H . J . Ryser conjectured ( Oberwolfach, 1967 ) that every Latin square of odd order has one.
32.
The problem of determining if a partially filled square can be completed to form a Latin square is NP-complete.
33.
:: : : No, the base heights do form a Latin square, as opposed to Graeco-Latin.
34.
The parallel class structure of an affine plane of order may be used to construct a set of mutually orthogonal latin squares.
35.
Wouldn't a solution to the 36 cube be an order 6 Graeco-Latin square, which doesn't exist.
36.
Actually it is restricted from the general Graeco-Latin square, since the bases already determine a Latin square on the heights.
37.
Actually it is restricted from the general Graeco-Latin square, since the bases already determine a Latin square on the heights.
38.
This construction is completely reversible and so strength 2, index 1 orthogonal arrays can be constructed from sets of mutually orthogonal latin squares.
39.
In the 1780s Euler demonstrated methods for constructing Graeco-Latin squares where " n " is odd or a multiple of 4.
40.
Whether or not that bit about the Latin squares " means " the fact you stated, I can't be sure.
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