Motion Planning CS 6160, Spring 2010 By Gene Peterson 5/4/2010

Motion PlanningCS 6160, Spring 2010

By Gene Peterson

5/4/2010

What is Motion Planning?

• Given a set of obstacles, a start position, and a destination position, find a path from start to destination that doesn’t go through any obstacles

• This project seeks shortest path• Many variations on the basic problem

Program

• Two parts: Course builder and simulator• Course Builder: made in C# using XNA and

Windows forms• User-friendly interface• Simulator: approximately 2000 lines of code• OpenGL rendering

Potential Project Goals

• Point object path finding• Polygonal object finding• Robot rotation• Real-time calculation• Non-convex obstacles• Moving obstacles• Curves• No global vision• 3D problem• Path weighting• Group travel

Procedure

• Define input: obstacles, “robot”, starting position, and destination

• Create a visibility graph• Find the shortest path from start to

destination in the graph

Required Geometry

• Point• Vector• Line• Line Segment• Ray• Poly Line• Polygon

Defining a Polygon

• Ordered set vertices• Edges implicitly defined by adjacency in the

vertex list• May be clockwise or counter-clockwise wound

Open or Closed Polygons?Open Closed

Does it matter?

Open vs Closed Polygon

No Intersection

Intersection

Open vs Closed Polygons

• Cannot achieve shortest possible path with closed polygons

• Always can get slightly closer

• Open polygons cannot get any closer to the polygon without penetrating it

Course Builder

• Create 2D polygons and lay them out as a course

• Saved to file, loaded by simulator

Convex Hull Algorithm #1

• Trivial O(n^3) algorithm• Every pair of vertices is checked to see if it can

be an edge of the convex set

Convex Hull Algorithm #2

• Gift wrapping technique• O(n*h)• n: # of vertices• h: # of vertices in the convex hull

Convex Hull Algorithm #3

• Graham-Schmidt algorithm• O(n log n)• n: # of vertices• Sorts the points• Didn’t attempt algorithm #4: mix of gift

wrapping with Graham-Schmidt

Polygonal Robot

• Instead of a point robot, use a polygon• By calculating the Configuration Space of the

polygonal robot, the robot can be represented by a point

• C-Space is the areas that the point robot cannot move

Configuration Space

• Calculate C-Space version of each obstacle using Minkowski sum of the polygon and the robot

Minkowski Sum Algorithm #1

• Trivial algorithm - O(n^3)• For each vertex a in polygon A

For each vertex b in polygon BAdd point (a+b) to P

• Minkowski sum is the convex hull of points P

Minkowski Sum Algorithm #2

• Sweep method – O(n + m)• n: # of vertices in polygon A• m: # of vertices in polygon B• Sorts the points in each polygon in counter-

clockwise order• Next point is the point with smallest angle

Visibility Graph

• Graph vertices represent vertices of the C-Space obstacles and the start and destination points

• Edges represent possible paths• If a line segment from any vertex in the graph to any

other vertex in the graph doesn’t penetrate an obstacle, there’s an edge between those two vertices

• Edge weights represent distance between points• Shortest path in the graph is shortest path through

the obstacles

Graph

Line Segment-Polygon Intersection

Visibility Graph Algorithm #1

• O(n^3)• Create every possible line segment from pairs

of vertices in the graph• Any line segment that doesn’t penetrate an

obstacle gets added as an edge in the graph

Visibility Graph Algorithm #2

• O(n^2 log n)• Sweep algorithm• Uses the fact that some edges will obscure

other edges from being visible• Requires a line segment search tree data

structure

Ray-Line Segment AVG Tree

• Balanced binary search tree• Given a ray, find the line segment in the tree

that it will intersect first

Shortest Path

• Now that we have an obstacle course and a visibility graph for it, we can calculate the shortest path using graph algorithms

Dijkstra’s Algorithm

• Breadth first search with weights• Only non-negative edges• Requires priority queue

Path Finding in Video Games

• Generally discretised (grid based)• 2D path finding useful in many 3D games• Speed important (1/60th of a second/frame)• Accuracy less important• Much of the environment is static• Precomputation often available

Precomputation

• Precalculate visibility graph of C-Space obstacles only (don’t include any start or end point)

• To create the full visibility graph, insert start and destination points (calculating visibility for each polygon vertex for them)

• Use a copy of precalculated visibility graph so it can be reused

• Needs to be recomputed as often as the environment changes

Statistics

• TO FILL IN WHEN I HAVE THE FINAL STATISTICS AVAILABLE

More Speed?

• Multi-threading• CUDA• Performance optimizations (length squared,

etc)• Minimizing geometry where possible• Sacrifice accuracy• Amortize path finding cost over time

Non-Convex Polygons

Robot Rotation

• Unlike translation, rotating the robot changes the C-Space

• Even if rotation angles are discretized, greatly enlarges the graph size, and thus the computation time

• Gave up on this problem to pursue other aspects of motion planning

Goals Revisited

• Point object path finding• Polygonal object finding• Robot rotation• Real-time calculation• Non-convex obstacles• Moving obstacles• Curves• No global vision• 3D problem• Path weighting• Group travel

Too computationally expensive for real-time

Not as useful in video games, slow

Didn’t get to it

Beyond the scope of this project

AI Problem

AI Problem

Further Exploration

• Rotation• Better precomputation• Non-convex performance• Sacrificing accuracy for performance• Regional partitioning

Implicit Line Equation from 2 Points

• Given two points on the line, calculate a, b, c

Pick any a and b that make this true

Use any point (x,y) on the line

* Any multiple k ≠ 0 applied to (a, b, c) will result in the same line

Half Space (2D)

• Input: Line and a Point• Output: Real (+,-,0)• Geometric Interpretation:

0 means the point is on the line+ means on one side of the line- means on the other side

• Plug point into implicit line equation• Result is the half space that contains the point

(+,-,0)

Half Space (2D) cont.

+

-

Line-Line Intersection (2D)

• Line 0:• Line 1: • Solve for (x,y): two equations, two unknowns• Three cases:

- No intersection (parallel lines)- Always intersect (same line)- One intersection

Line-Line Intersection (2D) cont.

• No intersection (parallel lines)

• Always intersect (same line)

• Otherwise, solve for (x,y)• Constant time

Slope of the two lines are equal:

For k ≠ 0:

But not the same line:

Line Segment-Line Segment Intersection

• Find intersection point using line-line intersection (if it exists, if it doesn’t the line segments don’t intersect)

• Create a ray for each line segment• Ray ranges from 0 to 1 along the line segment• Calculate t of the intersection point for each

line segment• If both t’s are in [0,1], line segments intersect

Point-Line Intersection

• Calculate the halfspace the point lies in• If halfspace equals 0, point is on the line• Floating point error may play into this• Use an epsilon, point is near close enough to

being on the line

Point-Polygon Intersection

• Calculate the half space the point lies in for each edge of the polygon

• If the polygon is clockwise wound, the point must always be in - half space

• If the polygon is counter-clockwise wound, the point must always be in + half space

• O(# of vertices in the polygon)

+

+

+

+ -

-

-

-

ClockwiseWinding

+

++

+-

-

-

-

Counter-ClockwiseWinding

Polygon-Polygon Intersection

• Possibly intersecting or fully contained• Check each line segment in polygon A for

intersection with each line segment in polygon B

• Check for containment: pick a point from A and see if it’s in B (and do same check to see if B is in A)