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