The New Sudoku Players' Forum

[r.e.s.]

Just thought I'd mention a nice variant that imo deserves more attention ...

The page http://www.essex.ac.uk/maths/misc_pages/CircularSudoku.htm has a few links to "circular sudoku"; e.g. a sample puzzle at http://www.essex.ac.uk/maths/misc_pages/Sudoku1.pdf is such that each of the digits 0-9 must occur in each of 5 concentric rings and in each of 10 pairs of adjacent pie-shaped sectors.

Personally, I find the circular layout unappealing (mostly due to the cells not having a fixed size), but a design that's entirely equivalent -- and imo more appealing -- is rectangular with rows that "wrap around" (so the leftmost & rightmost columns are considered adjacent). In such rectangular form, the puzzle equivalent to the one just mentioned is as follows ...

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Circular Sudoku in rectangular form with wrapping rows:

    . . . . . 8 2 . . .
    9 8 . . . . . 5 2 .
    6 . . . 2 . . 8 9 1
    . 3 6 . . 5 . . . .
    . . . 4 7 . . . . .



In this form, the digits 0-9 must occur in each row and in each of the ten pairs of adjacent columns. I enjoy this variant a great deal!

[Red Ed]

This is just an interleaved pair of 5x5 Latin squares: one square on odd columns, the other on even.

Barring typos, your puzzle is equivalent to the pair:

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...2.         ..8..
9...2         8..5.
6.2.9   and   ...81
.6...         3.5..
..7..         .4...

Ignoring all the symmetries and just doing a straight count, the total number of solution grids for circular sudokus of radius R works out as (2R)! (R-1)!^2 L(R)^2, where L(R) is the number of reduced Latin squares of order R. So, that's:

• 2 of radius 1
• 4 of radius 2
• 2880 of radius 3
• 23224320 of radius 4
• 6554832076800 of radius 5
• 610511815767490560000 of radius 6 ...
  

... and so on.

[r.e.s.]

This is just an interleaved pair of 5x5 Latin squares: one square on odd columns, the other on even. [...]

Indeed. For some reason I mistakenly thought the two squares were not only interleaved, but interdependent (in a way other than merely having the necessary distinct alphabets).

In a futile attempt to redeem the situation, I then thought maybe some interdependence could be introduced by using an odd-sized alphabet, say 1-9, and having strategically-placed cells blocked out as not to be used, like this (where X's mark the cells not to be used):

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   X . . . . . . . . .
   . . X . . . . . . .
   . . . . X . . . . .
   . . . . . . X . . .
   . . . . . . . . X .

Again the puzzle would use wrap-around, with units consisting of rows and adjacent pairs of columns, but again it reduces to two independent interleaved Latin squares. Oh well.