--- In

mathforfun@yahoogroups.com, "video_ranger" <video_ranger@...>

wrote:

>

> Correction: The second and third of these are equivalent to each

> other under the permutation 2<-->3 plus reflection across the main

> diagonal, so there are at most two (rather than three) distinct:

>

> 1243

> 3421

> 4312

> 2134

>

> 1243

> 3421

> 2314

> 4132

>

I notice in the above two arrays; the diagonals have two different

numbers and the diagonals in the second have three different numbers.

That is, the diagonals in the first are 1414, two different numbers,

and 3232, two different numbers. The diagonals in the second are

1412, three different numbers, and 3234 three different numbers.

Regardless what transformation I perform on the second there are

always three different numbers in each diagonal. On the first I can

get two different numbers or four different numbers in the diagonals

but never three different numbers.

If this is always true then these two solutions to 4 x 4 Sudoku must

be unique under all "legal" transformations.

John

> unless these two are equivalent also (in which case Peter's

> conjecture that there's only one) would hold for these

> simplified "sudoku"

>