Hi,
In this post I describe a technique wich is based on a simple idea. I consider a group of three candidates in two cells as (a, b, f) (a, b) lying in the region of a row, a column or a box. The group is called here Pair-Plus. In fact there exist two pairs of candidates a and b in two cells incorporating in one cell a single fellow f. Now, I consider that the first cell ( a, b, f) of the group consists of two parts: the subgroup [a, b] and the fellow f. It is true that only one of the two entities [a, b] and f being in the same cell must be true.
If the subgroup [a, b] of the first cell ([a, b], f) is true, by ignoring f, this subgroup cell together with the other cell (a, b) form the naked pair (a, b) (a, b) that acts in the region. If the fellow f is true, then the Pair-Plus scheme reduces to (f) (a, b), where f constitute a single candidate acting in the region. In case that a candidate is rejected by the action of [a, b] and also by f, it becomes obvious that there exists a contradiction meaning that both [a, b] and f being in the same cell are true. Therefore we come to the conclusion that the candidates which are rejected by both parts [a, b] and f must be removed.
From above, we ascertain that there are two processes:
1) One is due to action of the naked pair (a, b)(a, b)
2) The other is due to action of the single fellow f
If the two processes lead to common rejections of candidates, these candidates must be removed from the puzzle.
Based on the above explained technique for a Pair-Plus group the same procedure can also be applied to a Triple-Plus or a Quad-Plus schemes. Let us consider for example that a Triple-Plus group appears as (a, b, c) ([a, b, c], f ) (b, c ). We can proceed in the same way as before assuming that the fellow f is missing in the second cell. This allows the first cell (a, b, c) to joint with (a, b, c) (b, c) to form a naked triple acting in the region. Also, assuming that the subgroup [a, b, c] is missing in the second cell, the remaining group is (a, b, c) (f) (b, c), where the fellow f acts as a single in the region. The Triple-Plus and also Quad-Plus schemes are hard to spot and rarely give common rejections.
+---- ----------------------+---------------------------+-------------------------------------+
| 129 7 4 | 2^89* 19 1*2^8 | 1*2*^368* 5 1*2*^369* |
| 3 6 1289 | 4 1579 1258 | 1278 179 129 |
| 5 1289 1289 | 2789 6 3 | 1278 1479 1249 |
+--------------------------+---------------------------+--------------------------------------+
| 8 129 5 | 6 149 124 | 127 3 129 |
| 1279 4 1279 | 2359 8 125 | 1257 6 129 |
| 1269 3 1269 | 259 159 7 | 4 19 8 |
+ --------------------------+---------------------------+--------------------------------------+
| 4679 89 36789 | 1 2 468 | 36 4 5 |
| 146 18 1368 | 358 345 9 | 136 2 7 |
| 12467 5 12367 | 37 347 46 | 9 8 1346 |
+ --------------------------+---------------------------+--------------------------------------+
In the puzzle above there exists a Pair-Plus scheme (2, 1, 9) (1, 9) in cells A1 and A5.
1) The action of naked pair: Ignoring the fellow 2, the remaining naked pair
(1, 9) (1, 9) rejects 1 (note the symbol * ) from A6, A7 and A9, and also 9 from
A4, A9. Now, it “appeared” a naked pair (2, 8)(2, 8) in cells A4 and A6 which
rejects 2 in A7, A9 and 8 in A7.
2) The action of the fellow: The single fellow 2 in A1 rejects (note the symbol ^ )
the 2 in A4, A6, A 7, and A9.
3) There are common rejections of the candidate 2 in A7 and A9. These candidates
can be removed from the puzzle.
The above described technique is effective and easy to apply. The Pair-Plus schemes are very easy to spot in a puzzle and easy to check if they give common rejections. Although these schemes appear very often in a pazlle , few of them can give common rejections.
Any critisism and help on the subject would be welcome.