How to use the A* path finding algorithm on a grid less 2D plane?

不羁的心 提交于 2021-02-07 08:16:52

问题


How can I implement the A* algorithm on a gridless 2D plane with no nodes or cells? I need the object to maneuver around a relatively high number of static and moving obstacles in the way of the goal. My current implementation is to create eight points around the object and treat them as the centers of imaginary adjacent squares that might be a potential position for the object. Then I calculate the heuristic function for each and select the best. The distances between the starting point and the movement point, and between the movement point and the goal I calculate the normal way with the Pythagorean theorem. The problem is that this way the object often ignores all obstacle and even more often gets stuck moving back and forth between two positions. I realize how silly mu question might seem, but any help is appreciated.


回答1:


Create an imaginary grid at whatever resolution is suitable for your problem: As coarse grained as possible for good performance but fine-grained enough to find (desirable) gaps between obstacles. Your grid might relate to a quadtree with your obstacle objects as well.

Execute A* over the grid. The grid may even be pre-populated with useful information like proximity to static obstacles. Once you have a path along the grid squares, post-process that path into a sequence of waypoints wherever there's an inflection in the path. Then travel along the lines between the waypoints.

By the way, you do not need the actual distance (c.f. your mention of Pythagorean theorem): A* works fine with an estimate of the distance. Manhattan distance is a popular choice: |dx| + |dy|. If your grid game allows diagonal movement (or the grid is "fake"), simply max(|dx|, |dy|) is probably sufficient.




回答2:


Uh. The first thing that come to my mind is, that at each point you need to calculate the gradient or vector to find out the direction to go in the next step. Then you move by a small epsilon and redo.

This basically creates a grid for you, you could vary the cell size by choosing a small epsilon. By doing this instead of using a fixed grid you should be able to calculate even with small degrees in each step -- smaller then 45° from your 8-point example.

Theoretically you might be able to solve the formulas symbolically (eps against 0), which could lead to on optimal solution... just a thought.




回答3:


How are the obstacles represented? Are they polygons? You can then use the polygon vertices as nodes. If the obstacles are not represented as polygons, you could generate some sort of convex hull around them, and use its vertices for navigation. EDIT: I just realized, you mentioned that you have to navigate around a relatively high number of obstacles. Using the obstacle vertices might be infeasible with to many obstacles.

I do not know about moving obstacles, I believe A* doesn't find an optimal path with moving obstacles.

You mention that your object moves back and fourth - A* should not do this. A* visits each movement point only once. This could be an artifact of generating movement points on the fly, or from the moving obstacles.




回答4:


I remember encountering this problem in college, but we didn't use an A* search. I can't remember the exact details of the math but I can give you the basic idea. Maybe someone else can be more detailed.

We're going to create a potential field out of your playing area that an object can follow.

  1. Take your playing field and tilt or warp it so that the start point is at the highest point, and the goal is at the lowest point.

  2. Poke a potential well down into the goal, to reinforce that it's a destination.

  3. For every obstacle, create a potential hill. For non-point obstacles, which yours are, the potential field can increase asymptotically at the edges of the obstacle.

Now imagine your object as a marble. If you placed it at the starting point, it should roll down the playing field, around obstacles, and fall into the goal.

The hard part, the math I don't remember, is the equations that represent each of these bumps and wells. If you figure that out, add them together to get your final field, then do some vector calculus to find the gradient (just like towi said) and that's the direction you want to go at any step. Hopefully this method is fast enough that you can recalculate it at every step, since your obstacles move.




回答5:


Sounds like you're implementing The Wumpus game based on Norvig and Russel's discussion of A* in Artifical Intelligence: A Modern Approach, or something very similar.

If so, you'll probably need to incorporate obstacle detection as part of your heuristic function (hence you'll need to have sensors that alert your agent to the signs of obstacles, as seen here).

To solve the back and forth issue, you may need to store the traveled path so you can tell if you've already been to a location and have the heurisitic function examine the past N number of moves (say 4) and use that as a tie-breaker (i.e. if I can go north and east from here, and my last 4 moves have been east, west, east, west, go north this time)



来源:https://stackoverflow.com/questions/4054701/how-to-use-the-a-path-finding-algorithm-on-a-grid-less-2d-plane

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