Most puzzles have an abundance of Almost Locked Sets to consider. Many of these ALS's share overlapped cells. It turns out that you can consider two overlapping ALS's as one large structure, and the digits inside this structure have a potentially useful interrelationship.

You form a Combined Overlapping ALS structure by taking the union of 2 different ALS's that share at least one cell.

The useful relationship is as follows:

**Combined Overlapping ALS Rule**

For the cells that make up the entire combined ALS structure, either all the digits found in the overlap region are found in the structure, or all the digits not found in the overlap region are found in the structure, or both.

Lets show how you can use this information, first with a hypothetical example.

Code: Select all

```
*-------------------------------------------------------------------*
| . . . | . C12347 . | . D67 . |
| . . . | A125 . . | . . . |
| . . . | A125 . . | . . . |
+----------------------+------------------------+--------------------+
| . . . | . B345 . | . . . |
| . . . | . B345 . | . . . |
| . . . |AB56 . . | . -6* . |
+----------------------+------------------------+-------------------+
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
*--------------------------------------------------------------------*
```

Overlapping ALS B: r6c4|r45c5 containing digits 3456.

Combined ALS: r236c4|r45c5 (union of ALS A and B cells)

Overlap Cells: r6c4 (intersection of ALS A and B cells)

Digits in Overlap Cells: 5&6

Digits not in Overlap Cells: 1&2&3&4

CoALS relationship: ((5&6) = (1&2&3&4))r236c4|r45c5 a strong internal link.

Thus you can form this AIC...

((5&6) = (1&2&3&4))r236c4|r45c5 - ((1v2v3v4) = 7)r1c5 - (7=6)r1c8 => r6c8 <> 6

Here are some real world examples. First a simple one:

Code: Select all

```
+-------------------+-----------------+---------------------+
| 5 3468 2 | 367 78 9 | 146 347 1346 |
| -3678 3469 3679 | 2367 1 368 | 24679 5 23469 |
| 1 369 3679 | 4 25 356 | 8 2379 369 |
+-------------------+-----------------+---------------------+
| 36AB 356 8 | 1 259 4 | 69B 29 7 |
| 9 7 4 | 23 6 35 | 12 8 125 |
| 2 1 56 | 8 59 7 | 3 49 4569 |
+-------------------+-----------------+---------------------+
| 367A 3569 1 | 569 4 2 | 79C 379 8 |
| 78 259 5679 | 5679 3 68 | 2479 1 249 |
| 4 2389 379 | 79 78 1 | 5 6 239 |
+-------------------+-----------------+---------------------+
```

Overlap digits are 3&6, non-overlap digits are 7&9

Thus have ((3&6) = (7&9))r4c17|r7c1

One can either generate an AIC using r7c7 to show r4c17|r7c1 contains 3&6, or one can just observe that since all the 7s and 9s in the combined ALS see the same 79 cell in r7c7, the (7&9) premise must be false and the combined ALS must contain 3&6. Either way you come up with it, any candidate three seeing all the threes in the combined ALS can be removed.

Code: Select all

```
*---------------------------------------------------------------------------*
| 356 4 1356 | 689 1569 1689 | 578 2 567 |
| 9 578 567 | 2468 2456 68 | 1 4568 3 |
| 56 18 2 | 3 14 7 | 9 48 56 |
|--------------------------+-------------------------+----------------------|
| 23567 23579 3567 | 1 23689 4 | 23578 3568 2567 |
| 8 123579 13567 | 269 2369 369 | 2357 1356 4 |
| 2346 123 1346 | 7 2368 5 | 238 9 126 |
|--------------------------+-------------------------+----------------------|
|AB347 AB37 8 | 5 -13479 2 | 6 A13 A19 |
| 1 -2357 9 | 46 3467 36 | 2345 35 8 |
| B2345 6 B345 | 489 1349 1389 | 2345 7 C1259 |
*---------------------------------------------------------------------------*
```

Overlap digits are 3&4&7, non-overlap digits are 1&2&5&9

Thus have ((3&4&7) = (1&2&5&9))r7c1289|r9c13

Since all the 1&2&5&9 candidates in the combined ALS see the r9c9 1259 cell, that cannot be true, therefore r7c1289|r9c13 contains 3&4&7.

We can remove the sevens from r7c5 and r8c2 because they see all the sevens in the combined ALS.

Code: Select all

```
+-----------------+-------------------+--------------------+
| 4 78 39 | 368 1 3689 | 23679 2378 5 |
| 6 15 15 | 2 389 7 | 39 38 4 |
| 2 78 39 | 3468 3489 5 | 1 378 67 |
+-----------------+-------------------+--------------------+
| 357 4 18 | 9 2378A 238AB| 357 6 127 |
| 37 16- 2 | 1367 5 36AB| 8 4 9 |
| 357 9 168 | 16 278A 4 | 2357 12357 127 |
+-----------------+-------------------+--------------------+
| 9 2356 7 | 348 2348 238B | 2456 125 126 |
| 8 26D 46 | 5 279C 1 | 2467 279 3 |
| 1 235 45 | 347 6 239B | 2457 2579 8 |
+-----------------+-------------------+--------------------+
```

Overlap digits are 2&3&6&8, non-overlap digits are 7&9

Thus have

((2&3&6&8) = (7&9))r46c5|r4579c6 - ((7or9) = 2)r8c5 - (2=6)r8c2 => r5c6 <> 6

**Proof of the Combined Overlapping ALS (CoALS) rule**

Assume two overlapping ALS's.

A digit found in the overlap must be part of both ALS's

If an overlap digit is not present in the combined ALS, then all the remaining digits in both ALS's must be true. This means that all non-overlap digits must be present.

If a non-overlap digit is not present in the combined ALS, then for at least one ALS, all the remaining digits must be present. This has to include all overlap digits, therefore all overlap digits must be present.

Therefore we can state that either all the overlap digits can be found in the combined overlapping ALS, or all the non-overlap digits can be found in the combined overlapping ALS.

This translates directly into a strong link between the ANDed overlap digits and the ANDed non-overlap digits over the entirety of the combined ALS cells.

Thanks to ronk and Mike Barker for discussions related to overlapping ALS which helped uncover this relationship.