I've come up with a solution which shows a similarity to exocet+sk-loop puzzles.
It follows the same structure as champagne's solution, but doesn't use uniqueness to determine the base pair.
First, I will use a more sk-loop-shaped rank0 pattern (with the same elims as the MSLS):
- Code: Select all
4679 4679 \679
,---------------------,------------------------,-----------------------,
| 35678 34589 *4679–3|*4679 5689 4578 |#679 1 2 |
| 15678 14589 *4679 |*4679–12 125689 124578 |#679 56789 3 |
| 15678 1589 2 | 3 15689 1578 | 4 56789 6789 |
:---------------------+------------------------+-----------------------:
| 237 2349 1 | 8 236 234 | 23679 *4679–23 5 | 4679
| 235 6 349 | 124 7 1235–4 | 8 2349 149 | \479b4
| 2358–7 2358–4 347 | 1246 1235–6 9 | 123–67 23467 1467 | \46b5
:---------------------+------------------------+-----------------------:
| 1236 123 8 | 5 1239 1237 | 123679 *4679–23*4679–1| 4679
| 9 123 36 | 127 4 1238–7 | 5 23678 1678 | \6b7
| 4 7 5 | 129 1238–9 6 | 123–9 2389 189 | \79b8
'---------------------'------------------------'-----------------------'
I like to think about it in terms of of fitting 4679 into r47:
There can be two of 4679 in r47c12, since they are forced into r12c3,
there can be two of 4679 in r47c56, since they are forced into r12c4,
there can be one of 4679 in r47c7 and three in r47c89.
Combined, that is eight ways to place them, so we need all of them.
That is, two in the left stack, two in the middle stack, and four in the right stack.
An important feature of the exocet+sk-loop combo is a so-called PLQ (pattern-locked quad).
Here, we can get something similar:
- Code: Select all
,--------------------,----------------------,---------------------,
| 35678 34589 4679 | 4679 5689 4578 |B679 1 2 |
| 15678 14589 4679 | 4679 125689 124578 |B679 56789 3 |
| 15678 1589 2 | 3 15689 1578 | 4 56789 6789 |
:--------------------+----------------------+---------------------:
| 237 2349 1 | 8 236 234 | 23679 T679–4 5 |
| 235 6 349 | 124 7 1235 | 8 2349 149 |
| 2358 2358 347 | 1246 1235 9 | 123 23467 1467 |
:--------------------+----------------------+---------------------:
| 1236 123 8 | 5 1239 1237 | 123679 T4679 T4679 |
| 9 123 36 | 127 4 1238 | 5 23678 1678 |
| 4 7 5 | 129 1238 6 | 123 2389 189 |
'--------------------'----------------------'---------------------'
The base digits appear once each in r4c8+r7c89, together with 4r7.
The non-base 679 is then the digit which takes c7.
Each of 4679 appears once in r47c789 (but we don't know exactly which cell in c7 is part of the quad).
That means that each of 4679 appears exactly once in r47c1256.
Another key feature of a variety of sk-loops is the STP (single-truth property).
Again, it is in some form present here, too:
- Code: Select all
,--------------------,----------------------,---------------------,
| 35678 34589 4679 | 4679 5689 4578 | 679 1 2 |
| 15678 14589 4679 | 4679 125689 124578 | 679 56789 3 |
| 15678 1589 2 | 3 15689 1578 | 4 56789 6789 |
:--------------------+----------------------+---------------------:
|a237 a2349 1 | 8 b236 234 | 23679 c679 5 |
| 235 6 349 | 124 7 1235 | 8 2349 149 |
| 2358 2358 347 | 1246 1235 9 | 123 23467 1467 |
:--------------------+----------------------+---------------------:
|B1236 123 8 | 5 A1239 A1237 | 123679 C4679 C4679 |
| 9 123 36 | 127 4 1238 | 5 23678 1678 |
| 4 7 5 | 129 1238 6 | 123 2389 189 |
'--------------------'----------------------'---------------------'
Suppose 79r4c12. The left stack is full, so 6 must take the middle stack, so 6r4c5.
But then r4c8 has no value left, i.e. contra.
Suppose 79r7c56 instead. The middle stack is full, so 6 must take the left stack, so 6r7c1.
But then r7c89 have only 4 left, i.e. contra.
Therefore, 79 appear in different stacks. By complement, 46 also appear in different stacks.
This gives us three options for the left stack: 47, 67, and 69.
In the 47 case, we get 69 in the middle stack. Filling in 4679b4578 and r4c8 gets us here:
- Code: Select all
,----------------,-------------------,--------------------,
| 3568 3589 47 | 469 58 4578 | 679 1 2 |
| 1568 1589 47 | 469 1258 124578 | 679 5678 3 |
| 1568 1589 2 | 3 158 1578 | 4 5678 6789 |
:----------------+-------------------+--------------------:
| 7 4 1 | 8 6 *23 |*23 9 5 |
| 235 6 9 | 2–14 7 1235 | 8 4–23 14 |
| 2358 2358 3 | 124 1235 9 | 23–1 23467 1467 |
:----------------+-------------------+--------------------:
| 123 123 8 | 5 9 *123 | 12367 467 467 |
| 9 123 6 | 7 4 *1238 | 5 238 18 |
| 4 7 5 |*–12 *1238 6 | 1–23 238 189 |
'----------------'-------------------'--------------------'
23r4b8\c56r9 => –23r9c7
r46c7=23, r5c89=14, r5c4=2, no value left in r9c4
Therefore, we are left with 67 and 69.
We can fill in 6r7c1, 6r6c4, 4r4c6, 3r8c3. 12r78c2 eliminate other 12c2.
This reduces the skfr to 9.0 (7.3 if we use uniqueness to obtain the base pair).
The 67 case takes a bit longer to break.
singles 4679b4578, r4c2, r4c5, r5c4, r9c4, r5c9, r7c9, r6c9, r7c8, 7b6
- Code: Select all
,---------------,-----------------,----------------,
| 358 4589 67 | 49 568 578 | 679 1 2 |
| 158 4589 67 | 49 1568 2–1578| 679 5689 3 |
| 158 589 2 | 3 1568 7–158 | 4 5689 689 |
:---------------+-----------------+----------------:
| 7 3 1 | 8 2 4 | 69 69 5 |
| 25 6 9 | 1 7 *35 | 8 *23 4 |
| 258 58 4 | 6 *35 9 |*23 7 1 |
:---------------+-----------------+----------------:
| 6 12 8 | 5 9 *13 |*123 4 7 |
| 9 12 3 | 7 4 18 | 5 268 68 |
| 4 7 5 | 2 *38–1 6 | 1–3 *389 89 |
'---------------'-----------------'----------------'
singles 7r3, 2c6; 1c6\b8; single 1r9
3b5689 forms an impossible pattern
We are left with the 69 case.
9r4c2, 9r9c4, 7r7c6, 7r6c3, stte
