Hi ronk,
I think we may have reached a standoff?
I'd like to say more on the matter after all. Please let me return to your
original objection:
Good dig, your persistence has paid off. However, after obtaining the BUG grid,
I thought the objective was to eliminate a candidate in a BUG cell. If that's
not a requirement of the BUG principle, there is at least a certain "elegance"
in doing so.
(By "BUG cell" I'm pretty sure you mean one of the cells that
contains "extra candidates.")
My question is: why would there be such a restriction? If I can reason
deductively (using the BUG principle) for the removal of a particular
candidate, then I claim that is a valid and elegant move, no matter
where that candidate lies.
Earlier, I wrote this a little differently:
But I don't quite see why it might be preferable to remove a candidate from one
of the cells containing extra candidates, over removing a candidate from one of
the bivalue cells.
And you responded:
I'm just taking Jeff's Corollary 2 very literally.
Jeff wrote:Corollary 2: Any exclusion implied by all non-BUG candidates in the grid must
be valid.
But this corollary does not seem to indicate the preference described in my
quote above. I interpret "any exclusion" to mean the removal of any
candidate, anywhere in the grid.
In summary, I claim that the following argument from my earlier post is
not only correct (I think we agree on that already), but also
has good style:
- Code: Select all
*--------------------------------------------------*
| 69+5 8 45 | 46 1 59 | 2 3 7 |
| 7 46 2 | 3 48+6 68 | 5 1 9 |
| 59 1 3 | 57 79 2 | 6 8 4 |
|----------------+----------------+----------------|
| 1 2 7 | 45 34 35 | 9 6 8 |
| 48 5 6 | 9 78 1 | 47 2 3 |
| 48 3 9 | 67 2 68+7 | 47 5 1 |
|----------------+----------------+----------------|
| 2 9 1 | 8 6+7 7+6 | 3 4 5 |
| 3 7 8 | 2 5 4 | 1 9 6 |
| 56 46 45 | 1 39 39 | 8 7 2 |
*--------------------------------------------------*
By the BUG principle, at least one of the "extra candidates"
must be the correct final value for the related cell.
This implies that r2c6=8, as shown below:
r1c1=5 => r1c3=4 => r1c4=6 => r2c6=8
r2c5=6 => r2c6=8
r6c6=7 => r7c6=6 => r2c6=8
r7c5=7 => r7c6=6 => r2c6=8
r7c6=6 => r2c6=8