机器人世界

小飞侠xp

发布于 2018-10-15 11:47:59

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发布于 2018-10-15 11:47:59

与人类一样，机器人通过它的“感官”来感知世界。例如，无人驾驶汽车使用视频、雷达与激光雷达来观察周围的世界。随着汽车不断地收集数据，它们会建立起一个3D观察世界，在这个世界，汽车可以知道自己在哪里，其他物体（如树木、行人和其他车辆）在哪里，以及它应该去哪里！

机器人世界： 1-D

首先，假设你有一个生活在1-D世界中机器人。你可以将这个1-D世界看做是一个单车道道路

我们可以将这条道路视为一个阵列，并将其分解为很多网格单元，便于机器人理解。在这种情况下，这条道路是一个一维网格，有5个不同的空间。机器人只能向前或向后移动。如果机器人从网格上掉下来，它将环绕回到另一侧（这被称为循环世界）。

均匀分布

机器人有一张地图，因此它知道这个1D世界中只有5个空间，但是，它没有感觉到任何东西或没有发生移动。那么，对于这个5个单元格长度(5个值的列表)，机器人在这些位置中的任何一个位置的概率分布p是多少？由于机器人最初不知道它在哪里，所以在任何空间中的概率是相同的！这就是概率分布，因此所有这些概率的总和应该等于1，所以1/5 spaces = 0.2。所有概率相同（并且具有最大不确定性）的分布称为均匀分布。

# importing resources
import matplotlib.pyplot as plt
import numpy as np
# uniform distribution for 5 grid cells
# we use "p" to represent probability
p = [0.2, 0.2, 0.2, 0.2, 0.2]
print(p)

下面的函数display_map将会输出一个条形图，用于显示机器人在每个网格空间中的概率。对于概率范围，y轴的范围为0到1。在均匀分布中，这看起来像一条扁平线。如果你想将它们分开，可以选择每个条的宽度<= 1。

def display_map(grid, bar_width=1):
    if(len(grid) > 0):
        x_labels = range(len(grid))
        plt.bar(x_labels, height=grid, width=bar_width, color='b')
        plt.xlabel('Grid Cell')
        plt.ylabel('Probability')
        plt.ylim(0, 1) # range of 0-1 for probability values 
        plt.title('Probability of the robot being at each cell in the grid')
        plt.xticks(np.arange(min(x_labels), max(x_labels)+1, 1))
        plt.show()
    else:
        print('Grid is empty')

# call function on grid, p, from before
display_map(p)

机器人传感器

机器人通过摄像头和其他传感器来感知世界，但这些传感器并不是完全正确。感知后的概率：机器人会感知它处于一个红色单元格中，并更新其概率。根据我们的例子：

机器人对颜色感知正确的概率是pHit = 0.6。
机器人对颜色感知不正确的概率是pMiss = 0.2，在这个示例中：它看到了红色，但实际上它在绿色单元格中

# given initial variables
p = initialize_robot(5)
pHit  = 0.6
pMiss = 0.2

# Creates a new grid, with modified probabilities, after sensing
# All values are calculated by a product of 1. the sensing probability for a color (pHit for red)
# and 2. the current probability of a robot being in that location p[i]; all equal to 0.2 at first.
p[0] = p[0]*pMiss
p[1] = p[1]*pHit
p[2] = p[2]*pHit
p[3] = p[3]*pMiss
p[4] = p[4]*pMiss

print(p)
display_map(p)

# given initial variables
p=[0.2, 0.2, 0.2, 0.2, 0.2]
# the color of each grid cell in the 1D world
world=['green', 'red', 'red', 'green', 'green']
# Z, the sensor reading ('red' or 'green')
Z = 'red'
pHit = 0.6
pMiss = 0.2

## Complete this function
def sense(p, Z):
    ''' Takes in a current probability distribution, p, and a sensor reading, Z.
        Returns an unnormalized distribution after the sensor measurement has been made, q.
        This should be accurate whether Z is 'red' or 'green'. '''
    
    q=[]
    for i in range(len(p)):
        hit = (Z == world[i])
        q.append(p[i] * (hit * pHit + (1-hit) * pMiss))
    return q

q = sense(p,Z)
print(q)
display_map(q)

# given initial variables
p=[0.2, 0.2, 0.2, 0.2, 0.2]
# the color of each grid cell in the 1D world
world=['green', 'red', 'red', 'green', 'green']
# Z, the sensor reading ('red' or 'green')
Z = 'red'
pHit = 0.6
pMiss = 0.2

## Complete this function
def sense(p, Z):
    ''' Takes in a current probability distribution, p, and a sensor reading, Z.
        Returns a *normalized* distribution after the sensor measurement has been made, q.
        This should be accurate whether Z is 'red' or 'green'. '''
    
    q=[]
    
    ##TODO: normalize q
    
    # loop through all grid cells
    for i in range(len(p)):
        # check if the sensor reading is equal to the color of the grid cell
        # if so, hit = 1
        # if not, hit = 0
        hit = (Z == world[i])
        q.append(p[i] * (hit * pHit + (1-hit) * pMiss))
        
    # sum up all the components
    s = sum(q)
    # divide all elements of q by the sum
    for i in range(len(p)):
        q[i] = q[i] / s
    return q

q = sense(p,Z)
print(q)
display_map(q)

Move函数

使其返回一个新的分布q，向右运动（U）个单位。此函数下的分布应随着运动U的变化而变化,如果U = 1，你应该可以看到p中的所有值都向右移动1。

## TODO: Complete this move function so that it shifts a probability distribution, p
## by a given motion, U
def move(p, U):
    q=[]
    # iterate through all values in p
    for i in range(len(p)):
        # use the modulo operator to find the new location for a p value
        # this finds an index that is shifted by the correct amount
        index = (i-U) % len(p)
        # append the correct value of p to q
        q.append(p[index])
    return q

p = move(p,2)
print(p)
display_map(p)

不精确的运动

对于初始分布[0,1,0,0,0]和U = 2的运动，当考虑运动中的不确定性时，得到的分布是：[0,0,0.1,0.8,0.1]

修改move函数，使其能够容纳机器人超越或未到达预期目的地的附加概率此函数下的分布应随着运动U的变化而变化，但其中有一些未达到目的地或超越目的地的概率。对于给定的初始p，你应该可以看到一个U = 1的结果并且包含不确定性：[0.0, 0.1, 0.8, 0.1, 0.0]。

## TODO: Modify the move function to accommodate the added robabilities of overshooting or undershooting 
pExact = 0.8
pOvershoot = 0.1
pUndershoot = 0.1

# Complete the move function
def move(p, U):
    q=[]
    # iterate through all values in p
    for i in range(len(p)):
        # use the modulo operator to find the new location for a p value
        # this finds an index that is shifted by the correct amount
        index = (i-U) % len(p)
        nextIndex = (index+1) % len(p)
        prevIndex = (index-1) % len(p)
        s = pExact * p[index]
        s = s + pOvershoot  * p[nextIndex]
        s = s + pUndershoot * p[prevIndex]
        # append the correct, modified value of p to q
        q.append(s)
    return q

## TODO: try this for U = 2 and see the result
p = move(p,1)
print(p)
display_map(p)

多次不精确移动

# given initial variables
p=[0, 1, 0, 0, 0]
# the color of each grid cell in the 1D world
world=['green', 'red', 'red', 'green', 'green']
# Z, the sensor reading ('red' or 'green')
Z = 'red'
pHit = 0.6
pMiss = 0.2

pExact = 0.8
pOvershoot = 0.1
pUndershoot = 0.1

# Complete the move function
def move(p, U):
    q=[]
    # iterate through all values in p
    for i in range(len(p)):
        # use the modulo operator to find the new location for a p value
        # this finds an index that is shifted by the correct amount
        index = (i-U) % len(p)
        nextIndex = (index+1) % len(p)
        prevIndex = (index-1) % len(p)
        s = pExact * p[index]
        s = s + pOvershoot  * p[nextIndex]
        s = s + pUndershoot * p[prevIndex]
        # append the correct, modified value of p to q
        q.append(s)
    return q

# Here is code for moving twice
# p = move(p, 1)
# p = move(p, 1)
# print(p)
# display_map(p)
## TODO: Write code for moving 1000 times
for k in range(1000):
    p = move(p,1)

print(p)
display_map(p)

谷歌无人车定位功能基本就是以上代码实现，编写了两个部分：感知和移动

感知与移动循环

# importing resources
import matplotlib.pyplot as plt
import numpy as np

def display_map(grid, bar_width=1):
    if(len(grid) > 0):
        x_labels = range(len(grid))
        plt.bar(x_labels, height=grid, width=bar_width, color='b')
        plt.xlabel('Grid Cell')
        plt.ylabel('Probability')
        plt.ylim(0, 1) # range of 0-1 for probability values 
        plt.title('Probability of the robot being at each cell in the grid')
        plt.xticks(np.arange(min(x_labels), max(x_labels)+1, 1))
        plt.show()
    else:
        print('Grid is empty')

给定列表为motions=[1,1]，如果机器人首先感知到红色，则计算后验分布，然后向右移动一个单位，然后感知绿色，之后再次向右移动，从均匀的先验分布p开始。 motions=[1,1]意味着机器人向右移动一个单元格，然后向右移动。你会获得一个初始变量以及完整的sense 与move函数，如下所示。

# given initial variables
p=[0.2, 0.2, 0.2, 0.2, 0.2]
# the color of each grid cell in the 1D world
world=['green', 'red', 'red', 'green', 'green']
# Z, the sensor reading ('red' or 'green')
measurements = ['red', 'green']
pHit = 0.6
pMiss = 0.2

motions = [1,1]
pExact = 0.8
pOvershoot = 0.1
pUndershoot = 0.1

# You are given the complete sense function
def sense(p, Z):
    ''' Takes in a current probability distribution, p, and a sensor reading, Z.
        Returns a *normalized* distribution after the sensor measurement has been made, q.
        This should be accurate whether Z is 'red' or 'green'. '''
    q=[]
    # loop through all grid cells
    for i in range(len(p)):
        # check if the sensor reading is equal to the color of the grid cell
        # if so, hit = 1
        # if not, hit = 0
        hit = (Z == world[i])
        q.append(p[i] * (hit * pHit + (1-hit) * pMiss))
        
    # sum up all the components
    s = sum(q)
    # divide all elements of q by the sum to normalize
    for i in range(len(p)):
        q[i] = q[i] / s
    return q


# The complete move function
def move(p, U):
    q=[]
    # iterate through all values in p
    for i in range(len(p)):
        # use the modulo operator to find the new location for a p value
        # this finds an index that is shifted by the correct amount
        index = (i-U) % len(p)
        nextIndex = (index+1) % len(p)
        prevIndex = (index-1) % len(p)
        s = pExact * p[index]
        s = s + pOvershoot  * p[nextIndex]
        s = s + pUndershoot * p[prevIndex]
        # append the correct, modified value of p to q
        q.append(s)
    return q


## TODO: Compute the posterior distribution if the robot first senses red, then moves 
## right one, then senses green, then moves right again, starting with a uniform prior distribution.
for k in range(len(measurements)):
    # sense and then move, reading the correct measurements/motions at each step
    p = sense(p, measurements[k])
    p = move(p, motions[k])

## print/display that distribution
print(p)
display_map(p)
## print/display that distribution

关于熵值的说明

在更新步骤之后熵值将下降，并且在测量步骤之后熵值将上升。通常情况下，熵值可以用来测量不确定性的量。由于更新步骤增加了不确定性，因此熵值应该增加。测量步骤降低了不确定性，因此熵值应该减少。让我们看一看当前的这个例子，其中机器人可以处于五个不同的位置。当所有位置具有相等的概率 [0.2,0.2,0.2,0.2,0.2] 时，会出现最大的不确定性。

进行测量就会降低不确定性，从而降低熵值。假设在进行测量后，概率变为 [0.05,0.05,0.05,0.8,0.05]。现在熵值减少到0.338。因此，一个测量步骤会使熵值降低，而一个更新步骤则会使熵值增加。

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原始发表：2018.10.15 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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