April 3, 2006: Type 6 Unique Rectangle

[keith]

Today's Nightmare has an example of what Ruud has suggested should be a "Type 6 Unique Rectangle".

http://www.sudoku.com/forums/viewtopic.php?t=3709

I have suggested the name "Unique X-wing", in a different Forum.

After making the elementary reductions you get to:

Code: Select all

+----------------+----------------+----------------+
| 356  2    56   | 8    7    1    | 9    35   4    | 
| 1358 4    58   | 69   69   2    | 35   18   7    | 
| 18   9    7    | 45   3    45   | 6    128  28   | 
+----------------+----------------+----------------+
| 258  58   9    | 25   1    7    | 4    6    3    | 
| 256  1    3    | 2569 4    59   | 8    7    259  | 
| 4    7    256  | 3    69   8    | 1    25   259  | 
+----------------+----------------+----------------+
| 578  58   1    | 49   2    49   | 357  358  6    | 
| 9    6    258  | 7    58   3    | 25   4    1    | 
| 27   3    4    | 1    58   6    | 27   9    58   | 
+----------------+----------------+----------------+

How to find it:

1. There is a rectangle defined by a diagonal pair.
2. The rectangle is an X-wing.
3. The rectangle is potentially non-unique.

Then what? Well, the diagonal pair must be the same value as the X-wing.

The diagonal pair is <18> in R23C18. The X-wing is on <1>, the non-unique pattern is, of course, <18>. So, R2C8 and R3C1 must be <1>.

The proof is simple: If either of the diagonal pair is <8>, there is a non-unique rectangle through a forcing loop on <1>.

In this case, placing the <1>'s is not much help in solving the puzzle. (To progress, look at the rectangle in R47C12.)

Best wishes,

Keith

[David Bryant]

Hi, Keith!

From the position you posted two "doble-implication chains" will crack it wide open.

The first is alpha star at r2c3:

r2c3 = 8 ==> r2c8 = 1
r2c3 <> 8 ==> r8c3 = 8 ==> r9c5 = 8 ==> r7c8 = 8 ==> r2c8 = 1

Interestingly, this permits the same conclusion as your "unique x-wing."

The second DIC is alpha star at r2c7:

r2c7 = 3 ==> r7c8 = 3 ==> r9c9 = 8
r2c7 = 5 ==> r2c3 = 8 ==> {2, 5} pair in row 8 ==> r8c5 = 8 ==> r9c5 = 5 ==> r9c9 = 8

r9c9 = 8 is enough to crack it wide open. dcb

[keith]

I said:

In this case, placing the <1>'s is not much help in solving the puzzle.

which I withdraw.

There are a number of ways to use chains to solve this puzzle. As with David's solution, the Unique Rectangle replaces the deduction made using a chain.

Replace "not much use" with "just one step".

Keith