If on a 2d plane there are a no. of obstacles of all possible 2d shapes(circles, quadrilaterals, triangles, irregular shapes...) then how do you implement a mechanism to find the shortest path around the obstacles? I'm considering visual c++, as it provides many graphical classes to draw such figures.
I have come quite far
1) Firstly i'll be using A* search(A-star) to find the path with least cost
2) The path with the least displacement from the straight path will be considered for best path. (not really sure though)
3) The shortest path to get around a figure, for eg from the start, is a line from that point to :
a) the farthest vertex in case of a polygon/quadrilateral
b) a point on the circumference such that the line drawn would be tangential to the circle, in case of a circle or arc
c) (not sure about irregular figures)
Now coming back to the 2) point- least displacement between 2 or more paths can be determined by comparing perpendiculars from those lines to the farthest points of an object on their respective sides. (hope i've made myself understood) .
So then- how do we draw perpendiculars to the straight path?
x1,x2,y1,y2,k and l are known. We just have to find a,b.
Slope of the straight path * slope of it's perpendicular = -1
=> (y2-y1)/(x2-x1) * (b-l)/(1-k) = -1
hence, b = [(x1-x2)/(y2-y1) * (a-k)] + l
I've imagined that by using pythagoras theorem we can find the other equation in terms of the co-ordinates. The lengths of each line can be found by this way: dx = x1-x2 dy = y1-y2 dist = sqrt(dxdx + dydy)
And then by solving these 2 eqns we can find the correct values of a,b.
I can't think of anything further- any ideas or suggestions?
Is it possible that you use polygonal (i.e. straight line segment) approximations for all shapes? This would simplify the implementation of the algorithm a lot.
Assuming that this is indeed possible: If you want to use A* then you'll need a graph representation of the possible paths that you can take. The nodes of this graph are the combination of:
- all the vertices of all shapes[1], and
- the start and end points
- all intersection points between {the straight line segment between the start and end point} and all shapes.
The edges in this graph, then, are between each pair of nodes only if
- there exists a straight line between their corresponding two vertices
- that doesn't intersect any of the shapes[2].
The length of each edge in the graph is then simply the (euclidean) distance between the two vertices it represents, and the shortest path is always a subset of these edges (I think), which you can find by applying A* to this graph.
[1] - To reduce the number of vertices, you can make all concave shapes convex (unless this causes the start or end point to lie inside such a shape, then it should be kept concave).
[2] - You can use a variety of data structures to speed up these queries, such as kD or quad trees, or maybe use a sweep line algorithm (such as http://en.wikipedia.org/wiki/Bentley%E2%80%93Ottmann_algorithm) in combination with a doubly-connected edge list.
Well i am not pretty sure if this might help but anyways every irregular object can be split into a combination of regular objects like a circle just that the radius of the curve keeps changing.So you can consider it to be a combination of arcs corresponding to different radii.
For the second point if have the points (l,k). Consider two points located on the line(x1,y1),(x2,y2) which are equidistant from(l,k). So the perpendicular will be the combination of all points which are equidistant from (x1,y1)and (x2,y2).
来源：https://stackoverflow.com/questions/13238267/how-do-you-implement-a-program-to-find-the-shortest-path-in-a-2d-plane