## Hi im looking for a sudoku bigger than 81X81

Everything about Sudoku that doesn't fit in one of the other sections

### Re: Defining the product of two Sudoku grids (Latin squares)

dukuso wrote:No modulo is needed, that's the purpose of the 4-dim representation.

Oh!!
Because of the identity
x = n*int(x/n) + mod(x,n),
an ordinary row index x in {0..n^2-1} has a unique representation of the form
x = n*a+b, with a=int(x/n), b=mod(x,n);
so by taking a,b (which are now in {0..n-1}) as two of the four indices, and similarly for the column index, nothing needs to be recovered by int/mod operations. Neat! (Sorry, I guess it's old hat to many, but I had to say it in my own words.)

Thanks for all the feedback.
r.e.s.

Posts: 337
Joined: 31 August 2005

### Re: Defining the product of two Sudoku grids (Latin squares)

r.e.s. wrote:
dukuso wrote:No modulo is needed, that's the purpose of the 4-dim representation.

Oh!! .

sorry, _that_ modulo from the equation (26) in the linked paper
is somehow need. I avoided it by defining

Just the modulo with the blocks for the sudoku isn't needed.
dukuso

Posts: 479
Joined: 25 June 2005

### Defining the product of two Sudoku grids (Latin squares)

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.

My question is ...
How can one show -- or 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?

==
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.
==
r.e.s.

Posts: 337
Joined: 31 August 2005

>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
dukuso

Posts: 479
Joined: 25 June 2005

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