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AZ Matt
Joined: 03 Nov 2005 Posts: 63 Location: Hiding under my desk in Phoenix AZ USA

Posted: Fri Jul 14, 2006 4:44 pm Post subject: Anomaly? 


Okay, I have another question about a technique that I used to solve a puzzle, and I don't know if it is a XYZwing varient or an "invalid" trialanderror solve.
Here is were I got to:
Code:  3 157 57 1769 8 79 4 2 16
6 9 78 127 27 4 3 5 18
2 4 18 5 16 3 68 7 9
4 8 59 29 25 6 1 3 7
7 25 6 3 4 1 258 9 28
19 125 3 8 75 279 25 6 4
8 3 179 67 167 5 269 4 26
19 17 4 167 3 2 69 8 5
5 6 2 4 9 8 7 1 3

And this is my thought process:
You can see rather easily that the open cells in r3 and c9 are all binary and linked; i.e., a solve for one solves for all. There are five “buddy” cells (if that is appropriate usage in this context) at r2c35, and r7c357. Of these, only two cells – r7c57 – are potentially solved by the combined effects of the open cells in r3 and c9. And this is where I see what I would call an anomaly in the r7c5 cell:
If r3c5 = 1 and r7c9 = 6, then r7c5 must = 7.
But if that is true, r7c4 must also = 7.
Therefore, that proposition cannot be true.
So we test the proposition and find:
If r3c5 = 1, then r7c9 must = 6.
Therefore, r3c5 must = 6.
And the puzzle drops.
I suppose someone will find a simple XYwing that similarly drops the puzzle from this position (I tend to get impatient), but my question is: Suppose that was the only way to solve the puzzle; would it be a “valid” puzzle?
Or is this a standard technique, or variation thereof, that I am not recognizing?
PS: This is Brain Bashers’ Super Hard from yesterday, July 13:
http://www.brainbashers.com/sudoku.asp 

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ravel
Joined: 21 Apr 2006 Posts: 536

Posted: Fri Jul 14, 2006 5:52 pm Post subject: 


To your question:
The technique is called elimination by contradiction, other formulated:
r3c5=1 => r3c7=6 => r1c9<>6 => r7c9=6. Then r7c4=7 and (with r3c5=1) r7c5=7  contradiction.
=> r3c5=6.
[edit: corrected typos]
The other way to solve it: in r6c6 there cannot be 2 (you have it in r8c6), then there is a box/line intersection: the 7 in box 8 must be in column 4, therefore it cannot be in the rest of the column (r12c4). Then there is an xywing from r3c7: either r2c9 or r3c5 is 1, so r2c4 cannot be 1.
Last edited by ravel on Sat Jul 15, 2006 1:04 pm; edited 1 time in total 

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TKiel
Joined: 22 Feb 2006 Posts: 292 Location: Kalamazoo, MI

Posted: Fri Jul 14, 2006 9:52 pm Post subject: 


There is also a colouring chain (6's) that has r7c5 and r7c9 share a colour, which must be the false colour. 

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AZ Matt
Joined: 03 Nov 2005 Posts: 63 Location: Hiding under my desk in Phoenix AZ USA

Posted: Fri Jul 14, 2006 11:38 pm Post subject: Thanks 


Thank you.
I am sure there are any number of ways to solve this from puzzle from here. I tend to solve puzzles before I have finished even basic solving techniques because I am always looking for patterns in the numbers, or what I call "big solves"  solutions that wipe out a great number of cells (and what I will now call contradictions).
So my real question is, if we assume for the moment my solve was the only way to solve the puzzle (elimination by contradiction, which I have heard used interchangably with forcing chains though I don't know that I see that connection here), would a "purist" consider that to close to trialanderror for it to be a "valid" puzzle?
Again, I hope I am not beating a dead horse . I was doing sudoku for quite awhile before I ever read anything technical about it, so while I am apparently capable of solving even the most difficult puzzles, I am hopelessly lost with lexicon of the sport. 

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TKiel
Joined: 22 Feb 2006 Posts: 292 Location: Kalamazoo, MI

Posted: Sat Jul 15, 2006 12:09 pm Post subject: 


AZ Matt wrote:  So my real question is, if we assume for the moment my solve was the only way to solve the puzzle (elimination by contradiction, which I have heard used interchangably with forcing chains though I don't know that I see that connection here), would a "purist" consider that to close to trialanderror for it to be a "valid" puzzle?

Depends on how one defines 'purist', but my guess would be no. The hardest puzzles are only solvable by this technique.
Have you checked out the 'advanced solving techniques' forum on Sudoku.com (click this link)? I think there are some threads there that shed considerable light on the subject. 

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ravel
Joined: 21 Apr 2006 Posts: 536

Posted: Sat Jul 15, 2006 1:28 pm Post subject: Re: Thanks 


AZ Matt wrote:  ... contradiction, which I have heard used interchangably with forcing chains though I don't know that I see that connection here ... 
Forcing chains in the classical sense are the "positive" variants of elimination chains. They show, that starting from all candidates of a cell or all possible cells for a number in a unit the same result is obtained. For each elimination chain there is an equivalent forcing chain (maybe containing "case distinctions"). E.g. the one above could be expressed this way:
r7c4=6 => r7c9<>6 => r1c9=6 => r3c7<>6 => r3c5=6
r7c4=7 =>
{either r7c5=1 => r3c5=6
or r7c5=6 => r7c9<>6 => r1c9=6 => r3c7<>6 => r3c5=6}
(the last line alone also shows that r7c5 cannot be 6, therefore r3c5=6)
Contradiction chains like the above one with multiple inference (r7c9=6 and r3c5=1=> r7c5=7) are also referred to as forcing or error nets. 

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