Mixing
Minimum and maximum determinant of a sudoku-matrix
$begingroup$
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.
matrices eigenvalues-eigenvectors determinant recreational-mathematics sudoku
edited Oct 12 '16 at 9:41
Rodrigo de Azevedo
13.1k41960
asked Jul 21 '14 at 14:45
PeterPeter
48.9k1240137
$endgroup$
|
show 3 more comments
$begingroup$
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.
matrices eigenvalues-eigenvectors determinant recreational-mathematics sudoku
edited Oct 12 '16 at 9:41
Rodrigo de Azevedo
13.1k41960
asked Jul 21 '14 at 14:45
PeterPeter
48.9k1240137
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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.
matrices eigenvalues-eigenvectors determinant recreational-mathematics sudoku
edited Oct 12 '16 at 9:41
Rodrigo de Azevedo
13.1k41960
asked Jul 21 '14 at 14:45
PeterPeter
48.9k1240137
$endgroup$
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.
matrices eigenvalues-eigenvectors determinant recreational-mathematics sudoku
matrices eigenvalues-eigenvectors determinant recreational-mathematics sudoku
edited Oct 12 '16 at 9:41
Rodrigo de Azevedo
13.1k41960
asked Jul 21 '14 at 14:45
PeterPeter
48.9k1240137
edited Oct 12 '16 at 9:41
Rodrigo de Azevedo
13.1k41960
asked Jul 21 '14 at 14:45
PeterPeter
48.9k1240137
edited Oct 12 '16 at 9:41
Rodrigo de Azevedo
13.1k41960
edited Oct 12 '16 at 9:41
Rodrigo de Azevedo
13.1k41960
edited Oct 12 '16 at 9:41
Rodrigo de Azevedo
13.1k41960
13.1k41960
asked Jul 21 '14 at 14:45
PeterPeter
48.9k1240137
asked Jul 21 '14 at 14:45
PeterPeter
48.9k1240137
asked Jul 21 '14 at 14:45
PeterPeter
48.9k1240137
48.9k1240137
$begingroup$
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
|
show 3 more comments
$begingroup$
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
$begingroup$
Hah, interesting question :) Any observations on the eigenvalues yet?
$endgroup$
– rschwieb
Jul 22 '14 at 16:42
$begingroup$
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
$begingroup$
So far, I concentrated on the determinant. This is difficult enough.
$endgroup$
– Peter
Jul 22 '14 at 16:44
3
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
$begingroup$
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
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
$begingroup$
$pmatrix {9&8&3&4&5&2&7&1&6\4&5&2&7&1&6&9&8&3\7&1&6&9&8&3&4&5&2\8&3&4&5&2&7&1&6&9\5&2&7&1&6&9&8&3&4\1&6&9&8&3&4&5&2&7\3&4&5&2&7&1&6&9&8\2&7&1&6&9&8&3&4&5\6&9&8&3&4&5&2&7&1}$ has determinant $-929 587 995$!!
$endgroup$
– Peter
Aug 2 '14 at 12:56
1
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
$begingroup$
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
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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:
$$
pmatrix {1&2&4&3&8&9&5&6&7\3&9&8&6&7&5&2&1&4\6&5&7&1&4&2&9&3&8\2&7&1&9&3&4&8&5&6\5&8&6&2&1&7&4&9&3\9&4&3&5&6&8&7&2&1\4&1&9&8&5&3&6&7&2\7&6&2&4&9&1&3&8&5\8&3&5&7&2&6&1&4&9}
$$
edited Mar 26 '17 at 20:45
Anton Grudkin
2,251719
answered Jun 13 '15 at 15:38
JPFJPF
314
$endgroup$
add a comment |
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1 Answer
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$begingroup$
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:
$$
pmatrix {1&2&4&3&8&9&5&6&7\3&9&8&6&7&5&2&1&4\6&5&7&1&4&2&9&3&8\2&7&1&9&3&4&8&5&6\5&8&6&2&1&7&4&9&3\9&4&3&5&6&8&7&2&1\4&1&9&8&5&3&6&7&2\7&6&2&4&9&1&3&8&5\8&3&5&7&2&6&1&4&9}
$$
edited Mar 26 '17 at 20:45
Anton Grudkin
2,251719
answered Jun 13 '15 at 15:38
JPFJPF
314
$endgroup$
add a comment |
$begingroup$
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:
$$
pmatrix {1&2&4&3&8&9&5&6&7\3&9&8&6&7&5&2&1&4\6&5&7&1&4&2&9&3&8\2&7&1&9&3&4&8&5&6\5&8&6&2&1&7&4&9&3\9&4&3&5&6&8&7&2&1\4&1&9&8&5&3&6&7&2\7&6&2&4&9&1&3&8&5\8&3&5&7&2&6&1&4&9}
$$
edited Mar 26 '17 at 20:45
Anton Grudkin
2,251719
answered Jun 13 '15 at 15:38
JPFJPF
314
$endgroup$
add a comment |
$begingroup$
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:
$$
pmatrix {1&2&4&3&8&9&5&6&7\3&9&8&6&7&5&2&1&4\6&5&7&1&4&2&9&3&8\2&7&1&9&3&4&8&5&6\5&8&6&2&1&7&4&9&3\9&4&3&5&6&8&7&2&1\4&1&9&8&5&3&6&7&2\7&6&2&4&9&1&3&8&5\8&3&5&7&2&6&1&4&9}
$$
edited Mar 26 '17 at 20:45
Anton Grudkin
2,251719
answered Jun 13 '15 at 15:38
JPFJPF
314
$endgroup$
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:
$$
pmatrix {1&2&4&3&8&9&5&6&7\3&9&8&6&7&5&2&1&4\6&5&7&1&4&2&9&3&8\2&7&1&9&3&4&8&5&6\5&8&6&2&1&7&4&9&3\9&4&3&5&6&8&7&2&1\4&1&9&8&5&3&6&7&2\7&6&2&4&9&1&3&8&5\8&3&5&7&2&6&1&4&9}
$$
edited Mar 26 '17 at 20:45
Anton Grudkin
2,251719
answered Jun 13 '15 at 15:38
JPFJPF
314
edited Mar 26 '17 at 20:45
Anton Grudkin
2,251719
edited Mar 26 '17 at 20:45
Anton Grudkin
2,251719
edited Mar 26 '17 at 20:45
Anton Grudkin
2,251719
2,251719
answered Jun 13 '15 at 15:38
JPFJPF
314
answered Jun 13 '15 at 15:38
JPFJPF
314
answered Jun 13 '15 at 15:38
JPFJPF
314
314
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add a comment |
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$begingroup$
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