There are a couple tricks to this one.
After TH 168# => –168r9c9:
- Code: Select all
,---------------,-----------------,----------------,
| 346 59 1 | 4689 368 7 | 3568 468 2 |
| 2346 59 8 | 1469 136 b16–2 | 1356 146 7 |
| 2346 246 7 | 1468 5 1268 | 1368 9 b16–8|
:---------------+-----------------+----------------:
| 7 168 9 | 5 4 #168 |#168 2 3 |
| 5 168 4 | 2 #168 3 | 7 #168 9 |
| 1268 1268 3 |#168 7 9 | 4 5 #168 |
:---------------+-----------------+----------------:
|a18 3 2 |#168 9 5 |#168 7 4 |
| 9 7 5 | 3 #168 4 | 2 #168 a18–6|
| 148 148 6 | 7 2 a#18 | 9 3 #5 |
'---------------'-----------------'----------------'
Let
a be the digit in r8c9.
Suppose there exists a solution where
a is different from the digit in r9c6.
From this solution, remove all digits outside # and set
ar9c9.
That gives us a 3-colouring of TH, which is a contradiction.
Therefore, in all solutions,
a appears in r9c6 and then also in r7c1. –6r8c9
Consider c456789. Each of them contains one of 168 in each band.
Therefore, the permutations of 168 in c456 have the same parity. The same is true for c789.
Notice that c58 also have the same parity, as all their respective 168s are different:
The top digits cannot be both 1s or both 8s, because they would have to be in the same row.
They also cannot both be 6s because of r8.
The middle digits are in the same row and cannot be the same, which also applies to the bottom digits.
Therefore, c456789 all have the same parity.
Let
b be the digit in r3c9.
Since c69 have the same parity and their bottom digits are the same, their top digits are the same also.
Therefore, we get
br2c6, –2r2c6, –8r3c9.
- Code: Select all
,-------------,----------------,----------------,
| 346 59 1 | 4689 368 7 | 3568 468 2 |
| 2 59 8 | 1469 136 b1–6 | 1356 146 7 |
| 346 46 7 | 1468 5 2 | 1368 9 b1–6 |
:-------------+----------------+----------------:
| 7 168 9 | 5 4 168 | 168 2 3 |
| 5 168 4 | 2 168 3 | 7 168 9 |
|b1–68 2 3 |a18–6 7 9 | 4 5 168 |
:-------------+----------------+----------------:
|a18 3 2 | 168 9 5 | 168 7 4 |
| 9 7 5 | 3 168 4 | 2 168 a18 |
| 148 148 6 | 7 2 a18 | 9 3 5 |
'-------------'----------------'----------------'
In r6,
a must take r6c4 (–6r6c4) and
b (which is different from
a because of c9) must then take r6c1 (–8r6c1).
Now,
br3c9 and
br6c1 together see all candidates of 6b1. –6
b, stte