The New Sudoku Players' Forum

[Biscuit]

When I first posted on this board I mentioned that I thought Sudokus should yield to some sort of linear algebra strategy.... I haven't posted much since, but the thought has continued to dominate my thinking, more so than actually solving the puzzles themselves.

I have formulated several half baked ideas, none of which are particularly interesting, however today I have tried something new which quite fascinates me.

I have dubbed it the Sudoku Transform and it's quite simple.

In a completed puzzle, every location has a value. Say for instance, the grid location (1,1) has value 3. A new grid can be defined be swapping the last (or first, it works out similarly) co-ordinate with the value. So in this case we would get (1,3) = 1.

Does that make sense?

At first, I was very excited because I thought you got a new Sudoku from this transformation. Then I got even more excited because you don't. You get something else.

I haven't fully thought through exactly where this is going, but the reason I was thinking along these lines is because I was wondering if some puzzles which are, say, Fiendish, could under some transform, be rendered more difficult/easy.

Anyway, I am off. Sorry this post isn't really a fully thought out idea, but I wanted to share it in case anyone else has been thinking along these lines, or finds it of interest.

Also, if I get too drunk tonight it might come in handy to restore my train of thought tomorrow.

Cheers,

Biscuit

[sciguy47]

Actually, I thought of something else. What I was thinking was that every Sudoku (actually every 9x9 grid divided into 3x3 sections) was a 3x3x3x3 tesseract which was chopped into 3x3 "sheets" which were then arranged to make a 9x9 grid. I occured to me that the same tesseract could be chopped up and rearranged in different ways. But, all of these variants represent the same tesseract, even though they might not look the same.

[gfroyle]

I have dubbed it the Sudoku Transform and it's quite simple.

Consider a general Latin square of order 9 - in other words a 9x9 grid where each row and column contains each number from 1 to 9 exactly once. This is just a Sudoku grid without the extra constraint of the 3x3 boxes (so every Sudoku grid is a Latin square, but there are lots of Latin squares that are not Sudoku grids).

Code: Select all

1 7 4  2 8 5  3 9 6 
4 1 7  5 2 8  6 3 9 
7 4 1  8 5 2  9 6 3 

3 9 6  1 4 7  2 8 5 
6 3 9  7 1 4  5 2 8 
9 6 3  4 7 1  8 5 2 

2 8 5  3 9 6  1 4 7 
5 2 8  6 3 9  7 1 4 
8 5 2  9 6 3  4 7 1 


Then we can write it down by listing 81 triples, representing the row, column and symbol in that place... for the example above we have

1 1 1
1 2 7
1 3 4
1 4 2
...
9 8 7
9 9 1


Now suppose you write it down, go out and get drunk, and come back the next day and see the 81 triples. But you have forgotten whether each triple represents (row, column, symbol) in that order, or maybe (symbol, row, column) or some other order.

The key is that ANY way you read the triples, you will get a Latin square. So in fact, any set of triples gives you SIX related Latin squares (depending on whether you read the triples as r-c-s, r-s-c, c-r-s, c-s-r, s-r-c, s-c-r).

Now let's specialize to Sudokus and do the same trick. The problem is that most Sudokus do not transform to other Sudokus because the extra constraint of the 3x3 boxes is not preserved.

But SOME Sudoku grids do transform to other valid Sudoku grids under at least some of the transformations....

For example, if we take this Sudoku grid

Code: Select all

1 2 3  4 5 6  7 8 9 
4 5 6  7 8 9  1 2 3 
7 8 9  1 2 3  4 5 6 

3 1 2  5 6 4  8 9 7 
6 4 5  9 7 8  3 1 2 
9 7 8  3 1 2  6 4 5 

2 3 1  6 4 5  9 7 8 
5 6 4  8 9 7  2 3 1 
8 9 7  2 3 1  5 6 4


and transform it by writing down the triples in (row,column, symbol) order, and then reading them as though they were (row,symbol, column) we get

Code: Select all

1 2 3  4 5 6  7 8 9 
7 8 9  1 2 3  4 5 6 
4 5 6  7 8 9  1 2 3 

2 3 1  6 4 5  9 7 8 
8 9 7  2 3 1  5 6 4 
5 6 4  8 9 7  2 3 1 

3 1 2  5 6 4  8 9 7 
9 7 8  3 1 2  6 4 5 
6 4 5  9 7 8  3 1 2


(Just for a quick check on how the transform works, let's verify a couple of cases: in the original grid, the entry at R2, C4 is S7 ("symbol 7") and so in the transformed grid, we should have R2, S4, C7 which means that a 4 should appear at R2, C7 which is true. Similarly in the original grid, the entry at R9, C9 is S4 and so in the new grid we should have R9, S9, C4 and so the entry at R9, C4 should be a 9, which is true.)

The obvious questions are

- when does a Sudoku grid remain Sudoku under the various transforms?
- how many are there?

Cheers

Gordon

[sciguy47]

I thought if use a set of 4 coordinates (One each for the row of the 3x3 block the number is, the column it's in, the row of the number within the block and the column within the block.), each variable is from 0-2(1-3 wouldn't work). Then you take the number in the cell, subtract 1 from it and covert it to base 3 eg. 1 -> 0 -> 00, 6 -> 5 -> 12. The you can do something similar to what you did.