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New Solution Rule for Illogical at sudokusolver.co.uk

 
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renne

Joined: 28 Dec 2005
Posts: 2
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PostPosted: Sun Jan 01, 2006 8:34 pm    Post subject: New Solution Rule for Illogical at sudokusolver.co.uk Reply with quote

Logical Solver Stops here:
Code:

  1         2         3         4         5         6         7         8         9         
---+--------------------------------------------------------------------------------------------
A  |  13        2         6         349       35        7         149       59        8         
B  |  8         9         5         6         2         14        14        7         3         
C  |  13        7         4         39        8         15        1269      259       169       
D  |  4         5         7         1         9         3         8         6         2         
E  |  9         8         3         2         4         6         5         1         7         
F  |  6         1         2         5         7         8         39        39        4         
G  |  2         36        9         34        1         45        7         8         56       
H  |  5         4         8         7         36        9         1236      23        16       
I  |  7         36        1         8         356       2         369       4         569       

Principals of the new Rule

Split the Sudoku into 9 3x3 Blocks an mark rows form A to C, columns for 1 to 3 ->
Code:

   |  1         2         3   
---+------------------------
A  |  13        2         6
B  |  8         9         5
C  |  13        7         4

   |  1         2         3   
---+------------------------
A  |  349       35        7
B  |  6         2         14
C  |  39        8         15

   |  1         2         3   
---+------------------------
A  |  149       59        8         
B  |  14        7         3         
C  |  1269      259       169       

   |  1         2         3   
---+------------------------
A  |  4         5         7
B  |  9         8         3
C  |  6         1         2 

   |  1         2         3   
---+------------------------
A  |  1         9         3
B  |  2         4         6
C  |  5         7         8

   |  1         2         3   
---+------------------------
A  |  8         6         2
B  |  5         1         7
C  |  39        39        4

   |  1         2         3   
---+------------------------
A  |  2         36        9
B  |  5         4         8
C  |  7         36        1

   |  1         2         3   
---+------------------------
A  |  34        1         45
B  |  7         36        9
C  |  8         356       2

   |  1         2         3   
---+------------------------
A  |  7         8         56
B  |  1236      23        16
C  |  369       4         569

Now over all 9 blocks every number has to 3 times in every row and 3 times in every column !!!




Now I count how often 4 is in the position A1 over all 9 blocks -> 1 to 3 times .. so i don't know exaktly
now for A2 -> zero times for sure
A3 -> 0 – 1 times
B1 -> 0 – 1 times
B2 -> 2 times
B3 -> 0 – 1 times
C1 -> 0
C2 -> 1
C3 -> 2

write the the table

Code:

   |  1         2         3   
---+------------------------
A  |  1 - 3     0         0 - 1
B  |  0 - 1     2         0 - 1
C  |  0         1         1         

now i look a the main sudoku an can see that i have to recalibrate A1 because 4 can either be in A4 or A7 so ->

for A1 1 to 2 times ->
Code:

   |  1         2         3   
---+------------------------
A  |  1 - 2     0         0 - 1
B  |  0 - 1     2         0 - 1
C  |  0         1         1         

now every sum of each row and column has to be 3

the only solution left is:
Code:

   |  1         2         3   
---+------------------------
A  |  2         0         1
B  |  0         2         1
C  |  0         1         1         


now i have to map it back to the sudoku

for example there has to be one "4" on B3 – block-position -> that is G6 in the sudoku

and so on

now it is solveable with normal logic
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Ruud
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Joined: 17 Sep 2005
Posts: 708
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Location: Netherlands

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PostPosted: Sun Jan 01, 2006 9:02 pm    Post subject: Reply with quote

Hi renne,

A tip: To reach the intended audience, you should post topics like this in the Solving Sudoku forum.


I checked the candidate grid you posted.

Not all logical solvers would not stop at that point. My solver encounters an XY-Wing in the following cells:

x=3, y=5, z=1
xy=R1C5, xz=R1C1, yz=R3C6. Candidate 1 excluded from R3C1, leaving 3 as the remaining candidate.

Your technique looks interesting. Have you written a program for this?

Ruud.
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Nick70

Joined: 08 Jun 2005
Posts: 160
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PostPosted: Mon Jan 02, 2006 8:55 am    Post subject: Re: New Solution Rule for Illogical at sudokusolver.co.uk Reply with quote

Very interesting observation, requiring further thought.

One nitpick.
renne wrote:
Code:
   |  1         2         3   
---+------------------------
A  |  1 - 2     0         0 - 1
B  |  0 - 1     2         0 - 1
C  |  0         1         1         

You said that C3=2, so the table would actually be
Code:
   |  1         2         3   
---+------------------------
A  |  1 - 2     0         0 - 1
B  |  0 - 1     2         0 - 1
C  |  0         1         2         

giving
Code:
   |  1         2         3   
---+------------------------
A  |  2         0         1   
B  |  1         2         0   
C  |  0         1         2         
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Nick70

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PostPosted: Mon Jan 02, 2006 9:35 am    Post subject: Re: New Solution Rule for Illogical at sudokusolver.co.uk Reply with quote

renne wrote:
Now I count how often 4 is in the position A1 over all 9 blocks -> 1 to 3 times .. so i don't know exactly

Unfortunately this is not correct. 4 can be in A1 1 to 4 times, not 3.

I'm afraid this fatally invalidates the reasoning.
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renne

Joined: 28 Dec 2005
Posts: 2
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PostPosted: Mon Jan 02, 2006 9:55 pm    Post subject: Reply with quote

You are right

The reasoning is incorrect
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