View previous topic :: View next topic 
Author 
Message 
Marty R.
Joined: 12 Feb 2006 Posts: 5660 Location: Rochester, NY, USA

Posted: Mon Feb 20, 2006 1:31 am Post subject: This is ridiculous 


This puzzle is rated "Evil" by the site I got it from, but I usually solve them; the rating sounds much harder than the actual puzzle is, since I solve the majority of them. But this one I have tried at least six times. Every time when it looks like I'm nearing the end, I'm backed into a corner with a duplicate number.
That normally means I've made some sort of mechanical error, and I usually solve it with one or two more tries. But six times is unreal! When a puzzle is too difficult for me, the usual scenario is that I make some degree of progress, then reach an impasse and can't solve another cell.
I probably shouldn't embarrass myself by asking, but have you ever seen a published puzzle that was incorrect? I know the odds are very low, but having to erase everything and restart six times is unprecedented for me in my short Sudoku career.
Code:  458
85
753
14
86592
64
671
29
198 


Back to top 


David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Mon Feb 20, 2006 2:02 am Post subject: This puzzle is OK 


Hi, Marty!
There are some invalid puzzles floating around on the web. But this isn't one of them. It has a unique solution, and doesn't require anything tougher than finding a couple of triplets. dcb 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3287 Location: near Detroit, Michigan, USA

Posted: Mon Feb 20, 2006 2:16 am Post subject: The solution 


Marty,
The puzzle is valid, and does not require heroic efforts.
Here is the solution as given by Sudoku Susser:
* The puzzle seems to be in good shape. All solved squares are correct, and the remaining possibilities in the unsolved squares are accurate.
* R5C1 is the only square in row 5 that can be <4>.
* R7C4 is the only square in row 7 that can be <9>.
* R8C5 is the only square in row 8 that can be <1>.
* R9C9 is the only square in block 9 that can be <2>.
From this deduction, the following moves are immediately forced:
R3C9 must be <6>.
* R9C6 is the only square in row 9 that can be <5>.
* R6C9 is the only square in column 9 that can be <5>.
* R8C8 is the only square in column 8 that can be <5>.
From this deduction, the following moves are immediately forced:
R8C1 must be <3>.
* R4C8 is the only square in column 8 that can be <6>.
* Intersection of row 1 with block 2. The value <7> only appears in one or more of squares R1C4, R1C5 and R1C6 of row 1. These squares are the ones that intersect with block 2. Thus, the other (nonintersecting) squares of block 2 cannot contain this value.
R2C5  removing <7> from <2479> leaving <249>.
R2C6  removing <7> from <167> leaving <16>.
* Intersection of row 3 with block 3. The value <9> only appears in one or more of squares R3C7, R3C8 and R3C9 of row 3. These squares are the ones that intersect with block 3. Thus, the other (nonintersecting) squares of block 3 cannot contain this value.
R1C8  removing <9> from <1239> leaving <123>.
R2C7  removing <9> from <13479> leaving <1347>.
R2C8  removing <9> from <12349> leaving <1234>.
* Intersection of column 5 with block 5. The value <8> only appears in one or more of squares R4C5, R5C5 and R6C5 of column 5. These squares are the ones that intersect with block 5. Thus, the other (nonintersecting) squares of block 5 cannot contain this value.
R4C6  removing <8> from <378> leaving <37>.
R6C4  removing <8> from <2378> leaving <237>.
* Intersection of column 8 with block 3. The values <24> only appears in one or more of squares R1C8, R2C8 and R3C8 of column 8. These squares are the ones that intersect with block 3. Thus, the other (nonintersecting) squares of block 3 cannot contain these values.
R2C7  removing <4> from <1347> leaving <137>.
R3C7  removing <4> from <149> leaving <19>.
* Intersection of block 9 with column 7. The values <346> only appears in one or more of squares R7C7, R8C7 and R9C7 of block 9. These squares are the ones that intersect with column 7. Thus, the other (nonintersecting) squares of column 7 cannot contain these values.
R2C7  removing <3> from <137> leaving <17>.
R4C7  removing <3> from <3789> leaving <789>.
R6C7  removing <3> from <13789> leaving <1789>.
* A set of 3 squares form a comprehensive hidden triplet. R1C3, R1C6 and R1C8 each contain one or more of the possibilities <136>. No other squares in row 1 have those possibilities. Since the 3 squares are the only possible locations for 3 possible values, any additional possibilities these squares have (if any) can be eliminated. These squares now become a comprehensive naked triplet.
R1C3  removing <2> from <1236> leaving <136>.
R1C6  removing <7> from <167> leaving <16>.
R1C8  removing <2> from <123> leaving <13>.
* R4C6 is the only square in column 6 that can be <7>.
* R7C6 is the only square in column 6 that can be <3>.
From this deduction, the following moves are immediately forced:
R7C7 must be <4>.
R8C7 must be <6>.
R9C7 must be <3>.
* R7C2 is the only square in row 7 that can be <8>.
From this deduction, the following moves are immediately forced:
R8C2 must be <7>.
R8C3 must be <4>.
R9C2 must be <6>.
R8C4 must be <8>.
* R3C6 is the only square in row 3 that can be <8>.
* R5C9 is the only square in row 5 that can be <7>.
From this deduction, the following moves are immediately forced:
R2C9 must be <3>.
R1C8 must be <1>.
R1C6 must be <6>.
R5C8 must be <3>.
R2C7 must be <7>.
R3C7 must be <9>.
R4C7 must be <8>.
R4C5 must be <2>.
R6C7 must be <1>.
R5C2 must be <1>.
R6C8 must be <9>.
R1C3 must be <3>.
R2C6 must be <1>.
R6C5 must be <8>.
R6C4 must be <3>.
R6C2 must be <2>.
R4C3 must be <5>.
R4C1 must be <9>.
R7C3 must be <2>.
R6C3 must be <7>.
R2C2 must be <9>.
R7C1 must be <5>.
R2C3 must be <6>.
R3C3 must be <1>.
R2C5 must be <4>.
R4C2 must be <3>.
R1C1 must be <2>.
R2C8 must be <2>.
R9C5 must be <7>.
R3C4 must be <2>.
R3C8 must be <4>.
R1C4 must be <7>.
R9C4 must be <4>.
R1C5 must be <9>.
Best wishes,
Keith 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5660 Location: Rochester, NY, USA

Posted: Mon Feb 20, 2006 5:44 pm Post subject: 


Thanks David and Keith. Deep down I really knew that the odds of this being an invalid puzzle were extremely low, but as I mentioned, having to erase and restart six times is pretty unusual for me. I guess I'll go for seven!! 

Back to top 




You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot vote in polls in this forum

Powered by phpBB © 2001, 2005 phpBB Group
