Forcing Chains

Advanced methods and approaches for solving Sudoku puzzles

Postby RW » Mon Apr 09, 2007 7:14 am

daj95376 wrote:First, I found 13 bilocation cells.

Yes, you are very right. I shall not try to count bilocation cells in the middle of the night ever again...:)

RW
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Postby RW » Sun Apr 15, 2007 12:14 pm

This one gets a little closer to proving leon's conjecture wrong:
Code: Select all
1.......9.2..6..7...3...5...8.42.......61.......8.7.4...9...1...6..7..8.5.......3

No bivalue cells, only 8 bilocation cells. One of the bilocation cells solves the puzzles with singles contradiction nets, one more when allowing locked candidates and three more with subsets. Three of the possible guesses need more advanced techniques:

Singles: [r6c2]=1
Locked Candidates: [r5c7]=8
Subsets: [r3c8]=1, [r9c6]=6, [r7c1]=8
No solution: [r3c1]=6,[r7c9]=7, [r3c4]=7

I'm starting to believe that there is a puzzle that will prove leon's conjecture wrong!

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Postby leon1789 » Wed Apr 18, 2007 10:52 am

udosuk wrote:
Code: Select all
1.......2
.9.4...5.
..6...7..
.5.9.3...
....7....
...85..4.
7.....6..
.3...9.8.
..2.....1

RW wrote:
Code: Select all
1.......9.2..6..7...3...5...8.42.......61.......8.7.4...9...1...6..7..8.5.......3

Very difficult puzzles:!::!::!:
RW wrote:I'm starting to believe that there is a puzzle that will prove leon's conjecture wrong!

...but...
udosuk wrote:the conjecure survives... For now...:)


Some puzzles can't be solved with singles contradiction nets (in the 1-level recursion). But with locked sets, the conjecture always "runs":)
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