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Sudtyro Hooked
Joined: 16 Jan 2007 Posts: 49

Posted: Tue Apr 10, 2007 4:39 pm Post subject: 4/1/07 Nightmare 


The following grid appears after running through the basics, followed by removal of the 8's in r7c56 using two Turbot chains.
Code:  137 367 9  5 3678 2  17 78 4
147 5 8  69 4679 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 346  57 478 58
8 23 5  123 234 7  6 14 9
6 47 47  189 89 5  2 18 3

Could sure use a hint or two to advance beyond this point. 

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GreenLantern Regular
Joined: 30 May 2006 Posts: 14

Posted: Wed Apr 11, 2007 2:48 am Post subject: 


Look for a couple of Nice Loops in Boxes 13. 

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Sudtyro Hooked
Joined: 16 Jan 2007 Posts: 49

Posted: Wed Apr 11, 2007 11:38 pm Post subject: 


GreenLantern wrote:  Look for a couple of Nice Loops in Boxes 13. 
GL,
Thanks for the hint! Following the basics (and solving manually), I normally continue with some 3D coloring to reveal the AICs. However, this puzzle was my first attempt at using a b/b plot to search exclusively for Nice loops. In the resulting "wiring diagram" I failed to include the 7's weak link between r1c5 and r1c7, and that was key to finding at least one Nice loop in boxes 13.
Alternatively, 3D coloring of the bilocated 1's and 7's in those boxes quickly leads to:
(7)r2c5 = (71)r2c1 = (1)r1c1  (1=7)r1c7  (7)r1c5 = (7)r2c5 => r2c5 = 7.
Maybe I'd better stick with 3D coloring...
BTW, I'm always interested in any ALS rules or Wings that provide the same elimination. 

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Ron Moore Addict
Joined: 13 Aug 2006 Posts: 72 Location: New Mexico

Posted: Mon Apr 16, 2007 10:13 pm Post subject: 


Just a few minor points. Here's the grid before the elimination of the two "8's" that you mention.
Code: 
....
 137 367 9  5 3678 2  17 78 4 
 147 5 8  69 4679 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 23468 3468  57 478 *58 
 8 23 5  123 234 7  6 14 9 
 6 47 47  #189 89 5  2 18 3 
'''' 
You can view the eliminations of (8)r7c56 as coming from one single finned Xwing for digit 8, with the base Xwing in cells r37c49 (marked with "*") and r9c4 (marked with "#") as the fin.
There's certainly nothing wrong with the way you wrote your AIC, but it would be a bit more natural for me to use the following, with the same effect:(7=1)r1c7  (1)r1c1 = (17)r3 = (7)r2c5 => r1c5 <> 7 (and then r2c5 = 7) There is an ALS XZ rule elimination which gives this result, although I think it's harder to see. An AIC for this is:(7=1)r1c7  (1=24697)r2c45689 => r1c5 <> 7 I believe the other elimination suggested by GL can be found by considering the Almost Locked Sets r1c1278 and r3c9. However, I confess I didn't see this at the time.
What I did see turned out to be quite fruitful. It's a grouped AIC (only the first node is multicell) which starts out this way:(8=92)r59c5  (2)r5c9 . . . Followup from the resulting elimination immediately solves several cells with singles only. In fact it gives the eliminations of the two 8's from the finned Xwing if you missed those (and r7c4 as well). 

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Sudtyro Hooked
Joined: 16 Jan 2007 Posts: 49

Posted: Thu Apr 19, 2007 7:17 pm Post subject: 


Ron Moore wrote:  Just a few minor points. Here's the grid before the elimination of the two "8's" that you mention.
Code: 
....
 137 367 9  5 3678 2  17 78 4 
 147 5 8  69 4679 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 23468 3468  57 478 *58 
 8 23 5  123 234 7  6 14 9 
 6 47 47  #189 89 5  2 18 3 
'''' 
You can view the eliminations of (8)r7c56 as coming from one single finned Xwing for digit 8, with the base Xwing in cells r37c49 (marked with "*") and r9c4 (marked with "#") as the fin. 
Nice "catch"...wish I'd hooked that one.
Ron Moore wrote:  There's certainly nothing wrong with the way you wrote your AIC, but it would be a bit more natural for me to use the following, with the same effect:(7=1)r1c7  (1)r1c1 = (17)r3 = (7)r2c5 => r1c5 <> 7 (and then r2c5 = 7) 
Agreed...my chain would be the AIC equivalent of the full discontinuous Nice loop, which I'm trying to learn about.
Ron Moore wrote:  There is an ALS XZ rule elimination which gives this result, although I think it's harder to see. An AIC for this is:(7=1)r1c7  (1=24697)r2c45689 => r1c5 <> 7 
Yes...very hard to spot. I used the AST rule: Almost Saw That...
Ron Moore wrote:  I believe the other elimination suggested by GL can be found by considering the Almost Locked Sets r1c1278 and r3c9. However, I confess I didn't see this at the time. 
OK, I actually got that one, after first coloring the bilocated 6's in r1 and c2.
Ron Moore wrote:  What I did see turned out to be quite fruitful. It's a grouped AIC (only the first node is multicell) which starts out this way:(8=92)r59c5  (2)r5c9 . . . 
Now, this is pretty sweet! Grouped AICs are new to me, but I assume the chain ended with ... = (8)r1c8 => r1c5 <> 8. After that, it's easy.
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? 

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Ron Moore Addict
Joined: 13 Aug 2006 Posts: 72 Location: New Mexico

Posted: Sun Apr 22, 2007 12:45 am Post subject: Grouped AIC's 


As I used it here, by "grouped AIC" I meant a chain in which the nodes may possibly be multicell 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 multicell 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 nongrouped 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 multicell.(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 expressionmeans 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 asso 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: 
....
 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) singlecell AIC's. Of course, in general other nodes of the chain might be multicell. 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 withFortunately I did find paydirt fairly quickly, since other nodes in the chain were singlecell 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: 
**
 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 6129 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 = (85)r4c9 = (5)r4c5  (5=129)r4c6r56c4 => 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 = (68)r6c8 = (85)r4c9 = (5)r4c5 => r4c5 <> 6 (and then r6c5 = 6) 

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Sudtyro Hooked
Joined: 16 Jan 2007 Posts: 49

Posted: Mon Apr 23, 2007 9:13 pm Post subject: Re: Grouped AIC's 


Ron Moore wrote:  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: 
....
 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. 
RM,
Thanks mainly to you and Myth Jellies, I’m fairly comfortable with the Eureka notation for a (grouped) AIC corresponding to, say, an ALSXZ rule, which requires two linked ALSs. However, what struck me as unusual about this case, which I had not seen before, was that only one multicell ALS was involved in the chain, AND it wasn’t linked to another ALS! For some unknown reason, I had assumed that an ALS had to link to at least one other ALS (via the restricted common). But, after seeing your hint, it was easy to complete the AIC using this “new” technique as:
(8=92)r59c5 – (2)r5c9 = (26)r2c9 = (68)r3c9 = (8)r1c8 => r1c5 <> 8,
and no special ALS “rule” was needed.
It might also be interesting to now ask if a formal ALS rule could reproduce this same AIC. After staring at the grid awhile, the only thing I could dredge up was (maybe) an ALSXYWing [Edit: corrected the name] rule involving the three sets (ordered A,C,B below):
(8=92)r59c5 – (2=1586)r3567c9 – (6=12798)r1c78r2c89r3c7
As I understand the ALSXYWing rule, digit 2 is the restricted common for the first two sets (A and C), while 6 is the restricted common for the second and third sets (C and B). Set C is the pivot, digit 8 is common to sets A and B, and all 8’s in A and B can be seen by (8)r1c5.
Bottom line is that even if this ALSXY rule is correct, it was only (barely) obvious to me with 20/20 hindsight and would certainly not have been discovered initially in a patternbased search. Maybe one wouldn’t WANT to, since your mostly singlecell AIC is much simpler to follow and allows for easier interpretation of the fundamental link structures.
The grouped AIC concept is slowly becoming clear and logical, although admittedly my own understanding has been blurred by a learning history of first trying to use patternbased solving techniques (Wings, ALSXZ rule, etc.) without really understanding and therefore not looking for the underlying AIC structure. What’s now reassuring is that there seems to be an almost limitless variety of allowed combinations of single and multicell nodes in an AIC. The trick is to find the proper combinations that produce a quick solution to the puzzle. That would be the hard part.
Your ALS overview and explanations are extremely helpful. Many thanks for putting it all together!
Last edited by Sudtyro on Tue Apr 24, 2007 9:50 pm; edited 1 time in total 

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Ron Moore Addict
Joined: 13 Aug 2006 Posts: 72 Location: New Mexico

Posted: Tue Apr 24, 2007 1:56 am Post subject: 


Sudtyro,
Everything you say seems right on the mark. That was a good alternative chain to produce the same elimination.
In a few days I'll be posting my solution to last Sunday's Nightmare (22 April). This was the toughest I'd seen in a long time. I didn't find any quick solution, but I did improve somewhat on the Sudocue solver. My solution will have some ALS AIC's, some grouped AIC's for a single digit, and some mixed AIC's (with both types of grouped nodes), and some AIC rings (I'm not such a "nice" guy but I think this is the same as a continuous nice loop). Maybe you can find some good alternatives to what I found. 

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GreenLantern Regular
Joined: 30 May 2006 Posts: 14

Posted: Thu Apr 26, 2007 8:07 am Post subject: 


I've just posted my solution to the April 22 Nightmare as a followup to your posting. I would be interested to hear your comments on Nice Loops and AICs (my questions at the end of the post) if you have time as I'm considering learning about AIC's. I haven't bothered yet since I've been getting good results with just Nice Loops. 

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