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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Wed Sep 15, 2010 4:40 am Post subject: Puzzle 10/09/15: Extreme |
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Most of the reasonable puzzles are posted from my current collection. I'll generate more soon.
However, I have a number of difficult and extreme puzzles remaining. So, I'm clearing some out.
| Code: | +-----------------------+
| . 1 4 | 3 . 7 | 6 . . |
| 8 . 9 | 4 . . | 3 2 . |
| . . . | 9 . . | . . . |
|-------+-------+-------|
| 3 . 6 | . . 9 | 2 . . |
| . 5 . | . 8 . | . . . |
| 2 . . | 1 . 6 | . . . |
|-------+-------+-------|
| 7 . . | 6 . . | 9 . . |
| . 4 . | . . . | . 5 . |
| . . . | . . . | . . . |
+-----------------------+
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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 Sep 15, 2010 12:00 pm Post subject: |
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Not that easy to find a minimal set of steps of depth less than or equal to 4 ...
| Quote: | 3-SIS Grouped X Chain (1)R2C7R7 : => r78c6<>1
3-SIS M Wing (58) : (85)R9C4 5B7 8R7 : (8)r9c4=(8)r7c8 : => r9c78<>8
4-SIS XY Chain or ALS XY Wing (145-8) : (81)R8C7, (14)R9C7, (45=8)R9C46 : => r8c4<>8
3-SIS S Wing : 5R2 (54)R9C6 4C7 : (5)r2c9=(4)r3c7 : => r3c7<>5 |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Wed Sep 15, 2010 8:16 pm Post subject: |
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Three steps - first a bit mucky!
| Quote: | AIC (3=2)r7c2 - r7c9=(2-3*-7)r9c9=(7-3)r9c8=r7c89 ; r7c35<>3
(Not sure how to notate but essentially the strong link on 2 and 7 in b9 stop either r9c89 being 3...)
sashimi x-wing(1) ; r7c89<>1
xy-chain ; (5=1)r2c9 - (1=5)r2c6 - (5=4)r9c6 - (4=1)r9c7 - (1=8)r8c7 - (8=5)r6c7 ; r6c9<>5, r3c7<>5 |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Thu Sep 16, 2010 1:46 am Post subject: |
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Peter: How about applying your logic from right-to-left? (for easier interpretation)
| Code: | (3)r7c89 = (37-2)r9c89 = (2)r7c9 - (2=3)r7c2 - loop => r7c35<>3 (for starters)
+-----------------------------------------------------------------------+
| 5 1 4 | 3 2 7 | 6 9 8 |
| 8 7 9 | 4 6 15 | 3 2 15 |
| 6 23 23 | 9 15 8 | 145 147 1457 |
|-----------------------+-----------------------+-----------------------|
| 3 8 6 | 57 457 9 | 2 14 145 |
| 4 5 1 | 2 8 3 | 7 6 9 |
| 2 9 7 | 1 45 6 | 458 348 345 |
|-----------------------+-----------------------+-----------------------|
| 7 23 2358 | 6 135 1245 | 9 1348 1234 |
| 19 4 238 | 78 1379 12 | 18 5 6 |
| 19 6 2358 | 58 1359 1245 | 148 13478 12347 |
+-----------------------------------------------------------------------+
# 70 eliminations remain
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More eliminations possible because of the continuous loop _  _
I refuse to admit how long I stared at the AHP(37)r9c89 before realizing it could be put into play. _  _
Last edited by daj95376 on Thu Sep 16, 2010 2:04 am; edited 1 time in total |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Thu Sep 16, 2010 2:02 am Post subject: |
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| peterj wrote: | AIC (3=2)r7c2 - r7c9=(2-3*-7)r9c9=(7-3)r9c8=r7c89 ; r7c35<>3
(Not sure how to notate but essentially the strong link on 2 and 7 in b9 stop either r9c89 being 3...) |
Peter : you have written your first AIC as a "Transport" or "Transfer" Matrix :
| Code: | .---------------------.---------------------.---------------------.
| 5 1 4 | 3 2 7 | 6 9 8 |
| 8 7 9 | 4 6 15 | 3 2 15 |
| 6 23 23 | 9 15 8 | 145 147 1457 |
:---------------------+---------------------+---------------------:
| 3 8 6 | 57 457 9 | 2 14 145 |
| 4 5 1 | 2 8 3 | 7 6 9 |
| 2 9 7 | 1 45 6 | 458 348 345 |
:---------------------+---------------------+---------------------:
| 7 23 2358 | 6 135 1245 | 9 1348 1234 |
| 19 4 238 | 78 1379 12 | 18 5 6 |
| 19 6 2358 | 58 1359 1245 | 148 13478 12347 |
'---------------------'---------------------'---------------------'
(32)R7C2 : (3)r7c2 (2)r7c2
2B9 .... : ....... (2)r7c9 (2)r9c9
7B9 .... : ....... ....... (7)r9c9 (7)r9c8
3B9 .... : ....... ....... (3)r9c9 (3)r9c8 (3)r7c89
-> (3)r7c2=(3)r7c89 : => r7c35<>3 (Here, the first column identifies the SIS used in each row). |
This suggests writing down the AIC as :
4-SIS AIC : (32)R7C2 [2B9 7B9] 3B9 : (3=2)r7c2-r7c9=HP(27)r9c89-(3)r9c89=r7c89 : => r7c35<>3 Furthermore, the AIC is a continuous network disguised in a good looking continuous nice loop (no need of a 20 feet pole to tackle it ...  ). The weak links are thus strong : r7c36<>2, r9c89 contains 237 at most or r9c89<>148.
BTW, I had these last eliminations from the following 6-SIS AIC : NP(27)r9c89=(2)r7c9-(2=3)r7c2-r7c8=NT(148)B9 : => r9c89<>148. You did a better job with the HP(27)B9 
Further comments :
If the last entry in the Transport Matrix is written in the first column, one obtains a 4x4 Symmetric Pigeonhole Matrix : each row contains at least one truth and each column contains at most one truth => each column is a derived SIS. In details :
| Code: | SIS\WIS + 3r7..... 2r7..... *r9c9... *r9c8
---------+-------------------------------------
(32)R7C2 + (3)r7c2. (2)r7c2
2B9 .... + ........ (2)r7c9. (2)r9c9
7B9 .... + ........ ........ (7)r9c9. (7)r9c8
3B9 .... + (3)r7c89 ........ (3)r9c9. (3)r9c8
---------+-------------------------------------
=> Elim. + (3)r7c35 (2)r7c36 (14)r9c9 (148)r9c8 |
Regards, JC
[edit : typos corrected and addition of overlooked eliminations on 2 - Thanks Danny for checking !]
Last edited by JC Van Hay on Thu Sep 16, 2010 2:37 am; edited 3 times in total |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Thu Sep 16, 2010 6:41 am Post subject: |
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This puzzle earned its rating. Six eliminations on <1> that weaken it for a 5-SIS on two values across eight cells.
| Code: | r89c15 <19> UR Type 4.2244 r89c5<>1
c57\r37 Sashimi X-Wing <> 1 r7c89
r7 b8 Locked Candidate 2 <> 1 r89c6
+-----------------------------------------------------------------------+
| 5 1 4 | 3 2 7 | 6 9 8 |
| 8 7 9 | 4 6 a15 | 3 2 15 |
| 6 23 23 | 9 b15 8 | c145 147 1457 |
|-----------------------+-----------------------+-----------------------|
| 3 8 6 | 57 457 9 | 2 14 145 |
| 4 5 1 | 2 8 3 | 7 6 9 |
| 2 9 7 | 1 45 6 | d458 e348 345 |
|-----------------------+-----------------------+-----------------------|
| 7 23 g2358 | 6 135 145 | 9 f348 234 |
| 19 4 38 | 78 379 2 | 18 5 6 |
| 19 6 h2358 | 58 359 4-5 | 148 13478 12347 |
+-----------------------------------------------------------------------+
# 61 eliminations remain
(5)r2c6 = r3c5 - r3c7 = (5-8)r6c7 = r6c8 - r7c8 = (8-5)r7c3 = (5)r9c3 => r9c6<>5
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Translation for JC:
| Code: | 5B2 5C7 8R6 8R7 5C3 : (5)r2c6 = (5)r9c3 : => r9c6<>5
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With that, it's obvious that I need to get some sleep. Goodnight! |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Thu Sep 16, 2010 7:03 am Post subject: |
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Danny, JC, thanks for taking the time to look at my move!
There is something about AHPs that makes them particularly hard to see - actually even plain HPs I usually come to from the corresponding quad/quin.
It definitely had a 'network' feel to it - but I felt in mitigation that is more like a side-effect than a real branching of the chain!  The AHP/ANQ is much cleaner. |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Thu Sep 16, 2010 10:24 am Post subject: |
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Peter : HPs and AHPs are easily detected while pencilmarking if the bi-locals are tagged at the same time. For an HP, there are 2 cells containing the same pair of 2 digits carrying the same tags in a box (..), in a row (--) or in a column (||) and for an AHP, one of the 2 cells associated with the bi-locals is a "hub" cell.
BTW, your 3rd move is identical to a 3-SIS S Wing (45)R9C6. In this way, your solution becomes perfect (3 steps of depth <= 4).
Danny :  , your solution .... Thanks a lot for your enjoyable tailored puzzles They are of great help to make progress in solving even tougher puzzles.
Regards, JC |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Thu Sep 16, 2010 1:58 pm Post subject: |
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Peter, that was a great move
I found several steps but never one that started to unravel the puzzle. Maybe I will try finding some chains later.....
Ted |
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ttt
Joined: 06 Dec 2008
Posts: 42
Location: vietnam
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| Posted: Thu Sep 16, 2010 4:54 pm Post subject: |
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| tlanglet wrote: | | I found several steps but never one that started to unravel the puzzle. Maybe I will try finding some chains later..... |
On my experience, to solve sudoku puzzles (supported by SS solver & SudoCue solver – Ruud’s solver: to see bilocations) I always consider AHP first. Don’t know that is useful for you or not...
BTW, Danny: I don’t think we are on difference sudoku universe, we are the same sudoku universe...
Again, I had drunk too much... tonight
ttt |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Sep 16, 2010 5:37 pm Post subject: |
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Type 4 UR (19)
ER (1)
Type 1 UR (23)
Multi-coloring (1) |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Fri Sep 17, 2010 2:25 am Post subject: |
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| ttt wrote: | | tlanglet wrote: | | I found several steps but never one that started to unravel the puzzle. Maybe I will try finding some chains later..... |
On my experience, to solve sudoku puzzles (supported by SS solver & SudoCue solver – Ruud’s solver: to see bilocations) I always consider AHP first. Don’t know that is useful for you or not...
ttt |
Thanks ttt for the suggestion. In fact, based on studying solutions I found on Eureka, I have concluded AN(PTQ) and AH(PTQ) along with chains are the normal type of steps to unravel difficult puzzles. In this case, I did find a couple of interesting "almosts" but they did not seem to offer further usefulness. It is great fun to solve a difficult puzzle, but every good step is also a small victory.
Ted |
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