The New Sudoku Players' Forum

[Pi]

if you start with a finished puzzle (81 numbers, all filled)
can this always be reduced to a very hard puzzle?
Does anyone know?

[PaulIQ164]

As a complete guess with zero substantiating evidence, I'm going to say yes.

[gfroyle]

This is undoubtedly true, yet impossible to prove.

There are such vast numbers of puzzles contained within a single grid that the grid would have to be outstandingly special for every one of those puzzles to be easy. And this just won't happen.

But there is no way of ever proving it...

Gordon

[bennys]

I agree but still I would like to see a fiendish set of clues for that one.

Code: Select all

+-------+-------+-------+
| 1 2 3 | 4 5 6 | 7 8 9 |
| 4 5 6 | 7 8 9 | 1 2 3 |
| 7 8 9 | 1 2 3 | 4 5 6 |
+-------+-------+-------+
| 2 3 1 | 5 6 4 | 8 9 7 |
| 5 6 4 | 8 9 7 | 2 3 1 |
| 8 9 7 | 3 1 2 | 5 6 4 |
+-------+-------+-------+
| 3 1 2 | 6 4 5 | 9 7 8 |
| 6 4 5 | 9 7 8 | 3 1 2 |
| 9 7 8 | 2 3 1 | 6 4 5 |
+-------+-------+-------+


[dukuso]

here the hardest according to
http://magictour.free.fr/suexrat9.exe
from 1000 randomized minimal sudokus over this grid:
rating: 281

...4.....
..6.8.12.
.....3.5.
..1.6..97
...8.72..
8.73.2...
.12...9..
.4.......
..8...6..



I think, it could be feasable to proof that there is a fiendish over each
grid. Suppose e.g. you fill in 2 bands and only consider all puzzles
over the 3rd band. Just 416 possibilities, can they all be made fiendish ?
(I don't know how exactly fiendish is defined)

Or take a 3-rookery, can it be fiendish ? Well, there are lots of
nonisomorphic 3-rookeries, though.

Or how about B5689s ? Only 2865 grids to examine.


-Guenter

[bennys]

thanks