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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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 Posted: Wed Nov 10, 2010 6:17 pm Post subject: Vanhegan |
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Fiendish, 5-877276, rated 2.2.0.1
I normally easily solve puzzles in this rating range, but I'm hopelessly stuck here after a Finned X-Wing and chain, both of which didn't make a dent.
| Code: |
+-------------+-------+---------+
| 4 6 3 | 9 5 7 | 18 18 2 |
| 2 57 57 | 1 8 3 | 49 49 6 |
| 8 1 9 | 2 6 4 | 3 5 7 |
+-------------+-------+---------+
| 1 24 46 | 5 3 9 | 67 27 8 |
| 7 289 28 | 6 4 1 | 29 3 5 |
| 3 59 56 | 8 7 2 | 16 19 4 |
+-------------+-------+---------+
| 5 3 24 | 7 1 8 | 24 6 9 |
| 9 78 1 | 4 2 6 | 5 78 3 |
| 6 2478 2478 | 3 9 5 | 78 24 1 |
+-------------+-------+---------+
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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: Wed Nov 10, 2010 8:51 pm Post subject: |
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There's an xy-chain - I couldn't see any wings etc.
| Code: | | (7=5)r2c3 - (5=6)r6c3 - (6=1)r6c7 - (1=8)r1c7 - (8=7)r9c7 ; r9c3<>7 |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Nov 10, 2010 10:02 pm Post subject: |
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| peterj wrote: | There's an xy-chain - I couldn't see any wings etc.
| Code: | | (7=5)r2c3 - (5=6)r6c3 - (6=1)r6c7 - (1=8)r1c7 - (8=7)r9c7 ; r9c3<>7 |
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Thanks, that's all it needed.
Is there a good way to look for potentially helpful chains? All I can do is pretty much try to find them by testing, basically a trial-and-error way of finding them. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Wed Nov 10, 2010 10:57 pm Post subject: |
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This puzzle probably has an interesting DP, but I can't find it. However, what I did find seems interesting.
| Code: | (2)r5c2 = HP(24-78)r49c2 = HP(78-24)b7q59 ; BUG contradiction! => r5c2=2
+--------------------------------------------------------------+
| 4 6 3 | 9 5 7 | 18 18 2 |
| 2 57 57 | 1 8 3 | 49 49 6 |
| 8 1 9 | 2 6 4 | 3 5 7 |
|--------------------+--------------------+--------------------|
| 1 24 46 | 5 3 9 | 67 27 8 |
| 7 289 28 | 6 4 1 | 29 3 5 |
| 3 59 56 | 8 7 2 | 16 19 4 |
|--------------------+--------------------+--------------------|
| 5 3 24 | 7 1 8 | 24 6 9 |
| 9 78 1 | 4 2 6 | 5 78 3 |
| 6 2478 2478 | 3 9 5 | 78 24 1 |
+--------------------------------------------------------------+
# 30 eliminations remain
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Wed Nov 10, 2010 11:26 pm Post subject: |
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Danny, I am relatively new to this - but have not seen a move like this  (I had to step through in SS to understand the logic!) I looked quickly at a BUG+3 scenario but convinced myself that a BUG situation did not exist - interesting that your move relies on a contradiction that would create a BUG. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Thu Nov 11, 2010 1:48 am Post subject: |
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Peter: This was a totally new approach for me. As I said, I was looking really hard for a DP. I noticed a strange pattern existed if I forced r9c3=78. I tried to turn it into a DP, but failed.
Occassionally, I'd also look at r5c2 in conjunction with r9c23. I kept running into a wall because the latter cells had four candidates and I couldn't remember how they could be used in a BUG scenario. So, I decided to see which candidates I could remove and create a BUG. I started with r5c2<>2, and the HP() relationships jumped out at me. I nearly fell out of my chair!
Regards, Danny |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Thu Nov 11, 2010 2:37 pm Post subject: |
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I just read the posts and need to review them but here is a two step solution using a AUR and a BUG+1.
AUR(28)r59c23 internal SIS r5c2=9,r9c23=4,r9c23=7; r5c3=8
(9)r5c2-(8)r5c2=(8)r5c3;
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(4)r9c23-(4=2)r7c3-(2=8)r5c3;
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(7)r9c23-(7=8)r8c2-r5c2=(8)r5c3;
BUG+1; r9c2=7
Ted |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Thu Nov 11, 2010 4:06 pm Post subject: |
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Ted: when I (originally) tried the <28> UR, I ended up with:
| Code: | *--------------------------------------------------*
| 4 6 3 | 9 5 7 | 18 18 2 |
| 2 57 57 | 1 8 3 | 49 49 6 |
| 8 1 9 | 2 6 4 | 3 5 7 |
|----------------+----------------+----------------|
| 1 24 46 | 5 3 9 | 67 27 8 |
| 7 29 8 | 6 4 1 | 29 3 5 |
| 3 59 56 | 8 7 2 | 16 19 4 |
|----------------+----------------+----------------|
| 5 3 24 | 7 1 8 | 24 6 9 |
| 9 78 1 | 4 2 6 | 5 78 3 |
| 6 48+7 27+4 | 3 9 5 | 78 24 1 |
*--------------------------------------------------*
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And just now realized that I could derive: r9c2<>4 & r9c3<>7 |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Thu Nov 11, 2010 7:49 pm Post subject: |
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| Marty R. wrote: | | Is there a good way to look for potentially helpful chains? All I can do is pretty much try to find them by testing, basically a trial-and-error way of finding them. |
Ultimately for me it's pretty much trial-and-error - but I try to shortcut the process a little bit by making it more "victim" driven than chain driven i.e. think about what I want to achieve rather than just randomly follow bivalues.
1) pick a digit that is in the right sort of pattern - not a single x-wing pattern, not dozens of them everywhere
2) pick a victim cell that will be productive - a bivalue, a bilocal etc.
3) pick the pincers that would eliminate it
4) try to make a chain...
rinse - repeat
So trial and error - but directed trial and error |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Nov 12, 2010 12:06 am Post subject: |
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Thanks Peter, I'll try and make my method more efficient. Besides, DT&E sounds a lot better than T&E.  |
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