# Does the question contain enough information to ask on regular mathematics?

## Intro

I'm trying to prove that all valid $$n^2 \times n^2$$ Sudoku solutions follow a cyclic permutation pattern for a subset of $$n!$$ Sudoku grids.

But, I'm not sure on how to prove it.

$$S= {[n....x]}$$

Let $$n$$ be any input length with non-repeating elements.

For $$n$$ in range shift as follows,

$$n$$ = 123456789

123456789
456789123
789123456
234567891
567891234
891234567
345678912
678912345
912345678


Now, suppose I try a different input following the same cyclic pattern

$$n$$ = 987654321

987654321
654321987
321987654
876543219
543219876
219876543
765432198
432198765
198765432


for $$S = n>> ∞$$,

Given valid solved $$n^2 \times n^2$$,

$$X!$$ amount can be generated in $$O(n)$$ time given any input of non-repeating elements Do not confuse with O(1) time because the bigger the element the more bits it needs!!! Thus, the more time it takes to execute!

## Question

I'm not familiar with mathematics, and I'm looking to be pointed in the right direction. How would I prove that this works for ALL $$n^2\times n^2$$ Sudoku Solutions?

• Are you asking the question in the title, or the question in the body? – Gerry Myerson Jun 10 at 4:51
• Are you asking if you have a valid solution to a sudoku puzzle then if for example you replace the digits $(1,2,3,4,5,6,7,8,9)$ with a permutation of $(1,2,3,4,5,6,7,8,9)$ that you'll also get a valid solution? I don't really understand what $S$ is and I don't understand the 2 sentences defining $n$ as well. Also in $n^2\times n^2$ the $n$ seems to stand for something else (you probably don't mean $123456789^2\times123456789^2$.). – kingW3 Jun 10 at 14:15
• A sudoku solution is a form of latin square, and the mathematical term that is used is a permutation of symbols. Latin squares are preserved under some additional symmetries that don't preserve the specialized sudoku solutions, but they share equally in preservation by permutation of symbols. – hardmath Jun 10 at 23:49
• @hardmath Also, found out I can take any $n^2 x n^2$ general Sudoku Problem and reduce it into an algorithm in poly-time that generates that limited $n!$ subset of equivalent instances in $poly-time$. I also found out that solving the correct instances is NP-complete. In other words, creating an algorithm that generates valid arbitrary Sudoku pre-solved grids is NP-complete. Because, it first requires that the general Sudoku be solved!!! By the way the reduction actually works! :) – Travis Wells Jun 11 at 0:08
• @kingW3 I'm defining $n$ as an arbitrary input of 9 elements. Doesn't have to be just 1-9. As long as there is no repeating elements it will generate a valid Sudoku variation in poly-time! And, it will work for any $n^2 x n^2$ solved valid Sudoku puzzle! – Travis Wells Jun 11 at 0:24
• @TravisWells I've thought that was your question? So $n$ is not a number but you seem to use it as such in $n^2 x n^2$,also in your examples did you just switch each instance of 1 with 9,2 with 8,3 with 7, etc.? What exactly do the examples have to do with the question, also for what subset of $n!$ Sudoku grids are you talking about? It could just be me but the whole question seems confusing, and the notation strange. – kingW3 Jun 11 at 1:24
• @kingW3 I have three links. To explain. 1 ---cs.stackexchange.com/questions/107183/… 2 ---github.com/tbw1995/Sudoku-Solver-Constrained and 3---repl.it/repls/TubbyPrestigiousFirm – Travis Wells Jun 11 at 1:45
• @kingW3 Sorry, to bombard you with reading material. But, I just swapped instances of any elements that follow the same mapping to yield valid $n!$ grids, but only one at a time. – Travis Wells Jun 11 at 1:46