Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 30 / 4

A balanced incomplete block design has $t$ treatments and $b$ blocks, each containing $k$ distinct treatments with $1<k<t$. Every treatment occurs in $r$ blocks and every distinct treatment pair occurs together in $\lambda >0$ blocks. A symmetric balanced incomplete block design has $b=t$, which implies $r=k$ by the first counting identity below.

For the separate even-order condition, let $N$ be the square treatment-by-block incidence matrix of a set system of a symmetric design. Its diagonal overlap counts are $r$ and its off-diagonal counts are $\lambda$, givingThe counting identities proved in the next slots show that $r=k$ and $r+(t-1)\lambda =r^2$. Consequently the eigenvalues of $N{N}^T$ are $r^2$ on the constant vector and $r-\lambda$ on its $(t-1)$-dimensional orthogonal complement. Taking determinants givesIf $b=t$ is even, the exponent $t-1$ is odd. The exponent of every prime in $\det (N{)}^2$ and in $r^2$ is even; therefore its exponent in the positive integer $r-\lambda$ must be even. This proves the even-order symmetric design square obstruction:It is a necessary condition, not a sufficiency claim.

Count incidences of a treatment with a block containing it. Summing over blocks gives $bk$, while summing over treatments gives $tr$. Hence the balanced incomplete block design satisfies

Fix one treatment. Each of its $r$ blocks supplies $k-1$ companion treatments, giving $r(k-1)$ incidences. Alternatively, each of the other $t-1$ treatments accompanies it in exactly $\lambda$ blocks. Thus

Let $N$ be the $t\times b$ incidence matrix of a set system. The balanced incomplete block design overlap counts give $N{N}^T=(r-\lambda )I+\lambda J$. On the zero-sum subspace its eigenvalue is $r-\lambda =r(t-k)/(t-1)>0$; on the constant vector its eigenvalue is $r+(t-1)\lambda >0$. Therefore $N{N}^T$ has matrix rank $t$. Since a product cannot have larger rank than either factor,This proves Fisher's inequality for block designs directly. The incomplete, nontrivial hypotheses ensure the positive eigenvalue needed for this argument.

Symmetry forces $r=k$, and the pair-counting identity becomes $k(k-1)=2\cdot 45=90$. Its only positive integral solution is $k=r=10$. Since the number of blocks is even, the even-order symmetric design square obstruction requires $r-\lambda =8$ to be a square. It is not. No design with these parameters exists.

The identities require $r=k$ and $k(k-1)=2\cdot 3=6$, so $k=r=3$. Take treatments $1,2,3,4$ and the four blocksEach treatment is absent from exactly one block, hence occurs three times. Each pair is absent from the two blocks missing one of its members, hence occurs together in the other two. This is the required symmetric balanced incomplete block design; here $r-\lambda =1$ is indeed a square.