1/9/2024 0 Comments Sudoku strategy triplets![]() ![]() Not a cell was a'solving, nothing to forbid. Twas the Sudoku 'fore Christmas, when throughout the grid, Next, we extend this technique using Hidden Pairs, Triples and Quads Opinion is to just think of it inductively: No matter what a1 equals, in the first example, there The proof of extending this idea beyond two cells can be done many ways. ![]() In either case, one could present this idea in a proof as: Hopefully the reasoning for this is clear. One idea that bears mentioning here is that it is not necessary that all the possibilities be Thus, if for example we had:(the grid above does not illustrate the following example.) The same idea can be extended to any number of cells contained within a common container. The choice of proof presentation is not standard, so one must try to logically decipher each proof Other ways to present the same idea such as: Pair 59 at hi3 forbids df3,h2,hi1=5 and hi1=9.Īlthough the above presentation of proof lingo is my preferred style, one might also see.In a proof, this could be presented as follows: Since cells h3,i3 are both contained within row 3 and box h2, we can exclude 5's from d3,f3,h1,h2,i1. In this example,note that cells h3,i3 both have only 5,9 left as possibilities. The following is an example of Naked Pairs. Must show that cell X = candidate N violates the rules. To eliminate a candidate in the possibility matrix, one generally One may notice that most proofs of ideas are proofs by contradication. Since no candidate can appear twice within the same large container, this is forbidden. Then both of the original cells would be forced to equal the other candidate.Let that cell solve as one of two paired candidates.That cell is also within the paired cells' common large container(s).Those two candidates from all the rest of the cells within that large container(s). Same two candidates left as the only possibilities for these two cells. Suppose that two cells within at least one large container (box, column, row) have exactly the ![]()
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