To keep everything in one package, I'm also adding the notes from earlier.
The numbers 1-9 can be found on
every row, column and diagonal
in every 3x3 square whose top row is on the 1st, 4th or 7th row
in every 3x3 square whose left column is on the 1st, 4th or 7th column
If the 3x3 square fits both of the above categories (so it's one of the 9 "sudoku ...
Nice puzzle. After getting over the sudoku preconceptions, this may even be a bit easier, since two digits in a 1x3 block rule out 6 digits already, so the row- an columnwise sudoku solving gets pretty easy.
Since this was an introductory puzzle, it was probably designed to be quite easy; didn't have to resort to anything fancier than "this square must be ...
Here is another solution:
(There's not much to say about the method of solution. I tried putting something other than 5 in the (2,2) position; there was only one option; there were only two options for the next thing to the right, one of which quickly turned out impossible; after that it was just a matter of filling in the only possible number at each step. ...
There are 61479419904000 seven by seven Latin squares. This is from OEIS, and hard to calculate in general.
If you just want to generate Latin squares quickly, you have a few options. One is quick and lazy: take a few template squares, and then generate more by permuting the rows, columns, and/or numbers. If you're comfortable with programming, you can ...
Here is another one:
Note that for any permutation of the first 4 columns there are 6 matching permutations of the last 4 that give rise to another solution. And similar for rows. So this is actually a family of solutions.
We have 3 even numbers to insert.
Since the numbers in rows, columns and diagonals are coprime, it means that the even numbers must have different row, different column and different diagonal.
Also, we have to manage the 3 even sums; a even sum can be made with E+E+E or E+O+O (where E=even and O=odd). The E+E+E case is not possible, as we already stated ...
I don't know how to enumerate all the "perfect Latin squares", so I started off by enumerating all the possible templates: sets of rows satisfying the domino and Latin criteria. A template in itself creates a perfect Latin square if it is symmetric around either the center or the main diagonal, but it does not rule out the possibility that some ...
I've been looking for a while now, and it appears that such puzzles do not exist. However, evidence of absence etc., etc. So, rather than settle for that kind of argument, how about we try to show why such games would be unlikely.
The Online Encyclopedia of Integer Sequences is a repository of famous sequences of integers, such as the sequence Fibonocci ...
I do think I've seen a sort of this puzzle in some newspaper, namely a simple one where you are given a 5x5 grid and have to fill in a Graeco-Latin square over a set of 5 letters and 5 numbers, given some full or partial cells.
A quick google search for graeco-latin sudoku does provide some fruit supporting this assertion.
Among results are a Graeco-Latin ...
The diagonals form a 10's-complement:
as do the anti-diagonals, and the main diagonal and anti-diagonal are self-complementing.
And another $123456789$ block:
A skew grid with wrapround (adjacent grids work too):
Here are perfect latin squares of sizes 8,10,12,14 and 16 all as far as I can tell without obvious symmetries. Sorry about the formatting, at least it's copy-n-paste friendly (you wouldn't want to check them by visual inspection anyway, I assume).
How they were constructed and why this method doesn't work for odd sizes:
If I'm understanding the question correctly, any 6x6 Latin square can be paired with another using the substitutions (2<->5)(->6->4->3->1->). For example:
In fact, I suspect all solutions will be of this form, meaning there are exactly 1,625,702,400 solutions, two for ...
There seem to be no sudoku-type puzzles that use a Graeco-Latin square, probably for the reason you found, the rarity of those squares.
There are however some physical puzzles that use them. One rather clever puzzle is the 36 cube, which was designed by Derrick Niederman and made by ThinkFun:
It seems that the puzzle requires you to complete a 6x6 Graeco-...
Well, the thirty-six officers problem was a 6x6 Graeco-Latin square puzzle which has been puzzling mathematicians for more than a hundred years, until Gaston Tarry in 1901 proved it had no solution. It has been featured on Puzzling Stack Exchange as well.
Quite easy using a constraint solver.
For example Minizinc language and then using Gecode solver:
int: N = 8;
array[1..N,1..N] of var 1..N: p;
set of int: not_primes = array2set([4, 6, 8, 9, 10, 12, 14, 15, 16]);
constraint forall(n in 1..N)(
alldifferent([p[n,g] |g in 1..N]) /\ alldifferent([p[g,n] |g in 1..N])