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JV
Joined: 09 Jan 2011
Posts: 24
Location: Devon, England
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 Posted: Sat Feb 26, 2011 2:37 pm Post subject: Menneske 5772850 |
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| Code: |
+--------------+-----------+-------------+
| 1 2469 268 | 5 3 7 | 468 249 689 |
| 245 24 258 | 9 6 248 | 1 7 3 |
| 3 2469 7 | 1 48 248 | 468 249 5 |
+--------------+-----------+-------------+
| 7 8 4 | 6 9 3 | 5 1 2 |
| 69 16 3 | 2 5 148 | 7 49 689 |
| 269 5 126 | 7 48 148 | 468 3 689 |
+--------------+-----------+-------------+
| 245 234 25 | 38 1 6 | 9 58 7 |
| 8 7 9 | 4 2 5 | 3 6 1 |
| 56 136 156 | 38 7 9 | 2 58 4 |
+--------------+-----------+-------------+
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Play this puzzle online at the Daily Sudoku site
i have a feeling that I used more moves than necessary. Anyway, there's a nice little move with the 69 UR, and you might see a 6-cell DP later. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sat Feb 26, 2011 10:31 pm Post subject: |
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On my first pass, I did not get to the AUR(69); I saw the AUR(24) first.
#1: AUR(24)r27c12 sis r2c36=2,r7c3=2,r2c6=4; r2c1<>2
(2)r2c36
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(2)r7c3-r6c3=(2)r6c1
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(4)r2c6-(4=2)r2c2
#2: ANP(45=2)r27c1-(2=5)r7c3-r7c8=(5)r9c8; r9c1<>5
#3: coloring (6)
Ted |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sun Feb 27, 2011 6:32 am Post subject: |
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| Code: | +-------------------+-----------------+-----------------+
| 1 2469 268 | 5 3 7 | 468 249 689 |
| 245 24 258 | 9 6 248 | 1 7 3 |
| 3 249(6) 7 | 1 48 248 | 48(6) 249 5 |
+-------------------+-----------------+-----------------+
| 7 8 4 | 6 9 3 | 5 1 2 |
| 69 (16) 3 | 2 5 48-1 | 7 49 689 |
| 269 5 26-1 | 7 (48) (148) | (468) 3 689 |
+-------------------+-----------------+-----------------+
| 245 234 25 | 38 1 6 | 9 58 7 |
| 8 7 9 | 4 2 5 | 3 6 1 |
| 56 136 156 | 38 7 9 | 2 58 4 |
+-------------------+-----------------+-----------------+ |
#1...ALS(48)[(1)r6c6 = (6)r6c7] - (6)r3c7 = (6)r3c2 - (6=1)r5c2; r6c3 and r5c6 <> 1
#2...xy-wing {256} r9c1, r7c3, r6c3
r56c1 <> 6 |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Feb 27, 2011 6:55 am Post subject: |
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JV,
Please post the original puzzle. The validity of a DP deduction depends on the initial clues, not only on the current state of the puzzle.
Also, I would like to see the Menneske rating.
Best wishes,
Keith |
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Pat
Joined: 23 Feb 2010
Posts: 207
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| Posted: Sun Feb 27, 2011 8:32 am Post subject: |
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the original puzzle can be seen at
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Feb 27, 2011 9:18 am Post subject: |
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Ted had the correct UR, but insufficient eliminations.
| Code: | *) r2c1=2 => DP (as Ted derived)
*) r7c1=2 => DP (Hidden Unique Rectangle)
*) r6c1=2 follows ... and a Skyscraper <6> completes the job
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I've encountered similar UR patterns previously, and I'd love to use the notation:
| Code: | (*2)r2c1 - r123c2 = (*2-4)r7c2 = (*4)r7c1; DP => r2c1<>2
= (*4 )r2c2
+--------------------------------------------------------------+
| 1 2469 268 | 5 3 7 | 468 249 689 |
| 245 24 258 | 9 6 248 | 1 7 3 |
| 3 2469 7 | 1 48 248 | 468 249 5 |
|--------------------+--------------------+--------------------|
| 7 8 4 | 6 9 3 | 5 1 2 |
| 69 16 3 | 2 5 148 | 7 49 689 |
| 269 5 126 | 7 48 148 | 468 3 689 |
|--------------------+--------------------+--------------------|
| 245 234 25 | 38 1 6 | 9 58 7 |
| 8 7 9 | 4 2 5 | 3 6 1 |
| 56 136 156 | 38 7 9 | 2 58 4 |
+--------------------------------------------------------------+
# 60 eliminations remain
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Last edited by daj95376 on Sun Feb 27, 2011 9:51 am; edited 1 time in total |
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JV
Joined: 09 Jan 2011
Posts: 24
Location: Devon, England
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| Posted: Sun Feb 27, 2011 9:21 am Post subject: |
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Ted & Norm: I'm very impressed!. In my readings I noticed a comment of Dnny's, that times had moved on since he first posted puzzles, and I can see that this is true. The (apparently) casual use of ANPs and ALSs is new to me. Very interesting.
The move I saw was quite straightforward, but not easy for me to express in Eureka. To prevent the DP in UR69, r46c19 you need r6c1 = 2, or r57c9 = 8.
r6c1 = 2 gives a pair 45 in the column: r9c1 = 6; r6c1 <> 6
r46c9 = 8 makes r6c7 = 46, and now there's an xy-wing, pivot r5c8 that gives r6c1 <>6.
Keith: sorry, I've realised that another comparatively new idea is using DPs sort of retrospectively, but had forgotten anout that. As to the rating, well, OK, though I'm sure everyone agrees that Menneske ratings don't correlate closely with difficulty. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Feb 27, 2011 2:56 pm Post subject: |
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Norm, Wonderful als 
Danny, great observation. I only looked at the internal/external sis and failed to realized the hidden UR condition. This is particularly annoying to me because I usually do look for all possible deletions offered by a given pattern and have even posted some such events to show others the benefit of doing so. Thanks for the comment.
Ted |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Feb 27, 2011 3:35 pm Post subject: |
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JV,
I am still working to conquer the challenge of Eureka but have seen other posts that include some 'liberal" additives. In that manner, here is one possible notation for your pattern. The beginning part seems ok to me, but the back end is suspect.
als(2456)r279c1[(6)r9c1=(2)r27c1] -AUR(69)r56c19[(2)r6c1=(8)r56c8]-(8)r6c7=xy-wing(4-69)r6c7|r5c8|r5c1; r6c1<>6
I have also seen another variation where the end is notated as
..... xy-wing(4-69)[(6=4)r6c7-(4=9)r5c8-(9=6)r5c1]
Hopefully "those who know" will comment so that both you and I can benefit.
Ted
Last edited by tlanglet on Sun Feb 27, 2011 4:51 pm; edited 1 time in total |
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JV
Joined: 09 Jan 2011
Posts: 24
Location: Devon, England
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| Posted: Sun Feb 27, 2011 4:36 pm Post subject: |
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| Thanks Ted. I think I follow that. It's impressive anyway. |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Sun Feb 27, 2011 7:34 pm Post subject: |
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| tlanglet wrote: | I saw the AUR(24) first.
#1: AUR(24)r27c12 sis r2c36=2,r7c3=2,r2c6=4; r2c1<>2
(2)r2c36
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(2)r7c3-r6c3=(2)r6c1
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(4)r2c6-(4=2)r2c2 |
This is one of those cases where a mix of internal and external candidates can simplify the AUR chain. Using internal candidates in r2 and external candidates in r7:
AUR(24)r27c12:[(5)r2c1 = (2)r7c3] - (3)r5c3 = (2)r6c1 ==> r2c1<>2 |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Mon Feb 28, 2011 1:52 pm Post subject: |
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Ron,
Thanks for the reminder of mixed inferences. I seemed to have slipped into the mode of either internal or external and have forgotten the combination.
Ted |
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