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Ron Moore
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Joined: 13 Aug 2006
Posts: 72
Location: New Mexico
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 Posted: Tue Jun 26, 2007 1:40 am Post subject: 24 June 2007: Almost XYZ Wing |
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Start position for the 24 June 2007 Nightmare:
| Code: | 080000000002000836360000700000002300000900010709006005150000000000708900004003000
. 8 .|. . .|. . .
. . 2|. . .|8 3 6
3 6 .|. . .|7 . .
-----+-----+-----
. . .|. . 2|3 . .
. . .|9 . .|. 1 .
7 . 9|. . 6|. . 5
-----+-----+-----
1 5 .|. . .|. . .
. . .|7 . 8|9 . .
. . 4|. . 3|. . . |
In the "Almost Swordfish" thread for the 1 June, 2007 Nightmare, I mentioned that the approach there could be used with patterns besides fish. In this puzzle there is an "Almost XYZ Wing" pattern which can be used. I occasionally come across these, but in most cases they aren't really useful, either because they don't really advance the solution, or there is a more natural way to arrive at the same elimination. In this case, it turns out that there is an ALS XZ rule reduction which gives the same elimination, but the sets are fairly large and I didn't see it until I looked at the Sudocue solver's solution. The elimination is quite useful, so I'll consider this a reasonable example to illustrate the idea.
After initial basics and several naked or hidden subsets, the following position is reached:
| Code: | ·--------------------·----------------------·-------------------·
| 459 8 157 | 36 36 %4579 | 15 2459 1249 |
| %459 79 2 | %45 %4579 1 | 8 3 6 |
| 3 6 15 | 28 28 %459 | 7 459 149 |
·--------------------+----------------------+-------------------·
| 568 14 568 | 1458 14578 2 | 3 6789 *79-8 |
| *268-5 234 3568 | 9 34578 %457 | 246 1 #278 |
| 7 1234 9 | 1348 1348 6 | 24 #28 5 |
·--------------------+----------------------+-------------------·
| 1 5 78 | 246 2469 49 | 26 78 3 |
| 26 23 36 | 7 15 8 | 9 45 14 |
| 89 79 4 | 1256 1256 3 | 15 2678 #78+2 |
·--------------------·----------------------·-------------------· |
There are two independent AIC's in this position. The cells marked with "%" form a relatively simple grouped chain for digit 5:(5): r2c1 = r2c45 - r13c6 = r5c6 => r5c1 <> 5 Some might call this a "grouped" 2-string kite. The Sudocue solver finds this same elimination, much later in its solution path, from a Nishio (template check) step.
In the "Almost Swordfish" thread I used the term "spoiler candidate" to denote a candidate, which, if it were not present, would cause some familiar pattern to be created. That term reflects my perhaps unsophisticated intuition, while the term "surplus candidate for the pattern" sounds more conventional and polished, so I'll use that here. Consider the cells marked with "#" above -- r6c8, r59c9. With the annotation in r9c9, I've distinguished (2)r9c9 as we would a surplus candidate for a UR pattern, for example. Observe that if this surplus candidate is false, an XYZ wing pattern exists in the "#" cells. This XYZ wing would eliminate (8)r4c9. On the other hand, if (2)r9c9 is true, there is what could be called an "adjunct chain" to the XYZ wing pattern which also eliminates (8)r4c9.(2)r9c9 => not (2)r1c9 => (2)r1c8 => (8)r6c8 It's a bit nicer if we put all this in AIC form:(8)[XYZ wing r6c8|r59c9] = (2)r9c9 - (2)r1c9 = (2)r1c8 - (2=8)r6c8 => r4c9 <> 8 As with any other AIC, the elimination shown can be made because r4c9 sees all the digit 8 candidates in the first and last nodes of the chain.
We could also use a different adjunct chain:(8)[XYZ wing r6c8|r59c9] = (2)r9c9 - (2=6)r7c7 - (6=248)r56c7|r6c8 => r4c9 <> 8. By the way, the AIC for the ALS XZ rule elimination is: (8=2459)r1368c8 - (9=12478)r13589c9 => r4c9 <> 8
Now we have:
| Code: | ·------------------·--------------------·-------------------·
| 459 8 #157 | 36 36 #4579 | 15 2459 1249 |
| 459 79 2 | 45 4579 1 | 8 3 6 |
| 3 6 15 | 28 28 459 | 7 459 149 |
·------------------+--------------------+-------------------·
| 568 14 568 | 1458 14578 2 | 3 6789 79 |
| 268 234 3568 | 9 34578 #457 | 246 1 *28-7 |
| 7 1234 9 | 1348 1348 6 | 24 28 5 |
·------------------+--------------------+-------------------·
| 1 5 #78 | 246 2469 49 | 26 #78 3 |
| 26 23 36 | 7 15 8 | 9 45 14 |
| 89 79 4 | 1256 1256 3 | 15 2678 #278 |
·------------------·--------------------·-------------------· |
The previous elimination allows us to write this AIC:(7)r5c6 = (7)r1c6 - (7)r1c3 = (7)r7c3 - (7=8)r7c8 - (8)r9c9 = (8)r5c9 => r5c9 <> 7 After this, the rest is fairly routine -- an X wing for digit 8, and an XY wing.
Remark: In the Almost Swordfish thread, I did find an Almost XYZ wing in the first position I gave, but I didn't mention it because it didn't seem to advance the solution. For any readers who want to try their hand at spotting these patterns, you might see if you can spot an Almost XYZ wing in that position. Hint: Be sure to do this before the elimination of (7)r3c5 which I show there, as that destroys the potential XYZ wing. As in this puzzle, there may be alternative "adjunct chains" which can be used. |
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Para
Yokozuna
Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands
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| Posted: Tue Jun 26, 2007 4:38 am Post subject: |
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Hi Ron
I like the way you try to combine different techniques in AIC.
I think this move can be seen easier as an extension of a y-wing(which is basically an AIC), or a combination between ALS-xz and y-wing.
I once mentioned these extensions in this thread on another forum.
http://www.sudoku.com/boards/viewtopic.php?t=5327&postdays=0&postorder=asc&start=0 (yes i linked here before)
| Code: | ·--------------------·----------------------·-------------------·
| 459 8 157 | 36 36 4579 | 15 *2459 *1249|
| 459 79 2 | 45 4579 1 | 8 3 6 |
| 3 6 15 | 28 28 459 | 7 459 149 |
·--------------------+----------------------+-------------------·
| 568 14 568 | 1458 14578 2 | 3 6789 79-8 |
| 2685 234 3568 | 9 34578 457 | 246 1 B278 |
| 7 1234 9 | 1348 1348 6 | 24 A28 5 |
·--------------------+----------------------+-------------------·
| 1 5 78 | 246 2469 49 | 26 78 3 |
| 26 23 36 | 7 15 8 | 9 45 14 |
| 89 79 4 | 1256 1256 3 | 15 2678 B278 |
·--------------------·----------------------·-------------------· |
There are 2 ALS's:
Set A: R6C8: {28}
Set B: R59C9: {278}
Together they see all 2's in R1. So both sets can't contain 2, otherwise there would be no room for 2 in R1. Any 8 that sees both sets would force a 2 in both sets. Therefor we can eliminate 8 from R4C9.
I've been lookig for these types of eliminations for a while now, and they pop up more often than i expected. I find these easier than regular ALS-xz eliminations, mostly because here i have an idea where to look for them. I usually find these types of eliminations while scanning for skyscrapers and assciated eliminations. Always check if they share the same digit at the end. Empty rectangles tend to be very productive.
There is also this way to eliminate 2 from R9C9, and then you can just go ahead with the xyz-wing:
(2)R7C7 = (2)R56C7 - (2=8)R6C8 - (8)R79C8 = (8)R9C9 => R9C9<>2
This is one of the first times i try to write a elimination in this form. Did i do this correctly?
greetings
Para |
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Ron Moore
Addict
Joined: 13 Aug 2006
Posts: 72
Location: New Mexico
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| Posted: Tue Jun 26, 2007 6:38 am Post subject: |
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Para,
Very nice. Your AIC to eliminate (2)r9c9 is correct, and it's not really an easy one to start with.
I agree that your argument to eliminate (8)r4c9 is more natural than the almost XYZ wing. An AIC which expresses your reasoning is(8=2)r6c8 - (2)r1c8 = (2)r1c9 - (2=78)r59c9 => r4c9 <> 8 |
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Para
Yokozuna
Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands
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| Posted: Tue Jun 26, 2007 7:55 am Post subject: |
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| Ron Moore wrote: | | it's not really an easy one to start with. |
It looks easier in my head. It is easy to see that the R6C8 = 2 leads to R7C7 = 2 and R6C8 leads to R9C9 = 8. But to write it down as an AIC, it looks different.
Para |
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