# Solve Sudoku using Hidden Pairs Triples and Quads

Hidden pairs/triples/quads is really just a logical extension of Naked Pairs, Triples and Quads. Consider any large container. If it is subdivided into two or more parts by naked subsets, then the parts of that container disjoint from those naked subsets must be inhabited by hidden subsets.

Alternatively, one can consider the following logic:

• If a set of N candidates is limited to N cells within a large container.
• Then those cells can contain only those candidates in the set
• Thus one can safely exclude from those cells all candidates that are not members of the set.

The following is an example of a Hidden Pair . In this example, note that the 4's in box b5 are limited to only two locations: a6,b5. Note further that the 6's in box b5 are limited to the same two locations. We can therefor forbid everything that is not 4,6 from cells b5,a6. The observant amongst you may note that we also have a Naked Quintuple with candidates {12389} at cells abc3,c56.

This idea could be presented in a proof as follows:

• Hidden Pair 46 at a6,b5 forbids a6=123 and forbids b5=1389.

Again, the manner of presenting this step in a proof is also not uniform across provers, and sometimes not even uniform within a prover...

As with Naked Subsets, one can extend the idea to triplets, quads, etc. with Hidden subsets.

A caveat is required with both hidden and naked subsets, also called hidden and naked n-tuples. With all n-tuples, it is not required that all candidates appear in all the cells, as in the latter example listed above with candidates 1,2,3.

Hidden pairs are a favorite of mine to employ during set up. They are often easier to find before entering any possibilities.

### Combining techniques

The next example of a hidden pair illustrates how one can combine techniques into one step. Consider the following partial puzzle grid: Note that the 3's and the 9's in box e8 are limited to the two cells d9,f9. Thus we have hidden pair 39 at df9. Note however that b9=3 and hi9=9 are also still possibilities. Because of locked 3's at df9, and locked 9's at df9, we could also forbid b9=3, hi9=9. One could write:

• Locked 3's at df9 forbids b9=3
• Locked 9's at df9 forbids hi9=9
• Hidden pair 39 at df9 forbids d9=2,f9=12
But, it is more efficient to merely write (and think):
• Hidden pair 39 at df9 forbids b9=3, hi9=9, d9=2, f9=12

A complete rendering as to how this latter approach is more logical, in my opnion, will be presented later in this blog. 