"ALS nodes in Grouped Continuous Nice Loops are especially interesting: If we take all ALS candidates, subtract the candidates used to enter and to leave the ALS (they are normal weak links that become strong links as usual in continuous loops), the remaining candidates become locked within the ALS cells and can provide eliminations in all cells, that see all instances of one of the locked candidates within the ALS. Please note that this possible only in continuous loops." http://hodoku.sourceforge.net/en/tech_chains.php
In other words, the loop entry and exit digits of the ALS are treated like normal weak link end-points in continuous loops and provide their own eliminations, but the rest of the ALS digits can also produce eliminations because they get locked in the ALS cells no matter which way you traverse the loop. That can make these loops quite powerful eliminators, as we saw in the other thread.
I'm not sure how widely that rule is understood, though, and that's why I brought it up here just in case. For instance, it seems to be missing from this SudokuWiki example (last one): http://www.sudokuwiki.org/AIC_with_ALSs. Let's take a look at that:
- Code: Select all
+----------------+-----------------+----------------+
| 7 5 9 | 8 26 124 | 124 3 146 |
| 2 48 348 | 13459 36 1459 | 7 16 1456 |
| 34 1 6 | 345 7 2345 | 245 9 8 |
+----------------+-----------------+----------------+
| 138 9 7 | 2 5 13 | 14 168 146 |
| 6 24 145 | 7 9 8 | 15 12 3 |
| 138 28 1358 | 13 4 6 | 9 1278 157 |
+----------------+-----------------+----------------+
| 5 3 2 | 6 1 79 | 8 4 79 |
| 149 6 14 | 459 8 4579 | 3 17 2 |
| 1489 7 148 | 349 23 249 | 6 5 19 |
+----------------+-----------------+----------------+
Andrew's loop and eliminations translated into Eureka and rYcX format:
(3)r2c3 - (3=6)r2c5 - r2c8 = (6-8)r4c8 = r4c1 - (8=1=3)r6c14 - (3)r6c3 = (3)r2c3
=> -3 r2c4; -6 r2c9; -1 r4c8; -8 r6c23, r9c1
That's six eliminations, but they're all based on the normal weak links of the loop. The extra ALS eliminations aren't there. What would they be?
In this loop we have one ALS node r6c14{138}. Of those three digits 3 and 8 are the entry/exit digits, and their eliminations have been taken care of by the normal weak link rules (three 8s, no 3s available). We still have the digit 1 left, however, and that's a locked digit. Either way you travel the loop, the 1s are locked in the two ALS cells, and other 1s can be eliminated from the cells that see them both. In this case we can eliminate from the row only: -1 r6c389. That's still three more eliminations.
Note that if the ALS cells were packed in the box-row then we'd get two more eliminations in the box, but that's not possible here. Also, if it were a bigger ALS there would be more locked digits and they could all produce eliminations. (In the other thread our sample loop produced a total of 14 eliminations, of which 11 were by two locked ALS digits and only three by the normal weak link rules.)
Do we all agree on this observation? If so, was it already obvious to everyone or did this possibly clarify something? I just know it wasn't very clear to me even though I thought I had a somewhat decent grasp on chaining logic. If someone has more examples of these, that might be instructive.
It doesn't seem like a human-friendly solving method, though, so I suppose the practical value of this method is more as a mathematical proof and/or for computer-based solving? Or do you actually use it manually? 
.
