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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Tue Aug 31, 2010 4:10 am Post subject: Puzzle 10/08/31: C |
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| Code: | +-----------------------+
| . . . | . 5 . | 6 . . |
| . 4 5 | . . 9 | . 7 8 |
| . . . | 7 . . | . . . |
|-------+-------+-------|
| . . 4 | 1 . 7 | 8 . . |
| 1 . . | . 9 . | . . . |
| . 2 . | 3 . 5 | 1 . . |
|-------+-------+-------|
| 2 . . | . . 8 | . 9 1 |
| 9 6 . | . . . | 5 . . |
| . . . | . . . | . . . |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Tue Aug 31, 2010 6:50 am Post subject: |
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One step xy chain...
| Quote: | xy-chain with memory (3=6)r2c1 - (6=2)r2c4* - (2=4)r8c4 - (4=6)r5c4 - (6=4)r5c6* - (24=3)r1c6 ; r1c1<>3
(I imagine a chain with less SL can be written using the 6 SLs - but this is as found) |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Tue Aug 31, 2010 12:13 pm Post subject: |
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Same logic (target) as Peter's proof (just avoiding t-chain or Triangular Matrix or Kraken Cell for a while ...  ) ...
| Quote: | M Wing (26) : (26)R2C4 6B1 2C3 : => r1c6<>2
7-SIS M Chain (26) : (42)R8C4 (26)R2C4 6B1 (69)R6C3 (97)R6C9 7C7 7C5 : (4)r8c4=(7)r8c5 : => r8c5<>4
Note : Peter, your chain is made of 5-SIS : 4 cells and 1 bi-local, and 5 "internal" WIS (total of WIS : 7). So, the stars * account for 1 extra WIS. JC
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Tue Aug 31, 2010 6:55 pm Post subject: |
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AAIC with fin (2)r1c6
If AIC is true: (3=4)r1c6-(4=6)r5c6-r3c6=r3c3-(6-3)r2c1; r1c1<>3
If fin is true: (2)r1c6-(2=6)r2c4-(6=3)r2c1; r1c1<>3
My initial pass started with an ANP that deleted 2 from r1c6 to setup the AIC, but then I realized an "almost" AIC did the trick in one step.
ANP[(26)r3c36=(34)ls:r3c56]-(34=2)r1c6-r2c4=(2)r2c7; r3c89<>2
Ted |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Tue Aug 31, 2010 7:14 pm Post subject: |
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| Quote: | | just avoiding t-chain or Triangular Matrix or Kraken Cell for a while |
mind if I ask why? |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Tue Aug 31, 2010 7:43 pm Post subject: |
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| peterj wrote: | One step xy chain...
| Quote: | xy-chain with memory (3=6)r2c1 - (6=2)r2c4* - (2=4)r8c4 - (4=6)r5c4 - (6=4)r5c6* - (24=3)r1c6 ; r1c1<>3
(I imagine a chain with less SL can be written using the 6 SLs - but this is as found) |
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this is just a suggestion but might it be easier to see this as a AAIC?
a finned xy-wing to be exact?
so that the strong inference between the xy-wing and the 4 in r1c6 is spelled out a little more clearly.
[xy-wing:(3=6)r2c1 - (6=2)r2c4 - (2=3)r1c6] = (4)r1c6 - (4=5)r5c6... etc...
{xy-wing} = (4)
I don't want to take away from your brilliant find. |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Tue Aug 31, 2010 8:17 pm Post subject: |
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| storm_norm wrote: | | this is just a suggestion but might it be easier to see this as a AAIC? |
Norm, I am sure you are right. But I have a natural inclination to follow chains of bivalues - like a treasure hunt! - so this was the easiest route for me. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Tue Aug 31, 2010 8:21 pm Post subject: |
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yeah, I was thinking since you wrote it out from r2c1 around, that is how you envisioned it.
nice find. |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Wed Sep 01, 2010 1:20 am Post subject: |
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| storm_norm wrote: | | Quote: | | just avoiding t-chain or Triangular Matrix or Kraken Cell for a while |
mind if I ask why? |
Colouring the bivalues from box B1 leads to an empty cell in R1C6 in one of the colour => r1c1<>3. This is trial and error.
So, mentally supposing r1c1=3, one of the shortest paths to prove that the cell R1C6 will be emptied is the following ordered sequence of SIS :5-SIS t chain : (36)R2C1 (62*)R2C4 (24)R8C4 4C5[explicitly (4)r83*c5] (243)R1C6 : => r1c1<>3 When the Sudoku puzzle rules are applied, the strong inferences are done inside the SIS and the weak inferences are done between a SIS and (eventually all) the previous SIS.
The ordered sequence of SIS is a shorthand notation for a network (would say Danny), here a t-chain. The fully developped eureka notation is Peter's chain, where a * indicates the starting point of a weak inference to a SIS further away in the sequence (the final point is not indicated). To get a clearer view of the weak inferences, a Triangular Matrix may be written, but I will refrain of doing so as the notations are quite heavy.
Note : the t chain contains only one "t" candidate, (2)r2c4, and no "z" candidate.
From here, the interpretation of the exclusion may spread in different directions as the 5 SIS may be picked and eventually grouped in different logical orders.
For example :Norm's Finned XY Wing : {(36)R2C1 (62)R2C4 (23=4)R1C6} 4C5[explicitly (4)r38c5] (42)R8C4 (26)R2C4 (63)R2C1 => r1c1<>3
Kraken Cell (234)R1C6 : the t chain is here read in reverse order
(2)r1c6 (26)R2C4 (63)R2C1 [XY Wing]
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(3)r1c6 (63)R2C1 [XY Wing]
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(4)r1c6 4C5[explicitly (4)r38c5] (42)R8C4 (26)R2C4 (63)R2C1 [Fin] Ted's proof can also be analyzed in the same way as it is based on the
5-SIS t chain : (36)R2C1 6R3[explicitly (6)r3c36] (64)R5C6 (43*2)R1C6 (26)R2C4 : => r1c3<>3 where 3* stands for a "z" candidate (There is no "t" candidate).
His AAIC is a different way of writing a Kraken Cell (234)R1C6 : (2)r1c6 ... [Fin]
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(3)r1c6 [AIC]
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(4]r1c6 ... [AIC] Final comments : in this puzzle, eliminating (2)r1c6, for example through a m wing (26) , or considering (2)r1c6 as a fin for an AIC are equivalent  However the more concise interpretation of the exclusion of (3)r1c1 remains the 5-SIS t chain even if the level of difficullty of Danny's puzzles (SE less than 8.2) doesn't require it.
JC |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Sep 01, 2010 3:06 am Post subject: |
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| Code: | +--------------------------------------------------------------+
| 37 79 29 | 8 5 234 | 6 1 234 |
| 36 4 5 | 26 1 9 | 23 7 8 |
| 8 1 26 | 7 34 2346 | 9 235 2345 |
|--------------------+--------------------+--------------------|
| 5 39 4 | 1 2 7 | 8 36 369 |
| 1 8 37 | 46 9 46 | 237 235 2357 |
| 67 2 69 | 3 8 5 | 1 4 79 |
|--------------------+--------------------+--------------------|
| 2 37 37 | 5 6 8 | 4 9 1 |
| 9 6 18 | 24 347 1234 | 5 238 237 |
| 4 5 18 | 9 37 123 | 237 2368 2367 |
+--------------------------------------------------------------+
# 62 eliminations remain
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A different XY-Wing w/fin folded into a discontinuous loop
| Code: | **** XY-Wing ****
(3)r1c1 - [(3=24)r1c6/r2c4+r5c6] - (6)r3c6 = r3c3 - (6=3)r2c1 - (3)r1c1
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Regards, Danny
Note: I don't put a "=>" on a discontinuous loop because I consider it redundant.
(However, my solver is dumb and will. _ _) |
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