11 Combinatorial Designs
The preceding algorithm terminates and computes the GCD of \(a\) and \(b\) correctly.
Let \(n\) be an integer with \(n \geq 2\) and let \(a\) be a nonzero integer in \(Z_n = \{ O,1 ,\ldots , n - 1\} \). Then \(a\) has a multiplicative inverse in \(Z_n\) if and only if the GCD of \(a\) and \(n\) is \(1\). If \(a\) has a multiplicative inverse, then it is unique.
Let \(n\) be a prime number. Then each nonzero integer in \(Z_n\) has a multiplicative inverse.
In a BIBD, each variety is contained in
blocks.
In a BIBD, we have
In a BIBD, we have
In a BIBD, \(b \geq v.\)
Let \(B\) be a subset of \(k {\lt} v\) elements of \(Z_v\) that forms a difference set mod \(v\). Then the blocks developed from \(B\) as a starter block form an SBIBD with index
Let \(\mathcal{B}\) be a Steiner triple system with pammeters \(b, v, k = 3, r, \lambda \). Then
and
If the index is \(\lambda = 1\), then there is a nonnegative integer \(n\) such that \(v = 6n +1\) or \(v = 6n +3\).
If there are Steiner triple systems of index \(\lambda = 1\) with \(v\) and \(w\) varieties, respectively, then there is a Steiner triple system of index \(\lambda = 1\) with \(vw\) varieties.
Let \(n\) be a positive integer. Let \(A\) be the \(n\)-by-\(n\) array whose entry \(a_{ij}\) in row \(i\) and column \(j\) is
Then \(A\) is a Latin square of order \(n\) based on \(Z_n\).
Let \(n\) be a positive integer and let \(r\) be a nonzero integer in \(Z_n\) such that the GCD of rand \(n\) is \(1\). Let \(A\) be the \(n\)-by-\(n\) array whose entry \(a_{ij}\) in row \(i\) and column \(j\) is
Then \(A\) is a Latin square of order \(n\) based on \(Z_n\).
Let \(n\) be a prime number. Then \(L_n^1, L_n^2, \ldots , L_n^{n-1}\) are \(n - 1\) MOLS of order \(n\).
Let \(n = p^k\) be an integer that is a power of a prime number \(p\). Then there exist \(n - 1\) MOLS of order \(n\). In fact, the \(n - 1\) Latin squares (10.12) of order \(n\) constructed from a finite field with \(n = p^k\) elements are \(n - 1\) MOLS of order \(n\).
Let \(n \geq 2\) be an integer, and let \(A_1, A_2,\ldots , A_k\) be \(k\) MOLS of order \(n\). Then \(k \leq n - 1\); that is, the largest number of MOLS of order \(n\) is at most \(n - 1\).
\(N(n) \geq 2\) for each odd integer \(n\).
If there is a pair of MOLS of order \(m\) and there is a pair of MOLS of order \(k\), then there is a pair of MOLS of order \(mk\). More generally,
Let \(n \geq 2\) be an integer and let
be the factorization of \(n\) into distinct prime numbers \(p_1, p_2, \ldots , p_k\). Then
Let \(n \geq 2\) be an integer that is not twice an odd number. Then there exists a pair of orthogonal Latin squares of order \(n\).
Let \(n \geq 2\) be an integer. If there exist \(n-1\) MOLS of order \(n\), then there exists a resolvable BIBD with parameters
Conversely, if there exists a resolvable BIBD with parameters (10.18), then there exist \(n-1\) MOLS of order \(n\).
Let \(L\) be an \(m\)-by-\(n\) Latin rectangle based on \(Z_n\) with \(m {\lt} n\). Then \(L\) has a completion.
Let \(L\) be a semi-Latin square of order \(n\) and index \(m\), where \(m {\lt} n\). Then \(L\) has a completion.