# Minimum and maximum determinant of a sudoku-matrix

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 $238 615 470$.

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.

• Hah, interesting question :) Any observations on the eigenvalues yet? – rschwieb Jul 22 '14 at 16:42
• So far, I concentrated on the determinant. This is difficult enough. – Peter Jul 22 '14 at 16:44
• 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). – Pierre-Guy Plamondon Jul 23 '14 at 15:36
• $\pmatrix {9&8&3&4&5&2&7&1&6\\4&5&2&7&1&6&9&8&3\\7&1&6&9&8&3&4&5&2\\8&3&4&5&2&7&1&6&9\\5&2&7&1&6&9&8&3&4\\1&6&9&8&3&4&5&2&7\\3&4&5&2&7&1&6&9&8\\2&7&1&6&9&8&3&4&5\\6&9&8&3&4&5&2&7&1}$ has determinant $-929\ 587\ 995$!! – Peter Aug 2 '14 at 12:56
• 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. – Peter Aug 2 '14 at 13:02

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