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Mepham/Stuart Puzzle 11/05/07

 
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Mogulmeister



Joined: 03 May 2007
Posts: 1151
PostPosted: Thu May 17, 2007 10:02 pm    Post subject: Mepham/Stuart Puzzle 11/05/07 Reply with quote
Code:
+---+---+---+
|.1.|...|...|
|..2|1..|4..|
|6.4|...|2.8|
+---+---+---+
|.9.|6.4|.2.|
|7..|...|..5|
|.5.|8.9|.3.|
+---+---+---+
|1.9|...|6.2|
|..5|..2|7..|
|...|...|.8.|
+---+---+---+


Different routes - what worked for you ?
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Nwohio



Joined: 12 May 2007
Posts: 9
PostPosted: Fri May 18, 2007 1:34 am    Post subject: Reply with quote
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 X-wing 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 XY-chain?
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Mogulmeister



Joined: 03 May 2007
Posts: 1151
PostPosted: Fri May 18, 2007 9:20 pm    Post subject: Reply with quote
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 "what-if".

2)For an x-y 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 x-y'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: 1151
PostPosted: Fri May 18, 2007 10:03 pm    Post subject: Reply with quote
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: 1151
PostPosted: Fri May 18, 2007 10:16 pm    Post subject: Reply with quote
....and this is also a forcing chain........

Whatever the value in r7c2, r7c8 is forced to 4. Smile
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Nwohio



Joined: 12 May 2007
Posts: 9
PostPosted: Sat May 19, 2007 12:02 am    Post subject: Reply with quote
Quote:
2)For an x-y 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: 5770
Location: Rochester, NY, USA
PostPosted: Sat May 19, 2007 12:52 am    Post subject: Reply with quote
Nwohio wrote:
Quote:
2)For an x-y 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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