- Code: Select all
*--------------------------------------------------------------*
| 25 1267 8 | 4 257 2569 | 269-1 3 b19 |
| 245 1267 47 | 3 257 2569 | 269-1 1689 b189 |
| 9 236 23 | 68 28 1 | 5 7 4 |
|--------------------+--------------------+--------------------|
| 7 9 25 | 25 14 8 | 136 146 13 |
| 238 238 236 | 26 14 7 | 19 149 5 |
| 1 4 56 | 9 35 356 | 8 2 7 |
|--------------------+--------------------+--------------------|
| 348 378 9 | 158 6 345 |a137 18 2 |
| 6 5 1 | 7 238 23 | 4 89 b389 |
| 2348 2378 47 | 18 9 34 |a137 5 6 |
*--------------------------------------------------------------*
Suppose the 1's in r1c79 were both false. Then the other two digits in the first ALS must be True, so r79c7 would both be 37, a locked pair.
Now here is the big trick with ALS chains. It doesn't matter whether the 3 in that ALS is in r7c7 or r9c7, they can see all of the 3's in the second ALS. (In this example there is only one 3, in r8c9, but there are more complex examples).
Since the 3 has been removed from the second ALS all the other three digits in it must be in it somewhere. In particular there must be a 1 in r1c9 or r2c9.
So, what all this shows is that at least one of r79c7 or r12c9 must be 1. Since r2c7 can see all of these 4 cells, it can't be 1. The 1 in r1c7 can also be removed in my PM. Perhaps this had already been removed in your RCB solution.
The more common name for this move is ALS-XZ Rule , X = 3, Z = 1 : (1=37) r7c9 - (3=189) r128c9 => - 1 r12c7. This is often written more simply as ALS-XZ Rule , X = 3, Z = 1 : (1=3) r7c9 - (3=1) r128c9 => - 1 r12c7.
Z is the elimination digit and X, the linking digit, is called the restricted common digit. When 2 ALSs are linked in this way, all the X's in the first ALS must be separate from, but must see all the X's in the 2nd ALS.
You can read about the ALS -XZ Rule
here or
here.
Leren
