In the puzzle "Sudoku", it is a square with 9 boxes across, 9 boxes
down and within that figuration is 9 six box squares.
The gist of the puzzle is NOT to duplicate any number either down,
across, or within a given square using the numbers 1-9.

I am trying to figure out the entire equation OF empirical results
and/or the way to calculate them...

I have worked with 2 functions, the second seemingly more precise. IE:
((N-1)/N^N  as N approaches infinity) but produces the same results
without rounding.

Using L(N) represent the number of good 2 rows, following any given
1st row, for rows of length N.  It is obvious that L(1) = 0 and L(2) =
1.  (agreed ??)  Then for L = 3 and greater, the following seems true:
 L(N) = (N-1) x (L(N-2) = L(N-1)).

No rounding is necessary, but of course is a big disadvantage of this
method is that it requires calculating all the preceding L(N)
values...in any case when N gets large it becomes necessary to use
floating-point notation, which introduces rounding, anyway.

I do believe it builds on itself, sort of like a Fibonacci series. 
But I wonder if there is a way of determining the Nth value in a
Fibonacci series without knowing the preceding values ??!!

My 'assumption' is that the number of 9x9 squares meeting "Sudoku's"
row and column requirement would be the product of the "good" values
for each of the 9 rows, where the 1st row value is 9 factorial.

The next part would be how of those good row/column squares ALSO meet
the 3rd "Sudoku" requirement for the inner 3x3 boxes ??????

I am 'dying' to figure this out !! And like a dog with a bone, I find
it hard to let go...

I'd appreciate the answer that works beyond my, seemingly, limited caluculations !!

Thanks !!!

Zym

There is a variation of Sudoku called Hexadoku
- it works the same but has 16 'numbers' - eg: 256 cells

Recently someone asked me to crack it.

I tried brute force (permutations) and had a P3gh running for hours.

The solution is to 'crack' it methodically.
Each cell belongs in three logical groups of 9, 9 and 6
- row, column and box

Initially each cell has a list of 'possible' numbers 0 .. 9
By seeing which cells are already filled in, in the three logical
groups for each cell, one can trim down the 'possible list' for each
cell.

If you get two cells in one group that have just two possible numbers,
then that means that all /other/ cells in the group /cannot/ possibly
contain either of those numbers. So you can eliminate those two
numbers from the 'possible list' for the other cells.

Similarly three cells in a group with only the same three possible
numbers, means that those three numbers cannot be in the other cells
in that group.

Ditto four with four identical 'possible lists' etc.

As one eliminates numbers from the 'possible lists' you get cells with
just one possible number appearing.
Take those and repeat the process until all cells are filled in.

Hexadoku 'cracks' in about 22 passes.

Here is the program (exe) :-
http://www.jerryfrench.co.uk/utils/hexadoku.zip

If you run the solver then click on a cell you'll see the initial
'possible list' and be able to figure out how it got trimmed down.

The trick is generating the Groups, then inspecting the Groups from
the point of view of each cell. The cell does not care if the Group is
a row, col or box, to the cell it is just a 'Group'.

It is a matter of elimination.
When you have eliminated the impossible ...

I will only answer one of your questions.

Yes, it is possible to compute any member of a general Fibonacci
series without computing all the previous values. It is pretty easy to
show that the expression is just:

F(n) = A*P^n + B*Q^n

Where P=(1+sqrt(5))/2 is the golden ratio and Q=(1-sqrt(5))/2

A and B are constants that you can easily deduce for any values of
F(0) and F(1). For example, if F(0)=0 and F(1)=1, then:

A+B=0, so B=-A, and A*(P-Q)=1, so A=1/(P-Q)=1/sqrt(5), so the standard
Fibonacci series is given by:

F(n) = (1/sqrt(5)*(P^n - Q^n)

ANSWER:

In the year 2005 Bertram Felgenhauer developed a solution for this question.
There are 6.670.903.752.021.072.936.960 possible 9x9-Sudoku's. 

This number is computed by the expression 9! × 722 × 27 ×
27.704.267.971 , the last factor being a prime number. The number was
undependently approved by Ed Russell.

This number is still containing symmetries. According to Ed Russel und
Frazer Jarvis there are 5.472.730.538 possibilities with respect to
symmetries.

See http://www.afjarvis.staff.shef.ac.uk/sudoku/