Have you ever seen a naked N-tuple have eliminations different from its complementary hidden M-tuple
Have you ever seen a basic row/column N-fish have eliminations different from its complementary M-fish
AFAIK the answer to both is "no". Therefore it is reasonable to believe a doubly-linked ALS-xz should have the same eliminations as its hidden set complement.
hobiwan wrote:- Code: Select all
.--------------------------------.-------------------------.--------------------.
| -1346-7-89 -1346-7-8 6-7-89 | 246-7 146-78 124-7 | 158 157 1578 |
| 2 B1678 B678 | A67 15-6-78 15-7 | 3 4 9 |
| 5 B1478 B78 | A47 3 9 | 2 6 178 |
:--------------------------------+-------------------------+--------------------:
| 13678 9 25678 | 2346-7 467 2347 | 156 12357 12357 |
| 367 23567 4 | 1 67 237 | 9 8 2357 |
| 1367 -12367 267 | 5 9 8 | 16 1237 4 |
:--------------------------------+-------------------------+--------------------:
| 479 457 1 | 8 2 3457 | 45 359 6 |
| 4689 24568 25689 | 349 145 1345 | 7 12359 12358 |
| 4789 24578 3 | 4-79 1457 6 | 1458 1259 1258 |
'--------------------------------'-------------------------'--------------------'
Almost Locked Set XZ-Rule: A=r23c4 {467}, B=r2c23,r3c23 {14678}, X=4,6, Z=6 => r1c12,r6c2<>1, r2c5<>6, r1c123456,r2c56,r49c4<>7, r1c123<>8
ALSs r23c23 and r23c4 have complementary AHSs 3469b1 and 124568b2, respectively. [2 edits: Other complementary AHSs exist as well.]
There are 14 eliminations for this complementary doubly-linked AHS-xz. Since eliminations r6c2<>1 and r49c4<>7 do not exist, I consider these eliminations an "overstatement" for the original ALS-xz.
Of course, no one is going to look at both the ALS-xz and complementary AHS-xz to determine the common eliminations. However, I do expect posters and programmers who claim to be using the ALS-xz to use N cover sets for N base sets, not N+1 cover sets, not N+2 cover sets, etc. (stated as an over-simplification of superposition.)
So what exactly are the consequences of this? For hobiwan's ALS xz above, I would probably select 1b1 and 7b2 to be cover set members, because they have more eliminations than cover sets 1c2 and 7c4, respectively. This merely leaves 1b1\c2 and 7b2\c4 as eliminations for the simpler locked candidates technique.