1 . 1 2 | . 7 . | 3 5 .
2 3 . . | 2 . 1 | 4 . 7
3 4 . . | 5 . . | . 1 6
------------+---------------+------------
4 2 . . | . . . | . 7 .
5 1 6 . | . 8 . | . . 2
6 . 4 8 | . . . | 6 3 .
------------+---------------+------------
7 . . . | 8 . . | . . 4
8 . . . | 1 5 . | . . 3
9 . . . | . 4 3 | 1 2 .
a b c d e f g h i
New entries are marked by *
1 . 1 2 | . 7 . | 3 5 .
2 3 . . | 2 . 1 | 4 . 7
3 4 . . | 5 *3 . | *2 1 6
------------+---------------+------------
4 2 . . | . . . | . 7 .
5 1 6 . | . 8 . | . *4 2
6 . 4 8 | . . . | 6 3 .
------------+---------------+------------
7 . *3 *1 | 8 . . | . . 4
8 . *2 *4 | 1 5 . | . . 3
9 . . . | . 4 3 | 1 2 .
a b c d e f g h i
Multi-digit entries are understood to show all that is not ruled out yet.
1 . 1 2 | . 7 . | 3 5 *89
2 3 . . | 2 *69 1 | 4 *89 7
3 4 . . | 5 3 . | 2 1 6
------------+---------------+------------
4 2 . . | . . . | . 7 .
5 1 6 . | . 8 . | . 4 2
6 . 4 8 | . . . | 6 3 .
------------+---------------+------------
7 . 3 1 | 8 . . | . . 4
8 . 2 4 | 1 5 . | . . 3
9 . . . | . 4 3 | 1 2 .
a b c d e f g h i
if e2 = 6 must have a1 = 6
if e2 = 9 cannot have a1 = 9
therefore never a1 = 9, leaving 68
1 *68 1 2 | . 7 . | 3 5 89
2 3 . . | 2 69 1 | 4 89 7
3 4 . . | 5 3 . | 2 1 6
------------+---------------+------------
4 2 . . | . . . | . 7 .
5 1 6 . | . 8 . | . 4 2
6 . 4 8 | . . . | 6 3 .
------------+---------------+------------
7 . 3 1 | 8 . . | . . 4
8 . 2 4 | 1 5 . | . . 3
9 . . . | . 4 3 | 1 2 .
a b c d e f g h i
Very clever and ingenious trick / shameless cheat:
If a1 = 8 then b9 = 8, h2 = 8, g8 = 8, i4 = 8
the last forces e4 = 1 and we see that d1,f1,d4,f4 must be 2x4 + 2x6
If we are allowed to use the assumption that the puzzle has a unique
solution then this is a contradiction
Therefore a1 = 6
1 *6 1 2 | *4 7 *89 | 3 5 89
2 3 . . | 2 *6 1 | 4 89 7
3 4 . . | 5 3 *89 | 2 1 6
------------+---------------+------------
4 2 . . | *6 . *4 | . 7 .
5 1 6 . | . 8 . | . 4 2
6 . 4 8 | . . . | 6 3 .
------------+---------------+------------
7 . 3 1 | 8 . . | . . 4
8 . 2 4 | 1 5 . | . . 3
9 . . *6 | . 4 3 | 1 2 .
a b c d e f g h i
If f5 = 7 must have a6 = 7 and d9 = 7
leaving no room for 7 in bottom left box
Therefore f5 != 7 leaving 5
The rest is routine
1 6 1 2 | 4 7 *9 | 3 5 *8
2 3 *8 *5 | 2 6 1 | 4 *9 7
3 4 *7 *9 | 5 3 *8 | 2 1 6
------------+---------------+------------
4 2 *9 *3 | 6 *1 4 | *8 7 *5
5 1 6 *7 | *3 8 *5 | *9 4 2
6 *5 4 8 | *9 *2 *7 | 6 3 *1
------------+---------------+------------
7 *7 3 1 | 8 *9 *2 | *5 *6 4
8 *9 2 4 | 1 5 *6 | *7 *8 3
9 *8 *5 6 | *7 4 3 | 1 2 *9
a b c d e f g h i