I normally have a hard time finding ALS-XZ
Just as every naked set has an equivalent hidden set that makes the same deduction, it is also the case that every ALS will have an equivalent AHS that makes the same deduction. A relatively large ALS implies a relatively small AHS which may be easier to see, especially if you use coloring.
- Code: Select all
|-----------------+-----------------+-----------------|
| 389 7 4 | 1 2359 29 | 3589 2389 6 |
| 1 5 389 | 2389 7 6 | 389 4 239 |
| 389 2 6 | 389 359 4 | 1 7 359 |
|-----------------+-----------------+-----------------|
| 579 6 79 | 259 8 129 | 3579 129 4 |
| 5789 4 789 | 6 129 3 | 579 129 259 |
| 2 3 1 | 59 4 7 | 59 6 8 |
|-----------------+-----------------+-----------------|
| 34 18 5 | 349 6 19 | 2 389 7 |
|34a7b 18' 237B |2-34A9 1239' 5 | 6 389' 39' |
| 6 9 23 | 7 23* 8 | 4 5 1 |
|-----------------+-----------------+-----------------|
The AIC representation for your ALS deduction is
(3&1&8&9=2)r8c2589 - (2=3)r9c5 => r8c4 <> 3
From the hidden point of view, not only are there fewer digits in this case, but you have your coloring to aid you. It is simple from coloring to note that 4's and 7's are locked in r8c134. One other digit that is almost locked in those cells is 2. Thus we have...
(4&7&2)r8c134 = (2)r8c5 - (2=3)r9c5 => r8c4 <> 3
In this case one can break the hidden triple down into a greater amount of smaller hidden sets like...
(4&7)r8c14 = (7-2)r8c3 = (2)r8c45 - (2=3)r9c5 => r8c4 <> 3
...or using the locked sevens AH-pair instead of the locked fours AH-pair...
(4)r8c4 = (4&7-7&2)r8c13 = (2)r8c45 - (2=3)r9c5 => r8c4 <> 3
...or even three almost hidden singles...
(4)r8c4 = (4-7)r8c1 = (7-2)r8c3 = (2)r8c45 - (2=3)r9c5 => r8c4 <> 3
