Determinants of binary matrices
Clash Royale CLAN TAG#URR8PPP
$begingroup$
I was surprised to find that most $3times 3$ matrices with entries in $0,1$ have determinant $0$ or $pm 1$. There are only six out of 512 matrices with a different determinant (three with $2$ and three with $-2$) and these are
$$
beginbmatrix1 & 1 & 0\ 0 & 1 & 1\ 1 & 0 & 1endbmatrix
$$
and the different permutation of this.
Hadamard's maximum determinant problem for $0,1$ asks about the largest possible determinant for matrices with entries in $0,1$ and it is known that the sequence of the maximal determinant for $ntimes n$ matrices for $n=1,2,dots$ starts with 1, 1, 2, 3, 5, 9, 32, 56, 144, 320,1458 (https://oeis.org/A003432). It is even known that the number of different matrices realizing the maximum (not by absolute value) is 1, 3, 3, 60, 3600, 529200, 75600, 195955200, (https://oeis.org/A051752).
My question is: what is the distribution of determinants of all $ntimes n$ $0,1$-matrices?
Here is the data for (very) small $n$:
$$
beginarraylccccccc
n & -3 & -2 & -1 & 0 & 1 & 2 & 3\hline
1 & & & & 1 & 1 & & \
2 & & & 3 &10 & 3 & & \
3 & & 3 & 84 & 338 & 84 & 3 & \
4 & 60 & 1200 & 10020 & 42976 & 10020 & 1200 & 60
endarray
$$
linear-algebra matrices determinants
asked Jan 15 at 20:04
DirkDirk
7,49843367
$endgroup$
|
show 2 more comments
$begingroup$
I was surprised to find that most $3times 3$ matrices with entries in $0,1$ have determinant $0$ or $pm 1$. There are only six out of 512 matrices with a different determinant (three with $2$ and three with $-2$) and these are
$$
beginbmatrix1 & 1 & 0\ 0 & 1 & 1\ 1 & 0 & 1endbmatrix
$$
and the different permutation of this.
Hadamard's maximum determinant problem for $0,1$ asks about the largest possible determinant for matrices with entries in $0,1$ and it is known that the sequence of the maximal determinant for $ntimes n$ matrices for $n=1,2,dots$ starts with 1, 1, 2, 3, 5, 9, 32, 56, 144, 320,1458 (https://oeis.org/A003432). It is even known that the number of different matrices realizing the maximum (not by absolute value) is 1, 3, 3, 60, 3600, 529200, 75600, 195955200, (https://oeis.org/A051752).
My question is: what is the distribution of determinants of all $ntimes n$ $0,1$-matrices?
Here is the data for (very) small $n$:
$$
beginarraylccccccc
n & -3 & -2 & -1 & 0 & 1 & 2 & 3\hline
1 & & & & 1 & 1 & & \
2 & & & 3 &10 & 3 & & \
3 & & 3 & 84 & 338 & 84 & 3 & \
4 & 60 & 1200 & 10020 & 42976 & 10020 & 1200 & 60
endarray
$$
linear-algebra matrices determinants
asked Jan 15 at 20:04
DirkDirk
7,49843367
$endgroup$
3
$begingroup$
Will Orrick probably has the latest data. Miodrag Zivkovic did an analysis in 2005 for n up to 9. While lower determinant values are heavily weighted, it is not known how quickly the counts for off as the determinant value grows. You can find related questions here on MathOverflow. Gerhard "Search For Determinant Spectrum Problem" Paseman, 2019.01.15.
$endgroup$
– Gerhard Paseman
Jan 15 at 20:33
1
$begingroup$
Related MO questions: mathoverflow.net/questions/18636/… mathoverflow.net/questions/18547/… mathoverflow.net/questions/39786/…
$endgroup$
– Timothy Chow
Jan 16 at 3:07
$begingroup$
@WillOrrick Of course - fixed!
$endgroup$
– Dirk
Jan 16 at 13:51
$begingroup$
I think it would be good to clarify the statement of the question, since there are different possible interpretations. The way I interpreted it was, roughly speaking, to understand the limiting shape of the histograms that are implicit in the data shown for $n=1,2,3,4.$ With this interpretation, understanding the precise asymptotics of the number of $n times n$ matrices with a given determinant is probably irrelevant (unless that value occurs a positive percentage of the time). However, the data suggest that the most common determinant value is $0$, which occurs $o(2^n^2)$ times (right?)
$endgroup$
– Matt Young
Jan 16 at 14:40
2
$begingroup$
The limiting shape of the histogram is known: in 2011 Hoi H. Nguyen and Van Vu proved that the log of the absolute value of the determinant approaches a normal distribution with mean $frac12log (n-1)!$ and variance $frac12log n$. (Note how large the mean is and how sharp the peak is--data for sizes less than $10$ are misleading.) This result applies to random $-1,1$ matrices and to many other classes of random matrices. Using the mapping mentioned in my comment to Timothy Chow's answer, you can adapt this result to $0,1$ matrices.
$endgroup$
– Will Orrick
Jan 16 at 17:35
|
show 2 more comments
$begingroup$
I was surprised to find that most $3times 3$ matrices with entries in $0,1$ have determinant $0$ or $pm 1$. There are only six out of 512 matrices with a different determinant (three with $2$ and three with $-2$) and these are
$$
beginbmatrix1 & 1 & 0\ 0 & 1 & 1\ 1 & 0 & 1endbmatrix
$$
and the different permutation of this.
Hadamard's maximum determinant problem for $0,1$ asks about the largest possible determinant for matrices with entries in $0,1$ and it is known that the sequence of the maximal determinant for $ntimes n$ matrices for $n=1,2,dots$ starts with 1, 1, 2, 3, 5, 9, 32, 56, 144, 320,1458 (https://oeis.org/A003432). It is even known that the number of different matrices realizing the maximum (not by absolute value) is 1, 3, 3, 60, 3600, 529200, 75600, 195955200, (https://oeis.org/A051752).
My question is: what is the distribution of determinants of all $ntimes n$ $0,1$-matrices?
Here is the data for (very) small $n$:
$$
beginarraylccccccc
n & -3 & -2 & -1 & 0 & 1 & 2 & 3\hline
1 & & & & 1 & 1 & & \
2 & & & 3 &10 & 3 & & \
3 & & 3 & 84 & 338 & 84 & 3 & \
4 & 60 & 1200 & 10020 & 42976 & 10020 & 1200 & 60
endarray
$$
linear-algebra matrices determinants
asked Jan 15 at 20:04
DirkDirk
7,49843367
$endgroup$
I was surprised to find that most $3times 3$ matrices with entries in $0,1$ have determinant $0$ or $pm 1$. There are only six out of 512 matrices with a different determinant (three with $2$ and three with $-2$) and these are
$$
beginbmatrix1 & 1 & 0\ 0 & 1 & 1\ 1 & 0 & 1endbmatrix
$$
and the different permutation of this.
Hadamard's maximum determinant problem for $0,1$ asks about the largest possible determinant for matrices with entries in $0,1$ and it is known that the sequence of the maximal determinant for $ntimes n$ matrices for $n=1,2,dots$ starts with 1, 1, 2, 3, 5, 9, 32, 56, 144, 320,1458 (https://oeis.org/A003432). It is even known that the number of different matrices realizing the maximum (not by absolute value) is 1, 3, 3, 60, 3600, 529200, 75600, 195955200, (https://oeis.org/A051752).
My question is: what is the distribution of determinants of all $ntimes n$ $0,1$-matrices?
Here is the data for (very) small $n$:
$$
beginarraylccccccc
n & -3 & -2 & -1 & 0 & 1 & 2 & 3\hline
1 & & & & 1 & 1 & & \
2 & & & 3 &10 & 3 & & \
3 & & 3 & 84 & 338 & 84 & 3 & \
4 & 60 & 1200 & 10020 & 42976 & 10020 & 1200 & 60
endarray
$$
linear-algebra matrices determinants
linear-algebra matrices determinants
asked Jan 15 at 20:04
DirkDirk
7,49843367
asked Jan 15 at 20:04
DirkDirk
7,49843367
edited Jan 16 at 13:51
Dirk
asked Jan 15 at 20:04
DirkDirk
7,49843367
asked Jan 15 at 20:04
DirkDirk
7,49843367
asked Jan 15 at 20:04
DirkDirk
7,49843367
7,49843367
3
$begingroup$
Will Orrick probably has the latest data. Miodrag Zivkovic did an analysis in 2005 for n up to 9. While lower determinant values are heavily weighted, it is not known how quickly the counts for off as the determinant value grows. You can find related questions here on MathOverflow. Gerhard "Search For Determinant Spectrum Problem" Paseman, 2019.01.15.
$endgroup$
– Gerhard Paseman
Jan 15 at 20:33
1
$begingroup$
Related MO questions: mathoverflow.net/questions/18636/… mathoverflow.net/questions/18547/… mathoverflow.net/questions/39786/…
$endgroup$
– Timothy Chow
Jan 16 at 3:07
$begingroup$
@WillOrrick Of course - fixed!
$endgroup$
– Dirk
Jan 16 at 13:51
$begingroup$
I think it would be good to clarify the statement of the question, since there are different possible interpretations. The way I interpreted it was, roughly speaking, to understand the limiting shape of the histograms that are implicit in the data shown for $n=1,2,3,4.$ With this interpretation, understanding the precise asymptotics of the number of $n times n$ matrices with a given determinant is probably irrelevant (unless that value occurs a positive percentage of the time). However, the data suggest that the most common determinant value is $0$, which occurs $o(2^n^2)$ times (right?)
$endgroup$
– Matt Young
Jan 16 at 14:40
2
$begingroup$
The limiting shape of the histogram is known: in 2011 Hoi H. Nguyen and Van Vu proved that the log of the absolute value of the determinant approaches a normal distribution with mean $frac12log (n-1)!$ and variance $frac12log n$. (Note how large the mean is and how sharp the peak is--data for sizes less than $10$ are misleading.) This result applies to random $-1,1$ matrices and to many other classes of random matrices. Using the mapping mentioned in my comment to Timothy Chow's answer, you can adapt this result to $0,1$ matrices.
$endgroup$
– Will Orrick
Jan 16 at 17:35
|
show 2 more comments
3
$begingroup$
Will Orrick probably has the latest data. Miodrag Zivkovic did an analysis in 2005 for n up to 9. While lower determinant values are heavily weighted, it is not known how quickly the counts for off as the determinant value grows. You can find related questions here on MathOverflow. Gerhard "Search For Determinant Spectrum Problem" Paseman, 2019.01.15.
$endgroup$
– Gerhard Paseman
Jan 15 at 20:33
1
$begingroup$
Related MO questions: mathoverflow.net/questions/18636/… mathoverflow.net/questions/18547/… mathoverflow.net/questions/39786/…
$endgroup$
– Timothy Chow
Jan 16 at 3:07
$begingroup$
@WillOrrick Of course - fixed!
$endgroup$
– Dirk
Jan 16 at 13:51
$begingroup$
I think it would be good to clarify the statement of the question, since there are different possible interpretations. The way I interpreted it was, roughly speaking, to understand the limiting shape of the histograms that are implicit in the data shown for $n=1,2,3,4.$ With this interpretation, understanding the precise asymptotics of the number of $n times n$ matrices with a given determinant is probably irrelevant (unless that value occurs a positive percentage of the time). However, the data suggest that the most common determinant value is $0$, which occurs $o(2^n^2)$ times (right?)
$endgroup$
– Matt Young
Jan 16 at 14:40
2
$begingroup$
The limiting shape of the histogram is known: in 2011 Hoi H. Nguyen and Van Vu proved that the log of the absolute value of the determinant approaches a normal distribution with mean $frac12log (n-1)!$ and variance $frac12log n$. (Note how large the mean is and how sharp the peak is--data for sizes less than $10$ are misleading.) This result applies to random $-1,1$ matrices and to many other classes of random matrices. Using the mapping mentioned in my comment to Timothy Chow's answer, you can adapt this result to $0,1$ matrices.
$endgroup$
– Will Orrick
Jan 16 at 17:35
3
3
$begingroup$
Will Orrick probably has the latest data. Miodrag Zivkovic did an analysis in 2005 for n up to 9. While lower determinant values are heavily weighted, it is not known how quickly the counts for off as the determinant value grows. You can find related questions here on MathOverflow. Gerhard "Search For Determinant Spectrum Problem" Paseman, 2019.01.15.
$endgroup$
– Gerhard Paseman
Jan 15 at 20:33
$begingroup$
Will Orrick probably has the latest data. Miodrag Zivkovic did an analysis in 2005 for n up to 9. While lower determinant values are heavily weighted, it is not known how quickly the counts for off as the determinant value grows. You can find related questions here on MathOverflow. Gerhard "Search For Determinant Spectrum Problem" Paseman, 2019.01.15.
$endgroup$
– Gerhard Paseman
Jan 15 at 20:33
1
1
$begingroup$
Related MO questions: mathoverflow.net/questions/18636/… mathoverflow.net/questions/18547/… mathoverflow.net/questions/39786/…
$endgroup$
– Timothy Chow
Jan 16 at 3:07
$begingroup$
Related MO questions: mathoverflow.net/questions/18636/… mathoverflow.net/questions/18547/… mathoverflow.net/questions/39786/…
$endgroup$
– Timothy Chow
Jan 16 at 3:07
$begingroup$
@WillOrrick Of course - fixed!
$endgroup$
– Dirk
Jan 16 at 13:51
$begingroup$
@WillOrrick Of course - fixed!
$endgroup$
– Dirk
Jan 16 at 13:51
$begingroup$
I think it would be good to clarify the statement of the question, since there are different possible interpretations. The way I interpreted it was, roughly speaking, to understand the limiting shape of the histograms that are implicit in the data shown for $n=1,2,3,4.$ With this interpretation, understanding the precise asymptotics of the number of $n times n$ matrices with a given determinant is probably irrelevant (unless that value occurs a positive percentage of the time). However, the data suggest that the most common determinant value is $0$, which occurs $o(2^n^2)$ times (right?)
$endgroup$
– Matt Young
Jan 16 at 14:40
$begingroup$
I think it would be good to clarify the statement of the question, since there are different possible interpretations. The way I interpreted it was, roughly speaking, to understand the limiting shape of the histograms that are implicit in the data shown for $n=1,2,3,4.$ With this interpretation, understanding the precise asymptotics of the number of $n times n$ matrices with a given determinant is probably irrelevant (unless that value occurs a positive percentage of the time). However, the data suggest that the most common determinant value is $0$, which occurs $o(2^n^2)$ times (right?)
$endgroup$
– Matt Young
Jan 16 at 14:40
2
2
$begingroup$
The limiting shape of the histogram is known: in 2011 Hoi H. Nguyen and Van Vu proved that the log of the absolute value of the determinant approaches a normal distribution with mean $frac12log (n-1)!$ and variance $frac12log n$. (Note how large the mean is and how sharp the peak is--data for sizes less than $10$ are misleading.) This result applies to random $-1,1$ matrices and to many other classes of random matrices. Using the mapping mentioned in my comment to Timothy Chow's answer, you can adapt this result to $0,1$ matrices.
$endgroup$
– Will Orrick
Jan 16 at 17:35
$begingroup$
The limiting shape of the histogram is known: in 2011 Hoi H. Nguyen and Van Vu proved that the log of the absolute value of the determinant approaches a normal distribution with mean $frac12log (n-1)!$ and variance $frac12log n$. (Note how large the mean is and how sharp the peak is--data for sizes less than $10$ are misleading.) This result applies to random $-1,1$ matrices and to many other classes of random matrices. Using the mapping mentioned in my comment to Timothy Chow's answer, you can adapt this result to $0,1$ matrices.
$endgroup$
– Will Orrick
Jan 16 at 17:35
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
This question is too hard, because easier questions are already known to be hard.
The middle column is A046747 in the OEIS, which is essentially equivalent to A057982, the number of singular ±1-valued matrices. The asymptotic behavior of this number was a longstanding open problem that was just recently solved by Tikhomirov.
As you mentioned yourself, the Hadamard maximal determinant problem is unsolved, with order 15 (for the ±1 version of the problem) being difficult already. The determinant spectrum problem, mentioned by Gerhard Paseman, is even harder, and is easier than your question, since it's just asking for which values are zero.
answered Jan 15 at 22:24
Timothy ChowTimothy Chow
34.7k11184311
$endgroup$
$begingroup$
Thanks for the pointers! That's somehow what I expected. One thing: As you say, there are some results for $pm1$-matrices. I fail to see how these results help for $0,1$-matrices. Or do you mean to say that $0,1$ matrices are harder than $pm1$-matrices (and also I here I don't see why this should be so).
$endgroup$
– Dirk
Jan 16 at 9:24
$begingroup$
Maybe it could be interesting to ask just about the expected value of the determinant of a random binary matrix. AFAICT, the other open problems that were mentioned do not reduce to this problem.
$endgroup$
– Squark
Jan 16 at 12:40
5
$begingroup$
See the Wikipedia article on Hadamard's maximal determinant problem for the mapping between $0,1$ matrices and $-1,1$ matrices. For every $ntimes n$ $0,1$ matrix with determinant $D$, there are $2^2n+1$ $-1,1$ matrices with determinant $2^nD$. The case $n=1$ is slightly different.
$endgroup$
– Will Orrick
Jan 16 at 13:30
$begingroup$
@WillOrrick Thanks! That's indeed helpful.
$endgroup$
– Dirk
Jan 16 at 13:53
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
This question is too hard, because easier questions are already known to be hard.
The middle column is A046747 in the OEIS, which is essentially equivalent to A057982, the number of singular ±1-valued matrices. The asymptotic behavior of this number was a longstanding open problem that was just recently solved by Tikhomirov.
As you mentioned yourself, the Hadamard maximal determinant problem is unsolved, with order 15 (for the ±1 version of the problem) being difficult already. The determinant spectrum problem, mentioned by Gerhard Paseman, is even harder, and is easier than your question, since it's just asking for which values are zero.
answered Jan 15 at 22:24
Timothy ChowTimothy Chow
34.7k11184311
$endgroup$
$begingroup$
Thanks for the pointers! That's somehow what I expected. One thing: As you say, there are some results for $pm1$-matrices. I fail to see how these results help for $0,1$-matrices. Or do you mean to say that $0,1$ matrices are harder than $pm1$-matrices (and also I here I don't see why this should be so).
$endgroup$
– Dirk
Jan 16 at 9:24
$begingroup$
Maybe it could be interesting to ask just about the expected value of the determinant of a random binary matrix. AFAICT, the other open problems that were mentioned do not reduce to this problem.
$endgroup$
– Squark
Jan 16 at 12:40
5
$begingroup$
See the Wikipedia article on Hadamard's maximal determinant problem for the mapping between $0,1$ matrices and $-1,1$ matrices. For every $ntimes n$ $0,1$ matrix with determinant $D$, there are $2^2n+1$ $-1,1$ matrices with determinant $2^nD$. The case $n=1$ is slightly different.
$endgroup$
– Will Orrick
Jan 16 at 13:30
$begingroup$
@WillOrrick Thanks! That's indeed helpful.
$endgroup$
– Dirk
Jan 16 at 13:53
add a comment |
$begingroup$
This question is too hard, because easier questions are already known to be hard.
The middle column is A046747 in the OEIS, which is essentially equivalent to A057982, the number of singular ±1-valued matrices. The asymptotic behavior of this number was a longstanding open problem that was just recently solved by Tikhomirov.
As you mentioned yourself, the Hadamard maximal determinant problem is unsolved, with order 15 (for the ±1 version of the problem) being difficult already. The determinant spectrum problem, mentioned by Gerhard Paseman, is even harder, and is easier than your question, since it's just asking for which values are zero.
answered Jan 15 at 22:24
Timothy ChowTimothy Chow
34.7k11184311
$endgroup$
$begingroup$
Thanks for the pointers! That's somehow what I expected. One thing: As you say, there are some results for $pm1$-matrices. I fail to see how these results help for $0,1$-matrices. Or do you mean to say that $0,1$ matrices are harder than $pm1$-matrices (and also I here I don't see why this should be so).
$endgroup$
– Dirk
Jan 16 at 9:24
$begingroup$
Maybe it could be interesting to ask just about the expected value of the determinant of a random binary matrix. AFAICT, the other open problems that were mentioned do not reduce to this problem.
$endgroup$
– Squark
Jan 16 at 12:40
5
$begingroup$
See the Wikipedia article on Hadamard's maximal determinant problem for the mapping between $0,1$ matrices and $-1,1$ matrices. For every $ntimes n$ $0,1$ matrix with determinant $D$, there are $2^2n+1$ $-1,1$ matrices with determinant $2^nD$. The case $n=1$ is slightly different.
$endgroup$
– Will Orrick
Jan 16 at 13:30
$begingroup$
@WillOrrick Thanks! That's indeed helpful.
$endgroup$
– Dirk
Jan 16 at 13:53
add a comment |
$begingroup$
This question is too hard, because easier questions are already known to be hard.
The middle column is A046747 in the OEIS, which is essentially equivalent to A057982, the number of singular ±1-valued matrices. The asymptotic behavior of this number was a longstanding open problem that was just recently solved by Tikhomirov.
As you mentioned yourself, the Hadamard maximal determinant problem is unsolved, with order 15 (for the ±1 version of the problem) being difficult already. The determinant spectrum problem, mentioned by Gerhard Paseman, is even harder, and is easier than your question, since it's just asking for which values are zero.
answered Jan 15 at 22:24
Timothy ChowTimothy Chow
34.7k11184311
$endgroup$
This question is too hard, because easier questions are already known to be hard.
The middle column is A046747 in the OEIS, which is essentially equivalent to A057982, the number of singular ±1-valued matrices. The asymptotic behavior of this number was a longstanding open problem that was just recently solved by Tikhomirov.
As you mentioned yourself, the Hadamard maximal determinant problem is unsolved, with order 15 (for the ±1 version of the problem) being difficult already. The determinant spectrum problem, mentioned by Gerhard Paseman, is even harder, and is easier than your question, since it's just asking for which values are zero.
answered Jan 15 at 22:24
Timothy ChowTimothy Chow
34.7k11184311
answered Jan 15 at 22:24
Timothy ChowTimothy Chow
34.7k11184311
answered Jan 15 at 22:24
Timothy ChowTimothy Chow
34.7k11184311
answered Jan 15 at 22:24
Timothy ChowTimothy Chow
34.7k11184311
34.7k11184311
$begingroup$
Thanks for the pointers! That's somehow what I expected. One thing: As you say, there are some results for $pm1$-matrices. I fail to see how these results help for $0,1$-matrices. Or do you mean to say that $0,1$ matrices are harder than $pm1$-matrices (and also I here I don't see why this should be so).
$endgroup$
– Dirk
Jan 16 at 9:24
$begingroup$
Maybe it could be interesting to ask just about the expected value of the determinant of a random binary matrix. AFAICT, the other open problems that were mentioned do not reduce to this problem.
$endgroup$
– Squark
Jan 16 at 12:40
5
$begingroup$
See the Wikipedia article on Hadamard's maximal determinant problem for the mapping between $0,1$ matrices and $-1,1$ matrices. For every $ntimes n$ $0,1$ matrix with determinant $D$, there are $2^2n+1$ $-1,1$ matrices with determinant $2^nD$. The case $n=1$ is slightly different.
$endgroup$
– Will Orrick
Jan 16 at 13:30
$begingroup$
@WillOrrick Thanks! That's indeed helpful.
$endgroup$
– Dirk
Jan 16 at 13:53
add a comment |
$begingroup$
Thanks for the pointers! That's somehow what I expected. One thing: As you say, there are some results for $pm1$-matrices. I fail to see how these results help for $0,1$-matrices. Or do you mean to say that $0,1$ matrices are harder than $pm1$-matrices (and also I here I don't see why this should be so).
$endgroup$
– Dirk
Jan 16 at 9:24
$begingroup$
Maybe it could be interesting to ask just about the expected value of the determinant of a random binary matrix. AFAICT, the other open problems that were mentioned do not reduce to this problem.
$endgroup$
– Squark
Jan 16 at 12:40
5
$begingroup$
See the Wikipedia article on Hadamard's maximal determinant problem for the mapping between $0,1$ matrices and $-1,1$ matrices. For every $ntimes n$ $0,1$ matrix with determinant $D$, there are $2^2n+1$ $-1,1$ matrices with determinant $2^nD$. The case $n=1$ is slightly different.
$endgroup$
– Will Orrick
Jan 16 at 13:30
$begingroup$
@WillOrrick Thanks! That's indeed helpful.
$endgroup$
– Dirk
Jan 16 at 13:53
$begingroup$
Thanks for the pointers! That's somehow what I expected. One thing: As you say, there are some results for $pm1$-matrices. I fail to see how these results help for $0,1$-matrices. Or do you mean to say that $0,1$ matrices are harder than $pm1$-matrices (and also I here I don't see why this should be so).
$endgroup$
– Dirk
Jan 16 at 9:24
$begingroup$
Thanks for the pointers! That's somehow what I expected. One thing: As you say, there are some results for $pm1$-matrices. I fail to see how these results help for $0,1$-matrices. Or do you mean to say that $0,1$ matrices are harder than $pm1$-matrices (and also I here I don't see why this should be so).
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– Dirk
Jan 16 at 9:24
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Maybe it could be interesting to ask just about the expected value of the determinant of a random binary matrix. AFAICT, the other open problems that were mentioned do not reduce to this problem.
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– Squark
Jan 16 at 12:40
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Maybe it could be interesting to ask just about the expected value of the determinant of a random binary matrix. AFAICT, the other open problems that were mentioned do not reduce to this problem.
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– Squark
Jan 16 at 12:40
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5
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See the Wikipedia article on Hadamard's maximal determinant problem for the mapping between $0,1$ matrices and $-1,1$ matrices. For every $ntimes n$ $0,1$ matrix with determinant $D$, there are $2^2n+1$ $-1,1$ matrices with determinant $2^nD$. The case $n=1$ is slightly different.
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– Will Orrick
Jan 16 at 13:30
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See the Wikipedia article on Hadamard's maximal determinant problem for the mapping between $0,1$ matrices and $-1,1$ matrices. For every $ntimes n$ $0,1$ matrix with determinant $D$, there are $2^2n+1$ $-1,1$ matrices with determinant $2^nD$. The case $n=1$ is slightly different.
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– Will Orrick
Jan 16 at 13:30
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@WillOrrick Thanks! That's indeed helpful.
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– Dirk
Jan 16 at 13:53
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@WillOrrick Thanks! That's indeed helpful.
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– Dirk
Jan 16 at 13:53
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Will Orrick probably has the latest data. Miodrag Zivkovic did an analysis in 2005 for n up to 9. While lower determinant values are heavily weighted, it is not known how quickly the counts for off as the determinant value grows. You can find related questions here on MathOverflow. Gerhard "Search For Determinant Spectrum Problem" Paseman, 2019.01.15.
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– Gerhard Paseman
Jan 15 at 20:33
1
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Related MO questions: mathoverflow.net/questions/18636/… mathoverflow.net/questions/18547/… mathoverflow.net/questions/39786/…
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– Timothy Chow
Jan 16 at 3:07
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@WillOrrick Of course - fixed!
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– Dirk
Jan 16 at 13:51
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I think it would be good to clarify the statement of the question, since there are different possible interpretations. The way I interpreted it was, roughly speaking, to understand the limiting shape of the histograms that are implicit in the data shown for $n=1,2,3,4.$ With this interpretation, understanding the precise asymptotics of the number of $n times n$ matrices with a given determinant is probably irrelevant (unless that value occurs a positive percentage of the time). However, the data suggest that the most common determinant value is $0$, which occurs $o(2^n^2)$ times (right?)
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– Matt Young
Jan 16 at 14:40
2
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The limiting shape of the histogram is known: in 2011 Hoi H. Nguyen and Van Vu proved that the log of the absolute value of the determinant approaches a normal distribution with mean $frac12log (n-1)!$ and variance $frac12log n$. (Note how large the mean is and how sharp the peak is--data for sizes less than $10$ are misleading.) This result applies to random $-1,1$ matrices and to many other classes of random matrices. Using the mapping mentioned in my comment to Timothy Chow's answer, you can adapt this result to $0,1$ matrices.
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– Will Orrick
Jan 16 at 17:35