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		zaks
 
 
  Joined: 25 Nov 2005 Posts: 13
 
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				 Posted: Tue Dec 13, 2005 7:14 pm    Post subject: 13 Dec 2005 very hard sudoku: solution | 
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				13 Dec (vh) sdk
 
 soln:
 
 
i use notations:
 
 
boxes (b1...b9):
 
 
box1 box 2 box3
 
 box4 box 5 box6
 
 box7 box 8 box9
 
 
columns (ca...ci) :
 
 
 c c c  c c c  c c c
 
 o o o o o o  o o o
 
 L L L  L L L L L L
 
 a b c  d e f  g h  i
 
 
rows (r1...r9):
 
 
row 9
 
row 8
 
row 7
 
row 6
 
row 5
 
row 4
 
row 3
 
row 2
 
row 1
 
 
moves: 
 
number-column-row (exactly as in chess)
 
e.g., 4b6 means "put number 4 in cell 6", etc
 
 
here's my solution (sorry without moves' numbers)
 
 
 1b8   1i6  1g1  1a3  1f2  ( finished with 1's )
 
 4c5  4h7  6i3  
 
 7i9  ( see row 9, col i, and box 2 =  see r9+ci+b2 )
 
 8g5 ( see r5 )
 
 7g4 ( see cg ) => 9g7  6g9  6d7  
 
 3i7  (see  r7 ) => 8h8
 
 5c3 ( in box 7, 5 can't be in cells a1, c1 because 
 
           in box 8, 5 must be in row 1)
 
 8c9  ( see box 1 and col c ) => now we fill box 1:
 
 5a7 7b7 => now fill row 7:
 
 8f7  8d4  8b6   ( finished with 8's )
 
 4e4  ( box 5 )
 
 6a6  ( box 4 )
 
 6b2  ( box 7 )  6e1  ( finished with 6's )
 
 7a5  ( box 4 )
 
 2i5   ( row 5 ) 
 
 9e5  ( row 5 )
 
 9i4  ( col 9 )
 
 5h4  ( row 4 ) 
 
 2b4   ( row 4 ) 
 
 3a4   ( row 4 )
 
 2a1   ( col a )  
 
 9c6  ( box 4 )
 
 3c1   ( col 3 )
 
 9b3  ( box 7 )
 
 9h2  ( box 9 )
 
 2h3  ( box 9 )
 
 3h6  ( box 6 )
 
 
the rest moves are more/less easy:
 
 
 2e2 
 
 3d2
 
 2f9
 
 3e8
 
 4d9
 
 5e9
 
 7e6
 
 5f6
 
 4f3
 
 7e3
 
 5d1
 
 9f1
 
 7f8
 
 9d8 x!
 
 
enjoy, zaks | 
			 
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		k-bizzle Guest
 
 
 
 
 
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				 Posted: Wed Dec 14, 2005 7:17 am    Post subject:  | 
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				| for the 5c3: why must the 5 in box 7 be in r1 and NOT in r3cd or r3cf? | 
			 
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		newbie Guest
 
 
 
 
 
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				 Posted: Thu Dec 15, 2005 5:14 am    Post subject: 13 Dec 2005 very hard sudoku: solution | 
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				5c3 is because of the 4's and 7's in rows 1 and 2 (ie Box 7 and Box 9 both contain 4's and 7's in rows 1 and 2 --> therefore box 8 must contain either a 4 or 7 in both r3cd and r3cf and there's already a 8 in r3ce --> hence there must be a 5 in row 1 of box  .  Couple the 5 in row 1 of Box 8 along with the 5 in r2cg of Box 9, that means the only place for the 5 in Box 7 is r3cc. | 
			 
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		alanr555
 
 
  Joined: 01 Aug 2005 Posts: 198 Location: Bideford Devon EX39
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				 Posted: Sat Dec 17, 2005 2:19 am    Post subject:  | 
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				 	  | Code: | 	 		  
 
It would appear that SamGJ is now compiling "Very Hard"
 
puzzles that are not amenable to full solution by "Mandatory
 
Pairs" methods - unlike a few weeks ago when the 'v.hard'
 
were often easier than the plain 'hard'.
 
 
This one was solved after "normalising" column 8 and row 7.
 
 
"Normalising" is the splitting of the candidate string for a row
 
or column into two or (rarely) three subgroups where each
 
subgroup portrays all possible candidates for a number of
 
cells equal to the number of candidates in the subgroup.
 
 
An example would be 26,378,237,26,38.
 
The candidate string for the line is 23678.
 
This MUST have five digits in it as five cells are involved.
 
Normalisation would spot the pair of 26 items
 
This gives (26)(2378) as subgroups BUT the second group has
 
a repeat of one of the digits in the first group - a real trigger
 
to paying attention!
 
Removing the '2' from the 237 cell gives the normalised set
 
of (26)(378). The second group now has three members (with
 
values 378,37,38) and covers three cells.
 
 
The original string 23678 is "congruent" for the line (as it has
 
five distinct digits and covers five cells) but it is not "normalised"
 
as it is capable of division into two congruent substrings (albeit
 
by - beneficial - modification of one of the cell profiles).
 
 
NB: The example values are NOT the same ones as in the puzzle!
 
 
Alan Rayner  BS23 2QT
 
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