JC, well it looks as if it's just you and me! Thanks for your solution which I've learned from.
We work in quite different ways and there's little point in arguing over our differences, but I can outline my approach to you so at least you can understand it.
When I devised the GEM marking system I found it could be used to find net-based deductions which made most ordinary puzzles too easy and took all the enjoyment out of solving them. I therefore disciplined myself only to use the linear AICs identified by GEM. Taken to extreme though, this would make the harder puzzles impossible to solve. My compromise was to allow recognisable pattern inferences to be used which depend on net-based analysis to prove. Ever since I've been enjoying trying to increase the difficulty of the puzzles I can solve. This suits me as I prefer to look on Sudoku puzzles more like a game of patience than a race.
For the opening move I can therefore use the recognisable SK loop pattern and apply it immediately, but it then needs a naked quad and an (8)XWing to reach the position after your first rank 0 logic step. (You didn't use the use the naked quad, but you didn't need it). So when I posted the opening post my solution was:
Phase 1
SK Loop (89=34)r13c2 - (34=78)r79c3 - (78=23)r8c13 - (23=67)r8c79 -
......... (67=25)r79c8 - (25=69)r13c8 - (69=45)r2c79 - (45=89)r2c13 - Loop
Eliminations: (89)r1c3, (4)r2c5, (4)r2c6, (69)r3c9, (2)r4c8, (34)r6c2, (5)r6c8, (78)r7c1, (2)r8c5, (67)r9c7
..(6789)NakedQuad:r2456c6 => r1c6 <> 89, r3c6<>69, r7c6<> 78, r9c6 <> 67
(8)2Fish:r17c25 => r28c5,r56c2 <> 8
Phase 2
(7)r2c5 = (7-8)r2c6 = (8)r1c5 - (8)r7c5 = (8-9)r8c4 = (9)r8c5 => r2c5 <> 9, r8c5 <> 7
(78)r2c56 = (8)r1c5 - (8)r7c5 = (8)r8c4 - (8=3)r8c3 - (3)r9c23 = (3-1)r9c7 = (1)r7c9 - (1)r7c1 = (1-5)r3c1
.. = (5)r2c13 - (5=6)r2c7 => r2c56 <> 6 (r2c5=7)
.. (7)boxline:b8c4 => r5c4 <> 7
.. (6)boxlines:b2r3 & c6b5 => r3c8 <> 6, r5c4 <> 6
(46)r2c79 = (4-1)r3c9 = (1-5)r3c1 = (5)r2c13 => r2c79 <> 5 (r2c7 = 6)
.. (5)boxline:b3r3 => r3c1 <> 5
At this point the grid is very close to yours where you use the weak and strong inference set matrix Phase 3
(1)r7c9 = (1-3)r9c7 = (3)r9c23 - (3=8)r8c3 - (8)r7c2 = (8)r1c2 - (58=4)r2c13 - (4)r2c9 = (4)r3c9 => r3c9 <> 1
.. Singles (1)r1c7,r3c1,r9c3,r7c9
(5)r2c3 = (5)r6c3 - (5)r6c7 = (5-3)r9c7 = (3)r9c2 - (3=8)r8c3 => r2c3 <> 8
(5)r2c3 = (5)r6c3 - (5)r6c7 = (5-3)r9c7 = (3)r9c2 - (3)r13c2 = (3)r1c3 - (3=89)r48c3 => r2c3 <> 9
.. (9)boxline:b1c2 => r56c2 <> 9
.. (67)NakedPair:r56c2 => r456c1,r79c2 <> 7
.. Singles (7)r7c8,r8c1,r9c4, (2)r7c1, (5)r7c6
.. (2)boxline:b7r9 => r9c78 <> 2, c8b3 => r3c9 <> 2
.. (3)linebox:c1b4 => r46c3 <> 3
.. (4)boxline:b7c2 => r13c2 <> 4
(9)r2c9 = (9)r2c6 - (9)r456c6 = (9)r5c4 => r5c9 <> 9
(5)r9c7 = (5)r6c7 - (5=6)r5c9 - (6)r8c9 = (6)r9c8 => r9c8 <> 5
Singles to the end.
REVISED VERSION
Phase 2
Translating your SWings into the same notation style (thank you), my phase2 can be replaced by:
(2)r4c9 = (2)r4c7 - (2=1) - (1)r3c9 = (1)r7c9 => r7c9 <> 2
(1)r3c1 = (1)r3c9 - (1=5)r7c9 - (5)r5c9 = (5)r5c1 => r3c1 <> 5
.. (5)linebox:r3b3 => r2c79 <> 5, (single (6)r2c7)
.. (6)linebox:c6b5 => r5c4 <> 6
(9)r2c9 = (29-5)r13c8 = (5-4)r3c9 = (4)r2c9 - loop => r3c9 <> 12
. . Singles (1)r1c7,r7c9,r9c3,r3c1
. . BoxLine:b3c8 => r79c8 <> 2
(The (7)r789c5 eliminations aren’t needed) Phase 3 (Colouring Version)
Subsequent analysis has shown there is a logical error in this procedure that leaves a loophole. The outcome is that the concept is OK but needs more eliminations to be made in the corner boxes before it can be applied. In this puzzle by the time these further eliminations are made, the puzzle is virtually solved, and needs only two more chains to finish so it becomes over-kill. Consequently it's unlikely that the method will ever be worthwhile, and my efforts have been wasted ****** ****! (expletives redacted)Repeating the SK Loop:
(89=34)r13c2 - (34=78)r79c3 - (78=23)r8c13 - (23=67)r8c79 -
(67=25)r79c8 -
(25=69)r13c8 - (69=45)r2c79 - (45=89)r2c13 - Loop
The coloured segment is now
(25=9)r13c8 - (69"=4')r2c79 following these reductions (shown below).
At parity I when 4' is true, every term in the loop must hold a single truth
At parity II when 9" is true, in the brackets the left terms will hold two truths and right ones will be false.
Following the loop around, where any digit in a left hand pair is restricted to single cell it can be marked (0") and its companion digit in the other cell can be marked (0=) which shows it must be true at p" but could possibly be true at p'.
This is the logical error. The digits restricted to a single cell can only be marked (0") if the companion digit is locked in the cell pair. Otherwise it should be marked (0=) as there is the chance that it could be also be true at p'.
The other digits in the SK loop boxes can be marked according to the legend below.
- Code: Select all
*---------------------------*---------------------------*---------------------------*
| <6> 3.4.8"9. 3"4' | <5> 2489 234 | 1 2=9. <7> |
| 4:58. <2> 4:58.9. | <1> 79 789 | 6 <3> 4'9" |
| 1 3.4.9= <7> | 69 2469 234 | <8> 2.5"9. 4"5' |
*---------------------------*---------------------------*---------------------------*
| 378 <1> 389 | <4> <5> 6789 | 237 6789 2369 |
| 578 679 <2> | 789 <3> 6789 | <4> <1> 569 |
| 34578 679 34589 | <2> <1> 6789 | 357 6789 3569 |
*---------------------------*---------------------------*---------------------------*
| 2"4' 4=7.8. <6> | <3> 2478 245 | <9> 2.5.7= 1 |
| 2'3.7"8. <5> 3.8= | 6789 679 <1> | 237. <4> 2:3:6' |
| <9> 3"4.7. 1 | 67 2467 245 | 2.3.5" 2.5.6"7. <8> |
*---------------------------*---------------------------*---------------------------*
GEM Marks: Par (0') (0") True at own parity false at the other
Super (0-) (0=) True at own parity possibly true at the other
Sub (0.) (0:) Possibly true at own parity false at the other
In box1 (4)r13c2 can't be true as at p' (4')r1c3 will be true and at p" (8"9=)r13c2 will be true
. . (4)linebox:c2b7 => r7c1 <> 4
Single (2)r7c1
This proves that p" is true and all (0') and (0.) digits are false, effectively solving the puzzle.
When I posted I felt that I must have missed an almost pattern that would do away with that 15 link chain in phase 2, but you've showed me I was looking in the wrong place.
The tempting pattern I mentioned in the opening post is that in the solution (6789)r2c56,r8c45 and (6789)r56c3,r46c8 are locked quads, but I can't prove that will always happen from the distribution of the givens.
I'm still considering the best way to notate the colouring deductions.
DPB
TAGdpbSKLoop
Edited 29th Oct when blunder discovered.
