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98.7..6..7..5...8...6.....48..3...9..2...........1...85.....9...379....5..8.5..3.;174173;GP;12_11

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| 9 8 12345 | 7 234 1234 | 6 125 # 123 | 12!

| 7 14 1234 | 5 2346 12346 | 123 8 9 |

| 123 15 6 | 128 2389 12389 | 12357 1257 4 | 12

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| 8 14567 145 | 3 2467 24567 | 12457 9 1267 |

| 1346 2 13459 | 468 46789 456789 | 13457 14567 1367 | 146

| 346 4567 3459 | 246 1 245679 | 23457 24567 8 | 246

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| 5 146 B 124 B | 12468 234678 1234678 | 9 12467 T 1267 |

| 1246 3 7 | 9 246 b 1246 b | 8 1246 T 5 |

| 1246 t 9 8 | 1246 T 5 12467 | 1247 3 1267 |

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S S S

Almost Double JExocet [a=r7c23,r8c8,r9c4][b=r8c46,r7c8,r9c1] => r8c1 <> 1246 (false)

SV's transformation of the DJE spoilers in r1c8 to Deadly Pattern guardians is absolutely right, but I couldn't find any kite patterns without resorting to branching.

Assuming the spoilers aren't so conveniently placed, I wanted to see what other deductions can be made in the JE tier when the DJE is clearly impossible because it would produce an empty cell.

In this puzzle there can only be a single spoiler digit, (1) or (2) r1c8, which consequently

** must be true in both pairs of base cells. (If there are two potential spoilers the options would probably be too numerous to be decisive.)

Here is tier 3 repeated and then re-labelled using (a) to represent the spoiler and (bcd) the other digits to show how thy must repeat in the mini-lines as shown in

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| 5 146 124 | 12468 234678 1234678 | 9 12467 1267 |

| 1246 3 7 | 9 246 1246 | 8 1246 5 |

| 1246 9 8 | 1246 5 12467 | 1247 3 1267 |

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| 5 ab ab | d8 d38-7 d38-7 | 9 c7 c7 | [abcd] = [1246]

| d 3 7 | 9 ac ac | 8 b 5 | [a] = [1 or 2]

| c 9 8 | b 5 7-abdc | ad-7 3 ad-7 |

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The non-spoiler digits (bcd) will repeat in the regular mini-line diagonal directions for the JE pattern. (a) will be true with one of (bcd) in each mini-row, while the other mini-rows will contain two non-base digits repeating in patterns forced by the givens. Here the cells seen by both sets of base cells can't contain (abc) but can contain (d). Because they will be limited to two truths in the partial fish cells, (b) and (c) must be true in at least one of the target cells, but could be true in two, and similarly for (d) in r7c4 & r8c1.

Note the targets for the r1c23 JE at r8c8,r9c4 must both contain (b) as there are no non-base digits available.

Now r9c6 must contain a non-base digit which can only be 7 which provides the eliminations shown.

Implementing these eliminations isn't enough for my methods to solve the puzzle, but adding the (5)r1c8 elimination leads to r7c2 = 6. So, as (a) = (1) or (2), (b)r7c2 = (6) and can be assigned in r8c8 & r9c4. This in turn leads to r7c4 = 1, and so (a)r7c3 must = 2 from which the solution falls out.

To avoid being labelled a hypocrite, I should register such impossible DJEs as a pattern that gives rise to the inferences I've used, but then there is the problem about how to notate the algebra involved.

**[Edit] On reviewing this I found I'd missed an important logical step, namely that neither component JE can be true.

If r7c23,r8c8,r9c4 were true, r8b9 would have to hold two base digits, and if r8c46,r7c8,r9c1 were true, this would hold for r9b7, neither of which is possible.