The theory of design of experiments came into being largely through the work of R.A.Fisher and F.Yates in the early 1930's. They were motivated by questions of design of careful field experiments in agriculture. Although the applicability of this theory is now very widespread, much of the terminology still bears the stamp of its origins.

Consider an agricultural experiment. Suppose it is desired to compare the yield of v different varieties of grain. It is quite possible that there would be an interaction between the environment (type of soil, rainfall, drainage, etc.) and the variety of grain which would alter the yields. So, b blocks (sets of experimental plots) are chosen in which the environment is fairly consistent throughout the block. In other types of experiments, in which the environment might not be a factor, blocks could be distinguished as plots which receive a particular treatment (say, are given a particular type of fertilizer). In this way, the classification of the experimental plots into blocks and varieties can be used whenever there are two factors which may influence yield.

The obvious technique of growing every variety in a plot in every block, may, for large experiments be too costly or impractical. To deal with this, one would use smaller blocks which did not contain all of the varieties. Now the problem is one of comparison, to minimize the effects of chance due to incomplete blocks, we would want to design the blocks so that the probability of two varieties being compared (i.e. are in the same block) is the same for all pairs. This property would be called balance in the design. Statistical techniques, in particular Analysis of Variance, could then be used to reach conclusions about the experiment.


A BIBD is a set X of v 2 elements called varieties or treatments and a collection of b > 0 subsets of X, called blocks, such that the following conditions are satisfied: The incomplete in the name of these designs refers to the condition v > k, i.e., the block size, k , is less than the total number of treatments, so no block contains all the varieties. If we allowed v = k, then all the conditions would be trivially satisfied and the resulting design would not be of much interest. Balanced refers to the constancy of the parameter. BIBD's are often referred to as (v,b,r,k,)- designs.

An example of a (7,7,3,3,1)-design is given by the set X consisting of the varieties, 1,2,3,4,5,6,7 and the following blocks:

{1,2,4} {2,3,5} {3,4,6} {4,5,7} {5,6,1} {6,7,2} {7,1,3}.
It is easily seen that the block size k is 3, and the repetition number r is also 3. Each pair of varieties appears together in only one block, so = 1.

A (4,4,3,3,2)-design is given by X = {1,2,3,4} with blocks:

{1,2,3} {2,3,4} {3,4,1} {4,1,2}.

A slightly larger example is the (8,14,7,4,3)-design on the set X = {1,2,3,4,5,6,7,8} with blocks:

{1,3,7,8} {1,2,4,8} {2,3,5,8} {3,4,6,8} {4,5,7,8} {1,5,6,8} {2,6,7,8} {1,2,3,6} {1,2,5,7} {1,3,4,5} {1,4,6,7} {2,3,4,7} {2,4,5,6} {3,5,6,7}.

Some elementary counting arguments show that there are necessary conditions that the parameters of a BIBD must satisfy.

Theorem VII.1.1 - Given a (v,b,r,k,)-design,

bk = vr
r(k-1) = (v-1).

Proof: Consider the set of pairs (x,B), where x is a variety and B is a block containing x. By counting this set in two ways we arrive at the first equation. There are v possible values for x, and since each appears in r blocks, vr will count the number of these pairs. On the other hand, there are b blocks and each contains k varieties, so bk also counts the number of these pairs.

The second equation is also obtained by counting. Fix a particular variety, say p, and count the number of pairs of varieties {p,y} where p and y appear in some block together and if the pair appears more than once it is multiply counted. There are v - 1 possible choices of y and each such pair will appear in blocks together, so there are (v-1) such pairs. On the other hand p appears in r blocks and can be paired with k - 1 other elements in such a block, thus r(k-1) =(v-1).

These conditions on the parameters imply that r and b can be calculated if v,k and are known.

Given a (v,b,r,k,)-design, we can represent it with a v × b matrix called the incidence matrix of the design. The rows are labeled with the varieties of the design and the columns with the blocks. We put a 1 in the (i,j)-th cell of the matrix if variety i is contained in block j and 0 otherwise. Each row of the incidence matrix has r 1's, each column has k 1's and each pair of distinct rows has common 1's. These observations lead to a useful matrix identity.

Theorem VII.1.2 - If A is the incidence matrix of a (v,b,r,k,)-design, then

AAt = (r-)I +J
where I is the v × v identity matrix and J is the v × v matrix of all 1's.

Proof: In the product, an off diagonal entry is the inner product of two distinct rows of A, which must be. A diagonal entry is the inner product of a row of A with itself, and so equals r.

We can use this result to prove another restriction on the parameters of a BIBD, due to Fisher.

Theorem VII.1.3 - [Fisher's Inequality] In a (v,b,r,k,)-design, b v.

Proof: Suppose that b < v and let A be the incidence matrix of the design. We can add v - b columns of 0's to A to get a v × v matrix B. Since these extra columns of 0's will not alter the inner products, we must have AAt = BBt. By taking determinants we see that,

det(AAt) = det(BBt) = (det B)(det Bt) = 0
since det B = 0 due to the columns of 0's. Now by Theorem VII.1.2 we have
det(AAt) = det

We can evaluate this determinant by subtracting the first column from each of the other columns and then adding each row to the first row to obtain the following:

det(AAt) = det
r + (v - 1) 000...0
r + (v - 1)00...0
0r + (v - 1)0...0
00r + (v - 1)...0
000...r + (v - 1)

So we have that,

det (AAt ) = [r + (v-1)](r-)v - 1.
But since r,v and are all positive, r + (v-1)> 0 and since k < v Thm.VII.1 implies that r >, so this product on the right can not equal 0 contradiction.

If in a BIBD we have v = b (and thus r = k), we say that the BIBD is symmetric or square (maybe, projective). Symmetric BIBD's are often referred to as (v,k,)-designs. For symmetric BIBD's, there is an additional constraint on the parameters.

Theorem VII.1.4 - [Bruck-Ryser-Chowla Theorem] The following conditions are necessary for the existence of a symmetric BIBD:

  1. If v is even, then k - is the square of an integer.
  2. If v is odd, then the following equation has a solution in integers x,y,z not all of which are 0:
    x2 = (k-)y22 + (-1)(v-1)/2z2
The proof of this result involves some very deep results from number theory and so we will not give it.

All of the results on the parameters of a BIBD that we have given have been necessary but not sufficient. That is, we can use them to rule out the existence of a BIBD for certain sets of parameters, but given a set of the parameters which satisfy all these conditions does not mean that there actually exists a BIBD with those parameters. There are many sets of possible parameters for which the existence question has not been settled.


We shall examine a number of families of BIBD's, concentrating on those which have relationships to other combinatorial structures.

VII.2.1 - k = 2, = 1

If a BIBD has the parameters k = 2 and = 1, then it is easily calculated that r = v - 1 and b = v(v - 1)/2. This means that the blocks of the design are just all possible pairs of varieties, i.e., the set of blocks is the set of all 2-subsets of X. If we interpret the varieties of the design as being vertices and the blocks as being edges, then a design with these parameters is a complete graph on v vertices. From the viewpoint of design theory, these are not very interesting designs even though they are an important class of graphs.

VII.2.2 - k = 3, = 1 - Steiner Triple Systems

When k = 3 and = 1, the BIBD is called a Steiner Triple System since the blocks are triples and each pair of varieties appears in exactly one triple. We can calculate that r = (v-1)/2 and b = v(v-1)/6. Since r is an integer, v must be odd. Since b is an integer and v is odd, either 3 divides v or 6 divides v-1, so v3 mod 6 or v1 mod 6. This necessary condition is also sufficient for the existence of a Steiner Triple System.

Theorem VII.2.2.1 - [Kirkman, 1847] There exists a Steiner Triple System of v varieties iff v3 and v1 or 3 mod 6.

Proof: We have already shown the necessity of these conditions. The sufficiency is proved by construction, but as this is rather long and messy we will not present it [details may be found in M. Hall's, Combinatorial Theory].

In 1853 J. Steiner posed the sufficiency of this theorem as a problem and it was proved in 1859 by M. Reiss. Neither mathematician was aware of the fact that the problem had been posed and solved by T.P. Kirkman in an 1847 article appearing in the Cambridge and Dublin Mathematical Journal. Indeed, in 1850 Kirkman went on to pose a more difficult but related problem. This problem, which appeared in "The Lady's and Gentleman's Diary" of 1850, has become to be known as Kirkman's Schoolgirl Problem and was presented as follows:

A teacher would like to take 15 schoolgirls out for a walk, the girls being arranged in 5 rows of three. The teacher would like to ensure equal chances of friendship between any two girls. Hence it is desirable to find different row arrangements for the 7 days of the week such that any pair of girls walk in the same row exactly one day of the week.
This problem, which in general asks for a Steiner Triple System on 6t +3 varieties whose blocks can be partitioned into 3t + 1 sets so that any variety appears only once in a set, had not been settled in general until 1971 (Ray-Chaudhuri and Wilson).

There are a number of constructions of STS's, we present only one of them.

Theorem VII.2.2.2 - If there is a Steiner Triple System on e varieties and a Steiner Triple System on f varieties then there is a Steiner Triple System on ef varieties.

Proof: Let A be the STS whose elements are a1 ,a2 ,...,ae and B the STS whose elements are b1 ,b2 ,...,bf . Let S consist of the ef elements cij, where i = 1, ...,e and j = 1,...,f. A triple {cir,cjs,ckt} is in the STS based on S if and only if one of the following conditions hold:

  1. r = s = t and {ai, aj, ak } is in A.
  2. i = j = k and {br, bs, bt } is in B.
  3. {ai, aj, ak } is in A and {br, bs, bt } is in B.
We leave it as an exercise to show that these triples on S form a STS.

It has been shown that there is only one STS of orders v = 3, 7 or 9. There are two nonisomorphic designs for v = 13 and 80 distinct STS's with v = 15. This work was done by hand but the v = 15 case was verified by computer. Some upper and lower bounds are known for other STS's but no other exact counts.

Of course, it is well known that Steiner Triple Systems are equivalent to idempotent totally symmetric quasigroups and so are connected to Latin Squares.

VII.2.3 - Hadamard 2-Designs

Hadamard matrices of order 4t can be used to create symmetric BIBD's, which are called Hadamard 2-Designs. The construction actually forms the incidence matrix of the BIBD, from which the design is easily obtained. The Hadamard designs have parameters v = 4t - 1, k = 2t - 1 and = t - 1, or v = 4t - 1, k = 2t, and = t. The construction, as we shall see, is reversible, so that BIBD's with these parameters can be used to construct Hadamard matrices.

Let H be an Hadamard matrix of order 4t. First normalize the matrix H (so that the first row and column are just +1's), then remove the first row and column. The 4t-1 × 4t-1 matrix which remains, say A, has 2t -1's in each row and column and 2t-1 +1's in each row and column, so the row and column sums are always -1 for A. The inner product of two distinct rows of A will be -1 and the product of a row with itself will be 4t-1. These statements are summarized by the matrix equations,

AJ = JAt = -J and AAt = 4tI - J
where I is the identity matrix and J is the all one matrix of the appropriate order. Now construct the matrix B = ½(A + J). B is a (0,1)-matrix, whose row and column sums are 2t-1, i.e., BJ = JB = (2t-1)J. Furthermore, the matrix equation,
BBt = tI + (t-1)J
is satisfied [verify]. Comparing this to the result of Thm. VII.1.2, we see that B is the incidence matrix of a symmetric (since B is a square matrix) BIBD with v = 4t-1, k = 2t-1 and = t-1. Similarly, if C = ½(J - A), C will be the incidence matrix of a (4t-1,2t,t)-design.

For example, let H be the 8 × 8 Hadamard matrix in the Hadamard Lecture Notes. In normalized form we have,

    + + + + + + + +
    + + - - + + - -       + - - + + - -         1 0 0 1 1 0 0
    + - + - + - + -       - + - + - + -         0 1 0 1 0 1 0
H = + - - + + - - +       - - + + - - +         0 0 1 1 0 0 1
    + + + + - - - -  A =  + + + - - - -    B =  1 1 1 0 0 0 0
    + + - - - - + +       + - - - - + +         1 0 0 0 0 1 1
    + - + - - + - +       - + - - + - +         0 1 0 0 1 0 1
    + - - + - + + -       - - + - + + -         0 0 1 0 1 1 0
So, labeling the rows of B with {1,2,...,7} we have the (7,3,1)-design whose 7 blocks are:
{1,4,5} {2,4,6} {3,4,7} {1,2,3} {1,6,7} {2,5,7} {3,5,6}
The matrix C is B with the 0's and 1's interchanged (a design obtained by interchanging the 0's and 1's of an incidence matrix is called the complementary design of the original) and its blocks form the (7,4,2)-design:
{2,3,6,7} {1,3,5,7} {1,2,5,6} {4,5,6,7} {2,3,4,5} {1,3,4,6} {1,2,4,7}.

Exercise: Prove that an Hadamard design (i.e. a symmetric BIBD with either of these sets of parameters) can be used to construct an Hadamard matrix.

VII.2.4 - Designs from MOLS

Complete sets of MOLS of order q can be used to construct two related families of BIBD's. The first family has the parameters:
v = q2      b = q(q + 1)     r = q + 1     k = q     = 1
and the second is a symmetric BIBD with parameters
v = q2 + q + 1     k = q + 1     = 1.
These designs are examples of finite geometries, the first family affine and the second projective (these will be studied in Chap. VIII).

Let L1, L2, ..., Lq-1 be a complete set of MOLS of order q and A a q × q matrix containing q2 distinct symbols. We define blocks of size q on this set of q2 symbols in the following way: There are q blocks which are the rows of A, and q blocks which are the columns of A. The remaining q2 - q blocks are formed by taking each Li in turn, superimposing it on A and taking as blocks the elements of A which correspond to a single symbol in the Li.

As an example consider the set of 3 MOLS of order 4:

                         1  2  3  4             1  2  3  4          1  2  3  4
                         2  1  4  3             3  4  1  2          4  3  2  1
                         3  4  1  2             4  3  2  1          2  1  4  3
                         4  3  2  1             2  1  4  3          3  4  1  2
Now, let A be the matrix,
                                         1   2   3   4
                                         5   6   7   8
                                         9  10  11  12
                                         13 14  15  16
The blocks of the design are:
{1,2,3,4} {5,6,7,8} {9,10,11,12} {13,14,15,16} {1,5,9,13} {2,6,10,14} {3,7,11,15} {4,8,12,16} - from the rows and columns, and {1,6,11,16} {2,5,12,15} {3,8,9,14} {4,7,10,13} {1,7,12,14} {2,8,11,13} {3,5,10,16} {4,6,9,15} {1,8,10,15} {2,7,9,16} {3,6,12,13} {4,5,11,14} - from the Latin squares.
The second design is obtained from the first as follows: To each block which was a row of A, a new element is added. To each block which was a column of A, a different new element is added. Also, other new elements are added to the blocks which come from the same Latin square. Thus, q + 1 new elements are added and each block now has size q + 1. One new block is added to the design, it contains all of the new elements. This modified design now has the required parameters.

Exercise: Construct two designs from the 2 MOLS of order 3.


There are a number of methods for constructing BIBD's. They are generally divided into two classes, direct and recursive. The direct methods usually give BIBD's with special restrictions on the parameters, while the recursive methods provide entire families of BIBD's. We will examine some of the direct methods and ignore the recursive ones.

We will first look at three ways to construct new BIBD's from a given symmetric BIBD. The restriction to symmetric BIBD's is a consequence of the following theorem.

Theorem VII.3.1 - In a (v,k,) - design, any two blocks have exactly elements in common.

Proof: Let A be the incidence matrix of the design. By Thm. VII.1.2 we have that

AAt = (r - )I + J
and from the proof of Thm. VII.1.3 that,
det (AAt ) = rk(r -)v-1.
Since (v-1) = r(k-1) and v > k, we must have r >, thus det (AAt ) > 0. Also since the design is symmetric, det(A) = det(At ) and we may conclude that det(A)0, so A-1 exists. Now, consider the calculation,
AAtA = ((r-)I + J)A = A((r-)I + J) = AAAt
since JA = AJ = kJ. We may multiply by A-1 to conclude that
AAt = AtA.
The (i,j)-th entry in the product on the left isif ij, however this entry in the product on the right is the inner product of columns i and j of A, i.e., it gives the number of elements in common in the two blocks which are represented by these columns.

Given a (v,k,)-design, D, we can form its dual design by interchanging the roles of the varieties and the blocks, that is, we number the blocks of D in any manner and the set of varieties of the dual design is the collection of numbers used, the blocks of the dual are associated to the varieties of D in the following manner, for each variety of D we form a block of the dual design by using all the numbers corresponding to the blocks of D which contain this variety. Theorem VII.3.1 then tells us that every pair of varieties in the dual design is contained in the same number of blocks of the dual design. It is clear that if A is the incidence matrix of the design D, then At is the incidence matrix of the dual design. It is also not hard to see that the dual of the dual design is the original design. The parameters of the dual design are the same as those of the original. It may turn out that the dual of a design is the original design, in which case we say that the design is self-dual. Consider the (4,3,2)- design at the beginning of this chapter, for clarity we will label the blocks with letters, so

A = {1,2,3} B = {2,3,4} C = {3,4,1} D = {4,1,2}.
The dual design is based on the set {A,B,C,D} and its blocks are
1 - {A,C,D} 2 - {A,B,D} 3 - {A,B,C} 4 - {B,C,D}.
It is not difficult to see that this dual design is actually the original design with A = 1, B = 2, C = 3 and D = 4, so this design is self-dual.

Homework: What should the concept of isomorphism mean for block designs? Using your definition, show that the (4,3,2)-design above is isomorphic to its dual design.

Another design that can be obtained from a (v,k,)-design is the residual design. Pick a block from the original design, remove it and remove all of its varieties from the remaining blocks, the result is the residual design. By theorem VII.3.1, each of the blocks of the original design has varieties in common with the selected block, so the new blocks all have size k -. Each variety that remains appears in as many blocks as it did before, that is in r = k blocks. Also, any pair of remaining varieties will still appear together in blocks. The new design then has v - k varieties and only one fewer block, so it is a (v-k, v-1, k, k-,) - design.

Consider the following (11,5,2)-design based on {0,1,2,...,10};

{1,3,4,5,9} {2,4,5,6,10} {0,3,5,6,7} {1,4,6,7,8} {2,5,7,8,9} {3,6,8,9,10} {0,4,7,9,10} {0,1,5,8,10} {0,1,2,6,9} {1,2,3,7,10} {0,2,3,4,8}.
We form the residual design obtained by removing the first block and all its elements to get;
{2,6,10} {0,6,7} {6,7,8} {2,7,8} {6,8,10} {0,7,10} {0,8,10} {0,2,6} {2,7,10} {0,2,8}
which is a (6,10,5,3,2) - design. It should be noted that by selecting different blocks to remove, one might obtain non-isomorphic residual designs.

Exercise: Given a design whose parameters are the parameters of a residual design, construct a symmetric design having the given design as one of its residual designs.

On the basis of this exercise, knowing that there is no (43,7,1)-design implies that there is no (36,42,7,6,1)-design.

Yet another design that is obtained from a (v,k,)-design is the derived design. To construct this design, we again select a block and remove it, and remove from all the remaining blocks all varieties which are not in the selected block. This will leave us with a (k,v-1,k-1,,-1)- design. In the previous example, again selecting the first block we obtain the derived design;

{4,5} {3,5} {1,4} {5,9} {3,9} {4,9} {1,5} {1,9} {1,3} {3,4}
Which is a (5,10,4,2,1) - design. Again it is possible to obtain non-isomorphic derived designs by selecting different blocks.


We will now consider a generalization of the concept of a BIBD, called a t-Design. Since t-designs are not as tied to statistical applications as BIBD's are, the terminology used is closer to mainstream mathematics.

A t-(v,k,) design is an ordered pair (S,B), where S is a set of cardinality v, and B is a family of k-subsets (called blocks) of S with the property that each t-subset of S is contained in precisely blocks of B. For nondegeneracy, we shall always assume that 0 < t < k < v. Clearly, a BIBD is a 2-(v,k,) design. While not much is known about general t-designs much work has been done in certain special cases. For each triple satisfying 0 < t < k < v there are many t designs that may be obtained trivially as follows. Let S be any v-set. Form C, the set of all possible k-subsets of S. The result (S,C) is a t-design that we call the full combinatorial design. In this design equals the binomial coefficient C(v-t,k-t). Let B be the family of k-subsets of S that includes each member of C exactly n times. (S,B) is a t-(v,k,n) design. In fact, given any t-(v,k,) design we can obtain a t-(v,k,n) design by replicating each block n times.

Nontrivial t-designs have been known for some time for each t, 0 < t < 6. However, it is only recently that it has been shown that t-designs other than the full combinatorial designs and their replications also exist for all t 6.

An example of a 3-(8,4,1) design on the set S = {1,2,...,8} is given by the blocks:

{1,2,5,6} {3,4,7,8} {1,3,5,7} {2,4,6,8} {1,4,5,8} {2,3,6,7} {1,2,3,4} {5,6,7,8} {1,2,7,8} {3,4,5,6} {1,3,6,8} {2,4,5,7} {1,4,6,7} {2,3,5,8}.
This is a particular example of an Hadamard 3-design. Let H be an Hadamard matrix of order > 4 which is standardized. Identify the set S with the columns of H and form the blocks in the following way: For each row of H other than the first form a block consisting of all the columns containing a +1 in this row and also form a block consisting of all the columns which contain a -1. If H is order n, the design formed will be a 3-(n, ½n, ¼n - 1) design.

Theorem VII.4.1 - Every t-(v,k,) design is a (t-1)-(v,k,*) design where * = (v - t + 1)/(k - t + 1).

Proof: Let D = (S,B) be a t-(v,k,) design. Let X be any (t-1)-subset of S, and L(X) the number of blocks of D that contain X. Clearly X is contained in v - t + 1 t-subsets of S. Let C(X) denote the collection of such subsets. Each of these t-subsets occurs in blocks of D, these being the L(X) blocks that contain X. Now each of these blocks contains k - t + 1 members of C(X). Thus, L(X)(k-t+1) = (v-t+1). Since L(X) does not depend on the particular (t-1)-subset chosen, the result follows with * = L(X).

Thus, our example of a 3-(8,4,1) design is also a 2-(8,4,3) design.

Corollary VII.4.2 - Let 0 = b the number of blocks and 1 = r the number of blocks containing a given element, and i the number of blocks containing a given i-subset, 0 i t. Then,

iC(k-i, t-i) = tC(v-i, t-i).

Proof: This follows from repeated application of Thm. VII.4.1.

Corollary VII.4.3 - Let u be an integer such that 0 u t; then the number of subsets in B intersecting a given t-subset in u elements is independent of the t-subset chosen. Proof: This is equivalent to Cor. VII.4.2.

Corollary VII.4.4 - The compliment of a t-design is a t-design.

Proof: A straightforward application of the principle of inclusion-exclusion shows that the complementary design of a t-(v,k,) design is a t-(v,v-k, *) design, where

* = C(v-k, t) / C(k, t).

Theorem VII.4.5 - The existence of a t-(v,k,) design implies the existence of a (t-i) - (v-i, k-i, ) design.

Proof: Let C(x) be the set of blocks of D = (S,B) that contain a given element x of S. Every (t-1)-subset of S-{x} occurs with x in exactly blocks of D, these blocks being those of C(x)). Thus, (S - {x},B'), where B' is obtained from the blocks of C(x) by removing x, is the required design for i = 1. Repeated application establishes the required result.

Corollary VII.4.6 - The existence of a t-(v,k,) design implies the existence of a (t-1)-(v-1,k, *) design, where * = - {the # of blocks containing a given t-1-subset}.

Proof: This design consists of the blocks other than those containing a given element x, that is, the set of blocks B - C(x) in Thm. VII.4.5.


We have barely scratched the surface in the study of block designs in this section. Some of the important references are:

M. Hall, Combinatorial Theory, Blaisdell, Waltham, Mass, 1967.

P.J.Cameron and J.H.van Lint, Graph Theory, Coding Theory and Block Designs Cambridge University Press, Cambridge, 1975. (Also, see the revised and updated conversion to a text by the same authors, Designs, Graphs, Codes and their Links, Cambridge University Press, 1991.)

I.F. Blake and R.C. Mullin, An Introduction to Algebraic and Combinatorial Coding Theory, Academic Press Inc, New York, 1976.

D. Raghavarao, Constructions and Combinatorial Problems in Design of Experiments, Wiley, New York, 1971.

P. Dembowski, Finite Geometries, Springer-Verlag, Berlin, 1968.

H. Ryser,Combinatorial Mathematics, MAA Carus Monographs #14, 1963.

D. Hughes and F.C. Piper, Design Theory, Cambridge University Press, Cambridge, 1985.

An encyclopedic work is:

Th. Beth, D. Jungnickel and H. Lenz, Design Theory, Cambridge University Press, Cambridge, 1986.