Solution to dml #2

Advanced methods and approaches for solving Sudoku puzzles

Solution to dml #2

Postby gurth » Thu Dec 14, 2006 10:43 am

Solution to dml #2

following on closely from dml #4, which it much resembles. That will serve to emphasise and reinforce the Emerging Strategies as described in my solution to dml #4.

Code: Select all
dml #2 : SR 10.5, gsfr 99983
 *-----------*
 |8..|..3|...|
 |.7.|.6.|.9.|
 |..4|5..|...|
 |---+---+---|
 |..5|...|..4|
 |.9.|..2|.7.|
 |1..|...|8..|
 |---+---+---|
 |...|..9|..3|
 |.2.|.1.|.6.|
 |...|7..|5..|
 *-----------*


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) ... Scanning the filtered candidates, strategy becomes clear : create the 5s conjugate chain, and eliminate the Candies; then force a selected candidate which will trigger collapse.
(To follow this you need to have studied my introduction to my solution to dml #4). First we do the accessible Candies. At this stage, those Candies which already place all the 5s correctly, even though the 5s chain is not yet fixed, are to be regarded as accessible. This is a much better approach than fixing the chain first.

(2) ?3b1...(?3k2...??)-3k2... (?3d8...??)-3d8... (?3d7...??)-3d7... (?3d5...??)-3d5, (?3d4...??)-3d4... (?4e4...??)-4e4...?? -3b1.

(3) ?1g2... (?2a7...??)-2a7, (?2a8...??)-2a8, (?2a9...??)-2a9, (?2a5...??)-2a5, (?2c8...??)-2c8, (?2c7...??)-2c7...?? -1g2.

(4) ?4g2... (?4b7...??)-4b7... (?4a7...??)-4a7...?? -4g2.

(5) ?6g2...?? -6g2. Now we need to get the 5s chain into shape.

(6) ?5f9...?? -5f9.

(7) ?5a9... (?9f5...??)-9f5, (?9d5...??)-9d5... (?6k2...??)-6k2, (?6d2...??)-6d2... (?8d2...??)-8d2...?? -5a9.

Chain fixed! Now we can do the rest of the Candies: 12a8, 6e9, 349f5. Start with the easier ones always, the ones withe the most consequences!

(8) ?4f5...?? -4f5. (A very easy Candy).

(9) ?9f5...?? -9f5. Another piece of cake. I expect the next to be a bit harder.

(10) ?3f5...?? -3f5. I was wrong. Looks like we are over the worst.

(11) ?6e9...?? -6e9. (That was much tougher than it looks.)

(12) ?2a8...?? -2a8. Surprisingly easy.

(13) ?1a8...?? -1a8. Very easy. I suppose every candidate eliminated at this stage is telling, reducing the difficulty for the next.

Now all the Candies are gone. It's time now to force one of the true candidates in cells which the Candies have left. (The false-5 cells). These true candidates are 4a8, 1e9, 7f5, 2b1, 8g2. Which one will be easiest to force?

4a8 can be forced by -4g8 and -4k8 (-4gk8).
7f5 can be forced by -7f36.
2b1 can be forced by -2cd1.
8g2 can be forced by -8dk2.
Out of these I choose to go for forcing 4a8:

(14) ?4k8...(?3e7...??)-3e7, (?3d7...??)-3d7, (?6c1...??)-6c1, (?6a3...??)-6a3... (?1k9...??)-1k9...?? -4k8.

So the next move will (I hope) force -4g8, 4a8 and then the whole puzzle should collapse! Let's see.

(15) ?4g8... (?6f2...??)-6f2, (?6d2...??)-6d2... (?2f3...??)-2f3... (?3e3...??)-3e3, (?4f2...??)-4f2...?? -4g8....final march of singles to the End interrupted by one cantankerous naked pair only.

Note: No sub-sub-nets this time; I used 3 on dml #4. Is this the difference between SR 10.6 and 10.5? Hard to tell. Maybe it's just practice makes perfect.

Next I plan to try Mauricio's tenner for a change of scenery. If anyone wants to see my solution to any other puzzle, I invite requests. Post them on my thread Gurth's Puzzles: I'm not likely to miss them there. (in General/Puzzle forum ).
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gurth
 
Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

Postby leon1789 » Mon Dec 18, 2006 9:52 am

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

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


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.

1) f6#5,
2) a4#4,
3) now if b1=5 (bilocation number) then k2#3, b3#1, g3#7, a3#1, c5#2, d7#3, f5#7, k2#6, d7#2, h3#7, c7#3, f2#3, a3#2, f2#4, and contradiction using singles and locked sets, so -5b1 (and then 5g1)
4) d5#7,
5) f9#2,
6) b4#8,
7) a4#9,
8) c1#9,
9) c6#1,
10) d5#3,
11) b3#1,
12) b4#1, and finish with singles and locked sets.
leon1789
 
Posts: 37
Joined: 15 November 2006


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