Contents
What is orthogonal Latin square design?
In combinatorics, two Latin squares of the same size (order) are said to be orthogonal if when superimposed the ordered paired entries in the positions are all distinct. A set of Latin squares, all of the same order, all pairs of which are orthogonal is called a set of mutually orthogonal Latin squares.
What is a diagonal Latin square?
It’s when the ordered pair{(aij,ai+1,j-1):1 ≤ i, j ≤ n}only shows up once in a latin square, then it’s left-diagonal latin square. Diagonal Complete Latin Square. We call a Latin Square diagonally complete latin square is when that latin square is both right-diagonal complete and left-diagonal complete.
What does mutually orthogonal mean?
Two vectors are said to “Mutually orthogonal” if the dot product of any pair of distinct vectors in the set is 0.
Why orthogonal array is used in Lean?
When an orthogonal array (OA) of n rows is used as the design matrix in an experiment, n is the number of runs. A lean design is a design matrix formed by deleting some rows and columns of an OA, which still allows efficient estimation of the effects of the factors corresponding to the remaining columns of the OA.
When are two Latin squares of the same order orthogonal?
In combinatorics, two Latin squares of the same size ( order) are said to be orthogonal if when superimposed the ordered paired entries in the positions are all distinct. A set of Latin squares, all of the same order, all pairs of which are orthogonal is called a set of mutually orthogonal Latin squares.
Can you superimpose a Latin square over a Greek square?
The two squares must be mutually orthogonal, meaning that each Greek-Latin pair occurs only once. Therefore the same square cannot be superimposed over itself, but we can use the two Latin squares above, switch one to Greek letters, and superimpose it over the other, to arrive at this Graeco-Latin square of order 3.
Can a group based Latin square be constructed?
Euler was able to construct Graeco-Latin squares of orders that are multiples of four, and seemed to be aware of the following result. No group based Graeco-Latin squares can exist if the order is an odd multiple of two (that is, equal to 4 k + 2 for some positive integer k ).
How did Brayton prove that Latin squares are self-orthogonal?
Full details can be found in the aforementioned paper, in a summary paper of Brayton, Coppersmith and Hoffman (1974), and also in Keedwell (2000). Brayton et al were able to prove that self-orthogonal latin squares exist of all orders n except n = 2, 3 and 6.