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# karpathy/nes.py

Last active October 23, 2023 17:50
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Natural Evolution Strategies (NES) toy example that optimizes a quadratic function
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 """ A bare bones examples of optimizing a black-box function (f) using Natural Evolution Strategies (NES), where the parameter distribution is a gaussian of fixed standard deviation. """ import numpy as np np.random.seed(0) # the function we want to optimize def f(w): # here we would normally: # ... 1) create a neural network with weights w # ... 2) run the neural network on the environment for some time # ... 3) sum up and return the total reward # but for the purposes of an example, lets try to minimize # the L2 distance to a specific solution vector. So the highest reward # we can achieve is 0, when the vector w is exactly equal to solution reward = -np.sum(np.square(solution - w)) return reward # hyperparameters npop = 50 # population size sigma = 0.1 # noise standard deviation alpha = 0.001 # learning rate # start the optimization solution = np.array([0.5, 0.1, -0.3]) w = np.random.randn(3) # our initial guess is random for i in range(300): # print current fitness of the most likely parameter setting if i % 20 == 0: print('iter %d. w: %s, solution: %s, reward: %f' % (i, str(w), str(solution), f(w))) # initialize memory for a population of w's, and their rewards N = np.random.randn(npop, 3) # samples from a normal distribution N(0,1) R = np.zeros(npop) for j in range(npop): w_try = w + sigma*N[j] # jitter w using gaussian of sigma 0.1 R[j] = f(w_try) # evaluate the jittered version # standardize the rewards to have a gaussian distribution A = (R - np.mean(R)) / np.std(R) # perform the parameter update. The matrix multiply below # is just an efficient way to sum up all the rows of the noise matrix N, # where each row N[j] is weighted by A[j] w = w + alpha/(npop*sigma) * np.dot(N.T, A) # when run, prints: # iter 0. w: [ 1.76405235 0.40015721 0.97873798], solution: [ 0.5 0.1 -0.3], reward: -3.323094 # iter 20. w: [ 1.63796944 0.36987244 0.84497941], solution: [ 0.5 0.1 -0.3], reward: -2.678783 # iter 40. w: [ 1.50042904 0.33577052 0.70329169], solution: [ 0.5 0.1 -0.3], reward: -2.063040 # iter 60. w: [ 1.36438269 0.29247833 0.56990397], solution: [ 0.5 0.1 -0.3], reward: -1.540938 # iter 80. w: [ 1.2257328 0.25622233 0.43607161], solution: [ 0.5 0.1 -0.3], reward: -1.092895 # iter 100. w: [ 1.08819889 0.22827364 0.30415088], solution: [ 0.5 0.1 -0.3], reward: -0.727430 # iter 120. w: [ 0.95675286 0.19282042 0.16682465], solution: [ 0.5 0.1 -0.3], reward: -0.435164 # iter 140. w: [ 0.82214521 0.16161165 0.03600742], solution: [ 0.5 0.1 -0.3], reward: -0.220475 # iter 160. w: [ 0.70282088 0.12935569 -0.09779598], solution: [ 0.5 0.1 -0.3], reward: -0.082885 # iter 180. w: [ 0.58380424 0.11579811 -0.21083135], solution: [ 0.5 0.1 -0.3], reward: -0.015224 # iter 200. w: [ 0.52089064 0.09897718 -0.2761225 ], solution: [ 0.5 0.1 -0.3], reward: -0.001008 # iter 220. w: [ 0.50861791 0.10220363 -0.29023563], solution: [ 0.5 0.1 -0.3], reward: -0.000174 # iter 240. w: [ 0.50428202 0.10834192 -0.29828744], solution: [ 0.5 0.1 -0.3], reward: -0.000091 # iter 260. w: [ 0.50147991 0.1044559 -0.30255291], solution: [ 0.5 0.1 -0.3], reward: -0.000029 # iter 280. w: [ 0.50208135 0.0986722 -0.29841024], solution: [ 0.5 0.1 -0.3], reward: -0.000009

### onerachel commented Oct 1, 2022

Wrt this: w = w + alpha/(npopsigma) * np.dot(N.T, A), my understanding is that the author optimizes over w directly using stochastic gradient ascent with the score function estimator: (NA)/(npop*sigma). Matrix N is transposed by N.T to multiply with A easier. The return value is the estimate average reward value over npop, then this value is multiplied with the step size alpha, the w is updated.