# cyberpython/ComputationalGeometry

Java implementations of Graham's Scan, Jarvis' March and the Bentley-Ottmann line-segments intersection algorithm
Java
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## ComputationalGeometry - Computational Geometry Algorithms in Java

This software was developed as an assignment for the "Special Topics On Algorithms" course at the Athens Univesity Of Economics And Business (Spring 2010).

## Screenshots

![Screenshot1][screenshot1] [screenshot1]: ComputationalGeometry/raw/master/screenshots/screenshot_th.png "Screenshot"

## Todo

###Segment Intersections:

• Bentley-Ottmann

###Convex Hull:

• Graham's Scan

## API

####aueb.geom.algorithms.Intersections public static List bentleyOttmann(List segments, List eventLog);

####aueb.geom.algorithms.ConvexHull public static List grahamsScan(List points, List events); public static List jarvisMarch(List points, List events);

The last parameter of all 3 methods is a List that is filled with events(algorithm steps) as the method executes. MUST NOT be null.

The Bentley-Ottmann implementation uses an ArrayList instead of a self-balancing binary search tree (e.g. a Red-Black tree) so its complexity is O((k*n+n^2)*log(n)) instead of O((k+n)*log(n)).

## Algorithms

###Bentley-Ottmann: ------------------------------------ BEGIN ------------------------------------- 1) Initialize a priority queue Q of potential future events, each associated with a point in the plane and prioritized by the x-coordinate of the point. Initially, Q contains an event for each of the endpoints of the input segments. 2) Initialize a binary search tree T of the line segments that cross the sweep line L, ordered by the y-coordinates of the crossing points. Initially, T is empty. 3) While Q is nonempty, find and remove the event from Q associated with a point p with minimum x-coordinate. Determine what type of event this is and process it according to the following case analysis:

``````	*	If p is the left endpoint of a line segment s, insert s into T.
Find the segments r and t that are immediately below and above s in T (if they exist)
and if their crossing forms a potential future event in the event queue, remove it.
If s crosses r or t, add those crossing points as potential future events in the event queue.

*	If p is the right endpoint of a line segment s, remove s from T.
Find the segments r and t that were (prior to the removal of s)
immediately above and below itin T (if they exist).
If r and t cross, add that crossing point as a potential future event in the event queue.

*	If p is the crossing point of two segments s and t (with s below t to the left of the crossing),
swap the positions of s and t in T. Find the segments r and u (if they exist) that are immediately
below and above s and t respectively. Remove any crossing points rs and tu from the event queue, and,
if r and t cross or s and u cross, add those crossing points to the event queue.
------------------------------------  END  -------------------------------------
``````

###Graham's Scan: ------------------------------------ BEGIN ------------------------------------- Three points are a counter-clockwise turn if ccw > 0, clockwise if ccw < 0, and collinear if ccw = 0 because ccw is a determinant that gives the signed area of the triangle formed by p1, p2, and p3. function ccw(p1, p2, p3): return (p2.x - p1.x)(p3.y - p1.y) - (p2.y - p1.y)(p3.x - p1.x)

``````	p0 = the point with min y‐coordinate;
<p1,p2,...,pm> : the remaining vertices in <Q>;
PUSH(p0,S);
PUSH(p1,S);
PUSH(p2,S);
for i=3 to m do
{ while ccw(N‐TOP(S),TOP(S),pi) <= 0
do POP(S);
PUSH(pi,S);
}
return S;
------------------------------------  END  -------------------------------------
``````

###Jarvis' March ------------------------------------ BEGIN ------------------------------------- Start from the lowest point p0 i=1; // for Right Chain REPEAT { find the point pi with the smallest polar angle with pi‐1; i=i+1; } UNTIL pi is the highest point // for Left Chain REPEAT { find the point pi with the smallest polar angle with pi‐1 for the negative axis; i=i+1; } UNTIL pi is the lowest point

``````------------------------------------  END  -------------------------------------
``````