David, even if you apply both of your 2 rules, as well as the "double latin square" concept (which, by the way, has an official name called "Euler Square"), you will still reach no fewer than two solutions:
- Code: Select all
+-------+-------+-------+
| 2 6 7 | 5 9 1 | 8 3 4 |
| 9 1 5 | 3 4 8 | 6 7 2 |
| 4 8 3 | 7 2 6 | 1 5 9 |
+-------+-------+-------+
| 3 4 8 | 6 7 2 | 9 1 5 |
| 7 2 6 | 1 5 9 | 4 8 3 |
| 5 9 1 | 8 3 4 | 2 6 7 |
+-------+-------+-------+
| 1 5 9 | 4 8 3 | 7 2 6 |
| 8 3 4 | 2 6 7 | 5 9 1 |
| 6 7 2 | 9 1 5 | 3 4 8 |
+-------+-------+-------+
+-------+-------+-------+
| 8 3 4 | 5 9 1 | 7 2 6 |
| 6 7 2 | 3 4 8 | 5 9 1 |
| 1 5 9 | 7 2 6 | 3 4 8 |
+-------+-------+-------+
| 3 4 8 | 6 7 2 | 9 1 5 |
| 7 2 6 | 1 5 9 | 4 8 3 |
| 5 9 1 | 8 3 4 | 2 6 7 |
+-------+-------+-------+
| 2 6 7 | 4 8 3 | 1 5 9 |
| 9 1 5 | 2 6 7 | 8 3 4 |
| 4 8 3 | 9 1 5 | 6 7 2 |
+-------+-------+-------+
So no, one can only reach a unique solution from my initial givens using my rule, which is also much simpler and easier to describe than any rule proposed by you or other solvers (in other forums).
But you certainly do deserve some credit for being the only player in this webspace to be gutsy enough to try it. Other members don't even have any heart to start it!