急Webots 单机器人路径规划，RRT 应用
请问有没有知道怎么在Webots 上实现单个机器人路径规划，用RRT，
求帮忙zuo一个我给你简单地图里面，epuck用RRT到达目的地，然后躲避障碍的模拟
4月31号前要，GitHub上面有类似的code,但是我看不懂，能做到的再来接
4月没31号
搞这个的人太少了吧，还得硬件配合调试
是什么语言的代码，你可以发出来看看
我就凑个热闹，反正也没31号
epuck用rrt能带机器人去4月31号#
so easy
https://cloud.tencent.com/developer/article/1691514
请用RRT带我去到4.31号，我恐怕不行。
可以采用最佳路径优先搜索算法
"""
Best-First Searching
@author: huiming zhou
"""
import os
import sys
import math
import heapq
sys.path.append(os.path.dirname(os.path.abspath(__file__)) +
"/../../Search_based_Planning/")
from Search_2D import plotting, env
from Search_2D.Astar import AStar
class BestFirst(AStar):
"""BestFirst set the heuristics as the priority
"""
def searching(self):
"""
Breadth-first Searching.
:return: path, visited order
"""
self.PARENT[self.s_start] = self.s_start
self.g[self.s_start] = 0
self.g[self.s_goal] = math.inf
heapq.heappush(self.OPEN,
(self.heuristic(self.s_start), self.s_start))
while self.OPEN:
_, s = heapq.heappop(self.OPEN)
self.CLOSED.append(s)
if s == self.s_goal:
break
for s_n in self.get_neighbor(s):
new_cost = self.g[s] + self.cost(s, s_n)
if s_n not in self.g:
self.g[s_n] = math.inf
if new_cost < self.g[s_n]: # conditions for updating Cost
self.g[s_n] = new_cost
self.PARENT[s_n] = s
# best first set the heuristics as the priority
heapq.heappush(self.OPEN, (self.heuristic(s_n), s_n))
return self.extract_path(self.PARENT), self.CLOSED
def main():
s_start = (5, 5)
s_goal = (45, 25)
BF = BestFirst(s_start, s_goal, 'euclidean')
plot = plotting.Plotting(s_start, s_goal)
path, visited = BF.searching()
plot.animation(path, visited, "Best-first Searching") # animation
if __name__ == '__main__':
main()
机器人导论课程作业｜Webots仿真｜RRT*算法｜Pure Pursuit算法
https://github.com/Lanly109/Webots-Homework/tree/master/RRT_planning
机器人路径规划之RRT算法
https://wenku.baidu.com/view/617a5a22cf1755270722192e453610661ed95afd.html
伪代码
Algorithm BuildRRT
Input: Initial configuration qinit, number of vertices in RRT K, incremental distance Δq)
Output: RRT graph G
G.init(qinit)
for k = 1 to K do
qrand ← RAND_CONF()
qnear ← NEAREST_VERTEX(qrand, G)
qnew ← NEW_CONF(qnear, qrand, Δq)
G.add_vertex(qnew)
G.add_edge(qnear, qnew)
return G
"←" denotes assignment. For instance, "largest ← item" means that the value of largest changes to the value of item.
"return" terminates the algorithm and outputs the following value.
这个源码看起来挺好的，下面这个是视频讲解
https://www.bilibili.com/video/BV1uL4y1s7yv?spm_id_from=333.999.0.0