Interesting. Once you make that deduction, you can then go further.
- Code: Select all
*-----------------------------------------------------------*
| 1 245 9 | 6 3 245 | 7 A25 8 |
|A235 8 35 | 7 1 9 |B25 4 6 |
| 6 245 7 | 245 8 245 | 3 1 9 |
|-------------------+-------------------+-------------------|
| 2345 25 B2345 | 245 9 7 | 6 8 1 |
| 245 7 1 | 3 6 8 | 245 9 B25 |
| 9 6 8 | 1 45 245 | 245 3 7 |
|-------------------+-------------------+-------------------|
|B245 3 6 | 9 45 1 | 8 7 A245 |
| 8 1 A245 | 45 7 6 | 9 B25 3 |
| 7 9 45 | 8 2 3 | 1 6 45 |
*-----------------------------------------------------------*
Both r4c1 and r5c1 now see an A and a B, so their twos can be removed
- Code: Select all
*-----------------------------------------------------------*
| 1 245 9 | 6 3 245 | 7 A25 8 |
|A235 8 35 | 7 1 9 |B25 4 6 |
| 6 245 7 | 245 8 245 | 3 1 9 |
|-------------------+-------------------+-------------------|
| 345 A25 B2345 | 245 9 7 | 6 8 1 |
| 45 7 1 | 3 6 8 |A245 9 B25 |
| 9 6 8 | 1 45 245 | 245 3 7 |
|-------------------+-------------------+-------------------|
|B245 3 6 | 9 45 1 | 8 7 A245 |
| 8 1 A245 | 45 7 6 | 9 B25 3 |
| 7 9 45 | 8 2 3 | 1 6 45 |
*-----------------------------------------------------------*
Then r6c7 sees both an A and a B. Killing that candidate 2 leaves a single two in row 6. Thus r6c6 = 2. Then r1c6, r3c6, and r4c4 <> 2. Then r3c4 = 2 and r3c2 <> 2. Leaving us with...
- Code: Select all
*-----------------------------------------------------------*
| 1 B245 9 | 6 3 45 | 7 A25 8 |
|A235 8 35 | 7 1 9 |B25 4 6 |
| 6 45 7 | 2 8 45 | 3 1 9 |
|-------------------+-------------------+-------------------|
| 345 A25 B2345 | 45 9 7 | 6 8 1 |
| 45 7 1 | 3 6 8 |A245 9 B25 |
| 9 6 8 | 1 45 2 | 45 3 7 |
|-------------------+-------------------+-------------------|
|B245 3 6 | 9 45 1 | 8 7 A245 |
| 8 1 A245 | 45 7 6 | 9 B25 3 |
| 7 9 45 | 8 2 3 | 1 6 45 |
*-----------------------------------------------------------*
The uniqueness rectangle on 45's in r1c2, r3c2, r1c6, & r3c6 means that the 2B color must be true. This basically solves the puzzle. All in all, that's pretty cool for a little bit of simple coloring.