The New Sudoku Players' Forum

[MrHamilton]

This is Epstein's puzzle from my thread on traveling pairs.
I may have set out to prove one theory and hit upon another.

Code: Select all

 
 *-----------------------------------------------------------*
 | 5     6     279   | 47    39    1     | 249   349   8     |
 | 2479  248   1279  | 5     39    478   | 6     1349  1249  |
 | 349   348   139   | 48    6     2     | 5     7     149   |
 |-------------------+-------------------+-------------------|
 | 367   9     37    | 2     47    5     | 1     8     467   |
 | 27    25    4     | 68    1     68    | 3     59    579   |
 | 16    15    8     | 3     47    9     | 47    2     56    |
 |-------------------+-------------------+-------------------|
 | 1234  7     6     | 9     8     34    | 24    145   1245  |
 | 1349  134   5     | 467   2     3467  | 8     1469  1479  |
 | 8     24    29    | 1     5     467   | 2479  469   3     |
 *-----------------------------------------------------------*

Theorem: In a unit (row, column, or box) consisting of double instances (i.e. no more than two occurances per box) of the various candidates except for a single cell which has an extra (third) instance of a candidate, that candidate can be excluded from that cell -- except when the candidate is also "extra" in 2 intersecting units! e.g. the same digit is a third instance simultaneously in a box, row, and column: then it is FORCED to be that value, by the BUG rules.)

IF this is valid we may proceed as follows:
exclude 7 from r4c1 (box 4)
exclude 4 from r2c6 (box 2)
exclude 7 from r4c9 (row four)
exclude 5 from r5c9 (row five)
exclude 4 from r8c4 (column four)

and the rest is solved by 1 instance of multi-colors followed by 48 instances of naked or hidden singles!

Comments, refutations?

Havard's excellent post on strong links may have been in the back of my mind here.

[Ruud]

Please try your technique on this grid:

Code: Select all

4     78    6      | 3     9     5      | 78    2     1     
37    9     25     | 8     1     26     | 57    36    4     
38    1     25     | 7     4     26     | 58    9     36   
-------------------+--------------------+-------------------
2     36    14     | 14    5     8      | 36    7     9     
16    78    378    | 16    2     9      | 4     5     38   
9     5     48     | 46    3     7      | 2     1     68   
-------------------+--------------------+-------------------
78    2     78     | 9     6     3      | 1     4     5     
5     4     9      | 2     8     1      | 36    36    7     
16    36    13     | 5     7     4      | 9     8     2     

Ruud.

PS. It can also be solved by the BUG technique, which is somewhat opposite to yours.

Ruud.

[MrHamilton]

Aaaargh

[MrHamilton]

And this one

Code: Select all

 *-----------*
 |298|...|31.|
 |745|213|698|
 |136|..9|27.|
 |---+---+---|
 |514|638|729|
 |629|...|843|
 |873|942|156|
 |---+---+---|
 |452|...|937|
 |381|597|462|
 |967|324|581|
 *-----------*

 
 *--------------------------------------------------*
 | 2    9    8    | 47   567  56   | 3    1    45   |
 | 7    4    5    | 2    1    3    | 6    9    8    |
 | 1    3    6    | 48   58   9    | 2    7    45   |
 |----------------+----------------+----------------|
 | 5    1    4    | 6    3    8    | 7    2    9    |
 | 6    2    9    | 17   57   15   | 8    4    3    |
 | 8    7    3    | 9    4    2    | 1    5    6    |
 |----------------+----------------+----------------|
 | 4    5    2    | 18   68   16   | 9    3    7    |
 | 3    8    1    | 5    9    7    | 4    6    2    |
 | 9    6    7    | 3    2    4    | 5    8    1    |
 *--------------------------------------------------*


Hmm, when the phenomenon occurs three different ways in a cell at an intersection of box, row, and column, you have to MAKE it 5 in c5r1!
Instead of excluding 5 that is.
Have we finally sorted it out then?

[Jeff]

Theorem: In a unit (row, column, or block) consisting of double instances of the various candidates EXCEPT for a single cell which has an extra instance of a candidate, that candidate can be excluded from that cell.

Hi MrHamilton, Don't quite understand yet. Correct me if I am wrong. I am just trying to make sure my interpretation is correct by equating terms which are commonly used in this forum.

block <=> box
double instances of the various candidates <=> bivalue cells
single cell which has an extra instance of a candidate <=> trivalue cell
c1r4 <=> r4c1 (you are the only one using cxrx notation, I am afraid)
left stack middle block <=> box 4
middle stack top block <=> box 2

Could you be kind enough to adopt the common terms making it easier for us to go through your posts?

(Edit: exception occurs when the candidate is also "extra" in 2 intersecting units! e.g. the same digit is "extra" simultaneously in a box, row, and column: then it is FORCED to be that value, by the BUG rules.)

This sounds like a BUG-Lite+1 case, except that the extra candidate could be in 3 intersecting units.

[MrHamilton]

block <=> box

Yes.

double instances of the various candidates <=> bivalue cells

Not precisely what I meant. A bivalue cell has two (any two) values. What I mean is a twice-occurring candidate. Sometimes, it amounts to the same thing; sometimes not. A box may have 3 instances of a certain candidate and still consist only of bivalue cells. An instance may occur once in a bivalue cell and once in a trivalue cell. The point is that, in my example, each candidate appears exactly twice in the box except the one I would like to exclude.

single cell which has an extra instance of a candidate <=> trivalue cell

See above. Again, not just any trivalue cell, but one that has a third instance of a candidate. Obviously if it has the only instance of a candidate, it is a hidden single and cannot be excluded.

c1r4 <=> r4c1 (you are the only one using cxrx notation, I am afraid)

Sorry about that. I will try to remember in future.

left stack middle block <=> box 4

Yes.

middle stack top block <=> box 2

Yes.

Could you be kind enough to adopt the common terms making it easier for us to go through your posts?

I will edit the original post.

[Jeff]

Theorem: In a unit (row, column, or box) consisting of double instances (i.e. no more than two occurances per box) of the various candidates except for a single cell which has an extra (third) instance of a candidate, that candidate can be excluded from that cell -- except when the candidate is also "extra" in 2 intersecting units! e.g. the same digit is a third instance simultaneously in a box, row, and column: then it is FORCED to be that value, by the BUG rules.)

Thanks MrHamilton for adjusting the description.

With the following grid, I applied the thorem to col 7 and tried to conclude that r2c7<>6, but actually r2c7=6. The extra 6 doesn't share with another unit either.

Code: Select all

 *-----------------------------------------------------------*
 | 2     4     368   | 9     36    1     | 7     68    5     |
 | 5     1368  9     | 4     36    7     | 126   268   136   |
 | 13    7     136   | 8     2     5     | 9     4     136   |
 |-------------------+-------------------+-------------------|
 | 4     268   268   | 5     1     3     | 26    9     7     |
 | 7     5     26    | 26    4     9     | 3     1     8     |
 | 9     13    13    | 7     8     26    | 5     26    4     |
 |-------------------+-------------------+-------------------|
 | 8     123   4     | 236   5     26    | 16    7     9     |
 | 13    123   7     | 236   9     8     | 4     5     16    |
 | 6     9     5     | 1     7     4     | 8     3     2     |
 *-----------------------------------------------------------*

[MrHamilton]

That's quite right...I could try to see why it's an exception; and by the time I'm done listing the exceptions there may be very little left of the rule

The rule as stated originally, and as clarified later, is wrong then.
It works nicely in Eppstein but for other reasons; or for no particular reason.
It's back to the drawing board.