Post the puzzle or solving technique that's causing you trouble and someone will help

If one of the ALS is a bi-value cell, then the other ALS cannot have a cell in the same box of the bi-value cell, is that correct?

If none of the ALS is a bi-value cell, and they both are in a row (or both in a column), can both these rows (or columns) appear in the same band (or stack)?
Marc
~~~<><~~~<><~~~<><~~~<><~~~
The Lunatic is on the grass...
...under the sun everything is in tune,
but the sun is eclipsed by the moon...

Lunatic
2015 Supporter

Posts: 16
Joined: 19 April 2008
Location: Ghent - Belgium

Hi Marc,

I think that there is is a definition(s) misunderstanding here.

A locked set should have x values in x cells. The word "PEER" is being used more frequently to describe a cell that can "SEE" another cell. When you use the correct definitions in their general form. Then the use of the terms "HOUSE", "ROW", "COLUMN","BOX" or "LINE" will be trivial. it will also allow easy description and transition to sudoku variants.

An almost locked set (ALS) falls short of fulfilling the "x values in x cells" ... an example would be X+1 values in X cells

The ALS-xz rule is something different. It is a technique the takes advantage of the interaction of more than one ALS allowing eliminations to happen.

Therefore with regards to your 1st question the answer is: No not correct. A cell from the 2nd ALS can share the same Box as the 1st ALS
Code: Select all
.  .  -z
zx .  -z
.  .  cx

.  .  bz
.  .  .
.  .  .

.  .  .
.  .  .
.  .  cz

For your 2nd question (& I'm sensing a programming style question here) the answer is: I dont have an example but I can't see why it should be a problem!!

tarek

Posts: 2699
Joined: 05 January 2006