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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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