www.argio-logic.net

from an idea of giorgio1957
in all the 9 sectors the position of a same digit CAN NOT be the same
that is: if in r1c1 we have 1, 1 cannot be present in r4c4, r4c7, r7c4, r7c7 (and, naturally, in r1c4, r1c7, r4c1, r7c1)
in our example ; in r3c2 we have 5, 5 can not be present in 56c5,r6c8,r9c5,r9c8

I am generating the puzzles considering only the elimination from cells of digits already present in the same position in other sector and without considering the 9 additional sectors created by the rule.

I could insert the algorithm of search of the singles, naked and hidden sets , blocking numbers on the additional sectors

doing this , however, will make this variant very difficulty, more than the 4 box (windoku) that also has 9 additional sectors , but how many times do you consider the 5 hidden sectors?

however we can think about making 2 version, one with the only elimination and the other with 9+9 sectors

what do you think about?

Is this the same as "disjoint groups" on fed sudoku?

I didn't know that this variant was already known.
the giorgio's idea was: let's make a puzzle where all digits don't stay in the same position in all the sectors.
and this brings to the same results of disjoint group sudoku (also known as offset sudoku)