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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Thu Aug 25, 2005 2:47 pm Post subject: Sudoku that has to be solved with advanced technique XYWing 


Hi,
Anyone interested in a couple of 19 numbers positions that can be solved only with the advanced technique XYWing? (or mybe some other even higher [guessing] technique)
Just let me know.
see u, 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Thu Aug 25, 2005 2:49 pm Post subject: 


Sorry, let's make it harder 17 cells Sudoku!
waiting for volunteers, 

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Katie
Joined: 05 Aug 2005 Posts: 13 Location: London

Posted: Thu Aug 25, 2005 3:46 pm Post subject: 


I'm always ready to try. Send me the details via the post message system, and I'll have a go (gulp) 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Fri Aug 26, 2005 7:43 am Post subject: 


This is just for Katie.
Nobody is allowed to solve it ... ;)
   1    7 
 4   9    
 2       
7   8 5    
    2  4  
3        
      2  1
   3   6  
5   7     
after 34 numbers it gets interesting.
see u, 

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Katie
Joined: 05 Aug 2005 Posts: 13 Location: London

Posted: Fri Aug 26, 2005 9:47 am Post subject: 


Hope this is right.
8 6 5 1 3 4 9 7 2
1 4 7 2 9 5 3 6 8
9 2 3 6 7 8 5 1 4
7 9 4 8 5 3 1 2 6
6 5 1 9 2 7 4 8 3
3 8 2 4 6 1 7 9 5
4 7 9 5 8 6 2 3 1
2 1 8 3 4 9 6 5 7
5 3 6 7 1 2 8 4 9
Did take me about 90 minutes though.
I'd enjoy more of these. Thanks 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Fri Aug 26, 2005 4:32 pm Post subject: 


Hi Katie,
Yup, the solution is Ok.
After the "forplay" you can find with the usual techniques, up to 34 numbers, getting to the position:
   1 3 4  7 2
 4 7 2 9 5 3  
 2 3 6 7 8 5  
7  4 8 5 3 1 2 
   9 2 7 4  3
3  2 4   7  5
4 7  5 8  2 3 1
2   3 4  6 5 7
5 3  7  2   
which is the Gpoint of the game.
Here you can continue with a XYwing combination starting from r2c1
and getting to eliminate 6 from r5c8.
If someone can to it better and quicker and nice, please don't keep it for yourself.
One more time, a big BRAVO for Katie. The lady is a Pro!
(Sorry, I had to correct the prev. message)
see u, 

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alanr555
Joined: 01 Aug 2005 Posts: 198 Location: Bideford Devon EX39

Posted: Tue Sep 06, 2005 12:24 am Post subject: 


Given that Katie has now solved this one, is it possible to explain?
I have worked through to the [30 to go] stage  leaving the following
solution grid (omitting resolved cells)
689; 568; 58;   ; 89 ;  
18;      ; 16; 68
19;      ; 149; 49
; 69;       ; 69
68; 15; 15;    ; 68; 
; 89;  ; 16; 16; ; 89;
 ; 69;  ; 69;   
; 18; 189;  ; 19;   
 , 16;  ; 16; ; 89; 49; 489
This is a "closed" set in that no digit appears without a mate in any row,
column or block and the number of unique digits in any row, column or
block is equal to the number of cells.
I am not proficient in "swordfish" but from my understanding of it, all
the chains have been used already in reaching the above  so that
A1/E1/E8/B8/B9/D9/D2/A2/A1 or G3/G6/F6/F5/J5/J3/G3 do not yield any
new information or lead to new exclusions.
How does XYWing from r2c1 lead to elimination of 6 from the 68 in r5c8?
I presume that "8" is the linking digit and notice that several rows and
columns contain more than two occurrences of digit 8.
r2c1, r2c8, r2c9 form a triplet set {18,16,68} and interesects with the
column 8 quintet set {16,149,68,89,49} with {16} as the link point.
The 6 in r5c8 can be eliminated if r2c8 (the link point) resolves to 1 from
the possibility set at 16. I do not understand, at present, why one would
be able to reach that conclusion. Enlightenment appreciated!!
Alan Rayner BS23 2QT 

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Guest

Posted: Thu Sep 15, 2005 7:58 am Post subject: 


hi alanr555
As someone_somewhere said using the Xwings to solve the problem:
r2c1 can be 1 or 8
if r2c1=1 > r2c8=6> r5c8 cannot be 6
if r2c1=8 > r5c1=6> r5c8 cannot be 6
so you can eliminated 6 in r5c8.
Hope this can help. 

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alanr555
Joined: 01 Aug 2005 Posts: 198 Location: Bideford Devon EX39

Posted: Thu Sep 15, 2005 11:20 pm Post subject: 


> As someone_somewhere said using the Xwings to solve the problem:
> r2c1 can be 1 or 8
> if r2c1=1 > r2c8=6> r5c8 cannot be 6
> if r2c1=8 > r5c1=6> r5c8 cannot be 6
> so you can eliminated 6 in r5c8.
Since writing in this forum, I have discovered how it works.
1) Identify a rectangle in the solution grid where three corners
involve only three digits and only two in each corner.
(here r2c1=18, r5c1=68, r2c8=68 involves only 1,6,8)
2) Consider the fourth corner. If it contains a digit that is NOT in the
opposite corner but IS one of the three digits above, then this
digit can be excluded.
(here the fourth corner is r5c8. It does not matter how many digits
it holds  whether they be 1,6,8 or others!
Digit 6 is part of the trio {1,6,8} but is NOT in the opposite corner
 which is r2c1 in this case. Thus 6 can be excluded from r5c8.
+++
Of course, the difficult part is identifying the rectangle!
I find that rewriting the solution grid with only the UNresolved cells
can assist in presenting a new view to the brain and allow the pattern
to be spotted that much more easily (I use a hyphen to represent the
space of a resolved cell). Of course this is useful only when one has
reduced the solution grid to a relatively small number of cells.
+++
I found that using the nomenclature of XYWing was confusing and
somewhat offputting. We need a new name. One suggestion might
be "rectangular trio".
Thanks for the assistance in clarifying the generality of it.
Alan Rayner BS23 2QT 

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keith
Joined: 19 Sep 2005 Posts: 3321 Location: near Detroit, Michigan, USA

Posted: Mon Sep 19, 2005 10:13 pm Post subject: 


Whoops!
I just solved this one without having to resort to swordfish or xwings  I just used my regular technique. I'll have to redo the puzzle and check my logic  maybe I made a logical error but got the correct answer.
What should I do? My current plan is to post every step and ask for challenges on the logic. It might take me a few days to document my solution.
Best wishes,
Keith 

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