I’ve been researching others’ techniques for generating sudoku puzzles, and I have come to the conclusion that nobody actually knows how to do it.
Seriously, if you Google for “generating sudoku puzzles”, most of the methods boil down to “randomly throw numbers at the board and see if it’s a sudoku”. Oh, there’s some cleverness you can do to try to intelligently preserve the randomness of a partially-complete sudoku puzzle (like not randomly putting a 9 into a column that already has one) but at the end of the day all of the solutions basically amount to bogosort.
I promised yesterday that I’d reveal the identity board math. I had planned a big reveal here with lots of mathy explanationing, but the math just isn’t that exotic. There’s a trick to it, but that’s about it. Here’s the math:
A Sudoku board of dimension n is n2xn2 digits, and has nxn sectors, each with n2 digits. Let’s write a function to find the number at position x, y on the board. The whole trick to my F(x,y) math is simply this: the numbers in every sector other than (0,0) are rotations or shifts of the numbers in sector (0,0).
Make sense? Here. The numbers in sector(0,0) are simply the digits 1..n2, arranged in a grid, so that’s easy:
And the math for every other sector is simply a selection into sector 0, 0. Specifically, for F(x,y) outside of sector (0,0), choose the F(x’,y’) IN sector 0,0 such that
x’ = ((x+y/n)%n
y’ = ((y+x/n)%n
So at the end of the day the whole function looks like this in ruby:
def idfx(x, y, n)
if x<n && y<n
idfx( ((x+y/n)%n, ((y+x/n)%n )
I should point out that I am operating under the belief that a sudoku puzzle is like a Rubik’s Cube: that a set of transformations can be performed on a valid puzzle that leave the puzzle changed but always valid, and that you can turn any sudoku puzzle into any other sudoku puzzle via these transformations. If I am right, I can take my identity puzzle board and shuffle it using these transformations, and just like scrambling a Rubik’s Cube I will be left with a unique and interesting puzzle that is, most importantly, still valid.
I have proven the first half of my assumption–there are transformations that will change a sudoku without affecting its validity. What I haven’t proven is the second half. I cannot yet transform puzzles arbitrarily. It may not be possible at all, but right now I’m simply hoping that I don’t know enough transformations yet.