) by using a perfect (k, )-grouping We also show that the inequality k(k 1) 4( + 1) holds in a PD(k( + 1), k, λ; ) This result confirs the non-existence of PD(k, k, λ; 1) for k 5 which was proved in [7] Throughout the paper, we shall use V = {0, 1,, v 1} and B = {B 1, B,, B b } for the point set and block set of a PD(v, k, λ; ) unless otherwise specified Whenever we consider a block B i = {b i1, b i,, b ik } B, we shall assue that the points are ordered as 0 b i1 < b i < < b ik v 1 Given an integer a and a block B i, by B i + a we shall denote the set {b i1 + a, b i + a,, b ik + a} where eleents are reduced odulo v if needed We shall also consider the differences x y (coputed odulo v) as well as the distances
δ(x, y) between the points x and y Let B i = {b i1, b i,, b ik }, i = 1,,, g, be g blocks of size k based on V = {0, 1,, v 1} Consider the gk(k 1) differences and call the ulti-set b ij b il, j, l {1,,, k}, j l, 1 i g, D = {b ij b il : j, l {1,,, k}, j l, 1 i g} the difference collection of G = {B 1, B,, B g } Notice that we use the set notation, the curly bracket, for any collection of objects whether it is a set or a ulti-set We also often allow a design to have repeated blocks We note that in a polygonal design PD(v, k, λ; ), the difference collection of the entire block set B consists of each integer 1 through v 1 equally B k(k 1) v 1 ties Definition 1 A collection G = {B 1, B,, B g } of k-subsets B i of a v-set V is called a generating set of a PD(v, k, λ; ) (V, B) if B = {B i + a : 0 a t i 1, 1 i g} where t i is the sallest positive integer such that B i + t i B i (od v) A polygonal design is cyclically generated if and only if it has a generating set Note that for ost blocks in the generating set t i will siply be v However, for blocks that are rotationally syetric, t i will be a proper divisor of v When the differences are coputed odulo v, each distance represented in a block, B i, will occur a ultiple of v/t i ties in that block Suppose all t i are equally v for a set G of blocks Then it follows that G generates a polygonal design if and only if each of the differences +1, +,, v (+1) are represented in the difference collection of G exactly λ ties This stateent can be odified slightly when G has a rotationally syetric block, that is, there is a t i that is a proper divisor of v We shall restrict our attention to cyclically generated designs We note that there are no known polygonal designs that are not cyclically generated It is obvious that whenever we have a polygonal design we can derive a cyclically generated polygonal design, perhaps, with a larger block set while keeping v, k and constant For given and k the polygonal designs PD(v, k, λ; ) can possibly exist only when v ( + 1)k and r = λ(v 1)/(k 1) When we look at the class of polygonal designs with v = ( + 1)k (in this case, we ust have b = ( + 1) λ and r = ( + 1)λ), the following lea, which is Corollary in [7] is very useful (For the proof of this lea we refer the reader to [7]) Lea 11 Let B = {b 1, b,, b k } be a block in a PD(( + 1)k, k, λ; ), where b 1 < b < < b k If 1 i < j k, then the difference b j b i lies in the set {(j i)( + 1), (j i)( + 1) ( 1),, (j i)( + 1) + }
Existence and construction of PD(6 +,, λ; ) In this section, we present a series of existence theores, whose proofs provide construction ethods for polygonal designs of block size k = on ( + 1) points with the exception of the case when (od ) and λ = 1 For this exceptional case a cyclically generated polygonal design does not exist For these constructions, we will be concerned with the distances (rather than differences), as by obtaining the distances of + 1, +,, + 1, we get all of the differences +1, +,, 5+ For polygonal designs PD(6+,, λ; ) we can write the possible sets of distances, contributed to the difference collection D of a generating set, by a single generating block in the for (d 1, d, d ) For this to be a possible triple of distances we ust have either d 1 + d + d = 6 + or d 1 + d = d In the case that we are exaining; ie, with k = and v = 6 +, d 1 + d = d is excluded as a possibility because d 1 + d ust be in { +, +,, 5 + } due to Lea 11 The following technical lea will be used repeatedly Lea 1 The integers 1,,,, n + 1 excluding n+1 can be partitioned into triples (Y 1, Y, Y ) such that Y 1 + Y = Y Proof: In the following we present two side-by-side tables having headers Y 1, Y, and Y Each row fro each table presents a valid Y 1, Y, Y triple These tables give the required partitions where each integer appears in exactly one triple Case 1: n is even Y 1 Y Y n n + 1 n + 1 n n + n n 4 n + n 1 n n + Y 1 Y Y n 1 n + n + 1 n n + n n 5 n + 4 n 1 5n 1 + 1 5n + Case : n is odd Y 1 Y Y n n + 1 n + 1 n n + n n 4 n + n 1 1 5n+1 5n+1 + 1 Y 1 Y Y n 1 n + 1 n n n + n 1 n 5 n + n n n + 4
Theore PD(6 +,, 1; ) exists for 1 (od ) Proof: To deonstrate this we partition the distances + 1, +,, + 1 into triples (d 1, d, d ) such that d 1 + d + d = 6 + and each distance is in exactly one triple First, rewrite the distances +1, +,, +1 as (+1), (+1) ( 1),, (+1)+ We use Lea 1 in order to partition the integers 1,,, with = n + 1 into triples (Y 1, Y, Y ) such that Y 1 +Y = Y Let (Y 1, Y, Y ) be one such triple in the partition of 1,,, We then see that the triples (( + 1) Y 1, ( + 1) Y, ( + 1) + Y ) and (( + 1) + Y 1, ( + 1) + Y, ( + 1) Y ) are both valid distance triples, as both su to 6 + Therefore, for each triple (Y 1, Y, Y ) of eleents fro {1,,, } we have two valid distance triples Now we observe that every distance +1 through +1 except for + 1, + 1, and + 1 + appears in exactly one of the distance triples However, the distances +1, +1 clearly for a final valid distance triple as they su to 6 + As all and + 1 + distances now appear in one distance triple we can for a generating set for the PD(6 +,, 1; ) by foring a generating block {1, d 1 + 1, d 1 + d + 1} for each distance triple (d 1, d, d ), showing that a PD(6 +,, 1; ) exists for all 1 (od ) Theore PD(6 +,, 1; ) exists for 0 (od ) Proof: In the following tables, triples (X 1, X, X ) are presented such that X 1 +X +X = 0 Fro each triple (X 1, X, X ) we siply for the distance triple (( + 1) + X 1, ( + 1) + X, ( + 1) + X ) Each of the necessary distances will appear once in the resulting triples and thus we can use the to for blocks which will generate PD(6 +,, 1; ) 5
Case 1: is odd X 1 X X X 1 X X ( + 1) + 1 ( 1) ( + 1) ( ) ( + ) ( ) ( + ) 1 ( 4) ( + ) 1 ( 5) ( + ) 1 +1 +1 + 1 ( 5+ 5+ 6 1) 6 + 1 +1 5+ 6 + 4 + ( 1) 5 Case : is even 1 + 1 + ( 1) + ( ) 5+ 1 6 ( 5+ +1 6 + 1) 1 ( +1 + 1) X 1 X X X 1 X X ( + 1) + 1 ( 1) ( + 1) ( ) ( + ) ( ) ( + ) 1 ( 4) ( + ) 1 ( 5) ( + ) + 1 5 6 5 6 + 1 + 1 ( 5 6 + 1) 1 + 1 + + ( 1) 4 + ( 1) 5 This copletes the proof + ( ) 5 6 1 ( 5 6 + 1) 1 ( + 1) Theore 4 PD(6 +,, ; ) exists for (od ) Proof: Case 1: is even Use Lea 1 to partition the integers 1,,, ( 1) and then for two distance triples as before fro each triple in this partition By repeating this a second tie we 6
will have every distance appearing twice in the resulting triples with the exception of the distances ( + 1), ( + 1), ( + 1) + 0, ( + 1) +, ( + 1) + which do not appear at all Now for the distance triples (( + 1) +, ( + 1) +, ( + 1) ), (( + 1), ( + 1), ( + 1) + ), and (( + 1), ( + 1) + 0, ( + 1) + ) This leaves us with every distance appearing twice except for + 1 which appears once However, ( + 1) = 6 + and so we can for the rotationally syetric block that only contributes a single distance of + 1 to the difference collection We now have each of the necessary distances represented twice and so we can for generating blocks for PD(6 +,, ; ) when (od ) Case : is odd Observe that by increasing the second and third values of each triple in Lea 1 by one we get a partition of the integers 1,,, excluding +1 and +1 into triples (X 1, X, X ) such that X 1 + X = X Using the previous partition of 1,,, 1 excluding 1 and this new partition each one to for distance triples as before, we are left issing one instance of ( + 1), ( + 1) + 1, ( + 1) 1, ( + 1) + 1, ( + 1) + + 1, ( + 1) + 1, ( + 1) + + 1, ( + 1) + and two instances of + 1 We now for the distance triples (( + 1), ( + 1) + 1, ( + 1) + + 1 ), (( + 1) +, ( + 1) 1, ( + 1) + 1 ), (( + 1) + 1, ( + 1) + 0, ( + 1) + 1 ) and (( + 1) + 0, ( + 1) + 0, ( + 1) + 0) Each distance is now represented exactly twice in this group of valid distance triples and so we can generate PD(6 +,, ; ) for all (od ) 7
Theore 5 PD(6 +,, ; ) exists for all Proof: Fro the following table we can for the distance triples (( + 1) + X 1, ( + 1) + X, ( + 1) + X ) The resulting distance triples contain each of the distances + 1, +,, + 1 inclusive exactly ties and so can be used to construct a PD(6 +,, ; ) X 1 X X 0 ( 1) 1 ( ) 4 0 1 1 ( 1) 1 ( 1) Reark 1 A PD(v, k, λ; ) can be fored fro PD(v, k, 1; ) by siply repeating each of the blocks in PD(v, k, 1; ) λ ties Siilarly, PD(v, k, λ; ) for λ can be fored fro a linear cobination of the blocks fro PD(v, k, ; ) and PD(v, k, ; ) We thus have that PD(6 +,, λ; ) exists for all cobinations of λ and except for when (od ) and λ = 1 Here, a cyclically generated PD(v, k, 1; ) cannot exist for (od ) Next, we conclude the current section by introducing a way to construct polygonal designs of block size using the λ-fold triple systes discussed in Chapter of [4] A λ-fold triple syste is a pair (V, B), where V is a finite set and B is a collection of -eleent subsets of V called triples such that each pair of distinct eleents of V belongs to exactly λ triples of B Theore 6 If there exists a cyclically generated λ-fold triple syste with N (N ) points, then PD(N( + 1),, λ; ) exists Proof: First take the distance triples (n 1, n, n ) fro the generating blocks for the λ-fold triple syste Observe that either n 1 + n + n = N or n 1 + n = n ust hold Fro each triple where n 1 + n + n = N for the + 1 distance triples of the for (n 1 ( + 1) + X 1, n ( + 1) + X, n ( + 1) + X ) 8
where the values of X 1, X, and X are taken fro the rows of the following table Observe that as (n 1 ( + 1) + X 1 ) + (n ( + 1) + X ) + (n ( + 1) + X ) = N( + 1) these will be valid distance triples X 1 X X 0 ( 1) 1 ( ) 4 0 1 1 ( 1) 1 ( 1) Siilarly, fro the triples where n 1 + n = n for the + 1 distance triples of the for (n 1 ( + 1) + Y 1, n ( + 1) + Y, n ( + 1) + Y ) where the values of Y 1, Y, and Y are taken fro the rows of the following table These will also be valid distance triples as (n 1 ( + 1) + Y 1 ) + (n ( + 1) + Y ) = (n ( + 1) + Y ) Y 1 Y Y 0 1 1 4 0 1 1 1 1 ( 1) 9
Recursive construction of new designs fro old In this section we present two ways to construct another polygonal designs fro given polygonal designs Both are iterative construction ethods and they produce designs over different size of point sets Theore 1 If a cyclic PD(k(+1), k, λ; ) exists then PD((k 1)(+1), k 1, λ ; ) exists where λ = (k 1)λ Proof: Take a block fro the generating set for PD(k(+1), k, λ; ) and for the k-tuple of first distances that appear in the block: {b b 1, b b,, v + b 1 b k } = {d 1, d,, d k } Replace all of the k possible pairs of adjacent distances, d i and d i+1 (or d k and d 1 ) by d i = d i+d i+1 (+1) to for the k-tuple {d 1, d,, d k } Now consider the (k 1)-tuples of ordered distances that can be obtained fro the k-tuple {d 1, d,, d k } by reoving a single distance d i for each i = 1,,, k (There are k such (k 1)-tuples) Fro each (k 1)-tuple, say {r 1, r,, r k 1 }, for the block {0, r 1, r 1 + r,, r 1 + r + + r k 1 } of size k 1 on (k 1)( + 1) points Repeat this process for each generating block The resulting blocks will for a generating set for PD((k 1)( + 1), k 1, λ ; ) This can be verified by counting the differences in the resulting blocks First, though, we use Lea 11 Fro this lea we know that when v = k(+1) the first differences (differences between adjacent points in a block) will be eleents of the set {( + 1), ( + 1) ( 1),, ( + 1) + }, the second differences will be eleents of the set {( + 1), ( + 1) ( 1),, ( + 1) + }, and ore generally that nth differences will be eleents of the set {n( + 1), n( + 1) ( 1),, n( + 1) + } Observe that for each tie a first difference appeared in a generating block for the original design, it will appear k ties in the generating blocks for the resulting design For each tie a second difference appeared in the original generating blocks, it will appear unchanged k ties in the new generating blocks and will be reduced by + 1 to the corresponding first difference once For each tie a third difference appeared in the original design, it will appear unchanged k 4 ties and will be reduced by + 1 to the corresponding second difference twice This pattern continues on up to the (k 1)th 10
differences which will be reduced all k ties by + 1 in the resulting blocks to the corresponding (k )th difference The result is that for every tie the differences + 1, +,, (k 1)( + 1) + appeared in the generating blocks for the original design, the differences + 1, +,, (k )( + 1) + will appear k 1 ties in the resulting generating blocks Theore If PD(v, k, λ; ) exists then PD(v + ( + 1), k, λ ; ) exists where λ = (v 1)λ Proof: The new design can be constructed by taking each of the blocks fro PD(v, k, λ; ) with points labeled 0, 1,,, v 1 and replacing each instance of v 1 with v +, each instance of v with v + 5, each instance of v with v + 8,, and each instance of v with v ( 1) These new blocks for a generating set for PD(v + ( + 1), k, λ ; ) This can be seen by observing that each of the differences + 1, +,, v ( + 1) appears vλ ties in the difference collection of the blocks of PD(v, k, λ; ) The replaceent described above will increase the difference, d, by 1,,, each λ ties The distance will be increased by + 1 exactly λ(d ) ties and will be unchanged the reaining λ(v d ) ties It is straightforward to check that each of the distances + 1, +,, (v + + 1) ( + 1) appears λ(v 1) ties in these new blocks These blocks thus can be used as a generating set to for PD(v + + 1, k, λ ; ) Reark 1 It follows that if there is a PD(v, k, λ; ), then there is a polygonal design with v + ( + 1)α points in blocks of size k for any positive α This iterative ethod has been provided by John Stufken [6] as a generalization of the result of Hedayat, Rao and Stufken [] for the case of = 1 4 Construction of PD(( + 1)k, k, λ; ) fro perfect (k, )- grouping In this section we introduce another technique that ay be useful in construction and in the study of polygonal designs Definition 41 A perfect (k, )-grouping is a collection of k-eleent ulti-sets having eleents fro {0, 1,, } such that each of {1,,, } appears as the distance between 11
two eleents of a ulti-set precisely N ties and 0 appears as the distance between two eleents of a ulti-set no ore than N/ ties The k-eleent ulti-sets that for a perfect (k, )-grouping will be called groups Exaple 41 (1) The collection {{0, 1, }} of the single group {0, 1, } is a perfect (, )- grouping with N = 1, while {{0, 1, 4, 6}} is a perfect (4, 6)-grouping with N = 1 () {{0, 0, }, {0, 1, }, {0, 1, }} is a perfect (, )-grouping with N = () The collections {{0, 0, 1, 1}} and {{0, 0, 1,, }} are perfect (4, 1)- and (5, )-groupings respectively with N = 4 (4) {{0,, 4}, {0, 1, 4}, {0, 1, 4}, {0, 1, }} is a perfect (4, 4)-grouping with N = Theore 41 PD(( + 1)k, k, λ; ) exists for soe λ if and only if there exists a perfect (k, )-grouping Proof: Suppose that PD(( + 1)k, k, λ; ) exists for soe λ According to Lea 11, given a point b i in an arbitrary block, the next point b i+1 ust be an eleent of the set {b i + ( + 1), b i + ( + ),, b i + ( + 1)} More generally, point b j ust belong to {b i + ( + 1)(j i), b i + ( + 1)(j i) ( 1),, b i + ( + 1)(j i) + } Now select the point b r in the block such that b r b i ( + 1)(r i) is iniized and apply Lea 11 again We thus have that an arbitrary point b j ust belong to the set {b r + ( + 1)(j r), b r + ( + 1)(j r) ( 1),, b r + ( + 1)(j r) + } However, as the value of b r b i ( + 1)(r i) was iniized by our selection of b r, b j cannot fall in the first half of these values with respect to b r and thus b j ust belong to the set {b r + ( + 1)(j r), b r + ( + 1)(j r) + 1,, b r + ( + 1)(j r) + } This allows us to write each point, b j, in the block in the for b r + ( + 1)(j r) + a j where a j takes on one of the values 0, 1,, By writing the points of each block in this for it follows that if a ulti-set {a 1, a,, a k } is fored for each block in the polygonal design then by collecting all such ulti-sets we can for a perfect (k, )-grouping Conversely, suppose we have a perfect (k, )-grouping Take a particular group {a 1, a,, a k } and for the (k 1)! blocks of the for {f(a 1 ), ( + 1) + f(a ),, (k 1)( + 1) + f(a k )} 1
where f is taken over all of the (k 1)! perutations on the group {a 1, a,, a k } such that f(a 1 ) = a 1 When this is done for each group in the perfect (k, )-grouping, all of the necessary differences + 1, +,, v ( + 1) with possible exception of (k 1) ultiples of + 1 will each be represented N(k )! ties in the difference collections fro these (k 1)! blocks All of the differences that are ultiples of + 1 can be added to the difference collections a single tie by the syetric block {0, + 1, ( + 1),, (k 1)( + 1)} Thus, we can add soe nuber of instances of this block to the previous blocks to ake all necessary differences +1, +,, v (+1) be represented in the difference collections for these blocks precisely N(k )! ties These blocks will thus for a generating set for PD(( + 1)k, k, λ; ) where λ = N(k )! Theore 4 k(k 1) 4( + 1) is a necessary condition for the existence of a perfect (k, )-grouping Proof: Suppose that there exists a perfect (k, )-grouping where k(k 1) > 4( + 1) If this is the case, then the average nuber of ties that each of the distances 1,,, appears per group of the grouping ust be greater than 4 We will develop a contradiction by deonstrating that if the distance appears on average ore than 4 ties per group, then the distance 0 will appear ore than half as any ties as the distance This is not allowed by the definition of a perfect (k, )-grouping We start by exaining the nuber of ties the distance appears in a given group Observe that can only appear as the distance between the integers 0 and We let the nuber of instances of 0 in a particular group be denoted x and the nuber of instances of in that group be denoted y The nuber of ties that appears as a distance in this group will then be siply xy Observe that the distance 0 will appear between two points in this group a iniu of 1 (x(x 1) + y(y 1)) ties We now find for what values of x and y the distance will appear in the group at least twice as any ties as the distance 0 by solving the inequality xy x(x 1) + y(y 1), or equivalently x + y xy (x y) It is straightforward to show that the only positive integral solutions (x, y) to this inequality are (, ), (1, ), (, 1), and (1, 1) With the exception of (x, y) = (1, 1) all of these solutions yield strict equality Thus, if the average nuber of ties the distance appears per group is greater than 4, there ust be a group in which xy > 4 and thus in which the values of 1
x and y are not solutions to our inequality This eans that the distance will appear in this group fewer than twice as any ties as the distance 0 As the only type of group in which the distance can possibly appear ore than twice as any ties as the distance 0 has (x, y) = (1, 1), we ust have (x y) + xy x y blocks of this for to copensate for the group with xy > 4 The average nuber of ties that the distance appears per group in these groups is thus xy + (x y) + xy x y 1 + (x y) + xy x y This ust still be greater than 4 for it to be possible for the average nuber of ties that appears as a distance to be greater than 4 However it is straightforward to show that the expression is never greater than 4 when x and y are positive integers with xy > 4 This contradicts the original assuption thus proving that a perfect (k, )-grouping does not exist for k(k 1) > 4( + 1) Exaple 4 This bound of the above result is sharp for sall k and as perfect groupings exist for (k, ) = (4, 1) and (5, ) both of which yield k(k 1) = 4( + 1) As we have seen in () of Exaple 41, each of these groupings consists of a single group of size k with N = 4, naely {{0, 0, 1, 1}} and {{0, 0, 1,, }} We note that the polygonal designs PD(1, 4, 4; 1) and PD(5, 5, 1; ) generated by and G = {{0,, 7, 10}, {0, 4, 6, 10}, {0, 4, 7, 9}} G = {{0, 5, 11, 17, }, {0, 5, 1, 16, }, {0, 5, 1, 17, 1}, {0, 6, 10, 17, }, {0, 6, 1, 15, }, {0, 6, 1, 17, 0}, {0, 7, 10, 16, }, {0, 7, 10, 17, 1}, {0, 7, 11, 15, }, {0, 7, 11, 17, 0}, {0, 7, 1, 15, 1}, {0, 7, 1, 16, 0}}, respectively are obtained fro these perfect groupings Notice that the exaples provided here uses the ethod presented in the proof of Theore 41 for converting between perfect groupings and polygonal designs, but we have reoved the redundant generating blocks; so the nuber of blocks and λ for each designs have been reduced by a half Reark 4 While the bound of the previous theore is sharp for sall k and as we have just seen in the above exaple, for larger values of k, this bound for is not the strongest possible bound We have a better bound that approaches the relationship k 48 for large k The proof of this is given in the appendix Reark 44 As a consequence of the previous two theores, PD(( + 1)k, k, λ; ) does not exists if k(k 1) > 4(+1) That is, for = 1, there exists no polygonal designs with 14
v = k if k 5 as it was proved in [7] For =, there exist no polygonal designs with v = 5k if k 6 Also it is an iediate consequence of the theores that PD(9,, 1; 1) and PD(5, 5, 1; ) are the only syetric polygonal designs with v = b = ( + 1)k Definition 4 A natural perfect (k, )-grouping is any perfect (k, )-grouping in which every group is a set, having no redundant eleents; ie, 0 does not appear as the distance between any two eleents of any one group in the grouping For exaple, the perfect groupings listed in (1) and (4) of Exaple 41 are natural groupings now the following theore shows that if we have a perfect natural grouping, we can have another perfect grouping Theore 4 If there exists a natural perfect (k, )-grouping then there exists a natural perfect (k, )-grouping of size for all Proof: We prove this by deonstrating that a natural perfect (k, + 1)-grouping can be fored fro a natural perfect (k, )-grouping This is done by taking every group in the natural perfect (k, )-grouping and foring new groups by adding one to every eleent of the group that is greater than or equal to n Do this for n = 1,,, + 1 to for + 1 new groups for each group in the natural perfect (k, )-grouping Each of the distances 1,,, will be represented an equal nuber of ties in the resulting groups and 0 will not appear at all For exaple, the perfect (4, 4)-grouping in (4) of Exaple 41 can be obtained fro the perfect (, )-grouping given in (1) of Exaple 41 in the anner described in the above proof Reark 45 For k = or 4 there exist natural perfect (k, )-groupings fored by a single k-eleent set These are the perfect (,)-grouping and (4,6)-grouping illustrated in (1) of Exaple 41 Unfortunately, there are no natural perfect (k, )-groupings consisting of a single (ulti-)set for any other cobinations of k and Natural perfect groupings with a single (ulti-)set are very difficult to construct and as a result we have not found any natural perfect (k, )-groupings for k 5 Reark 46 As we have found a natural perfect (, )-grouping and (4, 6)-grouping, it thus follows fro Theores 41 and 4 that PD((+1)k, k, λ; ) exists for soe λ when k = and or k = 4 and 6 Additionally, for k =, = 1, and k = 4, = 1,,, 5 we have anually confired the existence of such designs Thus, PD(( + 1),, λ; ) and PD(( + 1)4, 4, λ; ) exist for soe λ given any 15
Acknowledgent This research was perfored by the first two authors when they visited Iowa State University in 004 The authors would like to thank the Departent of Matheatics at Iowa State University for its support of the authors during the period that this research was perfored References [1] Colbourn, C J and Ling, A C(1998) A class of partial triple systes with applications in survey sapling Coun Statist Theory Meth, 7(1), 1009-1018 [] Colbourn, C J and Ling, A C(1999) Balanced sapling plans with block size four excluding contiguous units Australasian J of Cobin, 0, 7-46 [] Hedayat, A S, Rao, C R, and Stufken, J (1988), A sapling plan excluding contiguous units, J Stat Plann Infer 19, 159-170 [4] Lindner C C and Rodger, C A (1997) Design Theory CRC Press, Boca Raton, Florida [5] See, K, Song, S Y, and Stufken, J (1997) On a class of partially balanced incoplete block designs with applications in survey sapling Coun Statist Theory Meth, 6(1), 1-1 [6] Stufken, J (199) Cobinatorial and statistical aspects of sapling plans toavoid the selection of adjacent units J Cobin Infor Syste Sci 18, 81-9 [7] Stufken, J, Song, S Y, See, K, and Driessel, K (1999) Polygonal designs: Soe existence and non-existence results J Stat Plann Infer 77, 155-166 [8] Stufken, J and Wright, J (001) Polygonal designs with blocks of size K 10 Metrika 54, 179-184 16
Appendix Theore For given k, let the iniu value of for which a perfect (k, )-grouping exists be denoted k Then li k 1 k k 48 Proof: We shall prove the stateent only for the case when (od 4), as belongs to the other congruence classes odulo 4 can be treated siilarly Suppose there exists a perfect (k, )-grouping that consists of R groups of the for a 1, a,, a k with a 1 a a k Observe that the su of the distances between points in a group a 1, a,, a k will be (k 1)(a 1 + a k 1 ) + (k 4)(a + a k ) + + (nk n )(a n + a k n ) + ( (k ) ( ) ) 1 k 1 ( ) + k a ( k 1 ) + a ( k+1 ) where n = 1,,, k 1 We can rewrite this as ( (k 1 ) k ( ) ) k 1 k 1 a i i=1 k 1 ( (k ) ( ) 1 k 1 k (nk n )) (a n + a k n ) n=1 Suing up this over all groups and replacing k 1 ( (k 1 ) k ( ) ) k 1 R groups k 1 n=1 i=1 a i by (as k 1 ( (k ) 1 k i=1 a i ), we obtain ( ) k 1 (nk n )) (a n + a k n ) for an upper bound on the su of all distances in all groups Observe that when this expression achieves its axiu value it ust be the case that a n, a k n [ n+1 ] for n = 1,,, k 1 We thus have another upper bound RU for the su where U := ( (k 1 ) k ( ) ) k 1 k 1 ( (k ) ( ) 1 k 1 k (nk n )) n=1 [ n ] Note that U is an upper bound for the average su of the distances per group This ust be greater than or equal to the nuber of distances per group ultiplied by the iniu average distance which is k(k 1) We thus have the inequality (+1) +1 U k(k 1) ( + 1) + 1 17
Multiplying both sides of this inequality by + 1, expanding, siplifying, dividing by k and taking the liit as k approaches infinity yields k li k k 1 48 18