16
$\begingroup$

Let $A$ be a sudoku-matrix. Assume that its determinant is positive. What is the lowest, what the highest possible value for the determinant of $A$ ? $A$ must have the dominant eigenvalue $45$, but this does not seem to help establishing bounds.

My records so far :

$$\pmatrix{7&2&9&6&4&3&5&1&8 \\ 5&6&8&9&1&2&7&4&3 \\ 1&3&4&8&5&7&9&6&2 \\ 2&8&7&4&6&1&3&9&5 \\ 9&5&1&7&3&8&6&2&4 \\ 3&4&6&2&9&5&8&7&1 \\ 4&9&3&5&2&6&1&8&7 \\ 8&1&2&3&7&9&4&5&6 \\ 6&7&5&1&8&4&2&3&9}$$

leads to a sudoku-matrix with determinant $1215$. $$\pmatrix{4&3&1&9&7&5&2&6&8 \\ 6&7&2&3&8&1&9&5&4 \\ 8&9&5&6&4&2&7&1&3 \\ 5&4&9&1&6&8&3&2&7 \\ 7&1&3&4&2&9&6&8&5 \\ 2&8&6&5&3&7&4&9&1 \\ 1&5&4&7&9&6&8&3&2 \\ 9&2&7&8&5&3&1&4&6 \\ 3&6&8&2&1&4&5&7&9 }$$

leads to a sudoku-matrix with determinant $238 615 470$.

Additional question :

Can a sudoku-matrix have multiple eigenvalues and, even more interesting, be not diagonalizable or have a minimal polynomial different from the characteristic polynomial ?

I also found a singular sudoku matrix :

$$\pmatrix{6&5&3&9&4&7&8&1&2 \\ 9&8&7&1&6&2&4&3&5 \\ 4&2&1&3&5&8&6&7&9 \\ 5&3&8&4&2&6&1&9&7 \\ 2&7&4&5&9&1&3&8&6 \\ 1&9&6&7&8&3&2&5&4 \\ 8&6&5&2&1&9&7&4&3 \\ 3&1&9&6&7&4&5&2&8 \\ 7&4&2&8&3&5&9&6&1}$$

I found out that the determinant must be a multiple of $405$, so $405$ is a lower bound. I found a sudoku-matrix with determinant $405$ , so it remains to find the maximum.

$\endgroup$
  • $\begingroup$ Hah, interesting question :) Any observations on the eigenvalues yet? $\endgroup$ – rschwieb Jul 22 '14 at 16:42
  • $\begingroup$ So far, I concentrated on the determinant. This is difficult enough. $\endgroup$ – Peter Jul 22 '14 at 16:44
  • 3
    $\begingroup$ P.Newton and S.DeSalvo, in their paper The Shannon entropy of sudoku matrices (in Figure 2), have found sudoku-matrices with determinant up to 551 886 210 (in absolute value). $\endgroup$ – Pierre-Guy Plamondon Jul 23 '14 at 15:36
  • 2
    $\begingroup$ $\pmatrix {9&8&3&4&5&2&7&1&6\\4&5&2&7&1&6&9&8&3\\7&1&6&9&8&3&4&5&2\\8&3&4&5&2&7&1&6&9\\5&2&7&1&6&9&8&3&4\\1&6&9&8&3&4&5&2&7\\3&4&5&2&7&1&6&9&8\\2&7&1&6&9&8&3&4&5\\6&9&8&3&4&5&2&7&1}$ has determinant $-929\ 587\ 995$!! $\endgroup$ – Peter Aug 2 '14 at 12:56
  • 1
    $\begingroup$ This question is closely related to maximum determinant of a latin square. If my conjecture mentioned there is true, the given sudoku is best possible. $\endgroup$ – Peter Aug 2 '14 at 13:02
2
$\begingroup$

I found 6 non equivalent sudoku-matrix with a determinant equal to $929\,587\,995$.
Here they are in lexicographic form :

124359678539687241867214395248931756391765482675428913486192537753846129912573864
124389567398675214657142938271934856586217493943568721419853672762491385835726149
127345689534698217869271354245983761398716425671452938483169572752834196916527843
128379456397645821654182793273964185581237649946518372415823967769451238832796514
134278569569341827827695134298456371371982645645713298416837952783529416952164783
136259478529847631748631259295784163361925847487163925613592784874316592952478316 `

The sudoku-matrix given in Peter's note is equivalent to line 3. Here is the second line as example:
$$ \pmatrix {1&2&4&3&8&9&5&6&7\\3&9&8&6&7&5&2&1&4\\6&5&7&1&4&2&9&3&8\\2&7&1&9&3&4&8&5&6\\5&8&6&2&1&7&4&9&3\\9&4&3&5&6&8&7&2&1\\4&1&9&8&5&3&6&7&2\\7&6&2&4&9&1&3&8&5\\8&3&5&7&2&6&1&4&9} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.