The New Sudoku Players' Forum

[bluemoose]

The following trick that I have found may be useful. I have not been through all of the forums so I don't know if someone has already posted it. I apoligise if this trick has already been posted or if it is too obvious to waste time writing down.

There is one simple principle that I am introducing: When you know two squares are the same, you can apply the constraints given on one of the squares to the other.

The means by which you figure out where you can apply this is the hard part. Basically, all you need to do is find where there is the same set of unknown numbers in which all but one of the squares is in the same location. There should be two different squares that are equal.

I don't know if I expained it very well so here is an example:

++++++5++
++++++82+
++++++7++
++++++3++
++++++4++
+++1++A++
++++++B53
++++++C87
++++9+DE4

notice that both sets ABCD and BCDE are missing a 1, 2, 6, and 9.
Because no matter what, B, C, and D will be the same numbers, the remaining number will fill A and E. Because E can't be 9 or 2, A can't be 9 or 2. A can't be 1 either so A must be 6. Therefor, E must also be 6.

[bluemoose]

here is a better image of the example

[ . . . [ . . . [5 . . [
[ . . . [ . . . [8 2 . [
[ . . . [ . . . [7 . . [
+
[ . . . [ . . . [ 3 . . [
[ . . . [ . . . [ 4 . . [
[ . . . [1 . . [ A . . [
+
[ . . . [ . . . [ B 5 3 [
[ . . . [ . . . [ C 8 7 [
[ . . . [9 . . [ D E 4 [

[Shazbot]

Code: Select all

               |                | 5             
               |                | 8    2         
               |                | 7             
---------------+----------------+----------------
               |                | 3             
               |                | 4             
               | 1              | 269           
---------------+----------------+----------------
               |                | 1269 5    3   
               |                | 1269 8    7   
               | 9              | 126  16   4   


I'm not sure whether to read this with the assumption that other cells are filled in or not (esp in boxes 3 and 6) - I guess not, otherwise there wouldn't be multiple candidates for r6c7.

From a slightly different approach, I'd figure that candidates 2 and 9 in box 9 are locked to column 7, so they can't be in r6c7, which leaves 6 as the only possible candidate. This would force 6 to be a hidden single in r9c8.

But I understand where you're going, and I've thought once or twice along the same lines myself, but putting it into words and proving the theory hurt too much. I wonder if there's an application for this where locked candidates aren't a simple alternative?

[r.e.s.]

bluemoose,

What you're noticing can seen as an application of the <law of leftovers>: If a chosen collection of whole units is modified into a different collection of whole units by adding some cells and removing others, then the set of cells added and the set of cells removed must contain the same set of digits.
In your puzzle, box 9 is a 1-unit collection that can be modified into the 1-unit collection consisting of column 7, by "moving" 3,4,5,7,8,E from the box to the column, and doing so must make A=E. So whatever constraints apply to A must apply also to E, and vice versa.