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zaks
Joined: 25 Nov 2005 Posts: 13

Posted: Tue Dec 13, 2005 7:14 pm Post subject: 13 Dec 2005 very hard sudoku: solution 


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:
numbercolumnrow (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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kbizzle Guest

Posted: Wed Dec 14, 2005 7:17 am Post subject: 


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

Posted: Thu Dec 15, 2005 5:14 am Post subject: 13 Dec 2005 very hard sudoku: solution 


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: 194 Location: Bideford Devon EX39

Posted: Sat Dec 17, 2005 2:19 am Post subject: 


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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