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The squares $L_{1} (i, j) = i + j$ and $L_{2} (i, j) = 2 i + j$ on $F_{3}$ are orthogonal

Example

On $F_{3} = {0, 1, 2}$ , the two squares are $L_{1} = (\begin{array}{ccc} 0 & 1 & 2 \\ 1 & 2 & 0 \\ 2 & 0 & 1 \end{array}), L_{2} = (\begin{array}{ccc} 0 & 1 & 2 \\ 2 & 0 & 1 \\ 1 & 2 & 0 \end{array}) .$

Facts & Assumptions

Given: The two displayed arrays.

Verification

technique · direct

1.1

Reading the two arrays entrywise produces the nine ordered pairs $(0, 0), (1, 1), (2, 2), (1, 2), (2, 0), (0, 1), (2, 1), (0, 2), (1, 0)$ , which are exactly the elements of $F_{3}^{2}$ without repetition.

givenalgebra

2.1

Therefore the two arrays are orthogonal. This is the $q = 3$ instance of The linear Latin squares $L_{a} (i, j) = a i + j$ over $F_{q}$ are pairwise orthogonal.

step 1.1∎

Depends on

The linear Latin squares $L_a(i,j)=ai+j$ over $\mathbb F_q$ are pairwise orthogonal

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Deductive Press, Section 16.2: Latin Squares and MOLS (standard reference, not scraped)

The squares L1(i,j)=i+jL1​(i,j)=i+j and L2(i,j)=2i+jL2​(i,j)=2i+j on F3F3​ are orthogonal