This is a continuation of work started by Carcul, Bennys, Jeff, Ronk, and others. It turns out that Alternating Inference Chains (AICs) and Almost Locked Sets (ALS) work together in a harmonious and intuitive fashion (IMO). Most ALS can be described as a candidate premise which looks something like...
- Sa = S(b & c & ...) -
...Here, the '=' represents a strong inference, and the '-' represent a weak inference. The 'S' denotes the set of cells which make up an ALS. If digit 'a' is false in the ALS cells, then all of the remaining digits must be true in S. Also, if not all of the non-'a' digits are true in the ALS, then 'a' must be true. The relationship is actually a conjugate one. One could also form a weak link between 'a' and the remainder of the digits in S, but that does not tend to be as useful. Here is a real life example...
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*--------------------------------------------------------------------*
| 2379 6 49 | 5 28 48 | 3789 1 389 |
| 259 179 159 | 3 1268 168 | 589 79 4 |
| 35 14 8 | 7 9 14 | 2 356 356 |
|----------------------+----------------------+----------------------|
| 4 A139 7 | 2 1568 -1368 | 359 3569 3569 |
|A69 5 A169 | 146 7 B36 | 349 8 2 |
| 8 23 26 | 46 56 9 | 1 3456 7 |
|----------------------+----------------------+----------------------|
| 59 79 3 | 18 4 2 | 6 579 18 |
| 1 249 24569 | 689 36 7 | 34589 3459 3589 |
| 679 8 469 | 169 136 5 | 3479 2 139 |
*--------------------------------------------------------------------*
A3=A(1&6&9)-B6=B3; => all cells which see all A3's and B3 cannot be 3
If you look at the weak inference between the sixes, you see that we can form one because the six in B can see all of the sixes in A. In general, this is the rule. If you want to form a weak link between an outside candidate and an ALS candidate, the outside candidate needs to "see" every instance of that ALS candidate in the ALS. If the cells marked with 'A' are your ALS cells...
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|----------------------+----------------------+----------------------|
| 59 79 3 | 18 4 2 | 6 579 18 |
| 1 249 24569 |A689 A36 7 |A34589 A3459 A3589 |
| 679 8 469 | 169 136 5 | 3479 2 139 |
*--------------------------------------------------------------------*
...then you can obviously form weak links with all the ALS digits (345689) along the middle row. You can also form a weak link with 6's in the middle box, and weak links with 4 or 5 in the right box.
Exceptions to the general weak link rule: ALS Subsets in AIC's
Now, back in the day, when I used to play with the Pattern Overlay Method, I found that I got a huge number of good deduction equations when I had a series of cells in a group which looked something like this....
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| ab bc bcd | bcde bcdef . | . . . |
...Until recently, I hadn't ever really noted another advanced method that could make the same reductions that I could with POM on this structure. Then I combined AICs with ALSs and noted the following...
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| ab' bc' bcd' | bcde' bcdef' . | . . . |
AICs:
a=b' => a and/or b' is true
a=b'-b=c' => a and/or c' is true
a=b'-(b&c)=d' => a and/or d' is true
a=b'-(b&c&d)=e' => a and/or e' is true
a=b'-(b&c&d&e)=f' => a and/or f' is true
Take these inferences altogether, and you can form and use the following strong inference...
a=(b'&c'&d'&e'&f')
Now, I don't need to limit my weak links to the ALS via digit d to that row. I can use any d which "sees" d' as a weak link to the ALS. So, if I have the following...
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| ab' bc' bcd' | bcde' bcdef' . | . . . |
|----------------------+----------------------+----------------------|
| . . dg | . fg . | . . . |
I then have the AIC...
a=(b'&c'&d'&e'&f')-d=g-g=f-(b'&c'&d'&e'&f')=a
...which proves 'a' is true. Or if you'd rather, you have...
d'-d=g-g=f-f'
...and since the d and/or f is true, then the premise (d'&f') must be false, which also proves 'a' is true.
Another possibility might be...
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| ab' bc' bcd' | bcde' bcdef' . | . . . |
|----------------------+----------------------+----------------------|
| aX . . | ae . . | . . . |
...which has the AIC: a=(b'&c'&d'&e'&f')-e=a, eliminating the 'a' in the aX-cell.
Exceptions to the general weak link rule: ALS Subsets in AIC's
I mentioned ALS subsets above, but haven't defined them. ALS Subsets are a shortcut to seeing the combined strong inferences without forming all of the AIC's and then combining them together. You form a subset by eliminating one digit from the ALS candidates and then see if you have any naked singles or sets that you can isolate. You can then remove those digits and continue until there are no more revealed singles or sets. For example, if you are coming into the following ALS via some weak candidate 'a' link...
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| abc bc abcg | bcdef bcdef . | abceg . . |
...you can take away the 'a's from your ALS, leaving...
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| bc bc bcg | bcdef bcdef . | bceg . . |
...the naked bc pair eliminates all other 'b's and 'c's...
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| bc bc g | def def . | eg . . |
...the single 'g' and then the single 'e', and we are left with the relevent digits in this ALS...
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| bc bc g | df df . | e . . |
...putting primes on the relevent digits, we finally have...
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| ab'c' b'c' abcg' | bcd'ef' bcd'ef' . | abce'g . . |
...thus, the AIC can continue ...a-a=(b'&c'&d'&e'&f'&g'), and the primed candidates are the only ones we have to consider in our further weak links with the ALS.
Don't have any real world examples of this yet, but I will keep my eyes open.
