The New Sudoku Players' Forum

[Pep]

Example 1 with 6 cells:

Code: Select all

+---------------+---------------+---------------+
|  .    .    .  |  .    .    .  |  .    .    .  |
|  .    .    .  |  .    .    .  |  .    .    .  |
|  .    .    .  | ab    .    .  |  .    .    .  |
+---------------+---------------+---------------+
|  .    .    cd | abcd cdef  .  |  *c   *d   .  |
|  .    .    .  | abef  ef   .  |  .    .    .  |
|  .    .    .  |  .    .   *ef |  .    .    .  |
+---------------+---------------+---------------+
|  .    .    .  |  *a   .    .  |  .    .    .  |
|  .    .    .  |  *b   .    .  |  .    .    .  |
|  .    .    .  |  .    .    .  |  .    .    .  |
+---------------+---------------+---------------+

(ab) are restricted to c4, (cd) to r4, and (ef) to box5, but the six values are in the 6 cells. The starred candidates can be eliminated.


If 'a' and 'b' are candidates in r3c4, they must necessarily also be candidates elsewhere in row 3 and elsewhere in box 2. Therefore, by "(ab) are restricted to c4", you can't possibly mean 'a' and 'b' are candidates only in col 4 of the entire grid.

What I was trying to say was this: The set of 6 cells that are under consideration are {r3c4, r4c345, r5c45}, and within those 6 cells, (ab) occur only in c4. Of course, they may occur elsewhere in the full diagram. The word restricted was too strong.

[ChrisT]

Edit: Post removed, as it doesn't advance the discussion and my question has been answered. Thanks Carcul!

Chris

[Carcul]

Hi ChrisT.

Your reasoning is completely correct, the "2" can be eliminated from box 3 and from row 2. Ronk have already used this argument in a reply in this thread.

Regards, Carcul

[Pep]

A couple of days ago, it suddenly hit me that this thread is about native disjoint sets. We can of course define hidden disjoint sets with the obvious changes to the scheme I gave earlier. Then when there are N values, V, found in N cells, C, all occurrences of a value v within V are found only within the same unit, and nowhere else in that unit, then those N values form a hidden disjoint set. All other values, not in V, can be removed from C.

I really must find time to implement this and see what it does. My brain at the moment is a long way ahead of my fingers.

Pep

[ronk]

Example 2 with six cells:

Code: Select all

+---------------+---------------+---------------+
|  .    .    .  |  .    .    .  |  .    .    .  |
|  .    .    .  |  *4   .   *26 |  .    .    .  |
|  .    .    .  |  .    .    .  |  .    .    .  |
+---------------+---------------+---------------+
|  .    .    .  |  15   .   126 |  *1   .    .  |
|  .    .    .  |  45   .    56 |  .    .    .  |
|  .    .    .  |  .    *5   .  |  .    .    .  |
+---------------+---------------+---------------+
|  .    .    .  |  34   .   236 |  *3   .    .  |
|  .    .    .  |  .    .    .  |  .    .    .  |
|  .    .    .  |  .         .  |  .    .    .  |
+---------------+---------------+---------------+


Can anyone see anything wrong with this?

I guess your almost-almost-locked-set (n cells, n+2 candidates) is {r45c6} = {1256}, but I can't see it's a two-sector (or even three-sector) disjoint subset example. However, I believe it is with a few modifications.

Code: Select all

+---------------+---------------+---------------+
|  .    .    .  |  .    .    .  |  .    .    .  |
|  .    .    .  |  .    .  *236 |  .    .    .  |
|  .    .    .  |  .    .    .  |  .    .    .  |
+---------------+---------------+---------------+
|  .    .    .  |  .    45  126 |  .    .    .  |
|  .    .    .  |  14   .    56 |  .    .    .  |
|  .    .    .  |  .   *145  .  |  .    .    .  |
+---------------+---------------+---------------+
|  .    .    .  |       .   23  |  .    .    .  |
|  .    .    .  |  .    .   36  |  .    .    .  |
|  .    .    .  |  .    *3   .  |  .    .    .  |
+---------------+---------------+---------------+


While I expressed reservation in an earlier post, I now believe your example 1 is just fine.

Ron

[ronk]

Example 1 with 6 cells:

Code: Select all


+---------------+---------------+---------------+
|  .    .    .  |  .    .    .  |  .    .    .  |
|  .    .    .  |  .    .    .  |  .    .    .  |
|  .    .    .  | ab    .    .  |  .    .    .  |
+---------------+---------------+---------------+
|  .    .    cd | abcd cdef  .  |  *c   *d   .  |
|  .    .    .  | abef  ef   .  |  .    .    .  |
|  .    .    .  |  .    .   *ef |  .    .    .  |
+---------------+---------------+---------------+
|  .    .    .  |  *a   .    .  |  .    .    .  |
|  .    .    .  |  *b   .    .  |  .    .    .  |
|  .    .    .  |  .    .    .  |  .    .    .  |
+---------------+---------------+---------------+

(ab) are restricted to c4, (cd) to r4, and (ef) to box5, but the six values are in the 6 cells. The starred candidates can be eliminated.


I found one of these ... a Three-Sector Disjoint Subset pattern ... in puzzle #461 of the top1465. If the pattern wasn't very rare, I would've started a new thread on the topic. As it is rare, I'm only posting this here for those who might have an academic interest.

With basic techniques, the puzzle can be advanced to ...

Code: Select all

 19     3      256    | 24579  24789  4579   | 4568   45678  1678
 8      7      25     | 6      24     1      | 9     C45     3
 19     4      56     | 3579   3789   3579   | 568    2      1678
----------------------+----------------------+---------------------
 2      15     1347   | 8      3479   6      | 345   A4579  B79
 357   D68     9      | 2347   2347   347    | 1     A45678 A678
 37     68     347    | 1      5      3479   | 23468  46789  26789
----------------------+----------------------+---------------------
 4      19     137    | 37     367    2      | 68     689    5
 357    59     37     | 347    3467   8      | 26     1      269
 6      2      8      | 59     1      59     | 7      3      4

Sets: A = {r5c8,r6c8,r6c9} = {456789}
      B = {r4c9} = {79}
      C = {r2c8} = {45}
      D = {r5c2} = {68}

Excepting the cells of sets A, B, C, and D:
      7,9 may be eliminated from box 6  (4 elims)
      4,5 may be eliminated from col 8  (3 elims)
      6,8 may be eliminated from row 5  (0 elims)

Set A is an almost-almost-almost-disjoint set with three cells and six candidates. It shares two candidates with set B, two different candidates with set C, and an additional two different candidates with set D.

The following figure illustrates this relationship.

Code: Select all

                                       
            7 ------- B ------- 9     
            .                   .     
            .                   .     
      4 . . . . . 4       6 . . . . . 6
      |     .      \\   //      .     |
      |     .       \\ //       .     |
      C     7 ======= A ======= 9     D
      |             // \\             |
      |            //   \\            |
      5 . . . . . 5       8 . . . . . 8
                                       


In the final placement, the disjoint (locked) set A will contain exactly one each of the candidates from sets B, C, and D. Sets B, C, and D will contain the other candidate of their respective sets, of course. While we don't know the exact placements, we do know ...

1. Candidates 7 and 9 will be somewhere in box 6
2. Candidates 4 and 5 will be somewhere in col 8, and
3. Candidates 6 and 8 will be somewhere in row 5

As a result, elimations may be made as noted above. For large image, click thumbnail below.

[Obi-Wahn]

As I mentioned in Ruud's thread Sue De Coq revisited on the Sudoku Discussions forum, there is an inconsistency in the theory of Two Sector Disjoint Subsets overlooked so far that can cause a Sue De Coq to disappear if you apply other techniques first.

iii. The most general form of the pattern is as follows.

Consider the set of unfilled cells C that lies at the intersection of Box B and Row (or Column) R. Suppose |C|>=2. Let V be the set of candidate values to occur in C. Suppose |V|>= |C|+2. The pattern requires that we find |V|-|C| cells in B and R, with at least one cell in each, with candidates drawn entirely from V. Label the sets of cells CB and CR and their candidates VB and VR. Crucially, no candidate is allowed to appear in VB and VR. Then C must contain V\(VB U VR) [possibly empty], |VB|-|CB| elements of VB and |VR|-|CR| elements of VR. The construction allows us to eliminate the candidates V\VR from B\(C U CB) and the candidates V\VB from R\(C U CR).

The restriction highlighted by me isn't necessary. We can use candidates not present in the intersection as long as for every additional candidate we can find an additional cell.
The actual requirement for a TSDS is that in the combined box and line sectors we have N cells containing only N different candidates with every candidate appearing either in the line or the box only.

Consider the following puzzle:

Code: Select all

..5....7...918...2..............7...8..5......632....919...65.35....3.84..6......

. . 5|. . .|. 7 .
. . 9|1 8 .|. . 2
. . .|. . .|. . .
-----+-----+-----
. . .|. . 7|. . .
8 . .|5 . .|. . .
. 6 3|2 . .|. . 9
-----+-----+-----
1 9 .|. . 6|5 . 3
5 . .|. . 3|. 8 4
. . 6|. . .|. . .


After SSTS we get:

Code: Select all

.---------------.---------------.---------------.
| 246  18   5   | 36   236  49  | 489  7    16  |
| 346  7    9   | 1    8    5   | 34   36   2   |
| 26   138  48  | 7    26   49  | 3489 19   5   |
:---------------+---------------+---------------:
| 9    5    1   | 36   36   7   | 2    4    8   |
| 8    4    2   | 5    9    1   | 37   36   67  |
| 7    6    3   | 2    4    8   | 1    5    9   |
:---------------+---------------+---------------:
| 1    9    48  | 48   7    6   | 5    2    3   |
| 5    2    7   | 9    1    3   | 6    8    4   |
| 34   38   6   | 48   5    2   | 79   19   17  |
'---------------'---------------'---------------'

In this example we even have two TSDS that don't comply with Sue's definition.

In row 1 and box 1 we have 6 cells with 6 candidates were {1236} appear in the row only and {48} appear in the box only, but 3 isn't part of the intersection:

Code: Select all

.-----------------.----------------.----------------.
| (246) (18) 5    | (36) (236) 49  | 489  7    (16) |
| 3-46  7    9    | 1    8     5   | 34   36   2    |
| 26    13-8 (48) | 7    26    49  | 3489 19   5    |
:-----------------+----------------+----------------:
| 9     5    1    | 36   36    7   | 2    4    8    |
| 8     4    2    | 5    9     1   | 37   36   67   |
| 7     6    3    | 2    4     8   | 1    5    9    |
:-----------------+----------------+----------------:
| 1     9    48   | 48   7     6   | 5    2    3    |
| 5     2    7    | 9    1     3   | 6    8    4    |
| 34    38   6    | 48   5     2   | 79   19   17   |
'-----------------'----------------'----------------'

And in row 3 and box 3 we have 6 cells with 6 candidates were {489} appear in the row only and {136} appear in the box only, but 6 isn't part of the intersection:

Code: Select all

.----------------.----------------.------------------.
| 246  18   5    | 36   236  49   | 489    7    (16) |
| 346  7    9    | 1    8    5    | -34    (36) 2    |
| 26   13-8 (48) | 7    26   (49) | (3489) (19) 5    |
:----------------+----------------+------------------:
| 9    5    1    | 36   36   7    | 2      4    8    |
| 8    4    2    | 5    9    1    | 37     36   67   |
| 7    6    3    | 2    4    8    | 1      5    9    |
:----------------+----------------+------------------:
| 1    9    48   | 48   7    6    | 5      2    3    |
| 5    2    7    | 9    1    3    | 6      8    4    |
| 34   38   6    | 48   5    2    | 79     19   17   |
'----------------'----------------'------------------'

Afterwards the puzzle can be solved with SSTS.

If you use SudoCue with the default solving order it requires an XY-Chain at this point. If you move the Sue De Coq technique up behind the Singles techniques and reset the puzzle it can solve it using SSTS and 2 Sue De Coq only, no chain required.

Sorry for digging up this old thread, but it seems to be the reference for this technique.

[re'born]

As I mentioned in Ruud's thread Sue De Coq revisited on the Sudoku Discussions forum, there is an inconsistency in the theory of Two Sector Disjoint Subsets overlooked so far that can cause a Sue De Coq to disappear if you apply other techniques first.

iii. The most general form of the pattern is as follows.

Consider the set of unfilled cells C that lies at the intersection of Box B and Row (or Column) R. Suppose |C|>=2. Let V be the set of candidate values to occur in C. Suppose |V|>= |C|+2. The pattern requires that we find |V|-|C| cells in B and R, with at least one cell in each, with candidates drawn entirely from V. Label the sets of cells CB and CR and their candidates VB and VR. Crucially, no candidate is allowed to appear in VB and VR. Then C must contain V\(VB U VR) [possibly empty], |VB|-|CB| elements of VB and |VR|-|CR| elements of VR. The construction allows us to eliminate the candidates V\VR from B\(C U CB) and the candidates V\VB from R\(C U CR).

The restriction highlighted by me isn't necessary. We can use candidates not present in the intersection as long as for every additional candidate we can find an additional cell.
The actual requirement for a TSDS is that in the combined box and line sectors we have N cells containing only N different candidates with every candidate appearing either in the line or the box only.

An excellent observation Obi-Wahn. It follows immediately from the theory of Subset Counting (which, by the way, I think is a wonderful way to think about Sue De Coq to avoid memorizing what exclusions can be made from the pattern, in any of its forms.)

Consider the following puzzle:

Code: Select all

.---------------.---------------.---------------.
| 246  18   5   | 36   236  49  | 489  7    16  |
| 346  7    9   | 1    8    5   | 34   36   2   |
| 26   138  48  | 7    26   49  | 3489 19   5   |
:---------------+---------------+---------------:
| 9    5    1   | 36   36   7   | 2    4    8   |
| 8    4    2   | 5    9    1   | 37   36   67  |
| 7    6    3   | 2    4    8   | 1    5    9   |
:---------------+---------------+---------------:
| 1    9    48  | 48   7    6   | 5    2    3   |
| 5    2    7   | 9    1    3   | 6    8    4   |
| 34   38   6   | 48   5    2   | 79   19   17  |
'---------------'---------------'---------------'

If you use SudoCue with the default solving order it requires an XY-Chain at this point. If you move the Sue De Coq technique up behind the Singles techniques and reset the puzzle it can solve it using SSTS and 2 Sue De Coq only, no chain required.

It's off-topic, but I just can't help myself.

The potential deadly patterns in r14c45[36] and r13c15[26] imply that r1c5=2 and r1c1<>6, solving the puzzle.

Sorry for digging up this old thread, but it seems to be the reference for this technique.

It was worth it to me.

[ronk]

After SSTS we get:
In this example we even have two TSDS that don't comply with Sue's definition.

In row 1 and box 1 we have 6 cells with 6 candidates were {1236} appear in the row only and {48} appear in the box only, but 3 isn't part of the intersection:

Code: Select all

.-----------------.----------------.----------------.
| (246) (18) 5    | (36) (236) 49  | 489  7    (16) |
| 3-46  7    9    | 1    8     5   | 34   36   2    |
| 26    13-8 (48) | 7    26    49  | 3489 19   5    |
:-----------------+----------------+----------------:
| 9     5    1    | 36   36    7   | 2    4    8    |
| 8     4    2    | 5    9     1   | 37   36   67   |
| 7     6    3    | 2    4     8   | 1    5    9    |
:-----------------+----------------+----------------:
| 1     9    48   | 48   7     6   | 5    2    3    |
| 5     2    7    | 9    1     3   | 6    8    4    |
| 34    38   6    | 48   5     2   | 79   19   17   |
'-----------------'----------------'----------------'

Just like one can use naked triples or naked quads (instead of a naked pair) with a UR Type 3, I think this extension of Sue de Coq's formal definition has been known for some time.

Here are two interesting examples. They both have the larger set in a box instead of a line, and this set partially lies in the box-line intersection.

Code: Select all

#13 of top1465
6.9.....8...7.1...4............6...4.2.....3..3....5...1.5...7.8...9..........2..

After SSTS:
 6      57     9      | 234    2345   2345   |-1347  A1245   8
 23     58     23     | 7      458    1      |B469   B4569  B569
 4      578    1      | 69     2358   69     | 37    A25    -2357
----------------------+----------------------+---------------------
 157    9      8      | 123    6      2357   | 17    C12     4
 157    2      46     | 1489   1457   45789  | 16789  3      1679
 17     3      46     | 12489  1247   2479   | 5     -12689  12679
----------------------+----------------------+---------------------
 239    1      23     | 5      34     3468   | 34689  7      369
 8      46     57     | 12346  9      23467  | 1346  -1456   1356
 39     46     57     | 13468  1347   34678  | 2     -145689 13569

Sets: A = {r12c8} = {1245}
      B = {r2c789} = {4569}
      C = {r4c8} = {12}

Exclusions: r1c7<>4, r3c9<>5, r689c8<>1, r6c8<>2

Code: Select all

#25 of top1465
...9.31..6.7....8.2.........5....4......6..2..1.......8...7.......3..5.....4....9
 
After SSTS:
 45     48     458    | 9      2      3      | 1      67     67
 6      39     7      | 15     145    145    | 239    8      23
 2      39     1      | 67     8      67     | 39     45     45
----------------------+----------------------+---------------------
 379    5      26     | 1278   139    1279   | 4      13679  13678
 3479   478    3489   | 157    6      14579  | 37     2      1357
 3479   1      26     | 2578   3459   24579  | 3678   35679  35678
----------------------+----------------------+---------------------
 8     B246   -3459   | 1256   7      12569  | 236    1346   12346
A1479  B2467  A49     | 3     C19    -12689  | 5     -1467  -124678
-1357  B267    35     | 4      15     12568  | 23678  1367   9

Sets: A = {r8c13} = {1479}
      B = {r789c2} = {2467}
      C = {r8c5} = {19}

Exclusions: r8c689<>1, r8c6<>9, r7c3<>4, r9c1<>7

I think this technique has been pretty much subsumed by the ALS mutual exclusion rule, aka the doubly-linked ALS xz-rule.

[Obi-Wahn]

I think this technique has been pretty much subsumed by the ALS mutual exclusion rule, aka the doubly-linked ALS xz-rule.

In my POV it's not a chain and there's nothing "Almost" about it. It's just simply subset counting.

Example 2 with six cells:

Code: Select all

+---------------+---------------+---------------+
|  .    .    .  |  .    .    .  |  .    .    .  |
|  .    .    .  |  *4   .   *26 |  .    .    .  |
|  .    .    .  |  .    .    .  |  .    .    .  |
+---------------+---------------+---------------+
|  .    .    .  |  15   .   126 |  *1   .    .  |
|  .    .    .  |  45   .    56 |  .    .    .  |
|  .    .    .  |  .    *5   .  |  .    .    .  |
+---------------+---------------+---------------+
|  .    .    .  |  34   .   236 |  *3   .    .  |
|  .    .    .  |  .    .    .  |  .    .    .  |
|  .    .    .  |  .         .  |  .    .    .  |
+---------------+---------------+---------------+


Can anyone see anything wrong with this?

I guess your almost-almost-locked-set (n cells, n+2 candidates) is {r45c6} = {1256}, but I can't see it's a two-sector (or even three-sector) disjoint subset example.

I wonder if you meanwhile realized that this is a Five-Sector Disjoint Subset, Ron?
6 cells containing only 6 candidates and all cells sharing the same candidate seeing each other. Therefor none of the candidates can be true more than once in the set => Disjoint Subset.
All 1s appear in row 4, all 3s in row 7, all 4s in column 4, all 2s and 6s in column 6 and all 5s in box 5.
r47c46b5 => five sectors.

[ronk]

I think this technique has been pretty much subsumed by the ALS mutual exclusion rule, aka the doubly-linked ALS xz-rule.

In my POV it's not a chain and there's nothing "Almost" about it. It's just simply subset counting.

Ultimately, I think everything can be explained as subset counting. Problem is, other than chaining -- or networking -- ALSs no one has offered a technique to find the subsets ... especially for the manual solver.

I wonder if you meanwhile realized that this is a Five-Sector Disjoint Subset, Ron?

Not at the time, and I'm not sure I appreciate that POV even now.

[tarek]

I wonder if you meanwhile realized that this is a Five-Sector Disjoint Subset, Ron?

Not at the time, and I'm not sure I appreciate that POV even now.

The easiest way for me to spot these >3 sector DS is by tracing the snake, each sector hinges to the next over a corner (xz is 1 corner, xy are 2 corners)....... the elimination is where the head of the snake sees the tail.....

This is simplest form of poly-ALS rule elimination.... subset counting allows more complex eliminations.

tarek

[ronk]

Although I just got corrected by DonM in another thread I still refer to the original definition (plus the additions made in the "Sue de Co revisited" thread):

Consider the set of unfilled cells C that lies at the intersection of Box B and Row (or Column) R. Suppose |C|>=2. Let V be the set of candidate values to occur in C. Suppose |V|>= |C|+2. The pattern requires that we find |V|-|C| cells in B and R, with at least one cell in each, with candidates drawn entirely from V. Label the sets of cells CB and CR and their candidates VB and VR. Crucially, no candidate is allowed to appear in VB and VR. Then C must contain V\(VB U VR) [possibly empty], |VB|-|CB| elements of VB and |VR|-|CR| elements of VR. The construction allows us to eliminate the candidates V\VR from B\(C U CB) and the candidates V\VB from R\(C U CR).

I too like an accurate mathematical description. In the past you pointed out the inaccuracy of the red-higlighted portion above. What if we replaced "candidates drawn entirely from V" with "at least two candidates drawn from V"

If we change according to your proposal, we have to add something like "for every candidate in Cx but not in V Cx must contain an additional cell".

If CR and CB both contain candidates not in V, I wonder if any of those "extra" candidates can be the same.

I too like an accurate mathematical description. In the past you pointed out the inaccuracy of the red-higlighted portion above. What if we replaced "candidates drawn entirely from V" with "at least two candidates drawn from V"

I can't remember pointing that out. If I did, I was citing Obi-Wahn's comment (see above).

Sorry, I confused hobiwan with Obi-Wahn. I guess I need to get my hearing checked.

[hobiwan]

If CR and CB both contain candidates not in V, I wonder if any of those "extra" candidates can be the same.

I am not concerned about that. My problem: The definition explicitly states "|V|-|C| cells" for CR+CB, but for every additional candidate in CR or CB you have to find an additional cell in the same set.

Sorry, I confused hobiwan with Obi-Wahn. I guess I need to get my hearing checked.

Well, it's easy to mix them up (and you are not the first to do so)

[ronk]

If CR and CB both contain candidates not in V, I wonder if any of those "extra" candidates can be the same.

I am not concerned about that.

Here are three examples where both sets B (CB) and C (CR) contain the same digit.

Code: Select all

Ruud50k #2122
.2..6.......8...2...32..5.8..1...37.5.......4.87...6..4.6..97...7...1.......5..3.
 
After SSTS
 8     2     45    | 1579  6     357   |C149   149   37
 179   6     45    | 8     1379  357   |C149   2     37
 179   19    3     | 2     1479  47    | 5     6     8
-------------------+-------------------+------------------
 6     4     1     | 59    289   258   | 3     7     259
 5     3     29    | 1679  1279  267   | 28-19 189   4
 29    8     7     | 1459  12349 2345  | 6     159   1259
-------------------+-------------------+------------------
 4     5     6     | 3     28    9     | 7    B18   B12
 3     7     289   | 46    248   1     |A2489  459-8 569-2
 129   19    289   | 467   5     24678 |A2489  3     69-2

Sets: A = {r89c7} = {2489}, B = {r7c89} = {128}, C = {r12c7} = {149}

Elims: r5c7<>19, r8c8<>8, r89c9<>2

Code: Select all

top1465 #218
5.2....4.....81...6............3.7..4..5............9..97...3.....2...6..1....8..

After SSTS
 5     8     2     | 379   79    379   | 6     4     1
 79    347   349   | 6     8     1     | 5     237   237
 6     37    1     | 4     25    25    | 9     378   378
-------------------+-------------------+------------------
B19    2     58-9  | 189   3     4     | 7     158   6
 4    A367  A389   | 5    C16    79-68 | 2    C138  C38
B17    356-7 358   | 178   126   2678  | 4     9     358
-------------------+-------------------+------------------
 28    9     7     | 18    1456  568   | 3     25    245
 38    45    45    | 2     79    38    | 1     6     79
 23    1     6     | 379   45    3579  | 8     257   24579

Sets: A = {r5c23} = {36789}, B = {r46c1} = {179}, C = {r5c589} = {1368}

Elims: r4c3<>9, r6c2<>7, r5c6<>68

Code: Select all

top1465 #785
.......9.....3..51.7..418...13...4......5...37.8....2...1.64....9......2...5.....

After SSTS
 1    3   B26   | 678  278  5    | 27   9    4
A489 B68   49-26| 679  3    279  | 27   5    1
A29   7    5    | 29   4    1    | 8    3    6
----------------+----------------+---------------
 5    1    3    | 789  2789 2789 | 4    6    78
C69   26   29   | 4    5    78   | 1    78   3
 7    4    8    | 3    1    6    | 9    2    5
----------------+----------------+---------------
 28   5    1    | 279  6    4    | 3    78   789
C36   9    67   | 1    78   378  | 5    4    2
C34   28   47   | 5    279  2379 | 6    1    789

Sets: A = {r23c1} = {2489}, B = {r1c3,r2c2} = {268}, C = {r589c1} = {3469}

Elim: r2c3<>26

My problem: The definition explicitly states "|V|-|C| cells" for CR+CB, but for every additional candidate in CR or CB you have to find an additional cell in the same set.

There's no doubt that is true. Modifying rubylip's definition, while keeping the definition simple, is the problem.

[edit: 1) added 2nd and 3rd examples; 2) & 3) thanks to hobiwan, corrected elims for 3rd example]