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[karybakas]

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                                   On the detection of Hidden Triplets               
                                                                By Karybakas
Hello,
 Is it possible for a hidden triplet to provide information about its existence and location? If possible, data mining should be based on its characteristics such as the maximum and minimum number of digits in a cell and the maximum number of digits of the triplet.   
In this post I describe a process which can lead to revelation of a triplet hidden in the region of a row, a column or a box. The process is based on successive removal of candidates. As the number of candidates is reduced in each removal, if a triplet exists, it will appear in the final residue.
The following are the basic keys which are used in the steps of the process:
Step 1.  If there exist a digit in the region that occurs more than 3 times, that set of digits does not belong to a triplet, as the triplet cannot accommodate more than three digits. Therefore that set does not have any relation to a triplet and we ignore the set removing it.
Step 2. After the realization of step 1, if a digit appears to be alone in a cell, like 8 in the first cell in (8)(3481)(2148) without having a partner, this digit  does not belong to a triplet, because each cell of the triplet must have two or three digits. As a result the 8 and the rest of 8’s in the above cells will not belong to triplet as well. Therefore all 8’s will be removed.
Step 3. If there exist a digit in only three cells in a region, this digit must be removed, unless these cells   constitute a triplet.
The following example is elaborated with details.
Example 1
Let us take the row
 (149)(23568)(249)(48)(37)(29)(567)(16)(1689)
 Each of the digits 6, 9 is displayed 4 times. Removing these digits from all cells, the row reduces to the residue Re1
 Re1      (14)(2358)(24)(48)(37)(2)(57)(1)(18)
 Now, it is seen that 2 and 1 are each alone in the corresponding cell, so 2 and 1 are removed from all cells. This results in the residue Re2
 Re2       (4)(358)(4)(48)(37)(  )(57)(  )(8)     
The digits 4 and 8 in Re2 appear to be alone. Removing these from all cells, the final residue Re3 results
 Re3      (  )(35)(  )(  )(37 )(  )(57)(  )(  )
revealing the triplet (35)(37)(57).       
The numbers 2, 6, 8 in cell2 and 6 in cell7 can be removed from the original row.
It is evident from above that it is not always necessary to proceed until to the final residue to see the result i.e. if there exist a triplet, because in each removable the number of candidates reduces, so the existence of the triplet can become apparent early.
Example 2
Shortly, in the row billow
 (5’6’’8)(5’9)[4](126’’)[7](236’’)(5’8)(135’9)(6’’9)
5’ and 6’’ appear four times. The existence of these digits make 8 and 9 in some cells to be alone. Hence removing 5’, 6’’, 8, and 9 we end up with the triplet (12)(23)(13).
Note: The process described above concerning the detection of hidden triplets can be used for finding of hidden quad and hidden pair as well. In order to explore an area for the existence of a hidden quad, a hidden triplet or a hidden pair we start from step1 removing all digits that occurs 4 or more times and proceed until the end.     





[creint]

What you describe:
Step 1 remove singles first
Step 2 keep removing singles
Step 3 incorrect

Take this example:
(123456) (456) (456789) (456) (456789) (123456) (456789) (123456) (456)
reducing gives you 2x3 hidden triples, or naked sets of size 3 and 2x6
(123) () (789) () (789) (123) (789) (123) ()
You still need a way to find the correct group to form your set.

Most digital solvers here are just doing all combinations which can slow a solver down.
A better way is just the following like my solver, for setsize n:
Locked sets: take cells in region with count <= n, use backtracking on sum digitcount > n to reduce combinations
Hidden sets: take digits in region with count <= n, use backtracking on cell digitcount > n to reduce combinations