a-star

问题 Is it possible to modify A* to return the shortest path with the least number of turns ? One complication: Nodes can no longer be distinguished solely by their location, because their parent node is relevant in determining future turns, so they have to have a direction associated with them as well. But the main problem I'm having, is how to work number of turns into the partial path cost (g). If I multiply g by the number of turns taken (t), weird things are happening like: A longer path with

问题 Blue-Walls Green highlighted cells = open list Red Highlighted cells = closed list Hello, can anyone tell me how can i implement backtracking in a a star search algorithm? I\'ve implemented the a star search according to wiki, but it does not backtrack, what i mean by backtrack is that the open list(green cells) contains 2,0 and 3,3 as shown in the picture, upon reaching 2,0 the current node would \"jump\" to 3,3 since the cost is now more than 3,3 and continue the search from there, how can

问题 I was looking at what the guys in the Mario AI Competition have been doing and some of them have built some pretty neat Mario bots utilizing the A* (A-Star) Pathing Algorithm. (Video of Mario A* Bot In Action) My question is, how does A-Star compare with Dijkstra? Looking over them, they seem similar. Why would someone use one over the other? Especially in the context of pathing in games? 回答1: Dijkstra is a special case for A* (when the heuristics is zero). 回答2: Dijkstra: It has one cost

问题 From http://ccl.northwestern.edu/netlogo/models/community/Astardemo, I coded an A* algorithm by using nodes in a network to define least-cost paths. The code seems to work but it is much too slow when I use it at large spatial scales.My landscape has an extent of 1000 patches x 1000 patches with 1 patch = 1 pixel. Even if I reduce it at 400 patches x 400 patches with 1 patch = 1 pixel, it is yet too slow (I can\'t modify my landscape below 400 patches x 400 patches). Here is the code: to find