My question is simply the title:
What is the maximum possible value of determinant of a matrix whose entries either 0 or 1 ?
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Mher
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Since the determinant is the sum of $,n!,$ products of elements of the matrix chosen in a particular way, I'd say the maximum value is $,n!,$ ...but I've no example to offer. – DonAntonio Jun 20 '13 at 13:09
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3$n!$ is impossible. – Mher Jun 20 '13 at 13:13
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There can't be any example for that – Mher Jun 20 '13 at 13:13
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1An upper bound that is smaller than $n!$ is $n^{n/2}$. Google Hadamard's inequality. – KCd Jun 20 '13 at 13:14
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I can offer an example for $n-1.$ Taking $a_{ii}=0$ and $a_{ij}=1$ if $i\neq j.$ – Mher Jun 20 '13 at 13:16
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But for your energetic claims, @MherSafaryan, I see no proof or even hint... – DonAntonio Jun 20 '13 at 14:12
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@DonAntonio, proof of what? – Mher Jun 20 '13 at 14:41
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Of any of your claims below my first comment, @MherSafaryan – DonAntonio Jun 20 '13 at 14:43
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Do you need some help to understand that, @DonAntonio? – Mher Jun 20 '13 at 14:48
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Oh, dear: not at all, @MherSafaryan...hehe. It's just that it was you who asked, I offered a first, rather gross bound, and you claim that that is impossible...just like that. – DonAntonio Jun 20 '13 at 14:53
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And no: it was not obvious, or better: as obvious as your question could have been. – DonAntonio Jun 20 '13 at 14:53
2 Answers
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Quoting my question in another thread:
In fact, I don't even know how large the determinant of a 0-1 matrix can be. The Hadamard's bound for the absolute determinant of an $n\times n$ 0-1 matrix is $\frac{(n+1)^{(n+1)/2}}{2^n}$ (online ref. 1 and ref. 2), and the bound is sharp if and only if there exists a Hadamard matrix of order $n+1$. Yet, to my knowledge, there is no known sharp upper bound for the absolute determinant of a general $n\times n$ 0-1 matrix.
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Hm, can someone explain (or direct me to someplace) to understand why this answer is a community wiki? I've not dealt with community wikis before, so just curious. – Calvin Lin Jun 20 '13 at 14:44
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@CalvinLin See an explanation here. In the present case I turned my answer into a community wiki one on purpose, because I didn't think simply quoting my own question verbatim deserves any reputation points. – user1551 Jun 20 '13 at 15:13
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A few examples of $\{0,1\}$-matrices (with the largest determinants $-$ according to OEIS-A003432):
$n=2$: $\quad\det\left( \begin{array}{cc} \bf{1} & 0 \\ \bf{1} & \bf{1} \\ \end{array} \right) =1;$
$n=3$: $\quad\det\left( \begin{array}{ccc} \bf{1} & 0 & \bf{1} \\ \bf{1} & \bf{1} & 0 \\ 0 & \bf{1} & \bf{1} \\ \end{array} \right) =2;$
$n=4$: $\quad\det\left( \begin{array}{cccc} \bf{1} & 0 & \bf{1} & 0 \\ \bf{1} & \bf{1} & 0 & \bf{1} \\ 0 & \bf{1} & \bf{1} & 0 \\ 0 & 0& \bf{1} & \bf{1} \\ \end{array} \right) =3;$
$n=5$: $\quad\det\left( \begin{array}{ccccc} \bf{1} & 0 & \bf{1} & 0 & 0\\ \bf{1} & \bf{1} & 0 & \bf{1} & 0 \\ 0 & \bf{1} & \bf{1} & 0 &\bf{1}\\ 0 & 0 & \bf{1} & \bf{1} & 0 \\ \bf{1} & 0 & 0 & \bf{1} & \bf{1} \\ \end{array} \right) =5;$
$n=6$: $\quad\det\left( \begin{array}{ccccc} \bf{1} & 0 & \bf{1} & 0 & 0 & 0\\ \bf{1} & \bf{1} & 0 & \bf{1} & 0 & 0 \\ 0 & \bf{1} & \bf{1} & 0 &\bf{1} & 0\\ 0 & 0 & \bf{1} & \bf{1} & 0 & \bf{1}\\ \bf{1} & 0 & 0 & \bf{1} & \bf{1} & 0\\ \bf{1} & \bf{1} & 0 & 0 & \bf{1} & \bf{1} \\ \end{array} \right) =9;$
$n=7$: $\quad\det\left( \begin{array}{ccccc} \bf1 & 0 & \bf1 & 0 & 0 & \bf1 & \bf1 \\ \bf1 & \bf1 & 0 & \bf1 & 0 & 0 & \bf1 \\ \bf1 & \bf1 & \bf1 & 0 & \bf1 & 0 & 0 \\ 0 & \bf1 & \bf1 & \bf1 & 0 & \bf1 & 0 \\ 0 & 0 & \bf1 & \bf1 & \bf1 & 0 & \bf1 \\ \bf1 & 0 & 0 & \bf1 & \bf1 & \bf1 & 0 \\ 0 & \bf1 & 0 & 0 & \bf1 & \bf1 & \bf1 \\ \end{array} \right) =32;$
$n=8$: $\quad\det\left( \begin{array} \bf1 & 0 & \bf1 & 0 & 0 & \bf1 & \bf1 & 0 \\ \bf1 & \bf1 & 0 & \bf1 & 0 & 0 & \bf1 & \bf1 \\ \bf1 & \bf1 & \bf1 & 0 & \bf1 & 0 & 0 & \bf1 \\ 0 & \bf1 & \bf1 & \bf1 & 0 & \bf1 & 0 & 0 \\ 0 & 0 & \bf1 & \bf1 & \bf1 & 0 & \bf1 & 0 \\ \bf1 & 0 & 0 & \bf1 & \bf1 & \bf1 & 0 & \bf1 \\ 0 & \bf1 & 0 & 0 & \bf1 & \bf1 & \bf1 & 0 \\ 0 & 0 & \bf1 & 0 & 0 & \bf1 & \bf1 & \bf1 \\ \end{array} \right) =56;$
Oleg567
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1Very nice! I was also looking for patterns of symmetry but didn't search too long... (I stopped after having brute-forced $n=4$ with $2^{16}$ variants of the matrix ... ;-)) – Gottfried Helms Jun 21 '13 at 09:47
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Yes, we can find first 8 matrices among striped matrices ($\searrow \searrow \searrow$). What about others $n$ ??? – Oleg567 Jun 21 '13 at 09:51
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Impressive! Were you looking only among the striped ones? For $n=9$ you might need $17$ stripes, so you did not reach the largest value from the list? – orangeskid Jul 06 '22 at 18:25
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1@orangeskid: I looked among the striped shapes just to reduce the search area. The striped matrices provide maximum value for determinant only for small $n$ ($n\le 8$): $$n=9: \qquad det_{max}(striped) =125 \textrm{ vs } det_{max}=144; $$ $$n=10: \qquad det_{max}(striped)=312 \textrm{ vs } det_{max} = 320;$$ $$n=11: \qquad det_{max}(striped)=1215 \textrm{ vs } det_{max}=1458;$$ – Oleg567 Jul 07 '22 at 13:40
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Interesting! A question: given a striped determinant with $0$, $1$, is it largest among striped determinants with entries in $[0,1]$? Since it is not linear in the entries anymore... – orangeskid Jul 07 '22 at 20:55
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@orangeskid:
small update: $n=10$ has $det_{max}(striped) = 315$; and $n=11$ admits striped (cyclic) solution: $$ n=11:\qquad det\left( \begin{array}{lllllllllll} 1&0&1&0&0&0&1&1&1&0&1\ 1&1&0&1&0&0&0&1&1&1&0\ 0&1&1&0&1&0&0&0&1&1&1\ 1&0&1&1&0&1&0&0&0&1&1\ 1&1&0&1&1&0&1&0&0&0&1\ 1&1&1&0&1&1&0&1&0&0&0\ 0&1&1&1&0&1&1&0&1&0&0\ 0&0&1&1&1&0&1&1&0&1&0\ 0&0&0&1&1&1&0&1&1&0&1\ 1&0&0&0&1&1&1&0&1&1&0\ 0&1&0&0&0&1&1&1&0&1&1 \end{array} \right) = 1458 $$
Regarding striped determinant with entries in $[0,1]$: it looks so, but I have no proof.
– Oleg567 Jul 08 '22 at 07:38 -
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