Generate Random Binary Matrix

8
votes

I wish to generate 10,000 random binary matrices which have the same number of 1s per row and per column as a given binary matrix.

The matrix is ~500 x ~10,000. There are about 2,000,000 1s. There are no zero rows or columns.

My current method converts the binary matrix into a bipartite adjacency matrix, and performs 1,000,000 random edge switches to guarantee randomness. This takes 13,000 seconds for 1 matrix. I'm coding in python, using a modified version of networkx's double_edge_swap function.

Is there a more efficient way to generate such matrices?

2
pythonmatrixrandom
I was looking for the name of this problem. It is the main problem of discrete tomography "which deals with the reconstruction of a binary image from its horizontal and vertical linesums" and for the case of 2 dimensions (pairwise nonparallel lattice directions), the problem is in P. It would be interesting to know what needs 10,000 randomly chosen possible reconstructions. - Dan D.
You should specify if you need a particular distribution, since different methods might give slightly different distributions. - Veedrac
It depends if you want to improve only efficient for generate matricies the good solution will be call c( function for generate matrix from python. - ElConrado

2 Answers

2
votes

I think you can first build a special case of such a matrix, and then use numpy.shuffle to shuffle it:

row_sum = 2
col_sum = 1
arr     = np.zeros((5, 10))
#generate a special case, with given row_sum and col_sum
for i in range(row_sum):
    arr.ravel()[i::arr.shape[1]+row_sum] = 1
arr

Out[84]: 
array([[ 1.,  1.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  1.,  1.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  1.,  1.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  1.,  1.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  1.,  1.]])

np.random.shuffle(arr)
#np.random.shuffle(arr.T) to shuffle the columns
arr
Out[89]: 
array([[ 0.,  0.,  0.,  0.,  1.,  1.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  1.,  1.],
       [ 0.,  0.,  1.,  1.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  1.,  1.,  0.,  0.],
       [ 1.,  1.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.]])

arr.sum(1) #row sums
Out[90]: array([ 2.,  2.,  2.,  2.,  2.])

arr.sum(0) #col sums
Out[91]: array([ 1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.])
0
votes

Tried this, and it works

np.mod(np.random.permutation(N*N).reshape(N,N),2)

Example:

>>> np.mod(np.random.permutation(4*4).reshape(4,4),2)  
array([[0, 0, 0, 1],
      [1, 1, 1, 0],
      [1, 0, 0, 1],
      [0, 1, 1, 0]])
>>> np.mod(np.random.permutation(4*4).reshape(4,4),2)  
array([[0, 0, 0, 1],
      [1, 1, 0, 0],
      [1, 1, 1, 1],
      [0, 0, 1, 0]])