Solution to dml #155

Advanced methods and approaches for solving Sudoku puzzles

Solution to dml #155

Postby gurth » Wed Dec 20, 2006 9:13 am

Solution to dml #155

Code: Select all
 *-----------*
 |1..|9..|...|
 |.2.|.5.|.3.|
 |..4|..6|...|
 |---+---+---|
 |6..|...|1..|
 |.3.|.8.|.2.|
 |..7|...|..4|
 |---+---+---|
 |...|3..|7..|
 |...|.2.|.5.|
 |.8.|..4|..9|
 *-----------*        SE 10.7


Note: This solution is designed to be followed using the Simple Sudoku program.
"..." means proceed as far as SS allows, following the hints given by SS in the order given.
This will facilitate rapid checking of this solution.

"?" introduces a move that will be disproved by contradiction. (These moves are only introduced when SS grinds to a halt, saying "no hint available".) In order to "play" this move, you will have to turn off the "block invalid moves" feature of the program. Then insert the move and follow all hints given until you see a contradiction. Once you find the contradiction, you must retrace (withdraw in reverse order) all moves made since the disproved move, by clicking the "undo" arrow repeatedly. Then "correct" the disproved move. EG if this move was "?3f6", then REMOVE the 3 at f6, because you have proved the 3 at f6 false.

SOLUTION:

(1) ...

(2) ?4g1...?? -4g1.

(3) ?3h1... (?4d4...??)-4d4... (?2k1...??)-2k1... (?7b1...??)-7b1... (?7c1...??)-7c1... (?9g1...??)-9g1...?? -3h1.

(4) ?4d4... (?3h3...??)-3h3... (?3d5...??)-3d5... (?3f6... ((?2g3...??))-2g3... ((?2a9...??))-2a9, ((?2f1...??))-2f1... ((?1b4...??))-1b4...??)-3f6... (?3c1...??)-3c1... (?6k3...??)-6k3...?? -4d4.

(5) ?4a7... (?3k7... ((?6k3...??))-6k3... ((?1k3...??))-1k3...??)-3k7... (?2k3...??)-2k3, (?2k1...??)-2k1... (?1k3...??)-1k3, (?6k3...??)-6k3... (?5k4...??)-5k4... (?2d6...??)-2d6, (?2f4...??)-2f4, (?2f6...??)-2f6...?? -4a7...

(6) ?3a5... (?3h3... ((?6k3...??))-6k3, ((?1k3...??))-1k3... ((?1k4...??))-1k4, ((?6k4...??))-6k4...??)-3h3... (?1k4... ((?9c2...??))-9c2... ((?9b3...??))-9b3...??)-1k4, (?6k4...??)-6k4, (?9c2...??)-9c2... (?7b1...??)-7b1... (?9b3...??)-9b3...?? -3a5.

(7) ?9b7... (?8c1...??)-8c1, (?8b1...??)-8b1...?? -9b7...

(8) ?8b7...?? -8b7.

(9) ?5d2...?? -5d2.

(10) ?5e4...?? -5e4.

(11) ?9h1... (?8a3...??)-8a3...?? -9h1.

(12) ?1e4...?? -1e4.

(13) ?7e4...?? -7e4.

(14) ?1g8...?? -1g8.

(15) ?6g8...?? -6g8.

(16) ?6b3...?? -6b3...

(17) ?6b9... (?3f6...??)-3f6, (?3d6...??)-3d6... -6b9.

(Note that so far I have not solved a single cell. All cells solved together as singles to end with:)

(18) ...
_____________________________________________________________
gurth
 
Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

Postby leon1789 » Thu Dec 21, 2006 10:07 am

An other method :
-- contradiction nets using singles and locked sets (special case of avanced ALS technique);
-- only one guess on a bivalue cell or a bilocation number.

Code: Select all
   123 456 789
  *-----------*
a |1..|9..|...|
b |.2.|.5.|.3.|
c |..4|..6|...|
  |---+---+---|
d |6..|...|1..|
e |.3.|.8.|.2.|
f |..7|...|..4|
  |---+---+---|
g |...|3..|7..|
h |...|.2.|.5.|
k |.8.|..4|..9|
  *-----------*


Notation : C#z means C=z is a disproved move by a contradiction net using singles and locked sets (so -zC is proved). Moreover, "bold" results are definitively true.

*) g1#4, c1#9
*) now, if d3=8 (bivalue cell) then c5#1, a3#6, a3#5, g1#5, a5#4, b3#6, a7#4, g3#2, c1#5 and contradiction using singles and locked sets, so -8d3 (and 2d3)
*) now c2#9, h3#3, h1#3, k1#2, g8#1, g8#6, b3#9, k1#7, d2#4, a2#7, a2#5, c1#3, e6#5, k4#5, d6#7, d9#3, c4#8 and finish with singles and locked sets.
leon1789
 
Posts: 37
Joined: 15 November 2006


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