The New Sudoku Players' Forum

[ghfick]

.---------.---------.---------.
| 9 8 . | 7 . . | 6 . . |
| . . 5 | . 4 . | . . . |
| . . . | . . 9 | . 3 . |
:---------+---------+---------:
| 4 . . | . . . | . . 8 |
| 2 5 . | . 1 . | . . . |
| . . 8 | . . . | . 1 . |
:---------+---------+---------:
| . . 2 | . 5 . | . . 4 |
| . . . | 6 . . | . 9 . |
| . . . | . . 3 | 7 . . |
'---------'---------'---------'
98.7..6....5.4.........9.3.4.......825..1......8....1...2.5...4...6...9......37..

Note: I cannot get anywhere with this one.

[Mauriès Robert]

Hi ghfick,
The difficulty level of this puzzle can be greatly reduced by a DFS that allows the elimination of 4r1c8.
The puzzle is then solvable with classical techniques.
Robert

[ghfick]

Hi Robert,
Interesting. So, if I understand you, a DFS algorithm can consider exclusions as well as placements.
The size of the backdoor for this puzzle is 2. One of the 14 backdoor pairs is r1c8=2, r8c6=8.
Placing r1c8=2 alone has nearly the same effect as excluding r1c8<>4. Either way, the puzzle is still tough but solvable with 'classic' techniques.
Gordon

[Hajime]

the puzzle is still tough but solvable with 'classic' techniques.

How (confused)?

[ghfick]

Placing r1c8=2 gives an SE=9.0 while excluding r1c8<>4 gives an SE=9.1. In both cases the puzzle's solution path does not require brute force. Very tough solution paths though.

[Mauriès Robert]

Hi Gordon,
Here is a possible DFS that validates the 3r1c5 then the 4r1c3... a piece of manual bravery!

(-3r1c5)->3r2c4->3r1c3->(4r1c8 & 2r1c5->24p6p7B8)->AP1->7r5c8->AP2->3r5c9->AP3->1r7c4->9r7c2->AP4->6r3c5->8r89c5->7r7c6->8r8c5->9r9c5->AP5->3r6c12->3r4c5-7r6c5->36r6c12->17r4c23->3r4c5->... incompatibility* => r1c5=3
with (I have not written all the candidates for the chains which are generally long):
AP1=(-7r5c8)->6r5c8->... incompatibility*
AP2=(-3r5c9)->3r5c7->... incompatibility*
AP3=(-1r7c4)->1r7c6->...->9r9c5->... incompatibility*
AP4=(-6r3c5)->8r3c5->... incompatibility*
AP5=(-3r6c12)->67r6c12->... incompatibility*
(*) incompatibility with sudoku rules

(-4r1c3)->1r1c3->4r1c8->1r4c2->AP1->6r7c8->7r5c8->AP2->13r7c7->AP3->3r7c7->125r189c9->7r3c9->(9r2c9 & 6r3c1->37r2c12->2r3c2->6r2c6->8r3c5)->36r56c9->AP4->2r89c5->AP5->9r9c5->2r8c5->9r7c2->... incompatibility* => R1C3=4 + 2 placements
with :
AP1=(-6r7c8)->8r7c8->... incompatibility*
AP2=(-13r7c7)->8r7c7->... incompatibility*
AP3=(-3r7c7)->1r7c7->... incompatibility*
AP4=(-2r89c5)->(7rc5 & 9r9c5)->18r7c46->... incompatibility*
AP5=(-9r9c5)->2r9c5->... incompatibility*

This gives a depth of 5, which seems a lot to me and certainly there are simpler DFS.
Furthermore, the way I presented these eliminations with two chains and subchains, makes me think that we should be able to solve with B5-Braids, or more precisely SkB5-SmBraid (?).
For the rest of the puzzle, eliminations by the chains are possible until complete resolution.
For example:
(-6r2c6)->6r3c5->8r3c4->8r5c6->... => -6r5c6
(-9L5C7,-9L7C4)->(3L5C7 & 9L7C2)->3L7C1->3L2C2->3L4C3->1L4C2->9L5C3->... => -9L5C4
Etc...
Robert

[ghfick]

Very brave, Robert!
I decided to have a look at each of the single placements to see if there were any 'close' to a backdoor. I think the lowest SE came with r8c6=8 with SE=8.4.

[DEFISE]

Furthermore, the way I presented these eliminations with two chains and subchains, makes me think that we should be able to solve with B5-Braids, or more precisely SkB5-SmBraid (?).
Robert

Yes because this puzzle is in B4B i.e. in T&E(B4,1).
But even in B5B, manual solving would probably be even more tedious than yours !

Happy New Year to all !