sudoku 数独 Rectangle Elimination

Rectangle Elimination - SudokuWiki.org
This is the first novel 'tough' strategy in a long time and replaces Empty Rectangles which I feel is an overly complicated pattern. Ken Reek from Denver, USA, sent me his explanation and it is included in his solver SudoKoach. I think it sufficiently distinct to warrant its own test and go near the start. The pattern can also be expressed with AICs usually with at least one grouped cell. For those curious I've kept the Empty Rectangle test on the solver so you can go back a step, untick Rectangle Elimination and compare.

I will use the original puzzle Ken sent to me to illustrate his pattern.

The pattern works on a single number - in this case 9.

 

 

 

The base of the pattern consists of a Hinge cell G2 connected to the one remaining 9 in the row (or column), in this case G6. This part has to be a strong either/or link. First 'wing' cell in orange.

From the hinge G2 we look for another 9 in the opposite orientation (the column) and in a different box that is weakly linked - more than two 9s in the unit. The second 'wing' cell.

Consider the weakly-linked A2. If it's ON, then the other wing cell G6 must also be ON. However, this would eliminate ALL the 9s in the 'fourth corner box' (box 2, which is the fourth corner of the rectangle). So A2 cannot be ON, i.e. we can eliminate 9 as a possibility from A2. Simple as.

 

This is an interesting and elegant strategy discovered by Ken Reek, which is indeed simpler than Empty Rectangles. Let me break down the Rectangle Elimination pattern from your example:

Key Elements:

1. Hinge cell: G2 (containing 9)
2. First wing cell: G6 (forms strong link with hinge horizontally)
3. Second wing cell: A2 (weakly linked to hinge vertically)
4. Fourth corner box: Box 2 (where the elimination of 9s would occur)

The logic flow:

1. If A2 were 9:
   • G2 couldn't be 9 (same column)
   • Therefore G6 must be 9 (strong link with G2)
   • But this would eliminate all 9s in box 2
2. Therefore A2 cannot be 9

The pattern requires:

1. A hinge cell
2. A strong link from the hinge in one direction (row/column)
3. A weak link from the hinge in the perpendicular direction
4. The resulting rectangle formation must threaten to eliminate all instances of the digit in the fourth corner box

This is indeed more straightforward than Empty Rectangles and seems like it would be more frequently applicable. Given its relative simplicity, it makes sense to place it earlier in the solving order, perhaps around difficulty level 3, after basic subset patterns but before more complex patterns like X-Wings.

 

 

 

作者:Chuck Lu    GitHub