Almost Locked Sets

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Almost Locked Sets

..84...3....3.....9....157479...8........7..514.9..827..9.6...26517..4......9..56

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Maybe someone can convert this expression to English and walk me through the reasoning of the elimination. RCB said:
16. Almost Locked Sets [r79c7]=3=[r128c9]
16a. Removed common candidate: 1 from r2c7
I see that [r79c7] and [r128c9] represent the two ALSs, and r2c7, which has candidate 1, can see all instances of possible 1s in both these sets, but I don’t understand the significance of the 3 or what the two cases are that both eliminate 1 from r2c7.
Yogi
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Re: Almost Locked Sets

Code: Select all
`*--------------------------------------------------------------*| 25    1267  8      | 4     257   2569   | 269-1 3    b19     || 245   1267  47     | 3     257   2569   | 269-1 1689 b189    || 9     236   23     | 68    28    1      | 5     7     4      ||--------------------+--------------------+--------------------|| 7     9     25     | 25    14    8      | 136   146   13     || 238   238   236    | 26    14    7      | 19    149   5      || 1     4     56     | 9     35    356    | 8     2     7      ||--------------------+--------------------+--------------------|| 348   378   9      | 158   6     345    |a137   18    2      || 6     5     1      | 7     238   23     | 4     89   b389    || 2348  2378  47     | 18    9     34     |a137   5     6      |*--------------------------------------------------------------*`

Suppose the 1's in r1c79 were both false. Then the other two digits in the first ALS must be True, so r79c7 would both be 37, a locked pair.

Now here is the big trick with ALS chains. It doesn't matter whether the 3 in that ALS is in r7c7 or r9c7, they can see all of the 3's in the second ALS. (In this example there is only one 3, in r8c9, but there are more complex examples).

Since the 3 has been removed from the second ALS all the other three digits in it must be in it somewhere. In particular there must be a 1 in r1c9 or r2c9.

So, what all this shows is that at least one of r79c7 or r12c9 must be 1. Since r2c7 can see all of these 4 cells, it can't be 1. The 1 in r1c7 can also be removed in my PM. Perhaps this had already been removed in your RCB solution.

The more common name for this move is ALS-XZ Rule , X = 3, Z = 1 : (1=37) r7c9 - (3=189) r128c9 => - 1 r12c7. This is often written more simply as ALS-XZ Rule , X = 3, Z = 1 : (1=3) r7c9 - (3=1) r128c9 => - 1 r12c7.

Z is the elimination digit and X, the linking digit, is called the restricted common digit. When 2 ALSs are linked in this way, all the X's in the first ALS must be separate from, but must see all the X's in the 2nd ALS.