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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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 Posted: Wed Jan 12, 2011 2:36 am Post subject: Vanhegan Fiendish |
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Puzzle 5-542666, rated 2.4.0.1
I'd be interested in others' solutions. Somehow I get the feeling there's something easy here but I couldn't see it. I did solve it with a very seldom-used move for me.
| Code: |
+----------+-----------+------------+
| 3 69 8 | 5 1 4 | 79 2 679 |
| 5 1 69 | 23 23 7 | 4 8 69 |
| 2 4 7 | 6 9 8 | 1 3 5 |
+----------+-----------+------------+
| 19 5 19 | 78 6 3 | 78 4 2 |
| 7 8 2 | 4 5 9 | 6 1 3 |
| 6 3 4 | 1 78 2 | 5 9 78 |
+----------+-----------+------------+
| 19 69 3 | 278 4 5 | 28 67 18 |
| 8 7 16 | 9 23 16 | 23 5 4 |
| 4 2 5 | 378 78 16 | 389 67 189 |
+----------+-----------+------------+
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Play this puzzle online at the Daily Sudoku site |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Wed Jan 12, 2011 7:32 am Post subject: |
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Either r9c4<>3, r1c9=9 (BUG+1), NP(18)R79C9
Or r9c4=3, (8)r9c9=(9)r1c9 (BUG+2), NP(18)R79C9.
Therefore r6c9<>8=7
Checking : (8)r6c9*-(8=7)r4c7-(7=9)r1c7**-(9)r1c2***=(9)r7c2-(9=1)r7c1-[(1=8)r7c9-(8=7)r6c9* OR/AND (1)r7c9=(1)r9c9-(9)r9c9=(9)r9c7***-(9=7)r1c7**]
__* (8)r6c9-(8)r6c9 or (8)r4c7=(8)r7c9 (5 SIS AIC) : => r6c9<>8
_** (7)r1c7=(7)r1c7 or (9)r1c7-(7)r4c7 (5 WIS AIC) : => r4c7<>7
*** (9)r1c2=(9)r9c7 (4 SIS AIC) : => r1c7<>9
Note : derived WIS exceptionally leads here to an elimination ! (r1c7=79) |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Jan 12, 2011 8:41 am Post subject: |
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| JC Van Hay wrote: | Either r9c4<>3, r1c9=9 (BUG+1), NP(18)R79C9
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Hello JC. Welcome back. I'm lost on how you get from r9c4<>3 to r1c9=9 (BUG+1) |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Wed Jan 12, 2011 9:29 am Post subject: |
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I think there are a lot of xy-chain solutions... here's one
| Code: | | xy-chain (8=7)r4c7 - (7=9)r1c7 - (9=6)r1c2 - (6=9)r7c2 - (9=1)r7c1 - (1=8)r7c9 ; r6c9<>8, r79c7<>8 |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Wed Jan 12, 2011 10:36 am Post subject: |
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| daj95376 wrote: | | JC Van Hay wrote: | Either r9c4<>3, r1c9=9 (BUG+1), NP(18)R79C9
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Hello JC. Welcome back. I'm lost on how you get from r9c4<>3 to r1c9=9 (BUG+1) |
Hello Danny. My mistake. I went too fast and I didn't completely check for hidden singles. So I overlooked the contradiction with r9c9=9!
Trying to solve the puzzle with a BUG+4 drew however my attention on (78)R6c9 as a starting point for chains.
Reworking the puzzle to prove that r9c4=3 ("to keep up appearances"), I got : 1R9 1R8 (19)R4C3 9R2 : => -9r9c9.
JC |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Jan 12, 2011 4:43 pm Post subject: |
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| I spent time looking for XY-Chains which I thought were there. My seldom-used move was what I know as an XYZ-Transport. Note the seemingly useless XYZ-Wing 278 pivoted in box 8. Transport the 8 from r7c7 to r4c4 and the 8 can be removed from r9c4. That sets up an easy four-cell coloring chain; r7c9<>8. |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Wed Jan 12, 2011 7:22 pm Post subject: |
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Marty, what an interesting first move!
The combination of the XYZ Wing and the bilocals (8)r4c47 is equivalent to an almost XWing (8)r9c5=XWing(8)r47c47 due to the ALS(278)R7C47 : => r9c47<>8.
In Eureka notation : (8=7)r9c5-(7)r7c4=loop[(8=2)r7c4-(2=8)r7c7-(8)r4c7=(8)r4c4] : => r9c47<>7
That is a weak link between an ALS and an Almost Continuous Nice Loop, similar to an ALS XZ-rule :
ALS(87)R9C5-ACNL(728)R47C47 : => r9c47<>8 After that, the puzzle simply falls down with an XWing(8) instead of a Kite(8).
JC |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Jan 12, 2011 10:35 pm Post subject: |
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Another perspective on Marty's first elimination:
| Code: | (78=2)r9c5,r7c4 - (2=8)r7c7 - r4c7 = (8)r4c4 => r9c4<>8
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Nice find Marty! |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Thu Jan 13, 2011 3:47 pm Post subject: |
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| Marty R. wrote: | | I spent time looking for XY-Chains which I thought were there. My seldom-used move was what I know as an XYZ-Transport. Note the seemingly useless XYZ-Wing 278 pivoted in box 8. Transport the 8 from r7c7 to r4c4 and the 8 can be removed from r9c4. That sets up an easy four-cell coloring chain; r7c9<>8. |
Marty,
Some time ago I was advised by "Luke451" that your step could also be referred to as a "Kraken cell". It is a fun move.
Ted |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Thu Jan 13, 2011 4:08 pm Post subject: |
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Here is another approach.........
ANP(69=7)r12c9-(7=8)r6c9-(8=1)r7c9-7c1=(1-6)r8c3=r2c3-(6=9)r2c9; r9c9<>9
Another sequence with the same result is
ANP(18=9)r79c9-(1)r9c9=r9c6-(1=6)r8c6-r8c3=r2c3-(6=9)r2c9 which results in a contradiction; r9c9<>9
Still a different solution using a ANP() is involves Marty's xyz-wing.
ANP(28=7)r7c47-(7=8)r4c4-r4c7=(8)r6c9; r7c9<>8=1
Ted |
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