Here is an interesting example of a type 3 AUR. The initial grid:

- Code: Select all
`6 . .| . . 3| . . 7`

. . 7| . . .| 9 . .

. 8 .| 7 5 .| . 6 .

------+------+-------

3 . .| . 6 .| 1 . .

. . 8| 3 . 4| 7 . .

. . 5| . 2 .| . . 3

------+------+-------

. 1 .| . 8 2| . 7 .

. . 9| . . .| 8 . .

8 . .| 1 . .| . . 2

and after some steps we get:

- Code: Select all
`6 2459 124 | 28 149 3 | 245 1458 7`

245 2345 7 | 268 14 68 | 9 13458 1458

249 8 1234| 7 5 19 | 234 6 14

----------------+--------------+-----------------

3 2479 24 | 59 6 5789| 1 2489 489

12 269 8 | 3 19 4 | 7 259 569

1479 4679 5 | 89 2 1789| 46 489 3

----------------+--------------+-----------------

45 1 346 | 4569 8 2 | 3456 7 4569

2457 23457 9 | 456 37 56 | 8 1345 1456

8 3457 46 | 1 37 569 | 456 3459 2

There is a slightly different type 3 AUR in cells r8c25/r9c25, which can be used to eliminate a "2" from r2c2 (and r1c2):

[r2c2]-2-[r8c2|r9c2|r7c1](-4-[r7c3])-4-[r9c3]-6-[r7c3]-3-[r3c3]=3=[r3c7]=2=[r1c7]-2-[r1c4]=2=[r2c4]-2-[r2c2], => r2c2<>2.

As an exercise, I invite the reader to complete the solving of this grid (from the beginning).

Carcul