by **dukuso** » Tue Dec 13, 2005 7:57 am

>OK, there seems to be some miscommunication ...

>Below is my understanding -- corrections welcome -- of

>(1) the "usual" (quasigroup) product of any two Latin squares

>(apparently presence/absence of int & mod operations is a red herring), and

>(2) what I'll call (dukuso's) "4-dim" product of two Latin squares of square

>orders. I'm going to restate everything in this one place for easy reference.

you can have the usual binary quaigroups or ternary,quaternary,... quasigroups.

Or call it latin squares, cubes, tesseracts,...

Products of these are defined in the same way.

>My question is ...

>How can one show -- or at least explain -- that the "4-dim" product

>of two sudokugrids *is* a sudokugrid

>(as evidenced by dukuso's examples), whereas the same is *not* true

>of the usual product?

In 4d it's a property on the whole set {1,2,3} or better on two

of the 4 components of {1,2,3}^4.

In 2d the blocks are a property of only a subset of {1,2,..,9}

>==

>Definition (quasigroup product of Latin squares):

>Let A,B be Latin squares of orders n,m respectively, with elements in

>{0..n-1}, {0..m-1} respectively.

>The product C = A*B is defined as follows ...

>For all p,q in {0..n-1} and for all r,s in {0..m-1},

>C(m*p+r, m*q+s) = m*A(p,q) + B(r,s).

>Or, equivalently, for all i,j in {0..n*m-1},

>C(i,j) = m*A(int(i/m), int(j/m)) + B(mod(i,m), mod(j,m)).

>

>Definition ("4-dim" representation of a Latin square of square order):

>A Latin square S of order n^2 with elements in {0..n^2-1} is

>specified uniquely by its 4-dim representation T defined by

>T(a,b,c,d) = S(n*a+b, n*c+d) for all a,b,c,d in {0..n-1}.

>

>Definition ("4-dim" product of Latin squares of square order):

>Let A,B be Latin squares of order N^2,M^2 respectively, with elements

>in {0..N^2-1}, {0..M^2-1} respectively.

>The "4-dim" product of A and B is defined as the Latin square

>whose 4-dim representation is Z, where Z is given by

>Z(M*x_+y_) = M^2 * X(x_) + Y(y_)

>for all vectors x_,y_ in {0..N^2-1}^4, {0..M^2-1}^4 respectively,

>where X,Y are the 4-dim representations of A,B respectively.

>==

more generally:

Definition:

let d,n be positive integers , let S be a subset of D*D.

A d-ary S-sudoku of order n is a map

T:{1,2,..,n}^d --> {1,2,..,n^2} such that

for all (s1,s2) in S and for all (x1..xd) in {1..n}^d

|{T(y1,.,yd) | yi in {1..n}, yi=xi for all i not in {s1,s2} }|=n*n

Examples:

normal sudokugrids are 4-ary {(1,2),(3,4),(2,4)} sudokus

latin squares are 4-ary {(1,2),(3,4)} sudokus

for other examples see the pictures in the

sudoku as a 6-dimensional problem" - thread

Definition:

if T1 and T2 are d-ary S-sudokus of sizes m and n respectively.

Then let T1*T2:{1..n*m}^d -->{1..n*m*n*m} be defined by

(T1*T2)(x_*n+y_)=T1(x_)*n*n+T2(y_)

Theorem:

given n,m,d,S.

if T1 is a d-ary S-sudoku of size m and T1 a d-ary S-sudoku of size n,

then T1*T2 is a d-ary S-sudoku of size m*n