How To Make Min-plus Matrix Multiplication In Python Faster?

May 29, 2024 Post a Comment

So I have two matrices, A and B, and I want to compute the min-plus product as given here: Min-plus matrix multiplication. For that I've implemented the following: def min_plus_pro

Solution 1:

Here is an algo that saves a bit if the middle dimension is large enough and entries are uniformly distributed. It exploits the fact that the smallest sum typically will be from two small terms.

import numpy as np

defmin_plus_product(A,B):
    B = np.transpose(B)
    Y = np.zeros((len(B),len(A)))
    for i inrange(len(B)):
         Y[i] = (A + B[i]).min(1)
    return np.transpose(Y)


defmin_plus_product_opt(A,B, chop=None):
    if chop isNone:
        # not sure this is optimal
        chop = int(np.ceil(np.sqrt(A.shape[1])))
    B = np.transpose(B)
    Amin = A.min(1)
    Y = np.zeros((len(B),len(A)))
    for i inrange(len(B)):
        o = np.argsort(B[i])
        Y[i] = (A[:, o[:chop]] + B[i, o[:chop]]).min(1)
        if chop < len(o):
            idx = np.where(Amin + B[i, o[chop]] < Y[i])[0]
            for j inrange(chop, len(o), chop):
                iflen(idx) == 0:
                    break
                x, y = np.ix_(idx, o[j : j + chop])
                slmin = (A[x, y] + B[i, o[j : j + chop]]).min(1)
                slmin = np.minimum(Y[i, idx], slmin)
                Y[i, idx] = slmin
                nidx = np.where(Amin[idx] + B[i, o[j + chop]] < Y[i, idx])[0]
                idx = idx[nidx]
    return np.transpose(Y)

A = np.random.random(size=(1000,1000))
B = np.random.random(size=(1000,2000))

print(np.allclose(min_plus_product(A,B), min_plus_product_opt(A,B)))

import time
t = time.time();min_plus_product(A,B);print('naive {}sec'.format(time.time()-t))
t = time.time();min_plus_product_opt(A,B);print('opt {}sec'.format(time.time()-t))

Sample output:

True
naive 7.794037580490112sec
opt 1.65810227394104sec

Solution 2:

A possible simple route is to use numba.

from numba import autojit
import numpy as np
@autojit(nopython=True)defmin_plus_product(A,B):
    n = A.shape[0]
    C = np.zeros((n,n))
    for i inrange(n):
        for j inrange(n):
            minimum = A[i,0]+B[0,j]
            for k inrange(1,n):
                minimum = min(A[i,k]+B[k,j],minimum)
            C[i,j] = minimum
    return C