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569 58 5679 | 1 2 4 | 57 3 789

1 37 4 | 5 38 389 | 279 6 27

3589 2 359 | 39 6 7 | 45 -4589 1

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3569 35 8 | 367 1 235 | 237 279 4

7 4 36 | 368 9 238 | 1 28 5

2 1 359 | 378 4 58 | 6 789 3789

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358 3578 2357 | 4 358 389 | 29 1 6

345 9 1 | 2 35 6 | 8 457 37

3458 6 25 | 389 7 1 | 2345 2459 239

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From here, one can readily develop the following Alternating Inference Chain (AIC): (4)r3c7=(4-3)r9c7=(3)r4c7-(3=5)r4c2-(5=8)r1c2-(8)r1c9=(8)r3c8 => r3c8<>4, which also places r3c7=4 and => r9c7<>4.

At this point, boxes b789 are shown below and can serve to illustrate some interesting alternative methods for the next elimination of interest, r9c8<>2. Methods are listed more or less in increasing order of difficulty (for me).

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358 3578 2357 | 4 358 389 | *29 1 6

345 9 1 | 2 35 6 | 8 457 37

3458 6 *25 | 389 7 1 | 235 -2459 *239

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Method 1: Grouped AIC (2=359)r9c379-(9=2)r7c7 => r9c8<>2, which is equivalent to a WXYZ-Wing with W=9 and Z=2. This method seems straightforward and elegant.

Method 2: Subset Counting.

The four cells (2359)r9c379\r7c7 form a locked subset (LS) with total multiplicity of 5 (the one extra count is from the 2-digits). The presence of (2)r9c8 would eliminate all 2's in the LS and thereby reduce the multiplicity below 4. Hence, r9c8<>2. This method is a little more complicated and involves choosing an appropriate subset, doing some counting and having an understanding of multiplicity.

Method 3: Aligned Pair Exclusion (APE).

Cell r9c3 plus box b9 look promising for an APE to eliminate (2)r9c8 using the row pair r9c78 (Andrew Stuart's solver confirmed this). Combination 2-2 is obviously not allowed, while the starred cells eliminate the remaining ordered combinations 3-2 and 5-2. Note that Almost Locked Set (ALS) (239)r7c7\r9c9 is needed to eliminate the 3-2 combination. This method is definitely tougher to follow and requires choosing a suitable aligned pair and also knowing how an ALS affects the exclusions. At this point, I stumbled across...

Method 4: Almost APE.

Now, except for the presence of (4)r9c7 in the original full grid, the previous APE provides for the desired elimination, r9c8<>2. That, in turn, makes (4)r9c7 a “spoiler” of sorts, so one can form an Almost APE in the original grid with the strong link [APE]=(4)r9c7. Searching now for a suitable chain in that grid, one then finally arrives at

[APE]=(4)r9c7-(4)r3c7=(4-8)r3c8=(8)r1c9-(8)r1c2=(8)r7c2-(8)r9c1=(8-9)r9c4=(9)r7c6-(9=2)r7c7 => r9c8<>2.

Well, this method works, but that Almost APE is one ugly monkey.

Apparently, one can “almost” almost anything.