org.locationtech.jts.index.chain.MonotoneChain Maven / Gradle / Ivy
/*
* Copyright (c) 2016 Vivid Solutions.
*
* All rights reserved. This program and the accompanying materials
* are made available under the terms of the Eclipse Public License v1.0
* and Eclipse Distribution License v. 1.0 which accompanies this distribution.
* The Eclipse Public License is available at http://www.eclipse.org/legal/epl-v10.html
* and the Eclipse Distribution License is available at
*
* http://www.eclipse.org/org/documents/edl-v10.php.
*/
package org.locationtech.jts.index.chain;
import org.locationtech.jts.geom.Coordinate;
import org.locationtech.jts.geom.Envelope;
import org.locationtech.jts.geom.LineSegment;
import org.locationtech.jts.geomgraph.index.MonotoneChainEdge;
/**
* Monotone Chains are a way of partitioning the segments of a linestring to
* allow for fast searching of intersections.
* They have the following properties:
*
* - the segments within a monotone chain never intersect each other
*
- the envelope of any contiguous subset of the segments in a monotone chain
* is equal to the envelope of the endpoints of the subset.
*
* Property 1 means that there is no need to test pairs of segments from within
* the same monotone chain for intersection.
*
* Property 2 allows
* an efficient binary search to be used to find the intersection points of two monotone chains.
* For many types of real-world data, these properties eliminate a large number of
* segment comparisons, producing substantial speed gains.
*
* One of the goals of this implementation of MonotoneChains is to be
* as space and time efficient as possible. One design choice that aids this
* is that a MonotoneChain is based on a subarray of a list of points.
* This means that new arrays of points (potentially very large) do not
* have to be allocated.
*
*
* MonotoneChains support the following kinds of queries:
*
* - Envelope select: determine all the segments in the chain which
* intersect a given envelope
*
- Overlap: determine all the pairs of segments in two chains whose
* envelopes overlap
*
*
* This implementation of MonotoneChains uses the concept of internal iterators
* ({@link MonotoneChainSelectAction} and {@link MonotoneChainOverlapAction})
* to return the results for queries.
* This has time and space advantages, since it
* is not necessary to build lists of instantiated objects to represent the segments
* returned by the query.
* Queries made in this manner are thread-safe.
*
* @version 1.7
*/
public class MonotoneChain {
private Coordinate[] pts;
private int start, end;
private Envelope env = null;
private Object context = null;// user-defined information
private int id;// useful for optimizing chain comparisons
public MonotoneChain(Coordinate[] pts, int start, int end, Object context)
{
this.pts = pts;
this.start = start;
this.end = end;
this.context = context;
}
public void setId(int id) { this.id = id; }
public int getId() { return id; }
public Object getContext() { return context; }
public Envelope getEnvelope()
{
if (env == null) {
Coordinate p0 = pts[start];
Coordinate p1 = pts[end];
env = new Envelope(p0, p1);
}
return env;
}
public int getStartIndex() { return start; }
public int getEndIndex() { return end; }
/**
* Gets the line segment starting at index
*
* @param index index of segment
* @param ls line segment to extract into
*/
public void getLineSegment(int index, LineSegment ls)
{
ls.p0 = pts[index];
ls.p1 = pts[index + 1];
}
/**
* Return the subsequence of coordinates forming this chain.
* Allocates a new array to hold the Coordinates
*/
public Coordinate[] getCoordinates()
{
Coordinate coord[] = new Coordinate[end - start + 1];
int index = 0;
for (int i = start; i <= end; i++) {
coord[index++] = pts[i];
}
return coord;
}
/**
* Determine all the line segments in the chain whose envelopes overlap
* the searchEnvelope, and process them.
*
* The monotone chain search algorithm attempts to optimize
* performance by not calling the select action on chain segments
* which it can determine are not in the search envelope.
* However, it *may* call the select action on segments
* which do not intersect the search envelope.
* This saves on the overhead of checking envelope intersection
* each time, since clients may be able to do this more efficiently.
*
* @param searchEnv the search envelope
* @param mcs the select action to execute on selected segments
*/
public void select(Envelope searchEnv, MonotoneChainSelectAction mcs)
{
computeSelect(searchEnv, start, end, mcs);
}
private void computeSelect(
Envelope searchEnv,
int start0, int end0,
MonotoneChainSelectAction mcs )
{
Coordinate p0 = pts[start0];
Coordinate p1 = pts[end0];
//Debug.println("trying:" + p0 + p1 + " [ " + start0 + ", " + end0 + " ]");
// terminating condition for the recursion
if (end0 - start0 == 1) {
//Debug.println("computeSelect:" + p0 + p1);
mcs.select(this, start0);
return;
}
// nothing to do if the envelopes don't overlap
if (! searchEnv.intersects(p0, p1))
return;
// the chains overlap, so split each in half and iterate (binary search)
int mid = (start0 + end0) / 2;
// Assert: mid != start or end (since we checked above for end - start <= 1)
// check terminating conditions before recursing
if (start0 < mid) {
computeSelect(searchEnv, start0, mid, mcs);
}
if (mid < end0) {
computeSelect(searchEnv, mid, end0, mcs);
}
}
/**
* Determine all the line segments in two chains which may overlap, and process them.
*
* The monotone chain search algorithm attempts to optimize
* performance by not calling the overlap action on chain segments
* which it can determine do not overlap.
* However, it *may* call the overlap action on segments
* which do not actually interact.
* This saves on the overhead of checking intersection
* each time, since clients may be able to do this more efficiently.
*
* @param searchEnv the search envelope
* @param mco the overlap action to execute on selected segments
*/
public void computeOverlaps(MonotoneChain mc, MonotoneChainOverlapAction mco)
{
computeOverlaps(start, end, mc, mc.start, mc.end, mco);
}
private void computeOverlaps(
int start0, int end0,
MonotoneChain mc,
int start1, int end1,
MonotoneChainOverlapAction mco)
{
//Debug.println("computeIntersectsForChain:" + p00 + p01 + p10 + p11);
// terminating condition for the recursion
if (end0 - start0 == 1 && end1 - start1 == 1) {
mco.overlap(this, start0, mc, start1);
return;
}
// nothing to do if the envelopes of these subchains don't overlap
if (! overlaps(start0, end0, mc, start1, end1)) return;
// the chains overlap, so split each in half and iterate (binary search)
int mid0 = (start0 + end0) / 2;
int mid1 = (start1 + end1) / 2;
// Assert: mid != start or end (since we checked above for end - start <= 1)
// check terminating conditions before recursing
if (start0 < mid0) {
if (start1 < mid1) computeOverlaps(start0, mid0, mc, start1, mid1, mco);
if (mid1 < end1) computeOverlaps(start0, mid0, mc, mid1, end1, mco);
}
if (mid0 < end0) {
if (start1 < mid1) computeOverlaps(mid0, end0, mc, start1, mid1, mco);
if (mid1 < end1) computeOverlaps(mid0, end0, mc, mid1, end1, mco);
}
}
/**
* Tests whether the envelopes of two chain sections overlap (intersect).
*
* @param start0
* @param end0
* @param mc
* @param start1
* @param end1
* @return true if the section envelopes overlap
*/
private boolean overlaps(
int start0, int end0,
MonotoneChain mc,
int start1, int end1)
{
return Envelope.intersects(pts[start0], pts[end0], mc.pts[start1], mc.pts[end1]);
}
}