Hi Champagne
I'm afraid I can't follow your posts very well, but because we work in such different ways, perhaps it doesn't matter.
I've been analysing the first few examples in your list which share some common characteristics that produce some nice short cuts.
For any Almost Double JE with 4 base digits we have the following Potential Eliminations:
1. The non base digits in the target cells
2. The base digits in the cells seen by both pairs of base cells
3. The base digits in the PE cells associated with the partial fish
When there are 4 target cells:
4. The base digits in any cell seen by all 4 target cells
When the partial fish for all the base digits has a line parallel to the JE band that must hold 3 base digits
5. The non-base digits in that line in the partial fish cells.
Now your opening examples have 4 target cells and 2 partial fish lines that must hold 3 base digits and when that happens there will be a cross-line that holds two target cells and two partial fish cells that are all confined to holding the base digits. Therefore the base digits must be false in the other cells in that cross line.
When all the partial fish cells occupy a 3x3 array (as in your opening examples) those eliminations will empty one of the cells leaving 8 cells to hold 8 digits and therefore the non-base digits in the two remaining partial fish cells can also be eliminated.
To illustrate using your 5th puzzle
..............1.23..2.345.....6.......3.27.5..8...37.....9.54....4.7...51....627.;759669;dob;12_12_19;156;1689 ;169
- Code: Select all
*-------------------------*-------------------------*-------------------------*
| 3456789 1345679 156789 | 2578 5689 289 | 1689B 14689 146789 |
| 456789 45679 56789 | 578 5689 <1> | 689B <2> <3> |
> | 6789 1679 <2> | 78 <3> <4> | <5> 1689T 16789 |
*-------------------------*-------------------------*-------------------------*
| 24579 124579 1579 | <6> 14589 89 | 1389 13489 12489 |
> | 469 1469 <3> | 148 <2> <7> | 1689T <5> 14689T |
| 24569 <8> 1569 | 145 1459 <3> | <7> 1469 12469 |
*-------------------------*-------------------------*-------------------------*
| 23678 2367 678 | <9> 18 <5> | <4> 1368 168 B |
> | 23689 2369 <4> | 1238 <7> 28# | 13689 13689T <5> |
| <1> 359 589 | 348 48 <6> | <2> <7> 89 B |
*-------------------------*-------------------------*-------------------------*
689 169 18 8!
The Almost Double JExocet is (1689)[a:r12c7,r5c9,r8c8],[b:r79c9,r3c8,r5c7]
The spoiler is (8)r8c6 which must be false for the pattern to comply with the partial fish requirements.
The partial fish for the 4 digits are confined to r358c124 with the candidates shown in the bottom margin.
For the DJE to be true the partial fish cells must hold 8 base digits and the target cells 4 base digits
The p fish cells r358c12 must therefore hold 6 base digits which will exclude all the non-base ones.
In row 5, c12 must contain 2 base digits and the targets in c79 must contain the other two.
Therefore r5c4 can't contain a base digit and 4 would have be true there for the DJE to be true
This requires r38c4 to hold the two remaining base digits in the p fish cells, ie (8)r3c4 & (1)r8c4.
Now (8)r8c4 = (8)r8c6 represents the cases when the DJE is true and false, and what we want to find is a contradiction caused when one or other of them is true. My analysis of the puzzles 1,2,4 & 5 (No 3 is a mistake I think) leads me to believe that generally this will involve the use of branched chains from the various PEs from the DJE pattern to the spoiler.
This becomes manageable for a human solver by marking all the PEs for the DJE and constructing chains using the alternating premise that either the DJE or the spoiler is true. This leads to a bunch of eliminations common to either case and eventually onto a contradiction. However this tactic is equivalent to using memory chains.
For example in box 7 r7c5,r8c45 <> 8 is easy to prove this way, but the different paths would be a bit of a pig to notate.
Perhaps some of this could be of use to you.
David
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