The New Sudoku Players' Forum

[Madolite]

Hey guys, I have a question regarding placement suppositions - I.e. when you suppose that a given number (9 in my case, in the example below) belongs to a given box. Whenever you do this, you can sometimes disprove your own supposition if your next placement (of 9a, 9b etc in my example) crashes with any other 9s in the puzzle, or crashes with any other value that you know MUST be in that box that you've supposed the the 9 also must be in. In these cases, you've just proven that the first 9 (the red one) cannot be placed in that box. As such, the 9 MUST belong to the other box and the 3 MUST belong to the box you just supposed that 9 belonged to.

However, as you can see in my example - 9a, 9b, 9c and 9d ALL fit perfectly into their necessary boxes, without ever crashing with anything and it left me with all 9s pseudo-solved. But I'm still left with the conclusion that, yes, the initial (red) 9 CAN be placed in that box, but it may still belong to the other box of that pair.

My question is, does there exist some underlying logic with Sudoku that dictates that, if you can place every number of a given value in a box without them crashing with each other, you can then conclude that the placement is correct? Or is it still not sufficient (because there might be other values that the 9, in my example, could potentially cause crashes with)?

In other words, can you use placement supposition (as I like to call it) ONLY to say "9 can/cannot belong to this box", OR can you also use it to say "9 MUST belong to this box"? Thanks

[m_b_metcalf]

Madolite,

Welcome to the forum.

The technique you're using here is sometimes called the Thread of Ariadne, also known as guessing. You make a guess and, if it leads to a contradiction, you backtrack and take another path. You certainly can place the 9s is the way you illustrate, but you finish up with a clash (not crash) for instance in box 2 (we refer to a 3x3 set of cells as a box), where there is no cell in which to place a 2.

Hope that helps,

Mike Metcalf

[SpAce]

My question is, does there exist some underlying logic with Sudoku that dictates that, if you can place every number of a given value in a box cell without them crashing with each other, you can then conclude that the placement is correct? Or is it still not sufficient (because there might be other values that the 9, in my example, could potentially cause crashes with)?

I think you answered your own question by noticing that you can't place 9 in the cell you originally assumed for it (row 2, column 3). While the pattern of 9s you described is valid by itself, it will clash with other digits, which proves that pattern false. Thus, the answer to your question is NO. Solving techniques like Templates or POM exist that look at all possible placement patterns for each digit and find which ones are possible and which are not, but they're mostly used by software solvers and not humans.

In other words, can you use placement supposition (as I like to call it) ONLY to say "9 can/cannot belong to this box", OR can you also use it to say "9 MUST belong to this box"? Thanks

Valid solving techniques always produce definite answers: either a candidate MUST be true or it MUST be false. Nothing can be concluded when you only know something CAN be true or false, unless you combine it with other information that produces a definite answer (possibly for another candidate). However, it's a valid technique to test all possibilities of a variable (for example, all possible placements of a 9 in a house) and see what happens -- if they all agree with a certain conclusion, then that must be true.

In your example it's easy to see that r2c3 can't be a 9, because r2c4 must be a 9:

ex.png (38.53 KiB) Viewed 835 times

Look at the 4 and 7 in column 4. Those digits must also go into the similarly colored cells in box 8 (a Locked Pair, just like the 3 and 9 in box 1, which you had noted). It leaves digits 2, 3, and 5 as the only possibilities for the bluish mini-column in box 8 (a Locked Triple). That in turn leaves the digits 1, 8, and 9 as the only possibilities for the rest of the column (reddish cells: a Hidden Triple), but there's a 1 and 8 in row 2, which leaves 9 as the only possibility for cell r2c4 (a Naked Single) so it MUST be a 9. Thus we also know that r2c3 can't be a 9 (and because of that r1c3 must be). (A more direct way of seeing the Hidden Triple (189) in column 4 would be noticing that 1, 8, and 9 are in box 8 blocking those digits from the blue cells and leaving the three red cells as the only place left for them. Not how I saw it, though.)

[Pat]

re the solution described by SpAce,
i would call it a "hidden" duo c4 {1,8}
which creates a "hidden single" c4 9

[SpAce]

re the solution described by SpAce,
i would call it a "hidden" duo c4 {1,8}
which creates a "hidden single" c4 9

A good point. Yes, that's probably the simplest way of describing it -- but not necessarily spotting it in practice. Line-based hidden pairs aren't that easy to see directly unless you're actually looking for them. At least for me it was much easier to see the locked-pair first and start with that as I described. To see the hidden pair (18) directly, you have to look at a total of four givens in two disconnected houses and see how they interact with a third. I only had to look at two givens in one column and the pattern of givens in a connected box to get something to work with, which comes quite naturally.

[Madolite]

Alright, thanks for the replies, guys, and a more thorough answer by SpAce. Suddenly Sudoku became 10 times more exciting, cause I'm 100% self-taught, so I've never thought of the "hidden pair" finding techniques at all. Also nice to know the proper naming conventions.

Thanks again!