What is the maximum determinant in an 8 by 8 matrix of zeros and ones

I shouldn't expect there to be exact results; compare the similar problem with matrices with entries $\pm1$. For an $n$-by-$n$ matrix with entries $\pm1$ one gets an upper bound for the determinant of $n^$. Taking some extra care one can ensure the first column of $A$ has a $1$ atop a column of zeros so that deleting the first row and column gives a smaller matrix with the same determinant. It's not clear to me whether this is the biggest possible.... As pointed out in Robin Chapman's answer and Frederio Poloni's comment thereto, there is a one-to-one map between normalized $n$-by- $n$ $(-1,1)$ matrices and $(n-1)$-by- $(n-1)$ $(0,1)$ matrices under which the determinant gets multiplied by $2^$: The optimal $\tilde G$ is $-J+4B+(n-3)I$ where $B$ is a block-diagonal matrix with all-ones blocks along the diagonal. These all-ones blocks should be as nearly equal in size as possible and their number should be five if $n=7$, five or six if $n=11$, six if $15 \le n \le 59$, and seven if $n \ge 63$. The bound on $\det \tilde G$ one gets from this is seldom a perfect square, and even when it is, Tamura showed that the existence of an $R$ such that $RR^T$ equals the optimal $\tilde G$ is ruled out by the Hasse-Minkowski theorem in many cases. The smallest $n$ for which the bound might be attainable is 511. Nevertheless, there exist matrices with determinant quite close to the bound. For example, when $n=19$, a matrix whose determinant attains 97.5% of the bound is known. Using a lo...

This is an interview question: given a boolean matrix find the size of the largest contiguous square sub-matrix, which contains only "true" elements. I found this My questions: • How to keep track of the rectangles in the algorithm? • Why do we move the diagonal line? What if we move a vertical/horizontal line or both? • How to calculate the algorithm complexity? I can't understand O(N^(3/2)*logN) algorithm where N=n*n is the number of elements in the original matrix.) For your largest square sub-matrix problem, there is a DP algorithm whose complexity is linear with the number of elements: Let the original matrix is A[n][n], we are trying to find a matrix B[n][n], where B[i][j] indicates the size of largest square sub-matrix whose bottom-right element is A[i][j]. So for (i = 0 ; i < n ; ++i) for (j = 0 ; j < n ; ++j) if (A[i][j] == 0) B[i][j] = 0; else And the largest B[i][j] is the answer. p.s. I didn't check the array range for simplification. You can just consider the out-of-range elements are zero. Thanks for contributing an answer to Stack Overflow! • Please be sure to answer the question. Provide details and share your research! But avoid … • Asking for help, clarification, or responding to other answers. • Making statements based on opinion; back them up with references or personal experience. To learn more, see our

Hello Sal. I know I'm wrong and the answer is probably staring me in the face! Where's the fallacy in my thinking: As I understand it, a square matrix whose determinant is not zero is invertible. Therefore, using row operations, it can be reduced to having all its column vectors as pivot vectors. That's equvialent to an upper triangular matrix, with the main diagonal elements equal to 1. If normal row operations do not change the determinant, the determinant will be -1. HELP! :) Sal was a bit unfortunate when he said that "row operations do not change the determinant". The proof he gave was that row operations of the form [Ri -c*Rj -> Ri] (that is, replacing row i with row i minus row j times a scalar c) do not change the determinant. But there are row operations of different kind, such as k*Ri -c*Rj -> Ri (That is, replacing row i with row i times a scalar k minus row j times a scalar c). What can be proved is that operations of this kind do change the determinant. In fact, they multiply the determinant by k. And when you put an invertible matrix in RREF (that is, you turn it into an identity matrix), you must do these kinds of operations that scale the determinant. And they always end up scaling the determinant back to one, which is the determinant of any identity matrix. It depends on the size of A and B. Multiplying a matrix by a scalar, is the same as multiplying every row of that matrix by that scalar, and note, that multiplying a single row by a scalar is equivalent...

first, before moving on to the solution. Approach The problem can be solved using a greedy approach. Place a zero at the right-bottom corner of the first square sub-matrix, i.e. the sub-matrix with coordinates (1, 1) and (x, x), and create the rest of the matrix symmetrically, we can get the minimum number of zeros, or, the maximum number of ones. Thus by observing, a common conclusion can be drawn that there are number of zeroes, in the minimum arrangement. The total number of cells available is in a NxN matrix. .