Sudoku solution method valid?

Factual Questions
1

I haven’t seen this solution method mentioned and would like to know if it’s valid.

If you have two blanks in a row, column, or square whose only solution is one of two numbers then no other blanks in that row, column or square can contain one of those numbers.

Now, suppose you discover two blanks whose only solution in one of two numbers. But those two blanks are not within the same row, column, or square.

However, the blank(s) at the “intersection(s)” of those two duo-solution blanks cannot contain one of those two numbers.

So what you can do at that blank at the intersection is eliminate those two numbers as a solution within that intersection blank.

2

I don’t follow you. What would that mean in this grid?



    A  B  C    D  E  F    G  H  I
 a  -  -  -    -  -  -    -  -  -
 b  3  4  5    6  7  8    9  -  -
 c  -  -  -    -  -  -    -  -  -

 d  -  -  -    -  -  -    -  -  -
 e  -  -  3    4  5  6    7  8  9
 f  -  -  -    -  -  -    -  -  -

 g  -  -  -    -  -  -    -  -  -
 h  -  -  -    -  -  -    -  -  -
 i  -  -  -    -  -  -    -  -  -
3

I think what he asking is, if you have two blanks that you have narrowed down to the same two possible values, and you think of those two blanks as corners of a rectangle, then can you remove those two possibilities from the two opposite corners?

As an example using Arnold’s grid: Let’s say you have narrowed bE and gH to values 1 or 2. The question is, can you assume that gE and bH cannot contain 1 or 2?

Answer: I don’t know but I’m certainly going to give it a shot next time the opportunity presents itself.

4

What’s to prevent the two sites from having the same number? Then, their “intersection” could be the other number.

5

Yes, exactly. It’s not a valid technique.

6

Exactly. As stated, you only know that (site 1 has value A OR B) AND (site 2 has value A OR B). Because the values at the two sites aren’t correlated by not being in the same row, column or square, you have no more information at the intersection point from knowing both squares have possible values A or B than you would if you knew only one of them was A or B.

7

Arnold Winkelried, that wouldn’t work then because there is no intersection.

But in this case:



    A  B  C    D  E  F    G  H  I
 a  8  -  -    -  -  -    -  -  -
 b  -  x  -    -  x  -    -  -  -
 c  3  -  -    -  -  -    -  -  -

 d  4  -  -    -  -  -    -  -  -
 e  5  -  3    4  -  6    7  8  9
 f  6  -  -    -  -  -    -  -  -

 g  7  -  -    -  -  -    -  -  -
 h  -  x  -    -  x  -    -  -  -
 i  9  -  -    -  -  -    -  -  -

At the x(s)
the numbers 1 and 2 can be eliminated as solutions.

It’s always worked.

8

You’ve been lucky. For example, bA and eB could both be 1, which means that bB could be 2. I suspect the reason you haven’t hit it is because it’s fairly improbable. Assuming random values, bA and eB will be 1 only 25% of the time, and in thtat case bB will be 2 only 17% of the time, for a total likelihood of around 4%. Not a bad gamble, but still a gamble.

9

Proof by counterexample from this morning’s sudoku puzzle:



  A B C  D E F  G H I
a   3
b   -             X
c   7

d   9
e   6
f - 5 7  9 6 3  8 - 2

g   2
h   8
i   -

Your method, if I’m understanding it correctly, would preclude (b,H) from containing ‘4’, but the correct solution does have a ‘4’ there.

10

With that being said, there are some clever things that can be done in the vein suggested by the OP:
see the sections “Advanced Steps” and “For The Addict” at this page
Solving Sudoku