- html - 出于某种原因,IE8 对我的 Sass 文件中继承的 html5 CSS 不友好?
- JMeter 在响应断言中使用 span 标签的问题
- html - 在 :hover and :active? 上具有不同效果的 CSS 动画
- html - 相对于居中的 html 内容固定的 CSS 重复背景?
我正在 Python 2.7.12 中实现双向 A*
算法,并在 Russell 和 Norvig 第 3 章的罗马尼亚 map 上进行测试。边具有权重,目的是找到两个节点之间的最短路径。
这是测试图的可视化:
我的双向 A* 失败的示例是起点为 'a'
且目标为 'u'
。这是我的实现找到的路径: ['a', 's', 'f', 'b', 'u']
的长度为 535
。
这是从 'a'
到 'u'
的实际最短路径: ['a', 's', 'r', 'p', 'b', 'u']
的长度为 503
。
正如我们所看到的,我的实现未能找到最短路径。我认为问题可能出在我的停止条件上,但我不知道。
这是我的 A* 实现(我使用欧几里德距离作为启发式)和其他一些帮助类和函数的 python 脚本:
from __future__ import division
import math
from networkx import *
import random
import pickle
import sys
import heapq
import matplotlib.pyplot as plt
class PriorityQueue():
"""Implementation of a priority queue"""
def __init__(self):
self.queue = []
self.node_finder = dict()
self.current = 0
self.REMOVED_SYMBOL = '<removed>'
def next(self):
if self.current >=len(self.queue):
self.current
raise StopIteration
out = self.queue[self.current]
self.current += 1
return out
def pop(self):
while self.queue:
node = heapq.heappop(self.queue)
nodeId = node[1]
if nodeId is not self.REMOVED_SYMBOL:
try:
del self.node_finder[nodeId]
except KeyError:
dummy=1
return node
def remove(self, nodeId):
node = self.node_finder[nodeId]
node[1] = self.REMOVED_SYMBOL
def __iter__(self):
return self
def __str__(self):
return 'PQ:[%s]'%(', '.join([str(i) for i in self.queue]))
def append(self, node):
nodeId = node[1]
nodePriority = node[0]
node = [nodePriority, nodeId]
self.node_finder[nodeId] = node
heapq.heappush(self.queue, node)
def update(self, node):
nodeId = node[1]
nodePriority = node[0]
node = [nodePriority, nodeId]
self.remove(nodeId)
self.node_finder[nodeId] = node
heapq.heappush(self.queue, node)
def getPriority(self, nodeId):
return self.node_finder[nodeId][0]
def __contains__(self, key):
self.current = 0
return key in [n for v,n in self.queue]
def __eq__(self, other):
return self == other
def size(self):
return len(self.queue)
def clear(self):
self.queue = []
def top(self):
return self.queue[0]
__next__ = next
def bidirectional_a_star(graph, start, goal):
if start == goal:
return []
pq_s = PriorityQueue()
pq_t = PriorityQueue()
closed_s = dict()
closed_t = dict()
g_s = dict()
g_t = dict()
g_s[start] = 0
g_t[goal] = 0
cameFrom1 = dict()
cameFrom2 = dict()
def euclidean_distance(graph, v, goal):
xv, yv = graph.node[v]['pos']
xg, yg = graph.node[goal]['pos']
return ((xv-xg)**2 + (yv-yg)**2)**0.5
def h1(v): # heuristic for forward search (from start to goal)
return euclidean_distance(graph, v, goal)
def h2(v): # heuristic for backward search (from goal to start)
return euclidean_distance(graph, v, start)
cameFrom1[start] = False
cameFrom2[goal] = False
pq_s.append((0+h1(start), start))
pq_t.append((0+h2(goal), goal))
done = False
i = 0
mu = 10**301 # 10**301 plays the role of infinity
connection = None
while pq_s.size() > 0 and pq_t.size() > 0 and done == False:
i = i + 1
if i % 2 == 1: # alternate between forward and backward A*
fu, u = pq_s.pop()
closed_s[u] = True
for v in graph[u]:
weight = graph[u][v]['weight']
if v in g_s:
if g_s[u] + weight < g_s[v]:
g_s[v] = g_s[u] + weight
cameFrom1[v] = u
if v in closed_s:
del closed_s[v]
if v in pq_s:
pq_s.update((g_s[v]+h1(v), v))
else:
pq_s.append((g_s[v]+h1(v), v))
else:
g_s[v] = g_s[u] + weight
cameFrom1[v] = u
pq_s.append((g_s[v]+h1(v), v))
if v in closed_t:
if g_s[u] + weight + g_t[v] < mu:
mu = g_s[u] + weight + g_t[v]
connection = v
done = True
else:
fu, u = pq_t.pop()
closed_t[u] = True
for v in graph[u]:
weight = graph[u][v]['weight']
if v in g_t:
if g_t[u] + weight < g_t[v]:
g_t[v] = g_t[u] + weight
cameFrom2[v] = u
if v in closed_t:
del closed_t[v]
if v in pq_t:
pq_t.update((g_t[v]+h2(v), v))
else:
pq_t.append((g_t[v]+h2(v), v))
else:
g_t[v] = g_t[u] + weight
cameFrom2[v] = u
pq_t.append((g_t[v]+h2(v), v))
if v in closed_s:
if g_t[u] + weight + g_s[v] < mu:
mu = g_t[u] + weight + g_s[v]
connection = v
done = True
if u in closed_s and u in closed_t:
if g_s[u] + g_t[u] < mu:
mu = g_s[u] + g_t[u]
connection = u
stopping_distance = min(min([f for (f,x) in pq_s]), min([f for (f,x) in pq_t]))
if mu <= stopping_distance:
done = True
#connection = u
continue
if connection is None:
# start and goal are not connected
return None
#print cameFrom1
#print cameFrom2
path = []
current = connection
#print current
while current != False:
#print predecessor
path = [current] + path
current = cameFrom1[current]
current = connection
successor = cameFrom2[current]
while successor != False:
path = path + [successor]
current = successor
successor = cameFrom2[current]
return path
# This function visualizes paths
def draw_graph(graph, node_positions={}, start=None, goal=None, path=[]):
explored = list(graph.get_explored_nodes())
labels ={}
for node in graph:
labels[node]=node
if not node_positions:
node_positions = networkx.spring_layout(graph)
edge_labels = networkx.get_edge_attributes(graph,'weight')
networkx.draw_networkx_nodes(graph, node_positions)
networkx.draw_networkx_edges(graph, node_positions, style='dashed')
networkx.draw_networkx_edge_labels(graph, node_positions, edge_labels=edge_labels)
networkx.draw_networkx_labels(graph,node_positions, labels)
networkx.draw_networkx_nodes(graph, node_positions, nodelist=explored, node_color='g')
if path:
edges = [(path[i], path[i+1]) for i in range(0, len(path)-1)]
networkx.draw_networkx_edges(graph, node_positions, edgelist=edges, edge_color='b')
if start:
networkx.draw_networkx_nodes(graph, node_positions, nodelist=[start], node_color='b')
if goal:
networkx.draw_networkx_nodes(graph, node_positions, nodelist=[goal], node_color='y')
plt.plot()
plt.show()
# this function calculates the length of the path
def calculate_length(graph, path):
pairs = zip(path, path[1:])
return sum([graph.get_edge_data(a, b)['weight'] for a, b in pairs])
#Romania map data from Russell and Norvig, Chapter 3.
romania = pickle.load(open('romania_graph.pickle', 'rb'))
node_positions = {n: romania.node[n]['pos'] for n in romania.node.keys()}
start = 'a'
goal = 'u'
path = bidirectional_a_star(romania, start, goal)
print "This is the path found by bidirectional A* :", path
print "Its length :", calculate_length(romania, path)
# visualize my path
draw_graph(romania, node_positions=node_positions, start=start, goal=goal, path=path)
# compare to the true shortest path between start and goal
true_path = networkx.shortest_path(romania, start, goal, weight='weight')
print "This is the actual shortest path: ", true_path
print "Its lenght: ", calculate_length(romania, true_path)
#visualize true_path
draw_graph(romania, node_positions=node_positions, start=start, goal=goal, path=true_path)
罗马尼亚的泡菜数据可以从here下载.
最佳答案
我更正了 PriorityQueue
和 bi Direction_a_star
中的一些错误。现在工作正常。
更正后的类和函数的代码如下:
class PriorityQueue():
"""Implementation of a priority queue"""
def __init__(self):
self.queue = []
self.node_finder = dict()
self.current = 0
self.REMOVED_SYMBOL = '<removed>'
def next(self):
if self.current >=len(self.queue):
self.current
raise StopIteration
out = self.queue[self.current]
while out == self.REMOVED_SYMBOL:
self.current += 1
out = self.queue[self.current]
self.current += 1
return out
def pop(self):
# TODO: finish this
while self.queue:
node = heapq.heappop(self.queue)
nodeId = node[1]
if nodeId is not self.REMOVED_SYMBOL:
try:
del self.node_finder[nodeId]
except KeyError:
dummy=1
return node
#raise KeyError('pop from an empty priority queue')
def remove(self, nodeId):
node = self.node_finder[nodeId]
node[1] = self.REMOVED_SYMBOL
def __iter__(self):
return self
def __str__(self):
return 'PQ:[%s]'%(', '.join([str(i) for i in self.queue]))
def append(self, node):
# node = (priority, nodeId)
nodeId = node[1]
nodePriority = node[0]
node = [nodePriority, nodeId]
self.node_finder[nodeId] = node
heapq.heappush(self.queue, node)
def update(self, node):
nodeId = node[1]
nodePriority = node[0]
node = [nodePriority, nodeId]
self.remove(nodeId)
self.node_finder[nodeId] = node
heapq.heappush(self.queue, node)
def getPriority(self, nodeId):
return self.node_finder[nodeId][0]
def __contains__(self, key):
self.current = 0
return key in [n for v,n in self.queue]
def __eq__(self, other):
return self == other
def size(self):
return len([1 for priority, node in self.queue if node!=self.REMOVED_SYMBOL])
def clear(self):
self.queue = []
def top(self):
return self.queue[0]
__next__ = next
def bidirectional_a_star(graph, start, goal):
if start == goal:
return []
pq_s = PriorityQueue()
pq_t = PriorityQueue()
closed_s = dict()
closed_t = dict()
g_s = dict()
g_t = dict()
g_s[start] = 0
g_t[goal] = 0
cameFrom1 = dict()
cameFrom2 = dict()
def euclidean_distance(graph, v, goal):
xv, yv = graph.node[v]['pos']
xg, yg = graph.node[goal]['pos']
return ((xv-xg)**2 + (yv-yg)**2)**0.5
def h1(v): # heuristic for forward search (from start to goal)
return euclidean_distance(graph, v, goal)
def h2(v): # heuristic for backward search (from goal to start)
return euclidean_distance(graph, v, start)
cameFrom1[start] = False
cameFrom2[goal] = False
pq_s.append((0+h1(start), start))
pq_t.append((0+h2(goal), goal))
done = False
i = 0
mu = 10**301 # 10**301 plays the role of infinity
connection = None
while pq_s.size() > 0 and pq_t.size() > 0 and done == False:
i = i + 1
if i % 2 == 1: # alternate between forward and backward A*
fu, u = pq_s.pop()
closed_s[u] = True
for v in graph[u]:
weight = graph[u][v]['weight']
if v in g_s:
if g_s[u] + weight < g_s[v]:
g_s[v] = g_s[u] + weight
cameFrom1[v] = u
if v in closed_s:
del closed_s[v]
if v in pq_s:
pq_s.update((g_s[v]+h1(v), v))
else:
pq_s.append((g_s[v]+h1(v), v))
else:
g_s[v] = g_s[u] + weight
cameFrom1[v] = u
pq_s.append((g_s[v]+h1(v), v))
else:
fu, u = pq_t.pop()
closed_t[u] = True
for v in graph[u]:
weight = graph[u][v]['weight']
if v in g_t:
if g_t[u] + weight < g_t[v]:
g_t[v] = g_t[u] + weight
cameFrom2[v] = u
if v in closed_t:
del closed_t[v]
if v in pq_t:
pq_t.update((g_t[v]+h2(v), v))
else:
pq_t.append((g_t[v]+h2(v), v))
else:
g_t[v] = g_t[u] + weight
cameFrom2[v] = u
pq_t.append((g_t[v]+h2(v), v))
if u in closed_s and u in closed_t:
if g_s[u] + g_t[u] < mu:
mu = g_s[u] + g_t[u]
connection = u
try:
stopping_distance = max(min([f for (f,x) in pq_s]), min([f for (f,x) in pq_t]))
except ValueError:
continue
if mu <= stopping_distance:
done = True
connection = u
continue
if connection is None:
# start and goal are not connected
return None
#print cameFrom1
#print cameFrom2
path = []
current = connection
#print current
while current != False:
#print predecessor
path = [current] + path
current = cameFrom1[current]
current = connection
successor = cameFrom2[current]
while successor != False:
path = path + [successor]
current = successor
successor = cameFrom2[current]
return path
关于python - 双向 A* 未找到最短路径,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42044901/
这个问题在这里已经有了答案: Integer summing blues, short += short problem (5 个答案) 关闭 7 年前。 版本:Visual Studio Prof
我尝试执行以下代码: public class Test5 { /** * @param args */ public static void main(String[] args) {
这是我的任务,我尝试仅使用简短的 if 语句来完成此任务,我得到的唯一错误是使用“(0.5<=ratio<2 )”,除此之外,构造正确吗? Scanner scn = new Scanner(
已关闭。此问题需要 debugging details 。目前不接受答案。 编辑问题以包含 desired behavior, a specific problem or error, and the
我有一个简单的类型 data Day = Monday | Tuesday | Wednesday | Thursday | Friday 我是haskell的新手,所以我写==如下。 (==) :
如何实现“简短”和“详细”两个按钮? “短”应该是默认值,并显示页面的一个版本。单击“详细”按钮后,应显示该页面的另一个版本。 由于这有点难以解释,或许可以看下面的例子。 示例页面: 别管内容 需要j
有没有一种方法可以在 C# 中执行此操作,而无需为现有的每个 var 类型创建一个新方法来重载? $box = !empty($toy) : $toy ? ""; 我能想到的唯一方法是: if (t
我想使用 setInterval 创建一个节拍器。我希望能够达到 300 bpm 这样的高 bpm。即使文件足够短,可以根据需要播放多次,它也很容易 打嗝。此外,许多浏览器都存在短音频文件的问题——S
我们现在有一个正在生产中的应用程序,它会将 IAP 收据发送到我们的服务器,这些收据显然太短,而且我们的服务器没有经过 apple 的验证。 Apple 正确验证的长收据长度为 3192。短收据长度均
例如,许多软件使用的许可证 key 。我曾想过对一个序列进行密码签名,所以我可能有 4 个字节用于 ID,8 个字节用于签名,但我找不到合适的算法。 我需要的是攻击者无法轻易生成,但存储在大约 20
作为一个学生项目,我们正在构建一个机器人,它应该跑完规定的路线并捡起一个木制立方体。它的核心是一台运行 debian 的单板计算机,配备 ARM9,频率为 250MHz。因此 Controller 的
在将 short 转换为字节数组时,我在网上找到了以下解决方案,但不太理解所涉及的逻辑。 //buffer is an array of bytes, bytes[] buffer[position]
如何在 PHP namespace 环境中检查对象的类而不指定完整的命名空间类。 例如,假设我有一个对象库/实体/契约(Contract)/名称。 以下代码不起作用,因为 get_class 返回完整
我有一个 View 范围的托管 bean,其托管属性绑定(bind)到查询字符串参数。 JSF 给了我熟悉的异常: javax.faces.FacesException: Property reset
根据 this post我已经修复了对象检查器。有时代码可以很好地运行 10 个条目,使它们全部正确,有时它可以运行 5 个条目。有时它会导致条目错误。 在获取元素的内部文本时总是会失败。当它的 Y/
我正在编写一组工具,其中 C++ 应用程序使用 AES 加密标准对数据进行编码,而 Java 应用程序对其进行解码。据我所知, key 长度必须为 16 个字节。但是当我尝试使用不同长度的密码时,我遇
我有以下代码: short num_short = 1; int possible_new_short = 1; valid = 1; while (valid) { poss
因此,作为 C 的新手,我遇到了我的第一个 SIGSEGV 错误。它出现在一个简短的 C 程序中,该程序旨在成为“猜数字”游戏。它由一个比较两个数字的自定义函数和一个带有输入的 do-while 循环
我不是严格意义上的初级程序员,但我没有接受过数学以外的正规教育 - 所以这纯粹是业余爱好,可能是业余的。 我最近自己开发了一个算法来解决这个问题,但我想知道是否有任何相对简单的算法明显更高效/更快?
我正在使用短条件来区分记录列表中显示的值。 例如,如果我希望强调 ( ) 标识符大于 100 的客户的姓名,请执行以下操作: {# Displays the identifier of the c
我是一名优秀的程序员,十分优秀!