CS 763 F16. Moving objects in space with obstacles/constraints.

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1 Moving objects in space with obstacles/constraints. Objects = robots, vehicles, jointed linkages (robot arm), tools (e.g. on automated assembly line), foldable/bendable objects. Objects need not be physical (e.g. fly-through animation). We will concentrate on moving from one position to another, though visiting a sequence of positions is also very interesting. Outline: - translational motion (one rigid object) - with rotations - linkage motion (robot arm)

2 Translational motion a polygon translating among polygonal obstacles. Start with a point moving among polygonal obstacles. Then we can use the shortest path algorithm from previous lecture. But we do not really need the shortest path

3 How to find if there is some path from point s to point t among polygonal obstacles. the blue graph is called a roadmap - construct trapezoidal map of space outside obstacles - construct dual graph = the roadmap vertex = trapezoid; edge = two trapezoids share a boundary - let T(s) = trapezoid containing s; T(t) = trapezoid containing t - there is a path s > t iff T(s) and T(t) are connected in the roadmap graph - time O(n log n)

4 An alternative roadmap: the Voronoi diagram of the obstacles. Voronoi Diagrams: Planning Find the point q* of the Voronoi Then, for a given route, the point stays as far as possible from the obstacles

5 A disc moving among polygonal obstacles. Model as a point (the center of the disc) moving among enlarged obstacles. disc radius = r the center of the disc must stay distance r from the obstacles the disc cannot follow this path O Rourke Enlarging the obstacles is captured more formally via Minkowski sum, A B

6 Minkowski sum

7 A polygon moving among polygons by translation only. R = moving polygon P = an obstacle polygon Replace R by a reference point at the origin. Enlarge P to compensate we need where can the reference point be when R touches P?

8 Can polygon R move (via translations) from initial to final position among polygonal obstacles? high level idea - compute the Minkowski sum P ( R) for each obstacle P - take the union, to obtain new polygonal obstacles - test if a point (the reference point) can move from initial to final position among the new enlarged obstacles

9 Theorem. If P and R are convex polygons with n and m edges, respectively, then P R has at most n+m edges and can be found in O(n+m) time. intuition for the proof: as we slide R around P, tracing the path of the reference point in R, each vertex of P and each vertex of R cause a vertex in the path

10 Complexity of Minkowski sum R P R and P convex O(n+m) R convex, P non-convex O(nm) R, P non-convex O(n 2 m 2 ) These bounds are tight: R, P non-convex R convex, P non-convex Figure 83: Minkowski sum of O(nm) complexity. Figure 82: Minkowski sum of O(n 2 m 2 ) complexity

11 Computing the union of the enlarged polygons. Complexity of the union is reduced due to the following: two convex polygons boundaries can intersect many times they form pseudodiscs if their boundaries intersect at most twice Note: need a more careful definition in case of shared boundary segments Theorem 1: Let P1 and P2 be disjoint convex polygons. Then P1 R and P2 R form pseudodiscs. Theorem 2: If Q1... Qk are pairwise pseudodiscs of total size n then their union has size O(n)

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14 Putting it all together: To move polygon R of constant size among polygonal obstacles of total size n. If an obstacle is non-convex, triangulate to get convex pieces (total size is still O(n)). Forbidden space = union of enlarged convex polygons = Free space = complement of forbidden space. Forbidden space has size (number of vertices) O(n) by Theorems 1 and 2. The forbidden space can be computed in O(n log n) time. (or via simpler O(n log 2 n) time divide and conquer algorithm)

15 Translational motion planning in higher dimensions in 2D O(n log n) by above method in 3D O(n^2 log^2 n) by similar method (for convex robot) general road map algorithm of Canny O(n^d log n) where d is the number of degrees of freedom this applies to rotational motion as well

16 Motion planning with rotations. Consider a line segment ( ladder ) moving among polygonal obstacles in the plane. There are 3 degrees of freedom. Choose an endpoint of the ladder as a reference point. The position of the ladder is specified by (x, y, θ) where (x,y) is the position of the reference point, and θ is the rotation angle

17 a ladder moving among 3 rectangular obstacles configuration space θ=0 configuration space θ=30 configuration space θ=60 figures from Devadoss & O Rourke

18 configuration space in 3D (3rd dimension is θ) The 3D configuration space has curved boundaries and is hard to deal with

19 Roadmap technique using cell decomposition

20 For fixed θ, use previous technique to find configuration space. Divide into equivalence regions: two positions of the ladder are equivalent if moving the ladder parallel to itself, it hits the same two obstacles

21 Now increase θ. The combinatorics of the regions changes only at critical angles in the direction through two vertices. There are O(n^2) critical directions

22 Best algorithm takes O(n^2) time

23 Robot Arm Motion (Linkages) The study of linkages is old, e.g. Peaucellier linkage to convert rotary motion to linear motion We will just look at a chain (not a general graph) a polygonal chain where the segments ( links ) have fixed lengths and the angles between successive links may change. Two models: as P moves on a circle, Q moves on a line - intersection of links allowed, e.g. above, where linkage is essentially planar, but each link is slightly higher (in 3rd dimension) than previous - intersections forbidden. e.g. protein folding, robot arm in 3D

24 We will study two problems: 1. Given a chain with one endpoint fixed, where can the other endpoint reach? Allow intersections. 2. Given a chain, can we go from any configuration to any other? Forbid intersections. different views not to scale

25 Theorem. outer radius = inner radius = John Hopcroft, Deborah Joseph, and Sue Whitesides. "On the movement of robot arms in 2-dimensional bounded regions."

26 reachability of a chain of links 2 links Minkowski sum of annulus and disc

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30 2. Given a linkage, can we go from any configuration to any other? Forbid intersections. Erik Demaine In 3D, the answer is not always. Biedl, T., Demaine, E., Demaine, M., Lazard, S., Lubiw, A., O'Rourke, J., Overmars, M., Robbins, S., Streinu, I., Toussaint, G. and Whitesides, S., Locked and unlocked polygonal chains in three dimensions Fig. 1. A locked, open chain K with long knitting needles at the ends. OPEN. Can a chain of unit length links be locked? OPEN. Find a polynomial time algorithm to test if a 3D polygonal chain is locked. (It is PSPACE-hard to test if we can get from one configuration to another.)

31 Theorem. In 2D, any chain can be straightened. Any closed chain can be made convex. Robert Connelly, Erik D. Demaine, and Günter Rote. "Straightening Polygonal Arcs and Convexifying Polygonal Cycles." 2003 (Erik Demain s PhD thesis work) This implies that a linkage can go from any configuration to any other. initial config. straight config. final config. idea of proof: they show that it suffices to use expansive motions the distance between any two vertices never decreases

32 A better way to go from initial to final configuration avoid going through intermediate straight/convex chain. Hayley Iben, James F. O Brien, and Erik D. Demaine. "Refolding planar polygons." Fig. 2 The top row demonstrates how using the vertex-position metric alone will, as expected, generate a sequence with self intersections. The bottom row illustrates how the collision-avoidance machinery alters the vertex motions to avoid self intersection. Computation times were less than one second