StrmCkr wrote:find one eventually, make one quickly plausible: meet all your request probably not...

Thanks, StrmCkr! Now I can at least form some structures out of those. However, you're right that I'd see simpler ways to achieve all the eliminations. Let's see the different ways we can combine those ALSs.

First, I can see two doubly-linked ALS XZs: AC and BC.

Doubly-linked ALS-XZ (AC, x=2,4):

- Code: Select all
`.-----------------------------.---------------.------------------.`

| -2 -4 | | |

| -79 A (279) A (479) | -79 -79 -79 | -79 -79 A (79) |

| -2 -4 | | |

:-----------------------------+---------------+------------------:

| -1 C (12) C (14) | -1 -1 -1 | -1 -1 -1 |

| -1 -12 -14 | | |

| -1 -12 -14 | | |

:-----------------------------+---------------+------------------:

| -2 -4 | | |

| -2 -4 | | |

| -2 -4 | | |

'-----------------------------'---------------'------------------'

(4={1}2)r4c23 - (2{79}=4)r2c293 - loop => -2 c2, -4 c3, -1 r4,b5; -79 r2 (except for AC cells)

Doubly-linked ALX-XZ (BC, x=1,2) (also a Sue de Coq):

- Code: Select all
`.------------------------------.------------------.------------------.`

| -359 | | |

| -359 | | |

| B (359) | | |

:------------------------------+------------------+------------------:

| -124359 C (12) C (14) | -4 -4 -4 | -4 -4 -4 |

| B (1235) -124 -124 | | |

| B (1235) -124 -124 | | |

:------------------------------+------------------+------------------:

| B (359) | | |

| -359 | | |

| -359 | | |

'------------------------------'------------------'------------------'

(2={4}1)r4c23 - (1{359}=2)r3756 - loop -> -124 b4, -124 r4, -359 c1 (except for BC cells) => +4r4c3

I think those are much simpler to see as separate, independent units, and as we can see, solving BC actually breaks A so that's all we need to do (XZ AC is unnecessary). But, let's still look at the ALS XY-Wing possibilities.

ALS XY-Wing(s) (ACB: x=2,4, y=1,2, z=9):

- Code: Select all
`.-----------------------------.------------------.------------------.`

| | | |

| -9 A (279) A (79-4) | | A (79) |

| B (359) | | |

:-----------------------------+------------------+------------------:

| C (12) C (14) | | |

| B (1235) | | |

| B (1235) | | |

:-----------------------------+------------------+------------------:

| B (359) | | |

| | | |

| | | |

'-----------------------------'------------------'------------------'

It's impossible for me to write a single chain that combines all those x,y possibilities, so I do it in two parts:

ACB-1 (x=2, y=1, z=9): (947=2)r2c392 - (2=41)r4c23 - (1=2359)r3567c1 => -9 r2c2

ACB-2 (x=4, y=1,2, z=9): (927=4)r2c293 - (4=12)r4c32 - (1|2=359)r3567c1 -> contradiction => -4 r2c3

As we can see, the latter combination produces a contradiction, which results in the same eliminations as XZ (BC) (which is simpler). Thus, no real need for the XY complication.

However, to me the simplest way to achieve the same result is this smaller Sue de Coq (or doubly linked ALS XZ):

- Code: Select all
`.---------------------------.------------------.------------------.`

| -359 | | |

| -359 | | |

| F (359) | | |

:---------------------------+------------------+------------------:

| -12359 E (12) 4-1 | | |

| D (1235) -12 -12 | | |

| D (1235) -12 -12 | | |

:---------------------------+------------------+------------------:

| F (359) | | |

| -359 | | |

| -359 | | |

'---------------------------'------------------'------------------'

Sue de Coq (D:1235 + E:12 + F:359):

(1=2) - (2={359}1)r3567c1 - loop => -12 b4, -359 c1 (except SDC cells themselves)

=> 4r4c3 => 79r2c39 => 2r2c2 => 1r2c2 => 2359r3567c1

For that we don't need ALS A at all and only a part of C.