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nj3h Gold Member
Joined: 10 Jul 2006 Posts: 111 Location: Virginia / USA

Posted: Mon Apr 16, 2007 1:29 am Post subject: Jigsaw 32 


The subject puzzle, after solving as far as I could caused me to hit a stumbling block.
I got a hint of Law of Leftovers eliminations in Row 12.
Uing this puzzle as an example can someone help me work through this technique? I have read the links listed in the other messages, but I do not understand how to apply in this situation.
Thanks,
George
..........
 23578  358  4  1  2356789  236  25789  235678  235678 
:++++++++:
 235789  358  258  25689  25689  4  1  23578  23578 
:++++++++:
 129  6  3  58  58  7  29  129  4 
:++++++++:
 2578  4  1  2678  2678  9  3  2567  2567 
:++++++++:
 123589  7  258  23468  123468  236  2589  123569  23568 
:++++++++:
 6  9  278  2378  12378  5  4  1237  2378 
:++++++++:
 345  2  9  3457  3567  8  57  34567  1 
:++++++++:
 3458  358  27  234578  234578  1  6  234578  9 
:++++++++:
 3458  1  6  279  279  23  2578  234578  23578 
'''''''''' 

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Glyn Major Major Major
Joined: 16 Jan 2007 Posts: 92 Location: London

Posted: Mon Apr 16, 2007 10:38 am Post subject: 


Hi George.
Rows 1 and 2 must contain a 2 complete sets of the digits 1 to 9. That means they must contain two digit 7s.
There are three jigsaw pieces (Nonets) which have cells in rows number 1 and 2. Only the left and right hand Nonets (N1 and N3) can provide the two digit 7s we require, as the middle Nonet (N2) already contains a 7 in R3C6.
In N1 and N3 we must take a 7 from either row 1 or row 2, which excludes R4C1 from being a 7.
Hope this helps.
Glyn _________________ I have 81 brain cells left, I think. 

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Para Yokozuna
Joined: 08 Nov 2006 Posts: 384 Location: The Netherlands

Posted: Mon Apr 16, 2007 10:50 am Post subject: Re: Jigsaw 32 


nj3h wrote:  The subject puzzle, after solving as far as I could caused me to hit a stumbling block.
I got a hint of Law of Leftovers eliminations in Row 12.
Uing this puzzle as an example can someone help me work through this technique? I have read the links listed in the other messages, but I do not understand how to apply in this situation.
Thanks,
George
Code:  ..........
 23578  358  4  1  2356789  236  25789  235678  235678 
:++++++++:
 235789  358  258  25689  25689  4  1  23578  23578 
:++++++++:
 129  6  3  58  58  7  29  129  4 
:++++++++:
 2578  4  1  2678  2678  9  3  2567  2567 
:++++++++:
 123589  7  258  23468  123468  236  2589  123569  23568 
:++++++++:
 6  9  278  2378  12378  5  4  1237  2378 
:++++++++:
 345  2  9  3457  3567  8  57  34567  1 
:++++++++:
 3458  358  27  234578  234578  1  6  234578  9 
:++++++++:
 3458  1  6  279  279  23  2578  234578  23578 
'''''''''' 

Hi
This is just the explanation for this particular move. I haven't checked if this solves the puzzle.
Code: 
Nonet numbers
111233333
112222333
114442263
144555266
148555269
448555669
748866699
777888899
777778999 
We draw a line Between Row 2 and 3. There then are 5 cells under the line from nonets 1 and 3 and 5 cells above the line from nonet 2.
R345C1 + R3C29 = R2C3456 + R1C4
R2C3456 + R1C4(the cells above the line) contain the digits 1,2,4,5,6,8 and 9. Or to say it otherwise, they don't contain the digits 3 or 7. So there can't be a 3 or a 7 in the cells under the line so R4C1 can't contain a 7 and R5C1 can't contain a 3.
greetings
Para 

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nj3h Gold Member
Joined: 10 Jul 2006 Posts: 111 Location: Virginia / USA

Posted: Tue Apr 17, 2007 1:26 am Post subject: 


Para and Glyn,
Thanks so much for taking the time to help me out with this puzzle.
I have to admit that I still am lost for the next hint from sumoCue. I used the same method to check Rows 89 and Column 12 and 89 in a similat fashion to the explation of Rows 12 above. I saw nothing that eliminated any other candidates.
The way I read the hint comment is that the elimination are in Rows 12. I guess the comment means that using Rows 12 there are candidate eliminations elsewhere. There were no eliminations in Rows 12.
Now I get a message that says "Law of Leftovers eliminations in row 89". As I indicated above, I can find no eliminations using Rows 89. I drew a line between Rows 7 and 8. How does one know where to draw such a line.
Below is the current state of the grid that yielded the LoL elim in Rows 89 message.
Please help me to grasp this technique.
Again, thanks for the help.
Regards,
George
..........
 23578  358  4  1  2356789  236  25789  235678  235678 
:++++++++:
 235789  358  258  25689  25689  4  1  23578  23578 
:++++++++:
 129  6  3  58  58  7  29  129  4 
:++++++++:
 258  4  1  2678  2678  9  3  2567  2567 
:++++++++:
 12589  7  258  23468  123468  236  2589  123569  23568 
:++++++++:
 6  9  278  2378  12378  5  4  1237  2378 
:++++++++:
 345  2  9  3457  3567  8  57  34567  1 
:++++++++:
 3458  358  27  234578  234578  1  6  234578  9 
:++++++++:
 3458  1  6  279  279  23  2578  234578  23578 
'''''''''' 

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sudokuEd Grandmaster
Joined: 19 Jun 2006 Posts: 257 Location: Sydney Australia

Posted: Tue Apr 17, 2007 8:27 am Post subject: 


nj3h wrote:  Now I get a message that says "Law of Leftovers eliminations in row 89". ...I drew a line between Rows 7 and 8. How does one know where to draw such a line.  Thanks for bringing up LoL George. I'm new to LoL and find SumoCue's hint confusing but think I can make sense of it.
Drawing the line after row 7 there are at least 3 different ways to divide up the 3 nonets into above and below the line so that the SAME number of INCOMPLETE nonet cells are above and below (the most important bit for where to draw the line it seems).
I can see two 5cell sets and one 8cell set of leftover cells for r89. All of them could potentially give LoL eliminations since they must all have the same digits.
1)8 cells above the line are r567c39 + r7c48. 8 cells below the line are r8c123+r9c12345
2)5 cells above the line are r567c3 + r7c14. 5 cells below the line are r8c89 + r9c789
3)5 cells above the line are r7c189 + r56c9. 5 cells below the line are r8c4567 + r9c6
Unfortunately, I can't see that any of them give LoL eliminations the way that Glyn and Para did.
How to make sense of the hint? Any of these leftover sets requires a 6 below the line > there must be a 6 'above the line'. This is only available in r7c8 or r5c9 > no 6 in r5c8 since it can 'see' both those cells. So the 'LoL elimination in r89' is 6 from r5c8.
Of course, a much easier way (for us!) to get this same elimination is to notice that these are the only 2 6's in nonet 9 > no 6 in r5c8.
Hope this makes sense of SumoCue.
Good Luck!
Ed
Code:  ..........
 23578  358  4  1  2356789  236  25789  235678  235678 
:++++++++:
 235789  358  258  25689  25689  4  1  23578  23578 
:++++++++:
 129  6  3  58  58  7  29  129  4 
:++++++++:
 258  4  1  2678  2678  9  3  2567  2567 
:++++++++:
 12589  7  258  23468  123468  236  2589  123569  23568 
:++++++++:
 6  9  278  2378  12378  5  4  1237  2378 
:++++++++:
 345  2  9  3457  3567  8  57  34567  1 
:++++++++:
 3458  358  27  234578  234578  1  6  234578  9 
:++++++++:
 3458  1  6  279  279  23  2578  234578  23578 
'''''''''' 


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Para Yokozuna
Joined: 08 Nov 2006 Posts: 384 Location: The Netherlands

Posted: Tue Apr 17, 2007 8:58 am Post subject: 


Hi
What Ed just explained is the second way to use LoL. This time the 6's are conveniently both in the same nonet but they don't have to be.
The first eliminations you can make with LoL. That is by simply comparing which digits are in the cells above and under the line.
The second type of eliminations are eliminations made by digits that are already placed above or under the line. If one digit is already placed in a cell above the line, we know it must also be in the cells under the line or vice versa. Then we determine which cells this digit can be placed in. Then we check if there are any cells that can see all the cells(above or under the line) and we can eliminate this digit from these cells(that see all cells which can contain the digit in question).
There's some nice examples of this second way in this thread:
http://www.sudocue.net/forum/viewtopic.php?t=653
greetings
Para 

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Para Yokozuna
Joined: 08 Nov 2006 Posts: 384 Location: The Netherlands

Posted: Tue Apr 17, 2007 9:15 am Post subject: 


Hi
I don't know if there's an easy way to explain where to check. I usually suggest to check every set of rows and columns. Just like when looking for singles or naked subsets.
For this puzzle i would first suggest Law of Leftovers on Column 789. This gives the best eliminations. Which gives you 4 cells left and 4 cells right of the line.
The problem with the hints from SumoCue is that it seems to be checking first all rows and then all columns and it shows all possible eliminations, even the more difficult ones. Usually there are easier eliminatinions somewhere else.
greetings
Para 

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Glyn Major Major Major
Joined: 16 Jan 2007 Posts: 92 Location: London

Posted: Tue Apr 17, 2007 4:00 pm Post subject: 


The elimination here seems to be based on a 6 being required in the 5 cells of Nonets 8 and 9 that lie above the line above row 8 which mirror the 5 cells of Nonet 7 which lie below that line.
Code: 
The Nonets are defined thus:
111233333
112222333
114442253
144666255
147666258
447666558
947755588
999777788
999997888 
The two candidate positions for the 6 are in cells R5C9 and R7C8. These form a pointing pair on 6 with R5C8 being the shared peer of both so R5C8<>6.
We could have got the elimination plus one more if we had used the strong link between digit 6 at R7C5 and R7C8, here there are two shared peers R45C8<>6. If you do this there is another pointing pair on 6 to look for straight away.
Have fun. _________________ I have 81 brain cells left, I think. 

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Para Yokozuna
Joined: 08 Nov 2006 Posts: 384 Location: The Netherlands

Posted: Tue Apr 17, 2007 4:29 pm Post subject: 


Glyn wrote:  The elimination here seems to be based on a 6 being required in the 5 cells of Nonets 8 and 9 that lie above the line above row 8 which mirror the 5 cells of Nonet 7 which lie below that line.
The two candidate positions for the 6 are in cells R5C9 and R7C8. These form a pointing pair on 6 with R5C8 being the shared peer of both so R5C8<>6.
We could have got the elimination plus one more if we had used the strong link between digit 6 at R7C5 and R7C8, here there are two shared peers R45C8<>6. If you do this there is another pointing pair on 6 to look for straight away.
Have fun. 
Sumocue calls a lot eliminations LoL eliminations because they can be made using Law of Leftovers as well (as Ed showed).
Quote:  Code: 
The Nonets are defined thus:
111233333
112222333
114442253
144666255
147666258
447666558
947755588
999777788
999997888 

Is there a general way to number nonets? I generally like to number nonets to their relative position to eachother, mostly because that way the numbering looks like the numbering of nonets in a regular sudoku.
That maybe is just a personal thing.
greetings
Para 

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nj3h Gold Member
Joined: 10 Jul 2006 Posts: 111 Location: Virginia / USA

Posted: Tue Apr 17, 2007 6:36 pm Post subject: 


Again thanks to all the folks that are helping myself and others to understand the LoL concept.
SudokuEd: I can see the 3 sets above and below the line (8 cell, 5 cell, and another 5 cell). What I am having trouble visualizing is why the above and below the line cells are complimentary to each other. I have to be dense, but I haven't hit the aha moment yet.
Further, can you please provide more explanation for the discussion in the "How to make..." paragraph. For some reason I can not see what is going on there. I just do not understand the logic in this paragraph. I think this must tie back to the fact that I am not grasping the above and below the line concept.
The next paragraph make sense, though, since the elimination in r5c8 of the 6 can be made without LoL techniques. I missed this logical exclusion going through the puzzle.
Thanks for your comments and future explanations, Ed.
Para: Thanks again for your help. I will check out the link that you provided. I suspect if I could understand the above and below the line theory, I could understand the concept. For some reason, this technique just isn't clicking yet.
Regards to all,
George 

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Para Yokozuna
Joined: 08 Nov 2006 Posts: 384 Location: The Netherlands

Posted: Tue Apr 17, 2007 9:31 pm Post subject: 


nj3h wrote:  What I am having trouble visualizing is why the above and below the line cells are complimentary to each other. I have to be dense, but I haven't hit the aha moment yet. 
Hi
Ok, this is basically based on the elementary rules of sudoku. I'll use these pictures to help explain.
Basic rules of Sudoku:
 Every Row, Column and Nonet has to contain the numbers 19 exactly once.
In this example we have 3 nonets almost completely in Row 1, 2 and 3. Nonets coloured yellow. In each of these nonets we have to place numbers 19 exactly once.
When we fill these nonets with 19, we have filled Row 1, 2 and 3 almost completely with 19. There are 2 cells empty in row 3. The missing digits are 1 and 3, which are also the digits under the line.
We can explain this by the fact that the 3 nonets contain 19 once each, so in total three times.
Also row 1, 2, and 3 contain the digits 19 once each, again in total three times.
So row 1, 2 and 3 contain exactly the same digits as 3 nonets. So all the digits that are in the 3 nonets but aren't in row 1, 2 and 3 are missing from row 1, 2 and 3.
In this case the a digit 1 and 3 in R4C19 are in the 3 nonets but not in row 1, 2 or 3. But they have to be in row 1, 2 and 3. So they go in the cells that are in row 1, 2 and 3 but aren't in the 3 nonets. In this case R3C28.
I hope this is clear. I always feel my explanations lack some clarity.
greetings
Para 

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Glyn Major Major Major
Joined: 16 Jan 2007 Posts: 92 Location: London

Posted: Tue Apr 17, 2007 9:56 pm Post subject: 


Hi Para
I think this algorithm would label the nonets in the same way that Ruud does for this puzzle.
For each nonet find the left edge of its' top row. If you then order the corners firstly by ascending row number, then by ascending column number then everything should work out.
The corners are at R1C1,R1C4,R1C5,R3C3,R3C8,R4C4,R5C3,R5C9,R7C1 and these define nonets 1 to 9 respectively.
I guess it can be confusing because one instinctively wants to call the bottom corner nonet 9, but the symmetry here scuppers that. I'm not sure if Jean Christophe uses the same system.
By the way great notes on LOL, perhaps Ruud will make it a 'sticky'. _________________ I have 81 brain cells left, I think. 

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Para Yokozuna
Joined: 08 Nov 2006 Posts: 384 Location: The Netherlands

Posted: Tue Apr 17, 2007 10:51 pm Post subject: 


Hi
Yeah i figured that was the way you numbered your nonets. It just seem a bit unnatural to me. But it is the easiest way to program.
That is why i always tend to add a nonetnumbering so people know which nonets i am talking about compared to what programs use.
I am just going to be stubborn. (must be a human thing)
greetings
Para 

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