This post steps through the most fiendish of the 10 puzzles preselected for the review of Brian Challenger’s Super Fiendish Sudoku, SF 67. , and then adds one of the bypass victims, SF 77.
Looking at the trace, you might think basic is easy,
but the long fill strings reveal a tough line marking.
Advanced solving starts with hidden unique rectangles of type 2b. The floor to ceiling slinks project a victim clue into a reversible rectangle of four clues.
Then, after an X-chain ANL,
a regular 
127-wing is extended into an inference chaining i127-wing.
The midfield action continues with a 4-BARN (Bent Almost Restricted 4-Set),
and an ungainly 
looking APE in which 5r4c9 cannot be combined with 1, 2 or 7 in r4c8 within sight of ALS 157 and ALS 2457.
The APE removal is telling,
leading to an XY railway of interconnected nice loops.
No removals result, but a nice loop generated cluster is expanded by a trap and wraps blue in r2 and c4, for the collapse of Super Fiendish 67.
In his comment to this post, forum member strmckr provided this quick resolution to SF 67 immediately after line marking. He terms it an ALS W-wing.
The restricted common is a forcing chain. Since one ALS will lose it’s 9’s. Any candidate seeing all of any other value in both ALS must see a true value. 1r1c9 does just that, and the removal collapses SF 67. With all the ALS in every grid, can you hope to spot such an event? Not really, but you can see the appeal of writing code that can find every instance of it.
Super Fiendish 77 might have presented similar advanced challenges, but was too vulnerable to Sysudoku Basic. It takes a while but the collapse is relentless.
Next week, the Super Fiendish review ends with 87 and 97. For the next review, get a copy of J.B. Nakamoto’s Sudoku Puzzle Book, Extreme Level.
The 67 and 77 solutions:
Systematic Sudoku 
.——————–.—————–.——————.
| 5 247 478 | 34 1 6 | 9 2347 2378 |
| 78 6 4789 | 349 2 3789 | 1 5 378 |
| 1278 3 14789 | 5 479 789 | 6 247 278 |
:——————–+—————–+——————:
| 1267 9 17 | 8 57 4 | 3 127 12567 |
| 12378 1247 1478 | 6 3579 379 | 457 127 12579 |
| 36 47 5 | 1 39 2 | 47 8 69 |
:——————–+—————–+——————:
| 179 17 2 | 349 349 5 | 8 6 13 |
| 4 8 3 | 2 6 1 | 57 9 57 |
| 19 5 6 | 7 8 39 | 2 13 4 |
‘——————–‘—————–‘——————‘
basics gets this grid above: then
Almost Locked Set W-Wing: A=r1c23,r2c13,r3c1 {124789}, B=r7c12 {179}, connect by 9r27c4 => r9c1 cannot = 1
– singles to the end.
Thanks for that. I added the two 1’s, and will amend the post to include it. The unlikelihood of human spotting needs to be said, but it’s a fine example.
Its not a forcing chain, its an extension of the w wing aic technique replacing the bivavles of the w wing with an als then linking them together by group or bilocal strong link the same as the w wing.
The bivavles of the w wing is the smallest als size (n cells with n+1 digit),
The example is found by hand.
For more information check out
http://forum.enjoysudoku.com/als-w-wings-rings-t36860.html
For those that are interested in it.
The restricted common in this case is a forcing chain, meaning an inference chain acting as a weak link between two candidates, which in this case are the value sets in what I call an ALS toxic pair. It’s not an extension of the w wing, it’s an essential element of it. You mean ‘bi-value’, which is an ALS, so instead of “replacing”, you can just extend the definition of the w-wing to include a second multi-value ALS.
Alternate solution shown on Reddit: https://www.reddit.com/r/Sudoku_meta/comments/fmh873/super_fiendish_67/
The idea that unique rectangle methods depend on an assumption of uniqueness is mistaken. A justified assumption is that no composer will publish a puzzle with a solution in which rectangle corners can be interchanged for a second solution.
what ‘standard basic’ gets 2r1c2 and 8r5c3 in Super Fiendish 67?
Uniqueness strategies by definition depend on the assumption of uniqueness. You may call the assumption justified, and that is certainly reasonable, but I just found a Paul Stephens puzzle that wasn’t unique, it had two solutions. (Mastering Sudoku, 2007, #39). It happens. (It was a BUG, not an NUR). We see a puzzle with more than one solution every few months on Reddit. They do get published. All this means is that authors or publishers can make mistakes. I’ll come back to answer your question. I’ll need to look at the solution path, I don’t document “standard basic details. Simultaneous Bivalue Coloring, in practice uses only basic strategies, ordinarily. If something more advanced is needed, that seed pair will be abandoned and an easier one found. In general, SBN resolves all ordinary sudoku with little or no fuss. I will check those specific resolutions, where they came from. Sometimes I make “lucky mistakes.” It would not make a significant difference, the most it could do is require an additional SBN seed, and if that 2 was unresolved, and following Gordonian cell order, it would have come first.
Please read the assumption again. It doesn’t assume a unique solution, as you and a host of other experts have declared in print. Your example from Paul Stephens is not relevant. To say it differently, the unique rectangle methods assumption is that no solution in the back of the book will allow a second solution by a diagonal interchange across a rectangle. That’s just too obvious an error. If composer makes that error, his work is discredited and the work on that puzzle is not relevant.
Something went awry with that solution. The initial SBN does work, but doesn’t resolve the puzzle as described. I.e., r1c3={34}, the three chain comes to a contradiction. I had the correct givens and don’t know how I came to the state shown there, the basics do not accomplish it. I will redo that. Thanks for asking. (I suspect that I tried another SBN, got those results, and came back to the puzzle and forgot what I had done.)
Okay, there was an error. I redid the solution for that puzzle and it appears that I missed the extra 2 in box 1, I almost did it again! However, the SBN coloring for r1c4={34} still resolves to 4 through contradiction of the 3 chain, and then, with singles and a naked pair in Box 1, that 2 re-appears, and the rest of the puzzle follows, singles to the end. Thanks for questioning that, and I write more of relevance in a new Reddit comment: