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


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

$n$ = 987654321


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!


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?

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    $\begingroup$ Are you asking the question in the title, or the question in the body? $\endgroup$ – Gerry Myerson Jun 10 '19 at 4:51
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    $\begingroup$ 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 $123456789^2\times123456789^2$.). $\endgroup$ – kingW3 Jun 10 '19 at 14:15
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    $\begingroup$ 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. $\endgroup$ – hardmath Jun 10 '19 at 23:49
  • $\begingroup$ @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! :) $\endgroup$ – Travis Wells Jun 11 '19 at 0:08
  • $\begingroup$ @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! $\endgroup$ – Travis Wells Jun 11 '19 at 0:24
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    $\begingroup$ @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. $\endgroup$ – kingW3 Jun 11 '19 at 1:24
  • $\begingroup$ @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 $\endgroup$ – Travis Wells Jun 11 '19 at 1:45
  • $\begingroup$ @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. $\endgroup$ – Travis Wells Jun 11 '19 at 1:46

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