Android examples for java.lang:Math Geometry
Check if two points are on the same side of a given line.
/******************************************************************************* * Copyright (c) 2011 MadRobot./*from ww w.j a v a 2 s. c o m*/ * All rights reserved. This program and the accompanying materials * are made available under the terms of the GNU Lesser Public License v2.1 * which accompanies this distribution, and is available at * http://www.gnu.org/licenses/old-licenses/gpl-2.0.html * * Contributors: * Elton Kent - initial API and implementation ******************************************************************************/ //package com.java2s; public class Main { /** * Check if two points are on the same side of a given line. * Algorithm from Sedgewick page 350. * * @param x0 * , y0, x1, y1 The line. * @param px0 * , py0 First point. * @param px1 * , py1 Second point. * @return <0 if points on opposite sides. * =0 if one of the points is exactly on the line * >0 if points on same side. */ private static int sameSide(float x0, float y0, float x1, float y1, float px0, float py0, float px1, float py1) { int sameSide = 0; float dx = x1 - x0; float dy = y1 - y0; float dx1 = px0 - x0; float dy1 = py0 - y0; float dx2 = px1 - x1; float dy2 = py1 - y1; // Cross product of the vector from the endpoint of the line to the // point float c1 = dx * dy1 - dy * dx1; float c2 = dx * dy2 - dy * dx2; if ((c1 != 0) && (c2 != 0)) { sameSide = (c1 < 0) != (c2 < 0) ? -1 : 1; } else if ((dx == 0) && (dx1 == 0) && (dx2 == 0)) { sameSide = !isBetween(y0, y1, py0) && !isBetween(y0, y1, py1) ? 1 : 0; } else if ((dy == 0) && (dy1 == 0) && (dy2 == 0)) { sameSide = !isBetween(x0, x1, px0) && !isBetween(x0, x1, px1) ? 1 : 0; } return sameSide; } /** * Return true if c is between a and b. */ private static boolean isBetween(float a, float b, float c) { return b > a ? (c >= a) && (c <= b) : (c >= b) && (c <= a); } }