New method to solve Sudoku

Advanced methods and approaches for solving Sudoku puzzles

Re: New method to solve Sudoku

Postby eleven » Sat Mar 01, 2014 9:56 pm

Hi harshgoel2k,

i tried to understand, what you are doing, but i don't.

E.g. you find a pair, where i can't see one (second puzzle):
"Similarly Consider B5N1 and B5N6, the values set are 44 and 66".
Why ? There is a 5 in B6N6 and 45, 65 are free.

[Edit:] Ah, i saw now that surbier already mentioned this mistake in 2010. Would you fix it please, before you announce the same stuff again ?
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Re: New method to solve Sudoku

Postby David P Bird » Sun Mar 02, 2014 1:03 pm

Harshgoel2k, here’s my take on your approach.

The inferences available for reducing the candidates in an unsolved cell will be the same whatever way the data is presented. We therefore want data representations that make it easy to recognise when the conditions exist to allow eliminations from different classes of logical arguments.

Your methods provide ways of recognising the conditions for elementary eliminations using your representation, but at first sight don’t seem to provide any insights into any of the more complex elimination classses that are described and discussed in this forum.

As a trial, I took the third example in your web page and found that, although it was described as “near impossible”, it solved very quickly using basic recognition methods using the conventional grid representation.

This is the starting grid where the eliminations have been made in every cell in sight of each given.
Example 3: ......7...4..2.8.3...4...12..1..9.28.3.1.5.7.79.6..1..62...1...9.8.7..5...3......
Code: Select all
 *-----------------------*-----------------------*-----------------------*
 | 12358  1568   2569    | 3589   135689 368     | <7>    469    4569    |
 | 15     <4>    5679    | 579    <2>    67      | <8>    69     <3>     |
 | 358    5678   5679    | <4>    35689  3678    | 569    <1>    <2>     |
 *-----------------------*-----------------------*-----------------------*
 | 45     56     <1>     | 37     34     <9>     | 3456   <2>    <8>     |
 | 248    <3>    246     | <1>    48     <5>     | 469    <7>    469     |
 | <7>    <9>    245     | <6>    348    2348    | <1>    34     45      |
 *-----------------------*-----------------------*-----------------------*
 | <6>    <2>    457     | 3589   34589  <1>     | 349    3489   479     |
 | <9>    1      <8>     | 23     <7>    2346    | 2346   <5>    146     |
 | 145    157    <3>     | 2589   45689  2468    | 2469   4689   14679   |
 *-----------------------*-----------------------*-----------------------*

Using basic elimination methods this reduces very quickly to:
Code: Select all
 *-----------------*-----------------*-----------------*
 | 2    58   69    | 389  1    368   | <7>  469  4569  |
 | 1    <4>  679   | 5    <2>  67    | <8>  69   <3>   |
 | 3    58   679   | <4>  69   678   | 569  <1>  <2>   |
 *-----------------*-----------------*-----------------*
 | 45   6    <1>   | 7    3    <9>   | 45   <2>  <8>   |
 | 8    <3>  2     | <1>  4    <5>   | 69   <7>  69    |
 | <7>  <9>  45    | <6>  8    2     | <1>  3    45    |
 *-----------------*-----------------*-----------------*
 | <6>  <2>  45    | 389  59   <1>   | 349  489  7     |
 | <9>  1    <8>   | 23   <7>  346   | 2346 <5>  46    |
 | 45   7    <3>   | 289  569  468   | 2469 4689 1     |
 *-----------------*-----------------*-----------------*

Now it’s easy to follow digit 4 to find that it must be true either in r6c9 or r8c6 so that it can be eliminated from r8c9
This also forces (9)r5c9 and reduces to:
Code: Select all
 *--------------*--------------*--------------*
 | 2   58  69   | 389 1   368  | <7> 469 45   |
 | 1   <4> 679  | 5   <2> 67   | <8> 69  <3>  |
 | 3   58  679  | <4> 69  678  | 59  <1> <2>  |
 *--------------*--------------*--------------*
 | 45  6   <1>  | 7   3   <9>  | 45  <2> <8>  |
 | 8   <3> 2    | <1> 4   <5>  | 6   <7> 9    |
 | <7> <9> 45   | <6> 8   2    | <1> 3   45   |
 *--------------*--------------*--------------*
 | <6> <2> 45   | 389 59  <1>  | 349 489 7    |
 | <9> 1   <8>  | 23  <7> 34   | 234 <5> 6    |
 | 45  7   <3>  | 289 569 468  | 249 489 1    |
 *--------------*--------------*--------------*

This time we follow digit 5 and find that for cells r3c7 and r7c5 one must hold a 5 and the other must hold a 9. Hence (9)r3c5 can be eliminated.
The forced assignments that follow now solve the puzzle.

I therefore think you shouldn’t trust that puzzle site to provide difficult examples and should favour sources that I believe have already been pointed out to you.

DPB
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Re: New method to solve Sudoku

Postby eleven » Sun Mar 02, 2014 2:44 pm

It's not, that i think, that other sudoku presentations never would have advantages over the classical one.
You might remember Denis' "Extended sudoku board", where (row/column) fish could be easier spotted (see here or here).

But this paper only showed me mistakes so far, no goodies.
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Re: New method to solve Sudoku

Postby David P Bird » Sun Mar 02, 2014 5:31 pm

eleven wrote:It's not, that i think, that other sudoku presentations never would have advantages over the classical one.
You might remember Denis' "Extended sudoku board", where (row/column) fish could be easier spotted (see here or here).

Yes I accept that point and it's all down to finned fish and almost locked sets being the mathematical duals of each other in different data 'spaces'.

Where the jury is out is in deciding if Harshgoel2k’s presentation actually simplifies identifying any of our existing elimination patterns or points a way to any new ones.

In the past I spent a lot of time representing the options for the digits that occur together in the mini-line diagonals in the different box bands (Braid Analysis) which is yet another way to present the data. It turned out that only in very rare circumstances did this highlight any deductions that weren‘t already easily available from a conventional pencil-marked grid. (That said, this knowledge can be useful when solving magazine puzzles without pencil marks.)

DPB
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Re: New method to solve Sudoku

Postby harshgoel2k » Mon Mar 03, 2014 1:33 pm

Learned members, Thanks for your kind feedback.

Kindly help me to provide the hardest puzzle which is not solved by this forum and ........ have a unique solution. I will try to solve it using my method,

My point here is not to prove that my method is best, but having a different model of same process helps to rediscover which sometime not possible using the same method.

I agree , I need to reformat the document and make it more readable...
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Re: New method to solve Sudoku

Postby JasonLion » Mon Mar 03, 2014 8:34 pm

There are puzzles that haven't been solved yet. However, there aren't any puzzles that have remained unsolved despite a concerted effort to solve them.

What ever happened with solving Fata Morgana, which StrmCkr suggested several posts ago?

One of the fascinating things about Sudoku is just how wide a range the difficulty varies over. The puzzles available in most books/web sites are in (roughly) the easiest tenth of the difficulty range, despite having names that sound quite impressive (like near impossible, diabolical, etc).
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Re: New method to solve Sudoku

Postby eleven » Mon Mar 03, 2014 9:20 pm

harshgoel2k wrote:Kindly help me to provide the hardest puzzle which is not solved by this forum and ........ have a unique solution.

First of all, there is no hardest sudoku. There is a list by champagne of meanwhile more than 1 mio puzzles here in ph_13_10.zip, sorted after the Explainer rating, which is not up-todate, but still the most accepted public rating.
I am very sure, that you cannot solve one of them (without lucky mistakes), so i recommend you to start with number 1096206:
Code: Select all
........1.....234...3.1.52...2..4..5.6.......7...8..6...5..1..4.8..4.9..9..87....                     
(10.30;10.30;10.30;dob;12_12_03;464056;24;*)
When you have found the first number, please let us know.

And don't forget to ensure, that you can solve the puzzle in your paper without the pair's mistake.
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