A determinant optimization challenge

Question

Consider 30 by 30 Toeplitz matrices all of whose entries are 0 or 1. This challenge is a simple optimization challenge to find the matrix with the largest determinant possible.

Input None

Output A 30 by 30 Toeplitz matrix all of whose entries are 0 or 1 along with its determinant.

Score The determinant of the matrix you output. If two people get the same score, the first answer wins.

Leading entries so far

65,455,857,159,975 in Matlab by Nick Alger (roughly (10^13.8)
65,455,857,159,975 in Python by isaacg (roughly 10^13.8)
39,994,961,721,988 in Mathematica by 2012rcampion ( roughly 10^13.6)
39,788,537,400,052 in R by Flounderer (roughly 10^13.6)
8,363,855,075,832 in Python by Vioz- (roughly 10^12.9)
6,984,314,690,903 in Julia by Alex A. (roughly 10^12.8)

Annoying additional constraints July 16 2015

If it is at all possible, please use arbitrary or high precision arithmetic to compute the final output determinant so that we can be sure what it really is (it should always be an integer). In python this should be helpful.

I'm surprised that this problem is not already solved. Is the answer known for circulant matrices? — xnor, Commented Jul 15, 2015 at 20:19
@NickAlger If the library is publicly available for everyone, you can use it. — orlp, Commented Jul 16, 2015 at 6:01
It's interesting that two independent methods have achieved a Toeplitz matrix with exactly the maximum circulant matrix determinant. I don't have any mathematical intuition as to why—is that determinant just common for binary Toeplitz matrices? — lirtosiast, Commented Jul 17, 2015 at 4:16
@Min_25 I should have the maximum up to 19 by tomorrow. Will get the code/values to you in the related question, Lembik. With heuristic algorithms, I have maxed out at exactly the same values for n=30 as the other two posters so far. Multiple times, with randomization involved. Also with circulant matrices as the result every time I reach that maximum, even though my search is not restricted to circulant matrices. BTW, another baffling (to me) fact: The maximum for n=15 is exactly 2^17. — Reto Koradi, Commented Jul 21, 2015 at 6:07

Nick Alger · Accepted Answer · 2015-07-17 06:49:16Z

Matlab, 65,455,857,159,975 (10^13.8159)

The method is gradient ascent in the interior of the cube [0,1]^59, with many random initial guesses, and rounding at the end to make everything zeros and ones.

Matrix:

0   1   1   1   0   0   0   0   1   0   1   1   0   1   0   0   1   0   0   0   1   0   1   1   1   0   1   1   1   0
0   0   1   1   1   0   0   0   0   1   0   1   1   0   1   0   0   1   0   0   0   1   0   1   1   1   0   1   1   1
1   0   0   1   1   1   0   0   0   0   1   0   1   1   0   1   0   0   1   0   0   0   1   0   1   1   1   0   1   1
1   1   0   0   1   1   1   0   0   0   0   1   0   1   1   0   1   0   0   1   0   0   0   1   0   1   1   1   0   1
1   1   1   0   0   1   1   1   0   0   0   0   1   0   1   1   0   1   0   0   1   0   0   0   1   0   1   1   1   0
0   1   1   1   0   0   1   1   1   0   0   0   0   1   0   1   1   0   1   0   0   1   0   0   0   1   0   1   1   1
1   0   1   1   1   0   0   1   1   1   0   0   0   0   1   0   1   1   0   1   0   0   1   0   0   0   1   0   1   1
1   1   0   1   1   1   0   0   1   1   1   0   0   0   0   1   0   1   1   0   1   0   0   1   0   0   0   1   0   1
1   1   1   0   1   1   1   0   0   1   1   1   0   0   0   0   1   0   1   1   0   1   0   0   1   0   0   0   1   0
0   1   1   1   0   1   1   1   0   0   1   1   1   0   0   0   0   1   0   1   1   0   1   0   0   1   0   0   0   1
1   0   1   1   1   0   1   1   1   0   0   1   1   1   0   0   0   0   1   0   1   1   0   1   0   0   1   0   0   0
0   1   0   1   1   1   0   1   1   1   0   0   1   1   1   0   0   0   0   1   0   1   1   0   1   0   0   1   0   0
0   0   1   0   1   1   1   0   1   1   1   0   0   1   1   1   0   0   0   0   1   0   1   1   0   1   0   0   1   0
0   0   0   1   0   1   1   1   0   1   1   1   0   0   1   1   1   0   0   0   0   1   0   1   1   0   1   0   0   1
1   0   0   0   1   0   1   1   1   0   1   1   1   0   0   1   1   1   0   0   0   0   1   0   1   1   0   1   0   0
0   1   0   0   0   1   0   1   1   1   0   1   1   1   0   0   1   1   1   0   0   0   0   1   0   1   1   0   1   0
0   0   1   0   0   0   1   0   1   1   1   0   1   1   1   0   0   1   1   1   0   0   0   0   1   0   1   1   0   1
1   0   0   1   0   0   0   1   0   1   1   1   0   1   1   1   0   0   1   1   1   0   0   0   0   1   0   1   1   0
0   1   0   0   1   0   0   0   1   0   1   1   1   0   1   1   1   0   0   1   1   1   0   0   0   0   1   0   1   1
1   0   1   0   0   1   0   0   0   1   0   1   1   1   0   1   1   1   0   0   1   1   1   0   0   0   0   1   0   1
1   1   0   1   0   0   1   0   0   0   1   0   1   1   1   0   1   1   1   0   0   1   1   1   0   0   0   0   1   0
0   1   1   0   1   0   0   1   0   0   0   1   0   1   1   1   0   1   1   1   0   0   1   1   1   0   0   0   0   1
1   0   1   1   0   1   0   0   1   0   0   0   1   0   1   1   1   0   1   1   1   0   0   1   1   1   0   0   0   0
0   1   0   1   1   0   1   0   0   1   0   0   0   1   0   1   1   1   0   1   1   1   0   0   1   1   1   0   0   0
0   0   1   0   1   1   0   1   0   0   1   0   0   0   1   0   1   1   1   0   1   1   1   0   0   1   1   1   0   0
0   0   0   1   0   1   1   0   1   0   0   1   0   0   0   1   0   1   1   1   0   1   1   1   0   0   1   1   1   0
0   0   0   0   1   0   1   1   0   1   0   0   1   0   0   0   1   0   1   1   1   0   1   1   1   0   0   1   1   1
1   0   0   0   0   1   0   1   1   0   1   0   0   1   0   0   0   1   0   1   1   1   0   1   1   1   0   0   1   1
1   1   0   0   0   0   1   0   1   1   0   1   0   0   1   0   0   0   1   0   1   1   1   0   1   1   1   0   0   1
1   1   1   0   0   0   0   1   0   1   1   0   1   0   0   1   0   0   0   1   0   1   1   1   0   1   1   1   0   0

Code:

% Toeplitz 0-1 determinant optimization

n = 30;
m = n + n-1;

toeplitz_map = @(w) toeplitz(w(n:-1:1), w(n:end));

objective = @(w) det(toeplitz_map(w));

detgrad = @(A) det(A)*inv(A)';

toeplitz_map_matrix = zeros(n^2,m);
for k=1:m
    ek = zeros(m,1);
    ek(k) = 1;
    M = toeplitz_map(ek);
    toeplitz_map_matrix(:,k) = M(:);
end

gradient = @(w) (reshape(detgrad(toeplitz_map(w)),1,n^2)*...
                 toeplitz_map_matrix)';

%check gradient with finite differences
w = randn(m,1);
dw = randn(m,1);
s = 1e-6;
g_diff = (objective(w+s*dw) - objective(w))/s;
g = gradient(w)'*dw;
grad_err = (g - g_diff)/g_diff

warning('off')
disp('multiple gradient ascent:')
w_best = zeros(m,1);
f_best = 0;
for trial=1:100000
    w0 = rand(m,1);
    w = w0;
    alpha0 = 1e-5; %step size
    for k=1:20
        f = objective(w);
        g = gradient(w);
        alpha = alpha0;
        for hh=1:100
            w2 = w + alpha*g;
            f2 = objective(w2);
            if f2 > f
                w = w2;
                break;
            else
                alpha = alpha/2;
            end
        end

        buffer = 1e-4;
        for jj=1:m
            if (w(jj) > 1)
                w(jj) = 1 - buffer;
            elseif (w(jj) < 0)
                w(jj) = 0 + buffer;
            end
        end
    end

    w = round(w);
    f = objective(w);
    if f > f_best
        w_best = w;
        f_best = f;
    end
    disp(trial)
    disp(f_best)
    disp(f)
end

M = toeplitz_map(w_best);

The math behind computing the gradient:

In the elementwise inner product (Ie., Hilbert-Schmidt inner product), the gradient of the determinant has Riesz representative G given by

G = det(A)A^(-*).

The map, J, from optimization variables (diagonal values) to toeplitz matrices is linear, so the overall gradient g is the composition of these two linear maps,

g = (vec(G)*J)',

where vec() is the vectorization operator that takes a matrix and unfolds it into a vector.

Interior gradient ascent:

After this all you have to do is pick an initial vector of diagonal values w_0, and for some small step sizes alpha iterate:

w_proposed = w_k + alpha*g_k
to get w_(k+1), take w_proposed and truncate values outside of [0,1] to 0 or 1
repeat until satisfied, then round everything to 0 or 1.

My result achieved this determinant after doing roughly 80,000 trials with uniform random initial guesses.

The OEIS link you gave was for circulant matrices, which are a special case of Topelitz matrices. So better is still possible. — isaacg, Commented Jul 16, 2015 at 9:09
Yes of course, I was incorrect about that. I have edited my post to fix it. — Nick Alger, Commented Jul 16, 2015 at 9:14
Yes, it got to that value at iteration 250 and stayed there for 100000 iterations. The vector defining the 18x18 toeplitz matrix with determinant 2994003 was [0,0,0,1,0,1,1,1,1,0,1,1,0,0,0,1,0,1,0,0,0,1,0,1,1,1,1,0,1,1,0,0,0,1,0], where the order goes from the bottom left to top right. — Nick Alger, Commented Jul 18, 2015 at 19:13
I awarded the win to you as you came up with a new idea and came up with the highest number first IIRC. Oh and this shows why your answer works math.stackexchange.com/questions/1364471/… . — user9206, Commented Jul 21, 2015 at 7:38

isaacg · Accepted Answer · 2015-07-16 13:17:34Z

Python 2 with Numpy, 65,455,857,159,975 ~= 10^13.8

This is hill climbing, as straightforward as can be. Final determinant calculation performed using SymPy to enusure an exact result. All matrices found with this determinant are circulant.

Matrices found with this determinant, given as value of diagonal from bottom left to upper right:

01000100101101000011100111011101000100101101000011100111011
01011101110011100001011010010001011101110011100001011010010
01100001000111011101001110100101100001000111011101001110100
01110100111010010110000100011101110100111010010110000100011
01011101110001000011010010111001011101110001000011010010111
01000101100010110100111101110001000101100010110100111101110
01000100101101000011100111011101000100101101000011100111011

The first one, as a matrix:

[[1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1]
 [1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1]
 [1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0]
 [0 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1]
 [1 0 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1]
 [1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1]
 [1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0]
 [0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0]
 [0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1]
 [1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1]
 [1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1]
 [1 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0]
 [0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0]
 [0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0]
 [0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1 0]
 [0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0 1]
 [1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1 0]
 [0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 0 1 1]
 [1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 0 1]
 [1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 0]
 [0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1]
 [1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0]
 [0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0]
 [0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1]
 [1 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0]
 [0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 0 0]
 [0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1 0]
 [0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 1]
 [1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0]
 [0 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1]]

Code:

import numpy as np
import sympy as sp
import random
import time
SIZE = 30

random.seed(0)

def gen_diag():
    return [random.randint(0, 1) for i in range(SIZE*2 - 1)]

def diag_to_mat(diag):
    return [diag[a:a+SIZE] for a in range(SIZE-1, -1, -1)]

def diag_to_det(diag):
    matrix = diag_to_mat(diag)
    return np.linalg.det(matrix)

def improve(diag):
    old_diag = diag
    really_old_diag = []
    while really_old_diag != old_diag:
        really_old_diag = old_diag
        for flip_at in range(SIZE * 2 - 1):
            new_diag = old_diag[:]
            new_diag[flip_at] ^= 1
            old_diag = max(old_diag, new_diag, key=diag_to_det)
    return old_diag

overall_best_score = 0
time.clock()
while time.clock() < 500:
    best = improve(gen_diag())
    best_score = diag_to_det(best)
    if best_score > overall_best_score:
        overall_best_score = best_score
        overall_best = best
        print(time.clock(), sp.Matrix(diag_to_mat(overall_best)).det(), ''.join(map(str,overall_best)))


mat = diag_to_mat(overall_best)

sym_mat = sp.Matrix(mat)

print(overall_best)
print(sym_mat.det())

The .227 is a little worrying. Do you think there is a way to be confident what the determinant really is? — user9206, Commented Jul 16, 2015 at 7:09
It seems that stackoverflow.com/questions/6876377/… might help to evaluate the final determinant? — user9206, Commented Jul 16, 2015 at 8:08

Alex A. · Accepted Answer · 2015-07-16 17:53:46Z

R, 39 788 537 400 052

Here is my attempt to do a genetic algorithm but only with asexual reproduction. I hope I understood the challenge correctly. Edit: sped it up a bit, tried a different random seed, and restricted to 100 generations.

    options(scipen=999)

toeplitz <- function(x){
# make toeplitz matrix with first row
# x[1:a] and first col x[(a+1):n]
# where n is the length of x and a= n/2
# Requires x to have even length
#
# [1,1] entry is x[a+1]

N <- length(x)/2
out <- matrix(0, N, N)
out[1,] <- x[1:N]
out[,1] <- x[(N+1):length(x)]
for (i in 2:N){
  for (j in 2:N){
    out[i,j] <- out[i-1, j-1]
  }
} 

out
}

set.seed(1002)

generations <- 100
popsize <- 25
cols <- 60
population <- matrix(sample(0:1, cols*popsize, replace=T), nc=cols)
numfresh <- 5 # number of totally random choices added to population

for (i in 1:generations){

fitness <- apply(population, 1, function(x) det(toeplitz(x)) )
mother <- which(fitness==max(fitness))[1]

population <- matrix(rep(population[mother,], popsize), nc=cols, byrow=T)
for (i in 2:(popsize-numfresh)){
  x <- sample(cols, 1)
  population[i,x] <- 1-population[i,x]
}
for (i in (popsize-numfresh +1):popsize){
  population[i,] <- sample(0:1, cols, replace=T)
}


print(population[1,])
print(fitness[mother])
print(det(toeplitz(population[1,]))) # to check correct

}

Output:

print(population[1, 1:(cols/2)]) # first row
print(population[1, (cols/2+1):(cols)]) # first column (overwrites 1st row)

to <- toeplitz(population[1,])

for (i in 1:(cols/2)) cat(to[i,], "\n")

1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 
0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 
1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 
0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 
0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 
0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 
1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 
1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 
1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 
1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 
0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 
1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 
1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 
1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 
0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 
0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 0 
0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 
0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 
1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 
0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 0 
0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 
1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 
0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 
0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 
0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 0 
0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 1 
1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 0 
0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 1 
1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 
1 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1

\$\begingroup\$ This is very nice. You are winning by a long way currently. \$\endgroup\$
– user9206
Commented Jul 16, 2015 at 0:35 — user9206, Commented Jul 16, 2015 at 0:35
\$\begingroup\$ Not any more :) \$\endgroup\$
– user9206
Commented Jul 16, 2015 at 8:51 — user9206, Commented Jul 16, 2015 at 8:51

Alex A. · Accepted Answer · 2015-07-15 21:51:33Z

3

Julia, 6,984,314,690,902.998

This just constructs 1,000,000 random Toeplitz matrices and checks their determinants, recording the maximum encountered. Hopefully someone will come up with an elegant analytical solution, but in the meantime...

function toeplitz(a, b)
    n = length(a)
    T = Array(Int, n, n)
    T[1,:] = b
    T[:,1] = a
    for i = 2:n
        T[i,2:n] = T[i-1,1:n-1]
    end
    T
end

d = 0
A = Any[]

for i = 1:1000000
    # Construct two random 0,1 arrays
    r1 = rand(0:1, 30)
    r2 = rand(0:1, 30)

    # Compute the determinant of a toeplitz matrix constructed
    # from the two random arrays
    D = det(toeplitz(r1, r2))

    # If the computed determinant is larger than anything we've
    # encountered so far, add it to A so we can access it later
    D > d && begin
        push!(A, (D, r1, r2))
        d = D
    end
end

M,N = findmax([i[1] for i in A])

println("Maximum determinant: ", M, "\n")
println(toeplitz(A[N][2], A[N][3]))

You can view the output here.

edited Jul 15, 2015 at 21:51

answered Jul 15, 2015 at 21:32

Alex A.

24.7k5 gold badges38 silver badges119 bronze badges

\$\begingroup\$ I wonder how precise the determinant calculation is. I figure that the underlying computation is done in double precision? Since the digits after the decimal point are .998, there's probably a good chance that the closest integer is still the correct determinant. Generally, you will start getting floating point precision issues when applying a general purpose determinant calculation, e.g. based on a standard LR decomposition, to these matrices once they get fairly large. \$\endgroup\$
– Reto Koradi
Commented Jul 15, 2015 at 21:59
\$\begingroup\$ @RetoKoradi Looks like it uses an LU decomposition to get the determinant. \$\endgroup\$
– Alex A.
Commented Jul 16, 2015 at 3:02

Add a comment |

2012rcampion · Accepted Answer · 2015-07-16 04:18:58Z

Mathematica, 39,994,961,721,988 (10^13.60)

A simple simulated annealing optimization method; no optimization or tuning yet.

n = 30;
current = -\[Infinity];
best = -\[Infinity];
saved = ConstantArray[0, {2 n - 1}];
m := Array[a[[n + #1 - #2]] &, {n, n}];
improved = True;
iters = 1000;
pmax = 0.1;
AbsoluteTiming[
 While[improved || RandomReal[] < pmax,
   improved = False;
   a = saved;
   Do[
    Do[
      a[[i]] = 1 - a[[i]];
      With[{d = Det[m]},
       If[d > best,
          best = d;
          current = d;
          saved = a;
          improved = True;
          Break[];,
          If[d > current || RandomReal[] < pmax (1 - p/iters),
           current = d;
           Break[];,
           a[[i]] = 1 - a[[i]];
           ]
          ];
        ;
       ],
      {i, 2 n - 1}];,
    {p, iters}];
   ];
 ]
best
Log10[best // N]
a = saved;
m // MatrixForm

Sample output:

{20.714876,Null}
39994961721988
13.602
(1  1   1   0   1   0   0   0   0   1   1   0   1   1   1   0   1   1   0   1   1   0   0   0   0   1   0   0   0   0
0   1   1   1   0   1   0   0   0   0   1   1   0   1   1   1   0   1   1   0   1   1   0   0   0   0   1   0   0   0
0   0   1   1   1   0   1   0   0   0   0   1   1   0   1   1   1   0   1   1   0   1   1   0   0   0   0   1   0   0
0   0   0   1   1   1   0   1   0   0   0   0   1   1   0   1   1   1   0   1   1   0   1   1   0   0   0   0   1   0
0   0   0   0   1   1   1   0   1   0   0   0   0   1   1   0   1   1   1   0   1   1   0   1   1   0   0   0   0   1
1   0   0   0   0   1   1   1   0   1   0   0   0   0   1   1   0   1   1   1   0   1   1   0   1   1   0   0   0   0
0   1   0   0   0   0   1   1   1   0   1   0   0   0   0   1   1   0   1   1   1   0   1   1   0   1   1   0   0   0
0   0   1   0   0   0   0   1   1   1   0   1   0   0   0   0   1   1   0   1   1   1   0   1   1   0   1   1   0   0
0   0   0   1   0   0   0   0   1   1   1   0   1   0   0   0   0   1   1   0   1   1   1   0   1   1   0   1   1   0
0   0   0   0   1   0   0   0   0   1   1   1   0   1   0   0   0   0   1   1   0   1   1   1   0   1   1   0   1   1
1   0   0   0   0   1   0   0   0   0   1   1   1   0   1   0   0   0   0   1   1   0   1   1   1   0   1   1   0   1
1   1   0   0   0   0   1   0   0   0   0   1   1   1   0   1   0   0   0   0   1   1   0   1   1   1   0   1   1   0
0   1   1   0   0   0   0   1   0   0   0   0   1   1   1   0   1   0   0   0   0   1   1   0   1   1   1   0   1   1
1   0   1   1   0   0   0   0   1   0   0   0   0   1   1   1   0   1   0   0   0   0   1   1   0   1   1   1   0   1
1   1   0   1   1   0   0   0   0   1   0   0   0   0   1   1   1   0   1   0   0   0   0   1   1   0   1   1   1   0
0   1   1   0   1   1   0   0   0   0   1   0   0   0   0   1   1   1   0   1   0   0   0   0   1   1   0   1   1   1
1   0   1   1   0   1   1   0   0   0   0   1   0   0   0   0   1   1   1   0   1   0   0   0   0   1   1   0   1   1
1   1   0   1   1   0   1   1   0   0   0   0   1   0   0   0   0   1   1   1   0   1   0   0   0   0   1   1   0   1
1   1   1   0   1   1   0   1   1   0   0   0   0   1   0   0   0   0   1   1   1   0   1   0   0   0   0   1   1   0
0   1   1   1   0   1   1   0   1   1   0   0   0   0   1   0   0   0   0   1   1   1   0   1   0   0   0   0   1   1
1   0   1   1   1   0   1   1   0   1   1   0   0   0   0   1   0   0   0   0   1   1   1   0   1   0   0   0   0   1
1   1   0   1   1   1   0   1   1   0   1   1   0   0   0   0   1   0   0   0   0   1   1   1   0   1   0   0   0   0
0   1   1   0   1   1   1   0   1   1   0   1   1   0   0   0   0   1   0   0   0   0   1   1   1   0   1   0   0   0
0   0   1   1   0   1   1   1   0   1   1   0   1   1   0   0   0   0   1   0   0   0   0   1   1   1   0   1   0   0
0   0   0   1   1   0   1   1   1   0   1   1   0   1   1   0   0   0   0   1   0   0   0   0   1   1   1   0   1   0
0   0   0   0   1   1   0   1   1   1   0   1   1   0   1   1   0   0   0   0   1   0   0   0   0   1   1   1   0   1
1   0   0   0   0   1   1   0   1   1   1   0   1   1   0   1   1   0   0   0   0   1   0   0   0   0   1   1   1   0
0   1   0   0   0   0   1   1   0   1   1   1   0   1   1   0   1   1   0   0   0   0   1   0   0   0   0   1   1   1
1   0   1   0   0   0   0   1   1   0   1   1   1   0   1   1   0   1   1   0   0   0   0   1   0   0   0   0   1   1
1   1   0   1   0   0   0   0   1   1   0   1   1   1   0   1   1   0   1   1   0   0   0   0   1   0   0   0   0   1

)

Kade · Accepted Answer · 2015-07-15 23:06:47Z

Python 2, 8 363 855 075 832

This has a very basic, nearly nonexistent strategy involved.

from scipy import linalg

start = 2**28
mdet  = 0
mmat  = []
count = 0
powr  = 1
while 1:
 count += 1
 v = map(int,bin(start)[2:].zfill(59))
 m = [v[29:]]
 for i in xrange(1,30):
     m += [v[29-i:59-i]]
 d = 0
 try: d = linalg.det(m, check_finite=False)
 except: print start
 if d > mdet:
     print d
     print m
     mdet = d
     mmat = m
     start += 1
     powr = 1
 else:
     start += 2**powr
     powr += 1
     if start>(2**59-1):
         start-=2**59-1
         powr = 1
 if count%10000==0: print 'Tried',count

Here is the best matrix it found after ~5,580,000 tries:

1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0 0 0 0 1 0 0 1 0 1 1 0 1 0
1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0 0 0 0 1 0 0 1 0 1 1 0 1
1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0 0 0 0 1 0 0 1 0 1 1 0
0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0 0 0 0 1 0 0 1 0 1 1
1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0 0 0 0 1 0 0 1 0 1
0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0 0 0 0 1 0 0 1 0
1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0 0 0 0 1 0 0 1
0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0 0 0 0 1 0 0
0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0 0 0 0 1 0
1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0 0 0 0 1
1 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0 0 0 0
0 1 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0 0 0
0 0 1 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0 0
0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 0
0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0
0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0
1 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1
0 1 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 1
1 0 1 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1
1 1 0 1 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0
1 1 1 0 1 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1
1 1 1 1 0 1 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1
1 1 1 1 1 0 1 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1
1 1 1 1 1 1 0 1 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0
1 1 1 1 1 1 1 0 1 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1
1 1 1 1 1 1 1 1 0 1 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 0 0 1
0 1 1 1 1 1 1 1 1 0 1 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 0 0
0 0 1 1 1 1 1 1 1 1 0 1 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1 0
1 0 0 1 1 1 1 1 1 1 1 0 1 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 1
0 1 0 0 1 1 1 1 1 1 1 1 0 1 0 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1

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A determinant optimization challenge

6 Answers 6

Matlab, 65,455,857,159,975 (10^13.8159)

Python 2 with Numpy, 65,455,857,159,975 ~= 10^13.8

R, 39 788 537 400 052

Julia, 6,984,314,690,902.998

Mathematica, 39,994,961,721,988 (10^13.60)

Python 2, 8 363 855 075 832

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A determinant optimization challenge

6 Answers 6

Matlab, 65,455,857,159,975 (10^13.8159)

Python 2 with Numpy, 65,455,857,159,975 ~= 10^13.8

R, 39 788 537 400 052

Julia, 6,984,314,690,902.998

Mathematica, 39,994,961,721,988 (10^13.60)

Python 2, 8 363 855 075 832

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