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Mogulmeister
Joined: 03 May 2007 Posts: 695

Posted: Thu May 17, 2007 10:02 pm Post subject: Mepham/Stuart Puzzle 11/05/07 


Code:  ++++
.1.......
..21..4..
6.4...2.8
++++
.9.6.4.2.
7.......5
.5.8.9.3.
++++
1.9...6.2
..5..27..
.......8.
++++

Different routes  what worked for you ? 

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Nwohio
Joined: 12 May 2007 Posts: 9

Posted: Fri May 18, 2007 1:34 am Post subject: 


Code: 
++++
 9 1 37  24 24 8  35 57 6 
 5 8 2  1 36 367  4 79 39 
 6 37 4  3579 39 357  2 1 8 
++++
 3 9 1  6 5 4  8 2 7 
 7 4 8  23 123 13  9 6 5 
 2 5 6  8 7 9  1 3 4 
++++
 1 37 9  3457 8 35  6 45 2 
 8 6 5  349 1349 2  7 49 139 
 4 2 37  579 169 1567  35 8 19 
++++

Play this puzzle online at the Daily Sudoku site
I was able to work it to here with somewhat standard eliminations, including a finned Xwing at one point. I then saw that a 5 in r7c8 resulted in a 7 in both r1c3 and r1c8, which of course meant that r7c8 must be 4. This feels like a guess, though, so I would appreciate any comments.
Looking at it from the opposite direction, I see that either a 5 or 7 in r1c8 results in the 4 that I placed in r7c8. Is this an example of an XYchain? 

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Mogulmeister
Joined: 03 May 2007 Posts: 695

Posted: Fri May 18, 2007 9:20 pm Post subject: 


Hi NW,
1)Your approach is very similar to George Woods who often uses an intuitive forcing chain to solve puzzles  basically you look at a bivalue square and you find that, irrespective of either value, causes the same outcome in a target cell.
You dicovered that a contradiction occurs when you put 5 into r7c8 so 5 is ruled out.
Be careful when you say Quote:  I see that either a 5 or 7 in r1c8 results in the 4 in r7c8  If you put a 7 in r1c8 and just look down that column this forces r2c8 to be 9,r8c8 to be 4 and makes r7c8 a 5 !!
Purists who don't want to use trial and error are sometimes dismissive of this technique but as George says, the art is finding them ! I tend to agree with George but prefer myself to use logical structures (wings/fish/etc) which prove an elimination without having to do a "whatif".
2)For an xy chain  you need two pincer squares (in green) which point at the candidate to be eliminated and the 5 in r7c8 does eliminate  see below:
Another example with xy's explained:
http://www.dailysudoku.com/sudoku/forums/viewtopic.php?t=1867&sid=97712ff1d23e09f28305cf749b96e605
Last edited by Mogulmeister on Fri May 18, 2007 10:05 pm; edited 1 time in total 

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Mogulmeister
Joined: 03 May 2007 Posts: 695

Posted: Fri May 18, 2007 10:03 pm Post subject: 


Notice also that this pattern is also an ALS. Using the same squares
Let set A be the pink. Set B the green. Locked common is 7 and therefore the 5 in blue can see both 5's in A & B and is eliminated.


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Mogulmeister
Joined: 03 May 2007 Posts: 695

Posted: Fri May 18, 2007 10:16 pm Post subject: 


....and this is also a forcing chain........
Whatever the value in r7c2, r7c8 is forced to 4. 

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Nwohio
Joined: 12 May 2007 Posts: 9

Posted: Sat May 19, 2007 12:02 am Post subject: 


Quote:  2)For an xy chain  you need two pincer squares 
Isn't that what I did, although with a longer chain? The pincers are are r1c8 and r7c6.
r1c8 (5) or
r1c8 (7), r1c3 (3), r3c2 (7), r7c2 (3), r7c6 (5)
Thanks for the input. 

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Marty R.
Joined: 12 Feb 2006 Posts: 5658 Location: Rochester, NY, USA

Posted: Sat May 19, 2007 12:52 am Post subject: 


Nwohio wrote:  Quote:  2)For an xy chain  you need two pincer squares 
Isn't that what I did, although with a longer chain? The pincers are are r1c8 and r7c6.
r1c8 (5) or
r1c8 (7), r1c3 (3), r3c2 (7), r7c2 (3), r7c6 (5)
Thanks for the input. 
If you continue that second chain and make r7c8 a 4, then continuing in column 8, you end up with duplicate 7s. Or if you start with r1c8=7 and work straight down in column 8, then r7c8 is a 5.
So they're not true pincers since we can't say that one or the other is forced to be a 5. 

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