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# karpathy/nes.py

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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

### mynameisvinn commented Nov 16, 2017

@bercikr aside from the most recent parameter vector, why would you need previous parameter vectors? as long as youve correctly defined the fitness function f, then your NES model will never revisit (ie regress) previous states.

### GoingMyWay commented Mar 29, 2018 • edited

@alirezamika, hi, may I ask you a question, in line 50 why this update rule works in NES?

w = w + alpha/(npop*sigma) * np.dot(N.T, A)


since in every iteration, N is randomly sampled from Gaussian distribution.

### sandorvasas commented Jun 26, 2018

After reading the article, I feel each couple of years we can bring back old algorithms from the past, which in their time were too computationally intensive, and discover they work excellent, even [almost] better than current cutting-edge algorithms.

How can this be? Taking it to the extremes: if we have infinite computational capacity, even a model guessing with rand() will converge to a perfect policy.

The point I'm trying to make is that besides all the fancy names and "innovative" algorithms in deep learning, and ES, I'm starting to think that most of the achievements can be credited not to the algorithms themselves, but the constantly improving computational performance.

### EliasHasle commented Jul 31, 2018

@holdenlee I am wondering about that too. I guess if it were a mistake, it could still go unnoticed for this simple example.

### heidekrueger commented Aug 29, 2019

@holdenlee, @EliasHasle, the sigma in the denominator is indeed correct and required to make the ES-gradient the same length as the true gradient in expectation. (To see why, you could replace $F(\theta + \sigma\varepsilon)$ in the ES-definition by its first-order taylor expansion and then solve the expectation.)

### 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.

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