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alanr555
Joined: 01 Aug 2005 Posts: 193 Location: Bideford Devon EX39

Posted: Fri Jan 06, 2006 1:19 am Post subject: McCarthy Stone competition 


Code: 
The following puzzle was set with twentyeight initial clues
and I have been able to progress it as far as finding some
twentyfour additional values as shewn.
Using trial and error I have determined that the remaining
twentynine values are linked to a unique solution but I do
not know how to demonstrate this by logic.
There is a hundred pound prize and I can submit my entry
with the values derived by trial and error but I would prefer
to know the logic that confirms this result.
400 580 100
108 306 705
560 900 200
709 405 602
204 600 950
651 298 374
305 709 800
876 150 409
902 860 507
The twentynine missing values are
34123612417388138417394236279
but I have not laid out them as a grid
to avoid portraying the solution.
++
The original puzzle was set as
400 080 100
100 006 705
000 900 200
009 405 602
004 600 000
601 208 300
005 009 000
806 100 009
002 060 007
++
The candidates for unresolved cells are
 29 37;   27;  369 36
 29  ;  24  ;  49 
  37;  17 147;  348 38
 38  ;  13  ;  18 
 38  ;  137 137;   18
  ;   ;   
 14  ;  24  ;  126 16
   ;   23;  23 
 14 ;   34;  18 
This gives quite a few "pairs" (three in column two!) and only
digits 1,3,7 appear more than twice in a row, column or region.
There is probably an 'implication' chain somewhere  but where
should one be looking to locate it? I am not interested very much
in the exact location but my interest is really in WHY one should be
looking for it in that very location (where, with hindsight, it exists).
Alan Rayner BS23 2QT



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

Posted: Fri Jan 06, 2006 9:47 am Post subject: 


Hi Alan,
Nice Sudoku puzle. You did it till the point where classic techniques can solve it.
From this point on, I apply the "double implication chain" technique, starting from the cells that contain a pair.
In our concrete case, even if there are a lot of pairs, I get a contradiction starting from the first one. So:
1. path: r1c2 = 2, r1c2 <> 9, r1c8 = 9, r1c8 <> 6, r7c8 = 6
2. path: r1c2 = 2, r1c6 <> 2, r8c6 = 2, c7c5 <> 2, r7c8 = 2
and in r7c8 can't hold both digits 6 and 2. This means that digit 2 can be excuded from r1c2.
The rest can be done with standard techniques.
Hope I could help.
see u, Alan, 

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keith
Joined: 19 Sep 2005 Posts: 3150 Location: near Detroit, Michigan, USA

Posted: Sun Jan 15, 2006 3:20 pm Post subject: BUG me again! 


Let's take another look at this one. Using standard techniques, it can be solved to this stage:
Code: 
4 29 37 5 8 27 1 69 36
1 29 8 3 24 6 7 49 5
5 6 37 9 17 147 2 48 38
7 38 9 4 13 5 6 18 2
2 38 4 6 137 17 9 5 18
6 5 1 2 9 8 3 7 4
3 14 5 7 24 9 8 126 16
8 7 6 1 5 23 4 23 9
9 14 2 8 6 34 5 13 7

Now, there is a forcing chain that involves the cells:
R5C6, R1C6, R1C2, R1C8, R1C9, R3C9, R5C9, and back to R5C6. If R5C6 = 7, the chain is in the order given. If R5C6 = 1, the chain is traced out in the reverse order. The starting cell does not matter.
Examining the "corners" of the chain, we see that:
One of R5C6 and R1C6 is 7, so R3C6 cannot contain 7.
One of R5C9 and R5C6 is 1, so R5C5 cannot contain 1.
Now, we have a BUG! Every square, except one, has only two possibilities. R7C8 can be <126>.
Take a look at the row, column, and block to which this square belongs. If the possibilities are 2 or 6, then each of these possibilities occurs twice (in the row, column or block), which is not allowed. So, R7C8 must be 1, and the rest is trivial.
Keith 

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