Maximal determinant of a matrix filled with $\pm 1$ Is there an algorithm to determine what is the maximal determinant you can get just by putting $1$ or $-1$ in a square matrix?
For example in a $3\times3$ matrix:
$$
\begin{bmatrix}1 && -1 && 1\\ 1 && 1 && -1 \\ -1 && -1 && 1\end{bmatrix}
$$
Or do you have to go and bruteforce every possibility?

Hadarmard's inequality
$$
|\det(A)|^{2} \le \prod_{j=1}^{n}\sum_{k=1}^{n} |a_{jk}|^{2}
$$
applied to the case $a_{jk} = \pm 1$ yields $|\det(A)| \le n^{n/2}$. In the $3\times3$ case, this gives $\sqrt{27} \approx 5.19$. If 
$$
A = 
\begin{pmatrix}
1 & -1 & -1 \\
1 &  1 & 1  \\
-1 & -1 & 1
\end{pmatrix}
$$
then $\det(A) = 4$, so this inequality isn't a terrible overestimate. No help if you need the exact value though.
