As I used it here, by "grouped AIC" I meant a chain in which the nodes may possibly be multi-cell Almost Locked Sets. I suppose I should note that this term might be used by some to include chains for a single digit involving grouped nodes, where a node may be a single cell or possibly a group of two or three aligned cells within some box. I understood and could work with these AIC's long before I became comfortable with the ALS AIC's (or chains involving both types of multi-cell nodes). Myth Jellies helped me to clear up some of my thinking in his response to my post in
this thread. (I'll try to hit the important points here so that you don't need to consult that thread). Still, it did take a while for me to become comfortable with their use, and I think this was because at the time I didn't discipline myself to document all my solutions with AIC's. From what I've seen of your posts, you seem to be proficient with non-grouped AIC's so this may well come more quickly for you, but don't be discouraged if it takes a while.
Actually, the chain which I gave for the ALS XZ rule elimination is a grouped AIC, with the final node being multi-cell.
- (7=1)r1c7 - (1=24697)r2c45689 => r1c5 <> 7
The ALS XZ rule is usually expressed in terms of two ALS's A and B, a restricted common digit X, and the eliminated digit Z. I didn't explicitly identify these sets and digits because the AIC itself compactly documents this information. The two ALS's are the two nodes in the chain -- r1c7 and r2c45689; the digit which is used to weakly link the two ALS's is the restricted common digit X (digit 1 in this case); the eliminated digit Z is the one which begins and ends the chain (digit 7 in this case).
The parenthesized expression preceding each node contains the candidate digits which appear in the ALS. The "=" symbol indicates a strong internal link within the node. One fact which is used, perhaps tacitly, in all Almost Locked Set arguments is this: If an ALS does not contain one of its candidate digits, then it must contain all other of its candidate digits. So the expression
means that r2c45689 either contains the digit 1 (in some unspecified position within the node), or it contains all of the digits 2, 4, 6, 9, and 7 (in some unspecified order within the node). Myth Jellies likes to emphasize the "all" by writing the above as
so you might want to do this if it helps your intuitive understanding.
In this case, what we're really interested in is the fact that if r2c45689 doesn't contain 1, it must contain 7. Digit 7 is is our target digit -- the one for which we wish to show that some elimination exists somewhere. That is why 7 is listed as the last of the digits to the right of the "=" symbol, and also as the leftmost digit in the starting node. All together, the chain tells us that the first or the last node of the chain (or possibly both) must contain digit 7 somewhere. Therefore any cell which sees all of the "7" candidates in the start and end nodes must see a 7 somewhere and therefore cannot itself contain 7.
You can place any candidate digit of an ALS to the left of the "=" (strong link symbol), and all others to the right of it. By conventions of Eureka notation, the restricted common digit (used to weakly link between two nodes) is written as the last digit of the first node's candidates, and the first digit of the second node's candidates.
Sudtyro wrote:Grouped AICs are new to me, but I assume the chain ended with ... = (8)r1c8 => r1c5 <> 8.
You're basically correct here.
Code: Select all
.------------------.------------------.------------------.
| 137 367 9 | 5 368 2 | 17 78 4 |
| 14 5 8 | 69 7 469 | 3 29 126 |
| 347 23467 2347 | 3689 1 34689| 79 5 68 |
:------------------+------------------+------------------:
| 2 9 6 | 4 5 1 | 8 3 7 |
| 345 34 134 | 7 289 89 | 1459 6 125 |
| 457 8 147 | 2369 2369 369 | 145 29 15 |
:------------------+------------------+------------------:
| 9 1 23 | 2368 2346 36 | 57 478 58 |
| 8 23 5 | 123 234 7 | 6 14 9 |
| 6 47 47 | 189 89 5 | 2 18 3 |
'------------------'------------------'------------------'
The AIC I used was
- (8=92)r59c9 - (2)r5c9 = (2)r6c8 - (2=9)r2c8 - (9=7)r3c7 - (7=8)r1c8 => r1c5 <> 8
You might equally well have used (2)r2c9 instead of (2)r6c8. The first node in the chain says that either r59c5 must contain 8 (in some unspecified position in the node), or it must contain 2. In the latter case, the 2 must be in r5c5, which means that r5c9 could not be 2, and thus the weak link to (2)r5c9. From there onward, the chain propogates onward like other (more familiar) single-cell AIC's. Of course, in general other nodes of the chain might be multi-cell. In total, this chain shows that either the start or end node must contain digit "8" somewhere. Since r1c5 sees all of the "8" candidates in the two nodes, it cannot contain 8.
Sudtyro wrote:But...how do you go about looking for the proper groups to use in the AIC? Can you supply a thread or two on the technique?
I don't have a good answer for your question. It comes with experience and with seeing examples of usage. My study of other sites has been somewhat haphazard, so I can't point you to any single thread on the topic, but there may well be something somewhere.
I can tell you that in this case I was looking for an XYZ wing. As you know, an XYZ wing is a special case of an ALS XZ rule position. Just for exercise, suppose that r6c4 contained candidates "29" only, instead of "2369". In that case we would have an XYZ wing which would eliminate (9)r6c5 -- so see if you can write the grouped AIC which expresses this.
Generally, to find XYZ wings we look for some trivalued cell with digits XYZ and some bivalued cell with digits XZ aligned with it (but outside of its box), and if successful we then hope to find a bivalued cell with digits YZ within the box of the trivalued cell. If this doesn't work, since the XYZ and the XZ cells form an ALS, we may be able to use them together as the start node of a longer grouped AIC, as was done here. (I should mention that it's also possible to begin with the XYZ and XZ cells in a common box.)
When I mentally began investigating the possibility of a useful chain, I wassn't sure which if either of the digits 8 or 9 would be the target digit. So at that point I was thinking of the chain as starting with
Fortunately I did find paydirt fairly quickly, since other nodes in the chain were single-cell nodes, and was able to identify "8" as the target digit, so I then wrote the chain as shown above. I suppose it's valid to leave the chain as starting with 89, but I think for clarity of purpose the chain should begin only with the digit of interest.
Now, in rare cases you might find some grouped AIC which eliminates two or more digits, in which case you should begin and end your chain with all eliminated digits grouped together. As an example (not highly practical, but curious), consider this position from the 4 March 2007 Daily Nightmare. Myth Jellies commented on this puzzle in this thread:
http://www.sudocue.net/forum/viewtopic.php?t=586
Code: Select all
*----------------------------------------------------------------------*
| 6 9 378 | 4 1237 1238 | 12357 12357 35 |
| 1 478 378 | 289 2379 5 | 2379 2367 3469 |
| 345 457 2 | 19 1379 6 | 8 137 349 |
|----------------------+------------------------+----------------------|
| 7 1268 168 | 3 1256 B12 | 4 9 568 |
| 249 1246 169 | B1259 8 7 | 1235 12356 356 |
|A289 3 5 |AB129 6-129 4 |A12 1268 7 |
|----------------------+------------------------+----------------------|
| 2359 1258 4 | 7 1235 1238 | 6 358 3589 |
| 3589 1578 13789 | 6 4 138 | 3579 3578 2 |
| 2358 25678 367 | 258 235 9 | 357 4 1 |
*----------------------------------------------------------------------*
In the diagram, the cells of the start and end nodes of the chain below are marked with A and B, respectively.
- (129=8)r6c147 - (8)r6c8 = (8-5)r4c9 = (5)r4c5 - (5=129)r4c6|r56c4 => r6c5 <> 1,2,9, i.e. r6c5 = 6
The chain shows that either set A or set B must contain a "129" triple, and r6c5 sees all cells of both sets, so 1, 2, and 9 can be eliminated from it. Myth Jellies' post shows an easier way to reach this result, in effect using the following AIC:
- (6)r6c5 = (6-8)r6c8 = (8-5)r4c9 = (5)r4c5 => r4c5 <> 6 (and then r6c5 = 6)