The New Sudoku Players' Forum

[ronk]

Two groups identically colored (amber) in the same unit (box 3) are a contradiction

In order for this to be a contradiction, I think you need to qualify that a little bit more and state, "Two non-intersecting groups...."

I don't see a problem with the nodes (groups) over-lapping. Referring to 2s candidates of the example of the opening post: If there were a candidate at r3c9, it would be eliminated by the contradiction even if it were a member of two different nodes. IOW r3c9 would just be colored amber twice.

In addition, you might as well have colored the remaining cells in box 6 amber

I had other 2s candidates colored at one point, but decided to focus on the x-cycle loops. After all, it's the loop that allows us to make the deduction that amber must represent the false candidates (and pink the true). Then all other 2s eliminations and placements follow from that deduction.

Ron

[ronk]

Using the original strict definition of conjugate pairs there are no conjugates in box 3, so the "type 4" rule seems to be the only way to exclude candidates in that box.

OK, I see. You wouldn't have even colored any candidates in box 3.

I'm not up on 'conjugate nodes' so I can't really comment there, but it seems a good way to differentiate your technique from conjugate pairs and avoid confusion.

I don't recall 'conjugate nodes' being using anywhere. However, links between nodes (of one or more cells per node) are being treated as strong links in a few threads on this forum ... so I proposed the term here.

Ron

[Jeff]

I'm not up on 'conjugate nodes' so I can't really comment there, but it seems a good way to differentiate your technique from conjugate pairs and avoid confusion.

Hi Angus, Ronk and MJ, The discussion has been very interesting. Allow me to interrupt.

In the forcing chain definition, conjugate node is one type of "strong nodes" that means bivalue cell. It is desirable not to create definitions in colouring that would confuse with those in forcing chains.

In the forcing chain definition, "strong link" is used to express the relationship between 2 cells or 2 nodes within a unit, where node means grouped cells.

"Conjugate pair" is a good collective term for colouring to express the relationship between 2 cells or 2 groups of cells in a unit, where "conjugate grouped pair" may be used to express the latter if a clear distinction between the 2 types is required.

[Myth Jellies]

I don't see a problem with the nodes (groups) over-lapping. Referring to 2s candidates of the example of the opening post: If there were a candidate at r3c9, it would be eliminated by the contradiction even if it were a member of two different nodes. IOW r3c9 would just be colored amber twice.

If you allow r3c9 to be open for a candidate 2, then the starred cells below represent just one way for the grid to be solved without using a single pink cell.

Code: Select all

 2*  .   .   |A2   .   .   | 2   2  A2
 2   .   .   |A2*  .   .   | 2   2  A2
 .   .   .   | .  a2   .   |A2  A2  A2*
-------------+-------------+-----------
 .   2*  .   |a2   .   .   | 2   .   .
 .   2   .   | .   .   .   | 2*  2   .
 .   .   .   | .  A2*  .   | .   .  a2
-------------+-------------+-----------
 .   .   .   | .   .   .   | 2   2*  .
 .   .   2   | .   .   .   | .   .   .
 .   .   .   | .   .   2   | .   .   .


If you allowed intersecting groups in your comparison, then the cell where they intersect can satisfy the truth condition in that unit for that color for both groups. Hence there is no contradiction in that case.

You can only use non-intersecting groups when checking for rule 2.

[Myth Jellies]

Rules 4 and 5 use the following definitions:

Let A and B represent two cells of the same color, X, that do not lie in the same unit.

Let A' represent all of the remaining cells in a unit containing A.
Let B' represent all of the remaining cells in a unit containing B.

Rule 4: If you can find an A' and a B' that lie entirely in the same unit, and that do not share any cells, then color X must be true.

Rule 5: If you can find an A' and a B' that lie entirely in the same unit, and they do share cells, then any cells they do not share must be false.

Heh, pulled some filet-o-colors logic on you for that last one.

[ronk]

In the forcing chain definition, conjugate node is one type of "strong nodes" that means bivalue cell. It is desirable not to create definitions in colouring that would confuse with those in forcing chains.

Reply here.

[ronk]

If you allowed intersecting groups in your comparison, then the cell where they intersect can satisfy the truth condition in that unit for that color for both groups. Hence there is no contradiction in that case.

You can only use non-intersecting groups when checking for rule 2.

I stand corrected and will edit the opening post.

[edit: Another viewpoint, and referring to the example of the opening post:

If a candidate existed at the intersection of row 3 and col 9, then the turbot fish ... the basis for the contradictory eliminations ... "degenerates" into an x-wing, meaning that basis for eliminations has been errr ... eliminated. ]

Thanks for the explanation, Ron

[Jeff]

In the forcing chain definition, conjugate node is one type of "strong nodes" that means bivalue cell. It is desirable not to create definitions in colouring that would confuse with those in forcing chains.

Reply here.

Reply of your reply here.

[Myth Jellies]

Actually, rule 2 would just be a special case of rule 4 where A' and B' are both just a single cell.

[ronk]

Rules 4 and 5 use the following definitions:

I invite you to post examples of "rules 4 and 5". In keeping with the title of this thread, please post the simplest of non-trivial examples, preferrably from actual puzzles.

TIA, Ron

[Myth Jellies]

I invite you to post examples of "rules 4 and 5". In keeping with the title of this thread, please post the simplest of non-trivial examples, preferrably from actual puzzles.

I'll assume that the puzzle you used at the start of this thread is simple enough . One can use it to illustrate rule 4, thusly

Code: Select all

 2   .   .   | 2   .   .   | 2   2   2B'
 2   .   .   | 2   .   .   | 2   2   2B'
 .   .   .   | .   2X  .   | 2A' 2A' .
-------------+-------------+-----------
 .   2   .   | 2   .   .   | 2   .   .
 .   2   .   | .   .   .   | 2   2   .
 .   .   .   | .   2x  .   | .   .   2X
-------------+-------------+-----------
 .   .   .   | .   .   .   | 2   2   .
 .   .   2   | .   .   .   | .   .   .
 .   .   .   | .   .   2   | .   .   .


x and X are your two colors; cell A (r3c5) and cell B (r6c9) are both the same color. There exists sets A' and B' consisting of the remaining cells of groups containing A and B respectively, that are non-intersecting, and that lie entirely within the same group (box 3). Therefore, color X must be true.

Note that if we alter this grid slightly...

Code: Select all

 2   .   .   | 2   .   .   | 2   2   2xB'
 2   .   .   | 2   .   .   | 2   2   . 
 .   .   .   | .   2X  .   | 2xA'.   .
-------------+-------------+-----------
 .   2   .   | 2   .   .   | 2   .   .
 .   2   .   | .   .   .   | 2   2   .
 .   .   .   | .   2x  .   | .   .   2X
-------------+-------------+-----------
 .   .   .   | .   .   .   | 2   2   .
 .   .   2   | .   .   .   | .   .   .
 .   .   .   | .   .   2   | .   .   .


...A' and B' could just take the color x, and color x would be declared false via rule 2. On the other hand, if you alter the grid in a slightly different way...

Code: Select all

 2   .   .   | 2   .   .   | 2   2   2B'
 2   .   .   | 2   .   .   | 2   2   2B'
 .   .   .   | .   2X  .   | 2A' 2A' 2A'B'
-------------+-------------+-----------
 .   2   .   | 2   .   .   | 2   .   .
 .   2   .   | .   .   .   | 2   2   .
 .   .   .   | .   2x  .   | .   .   2X
-------------+-------------+-----------
 .   .   .   | .   .   .   | 2   2   .
 .   .   2   | .   .   .   | .   .   .
 .   .   .   | .   .   2   | .   .   .


...here we satisfy all the requirements for rule 4, except that the sets A' and B' are intersecting. Thus we can use rule 5 and eliminate the candidate 2's from the cells marked with just A' or just B' (but not from the cells marked A'B').

This should segue nicely into your use of groups in coloring. A' and B' just become groups that get the color x, and you can use similar logic and make the same deductions.

Sorry, I don't have a real example for rule 5; however the illustration above should work nicely since it highlights the subtle difference between rules 4 and 5. This is especially true if you just filter on 2's in your diagrams.