daj95376 wrote:David P Bird wrote:The possible alternative (ab/cd) notation for SK Loops is because they are NOT AICs and the logic covers three divisions of the truths not two.

Strange ... and yet ronk presented it like an AIC with the OR(X,Y):(3) logic.

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`Steve Kurzhal's original presentation w/ RonK's labels:`

1.......2.9.4...5...6...7...5.9.3.......7.......85..4.7.....6...3...9.8...2.....1

1 A478 34578 | 3567 3689 5678 | 3489 I369 2

L238 9 L378 | 4 K12368 K12678 |J138 5 J368

23458 A248 6 | 1235 12389 1258 | 7 I139 3489

-------------------------+-------------------------+-----------------------

2468 5 1478 | 9 1246 3 | 128 H1267 678

234689 B12468 13489 | 126 7 1246 | 123589 H12369 35689

2369 B1267 1379 | 8 5 126 | 1239 4 3679

-------------------------+-------------------------+-----------------------

7 C148 14589 | 1235 12348 12458 | 6 G239 3459

D456 3 D145 |E12567 E1246 9 |F245 8 F457

45689 C468 2 | 3567 3468 45678 | 3459 G379 1

(27)r13c2=(27-16)r56c2=(16)r79c2-(16)r8c13=(16-27)r8c45=(27)r8c79-

(27)r79c8=(27-16)r45c8=(16)r13c8-(16)r2c79=(16-27)r2c56=(27)r2c13-loop

The SK-Loop doesn't work unless the first SL forces both of (27) false in r13c2 and both of (27) true in r56c2. We may not have a consensus on how to notate these terms, but it sure looks like an AIC loop in all other respects!

This is even more apparent in the V-Loop version:

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` (48=27)r13c2 - (27=38)r2c13 - (38=16)r2c79 - (16=39)r13c8 -`

(39=27)r79c8 - (27=45)r8c79 - (45=16)r8c13 - (16=48)r79c2 - loop

_

It's probably worth mentioning that for both of these "loops", the usual presentation is ambiguous, and so to some extent, meaningless.

Each one has two consistent AIC interpretations ... leading to two valid AICs, neither of which (on its own) can justify a single elimination.

To justify the usual eliminations, you need to combine bits of knowledge gained from each loop, and add a little bit of insight as well.

That point is rarely mentioned, and it seems to have been forgotten by some.

For the first loop, the two interpretations are:

(2|7)r13c2 = ((2&7)-(1|6))r56c2 = (1&6)r79c2 - (1|6)r8c13 = ((1&6)-(2|7))r8c45 = (2&7)r8c79 -

(2|7)r79c8 = ((2&7)-(1|6))r45c8 = (1&6)r13c8 - (1|6)r2c79 = ((1&6)-(2|7))r2c56 = (2&7)r2c13 - loop

(2&7)r13c2 = ((2|7)-(1&6))r56c2 = (1|6)r79c2 - (1&6)r8c13 = ((1|6)-(2&7))r8c45 = (2|7)r8c79 -

(2&7)r79c8 = ((2|7)-(1&6))r45c8 = (1|6)r13c8 - (1&6)r2c79 = ((1|6)-(2&7))r2c56 = (2|7)r2c13 - loop

For the second loop, they are are:

((4|8)=(2&7))r13c2 - ((2|7)=(3&8))r2c13 - ((3|8)=(1&6))r2c79 - ((1|6)=(3&9))r13c8 -

((3|9)=(2&7))r79c8 - ((2|7)=(4&5))r8c79 - ((4|5)=(1&6))r8c13 - ((1|6)=(4&8))r79c2 - loop

((4&8)=(2|7))r13c2 - ((2&7)=(3|8))r2c13 - ((3&8)=(1|6))r2c79 - ((1&6)=(3|9))r13c8 -

((3&9)=(2|7))r79c8 - ((2&7)=(4|5))r8c79 - ((4&5)=(1|6))r8c13 - ((1&6)=(4|8))r79c2 - loop

Some of us will have seen these "AIC pairs" before, on this forum, with #1's and #2's replacing '|'s and &'s ... ab#1 instead of (a|b), and ab#2 instead of (a&b).

For a review of how the eliminations are arrived at:

For the first pair of loops, if we break the loops at the "- loop" weak links at the end of the presentations, and make the standard claim for AICs, we get these:

from the 1st loop: one of { (2|7)r13c2, (2&7)r2c13 } must be true in any particular solution to the puzzle

from the 2nd loop: one of { (2&7)r13c2, (2|7)r2c13 } must be true in any particular solution to the puzzle

From those, we get the usual eliminations for 2's and 7's in b1, like this.

- For the first loop, if it's (2&7)r2c13 that was true, then 2's and 7's in r13c13 would be eliminated.
- For the seocnd loop, if it's (2&7)r13c2 that was true, then 2's and 7's in r13c13 would be eliminated.
- If neither of those cases applied, then from the first loop, (2|7)r13c2 would be true, and from the second loop, (2|7)r2c13 would be true. A little thought reveals that in that case, one of {2,7} would be true in r13c2, and the other would be true in r2c13, and again 2's and 7's in r13c13 would be eliminated.
- Summary: In all 3 cases, 2's and 7's in r13c13 would be eliminated.

Conclusion: it's safe to eliminate them "now" ... they can't appear in any solution to the puzzle.

To get the remaining eliminations we need to make that same argument, breaking the loops at a different weak link; keeping the breakpoints synchronized; and making a similar argument for each pair of breakpoints.

--

The bottom line here, is that no amount of trying to write "

the SK loop" with special symbols in it, is going to get around the fact that it's going to take

two valid AIC (loopable AIC's) and some side arguments, to justify the usual eliminations that are claimed for "it".

The best we can say, is that the usual presentation is "merely suggestive" ... that it has just two consistent interpretations (involving &'s and |'s), and that we need both interpretations to arrive at any conclusions.

Finally: The chance that a newbie is going to be able to make proper sense out of any presentation (eliminations included), is virtually nil ... even if both (proper) AICs were displayed.

Edited: minor rewording