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November 25, 2017 15:14
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Histogram filters, Monte-Carlo robot localization
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# From cs373, my implementation based on | |
# https://classroom.udacity.com/courses/cs373/lessons/48684821/concepts/487362110923 | |
# | |
# The function localize takes the following arguments: | |
# | |
# colors: | |
# 2D list, each entry either 'R' (for red cell) or 'G' (for green cell) | |
# | |
# measurements: | |
# list of measurements taken by the robot, each entry either 'R' or 'G' | |
# | |
# motions: | |
# list of actions taken by the robot, each entry of the form [dy,dx], | |
# where dx refers to the change in the x-direction (positive meaning | |
# movement to the right) and dy refers to the change in the y-direction | |
# (positive meaning movement downward) | |
# NOTE: the *first* coordinate is change in y; the *second* coordinate is | |
# change in x | |
# | |
# sensor_right: | |
# float between 0 and 1, giving the probability that any given | |
# measurement is correct; the probability that the measurement is | |
# incorrect is 1-sensor_right | |
# | |
# p_move: | |
# float between 0 and 1, giving the probability that any given movement | |
# command takes place; the probability that the movement command fails | |
# (and the robot remains still) is 1-p_move; the robot will NOT overshoot | |
# its destination in this exercise | |
# | |
# The function should RETURN (not just show or print) a 2D list (of the same | |
# dimensions as colors) that gives the probabilities that the robot occupies | |
# each cell in the world. | |
# | |
# Compute the probabilities by assuming the robot initially has a uniform | |
# probability of being in any cell. | |
# | |
# Also assume that at each step, the robot: | |
# 1) first makes a movement, | |
# 2) then takes a measurement. | |
# | |
# Motion: | |
# [0,0] - stay | |
# [0,1] - right | |
# [0,-1] - left | |
# [1,0] - down | |
# [-1,0] - up | |
def sense(colors, p, measurement, sensor_right): | |
# update with sense probability | |
for i in range(len(colors)): | |
for j in range(len(colors[0])): | |
if measurement == colors[i][j]: | |
p[i][j] *= sensor_right | |
else: | |
p[i][j] *= (1-sensor_right) | |
# normalize | |
s = sum(map(sum, p)) | |
for i in range(len(colors)): | |
for j in range(len(colors[0])): | |
p[i][j] /= s; | |
return p | |
def move(colors, p, motion, p_move): | |
x,y = motion | |
cols = len(colors[0]) | |
rows = len(colors) | |
q = [[0 for col in range(cols)] for row in range(rows)] | |
for i in range(rows): | |
for j in range(cols): | |
# y positive means right, x postive means downwards | |
# [0,0] - stay | |
# [0,1] - right | |
# [0,-1] - left | |
# [1,0] - down | |
# [-1,0] - up | |
q[i][j] = p_move * p[(i-x)%rows][(j-y)%cols] + (1-p_move)*p[i][j] | |
return q | |
def localize(colors,measurements,motions,sensor_right,p_move): | |
# initializes p to a uniform distribution over a grid of the same dimensions as colors | |
pinit = 1.0 / float(len(colors)) / float(len(colors[0])) | |
p = [[pinit for row in range(len(colors[0]))] for col in range(len(colors))] | |
# >>> Insert your code here <<< | |
for measurement, motion in zip(measurements, motions): | |
p = move(colors, p, motion, p_move) | |
p = sense(colors, p, measurement, sensor_right) | |
return p | |
def show(p): | |
rows = ['[' + ','.join(map(lambda x: '{0:.5f}'.format(x),r)) + ']' for r in p] | |
print '[' + ',\n '.join(rows) + ']' | |
############################################################# | |
# For the following test case, your output should be | |
# [[0.01105, 0.02464, 0.06799, 0.04472, 0.02465], | |
# [0.00715, 0.01017, 0.08696, 0.07988, 0.00935], | |
# [0.00739, 0.00894, 0.11272, 0.35350, 0.04065], | |
# [0.00910, 0.00715, 0.01434, 0.04313, 0.03642]] | |
# (within a tolerance of +/- 0.001 for each entry) | |
colors = [['R','G','G','R','R'], | |
['R','R','G','R','R'], | |
['R','R','G','G','R'], | |
['R','R','R','R','R']] | |
measurements = ['G','G','G','G','G'] | |
motions = [[0,0],[0,1],[1,0],[1,0],[0,1]] | |
p = localize(colors,measurements,motions,sensor_right = 0.7, p_move = 0.8) | |
show(p) # displays your answer |
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