Repeated addition would be a very inefficient way to multiply numbers, imagine multiplying 1298654825 by 85324154. Much quicker to just use long multiplication using binary. 1100100 0110111 ======= 0000000 -1100100 –1100100 —0000000 —-1100100 —–1100100 ——1100100 ============== 1010101111100 For floating point numbers scientific notation is used. 100 is 1 * 10^2 (10 to the power of … Read more
Add in an auxiliary variable, say x4, with constraints: x4 >= c1*x1 x4 >= c2*x2 x4 >= c3*x3 Objective += c4*x4
I managed to find a solution with objective value 1861.54 by stapling a couple ideas together. Form unordered color clusters of size 8 by finding a min-cost matching and joining matched subclusters, repeated three times. We use d(C1, C2) = ∑c1 in C1 ∑c2 in C2 d(c1, c2) as the distance function for subclusters C1 … Read more
While the SLSQP algorithm in scipy.optimize.minimize is good, it has a bunch of limitations. The first of which is it’s a QP solver, so it works will for equations that fit well into a quadratic programming paradigm. But what happens if you have functional constraints? Also, scipy.optimize.minimize is not a global optimizer, so you often … Read more
This constraint t[0] + t[1] = 1 would be an equality (type=”eq”) constraint, where you make a function that must equal zero: def con(t): return t[0] + t[1] – 1 Then you make a dict of your constraint (list of dicts if more than one): cons = {‘type’:’eq’, ‘fun’: con} I’ve never tried it, but … Read more
As far as off the shelf solutions, check out MAXLOADPRO for loading trucks. It may be able to be configured to load any rectangular volume, but I haven’t tried that yet. In general 3d bin-packing problems have the added complication that the objects can be rotated into different positions so for any object with a … Read more
I’m not very familiar with quadratic programming, but I think you can solve this sort of problem just using scipy.optimize‘s constrained minimization algorithms. Here’s an example: import numpy as np from scipy import optimize from matplotlib import pyplot as plt from mpl_toolkits.mplot3d.axes3d import Axes3D # minimize # F = x[1]^2 + 4x[2]^2 -32x[2] + 64 … Read more
Both approaches will save time, but the first one is very prone to integer overflow. Approach 1: This approach will generate result in shortest time (in at most n/2 iterations), and the possibility of overflow can be reduced by doing the multiplications carefully: long long C(int n, int r) { if(r > n – r) … Read more
I personally found GLPK better (i.e. faster) than LP_SOLVE. It supports various file formats, and a further advantage is its library interface, which allows smooth integration with your application.
At a local minimum (or maximum) x, the derivative of the target function f vanishes: f'(x) = 0 (assuming sufficient smoothness of f). Gradient descent tries to find such a minimum x by using information from the first derivative of f: It simply follows the steepest descent from the current point. This is like rolling … Read more