纯真时期的解析几何，谁人能够铭记？
遇到一个有问题，三顾大牛求辅佐！！
1.问题：
现有一条线段，已知线段的两个端点。
求出该线段的中点，以该中点半径为 smallRadius 的圆，求该线段中垂线与圆的两个交点。
2.按理说:
过圆中心做一条直线，必有2解（画图可以看到直线和圆有两个交点）
3.蛋疼的地方：
求解的过程中竟然时不时的出现 delta = b^2-4*a*c < 0 的情况，和2 严重矛盾。
4.我的测试方案：
我写了一个方法，传入线段的两个点，和以线段中点为圆心的一个圆。
我100%确定，该线段的长度大于上述圆的直径（必须保证有两个交点么~）。
返回一个点数组，点数组包含了2解。
5。尾声：
为什么出现 delta < 0 这种奇葩的情况，求指点。
如能点醒我这块朽木，100整分奉送，在线等！！高手素来~~~
代码如下：
- Java code
package org.bruce.vertices.controller.geometry;/** * @author Bruce Yang * 用于打印调试~ */public class CGDbg { public static final boolean DEBUG_MODE = true; // 方便进行调试信息的输出，开关~ public static void println(Object info) { if(DEBUG_MODE) { System.out.println(info); } } public static void print(Object info) { if(DEBUG_MODE) { System.out.print(info); } }}***********************************************
- Java code
package org.bruce.vertices.controller.geometry;/** * @author BruceYang * 这个是对通用一次直线方程 A*x + B*y + C = 0 的封装~ * 本来封装的是斜截式，不过发现当斜率k不存在的时候会比较麻烦，因此该用一般式 * 再个就是接着用一般式的演变方式 x + B/A*y + C/A = 0，但是考虑到可能存在x ＝＝ 0 的情况，因此又舍弃~ * * 娘的，一般式还是他妈的无济于事啊，改回斜截式，多提供两个成员变量： * 一个boolean表示k是否存在，一个额外的float表示k不存在的时候直线方程 x=***, *** 等于多少~ */public class CGLine { // 特别声明为public类型，免得到时候访问的时候麻烦，到时候直接点就行了 private boolean kExists; // 大部分情况下 k 都应该是存在的，因此提供一个 true 的默认值~ public float k = 77885.201314f; public float b = 13145.207788f; public float extraX = 52077.881314f; /** * 这是当 k 存在时的构造方法~ * @param k * @param b */ public CGLine(float k, float b) { this.kExists = true; this.k = k; this.b = b; } /** * 已知两点，求直线的方程~ * @param p1 * @param p2 */ public CGLine(CGPoint p1, CGPoint p2) { if((p1.x - p2.x) != 0) { CGDbg.println("y = k*x + b, k exits!!"); this.kExists = true; this.k = (p1.y - p2.y)/(p1.x - p2.x); this.b = (p1.y - p1.x * k); } else { CGDbg.println("y = k*x + b, k doesn't exists!!"); // 如果走进这个分支，表示直线垂直于x轴，斜率不存在，保留k的默认值~ this.kExists = false; this.extraX = p1.x; } CGDbg.print("过p1("+p1.x+", " +p1.y + "), p2("+p2.x+", "+p2.y+")两点的直线方程表达式为: "); if(kExists) { CGDbg.println("y = " + k + "*x + " + b); } else { CGDbg.println("x = " + extraX + "(垂直于x轴!)"); } } /** * 点斜式~ * @param p 某点 * @param k 过该点的直线的斜率 */ public CGLine(float k, CGPoint p) { /** * (y-y') = k*(x-x') * 变形成斜截式为： * y = k*x + y' - k*x' * k = k, b = y'-k*x' */ this.kExists = true; this.k = k; this.b = p.y - k * p.y; } /** * 这是当 k 不存在时的构造方法~ * @param extraX */ public CGLine(float extraX) { this.kExists = false; this.extraX = extraX; } @Override public String toString() { return "Line.toString()方法被调用，y = k*x + b斜截式, k=" + this.k + ", b=" + this.b +", kExists=" + this.kExists + ", extraX=" + this.extraX; } public boolean iskExists() { return kExists; } public void setkExists(boolean kExists) { this.kExists = kExists; }}***********************************************
- Java code
package org.bruce.vertices.controller.geometry;/** * @author BruceYang * 对点的抽象~ */public class CGPoint { public float x; public float y; public CGPoint() { } public CGPoint(float x, float y) { this.x = x; this.y = y; } @Override public String toString() { return "x=" + this.x + ", y=" + this.y; }}
***********************************************
- Java code
package org.bruce.vertices.controller.geometry;/** * @author BruceYang */public class CGGeometryLib { /** * @param p0 第一个点的坐标 * @param p1 第二个点的坐标 * @return 两个点之间的距离 * 计算出两点之间的距离 */ public static float getDistanceBetween2Points(CGPoint p0, CGPoint p1) { float distance = (float)Math.sqrt(Math.pow(p0.y - p1.y, 2) + Math.pow(p0.x - p1.x, 2)); return distance; } /** * 获取传入两点的中点~ * @param p1 * @param p2 * @return */ public static CGPoint getMiddlePoint(CGPoint p1, CGPoint p2) { return new CGPoint((p1.x + p2.x) / 2.0f, (p1.y + p2.y) / 2.0f); } /** * 封装一下 Math 的 pow 、sqrt 方法，调用起来方便一些~ * @param d1 * @param d2 * @return */ public static double pow(double d1, double d2) { return Math.pow(d1, d2); } public static double sqrt(double d) { return Math.sqrt(d); } public static double sin(double theta) { return Math.sin(theta); } public static double cos(double theta) { return Math.cos(theta); } /** * 传入线段的两个端点，获取中点，以该中点为圆心做半径为 radius 的圆， * 经过线段中点做线段的垂线，返回垂线与圆的两个交点~ * Objective-C 里面的结果有点儿问题，不知道是什么原因，来java 里面碰碰有运气~ * @param p1 线段端点1 * @param p2 线段端点2 * @param radius 圆半径 * @return 线段中垂线与圆的两个交点~ */ public static CGPoint[] getWhatIWanted(CGPoint p1, CGPoint p2, float radius) { CGPoint[] target = new CGPoint[2]; CGPoint pMiddle = getMiddlePoint(p1, p2);// float segLength = getDistanceBetween2Points(p1, p2); CGLine l1 = new CGLine(p1, p2); if(l1.iskExists()) { if(l1.k != 0) { float kOfNewLine = -1 / l1.k; CGLine newLine = new CGLine(kOfNewLine, pMiddle); // 经过数学运算，得出二元一次方程组的表达式 double A = pow(newLine.k, 2) + 1; double B = 2 * (newLine.k * newLine.b - newLine.k * pMiddle.y - pMiddle.x); double C = pow(pMiddle.x, 2) + pow((newLine.b - pMiddle.y), 2) - pow(radius, 2); double delta = pow(B, 2) - 4 * A * C; if(delta < 0) { // 经实践检验有一定几率走入该分支，必须做特殊化处理~ CGDbg.println("竟然会无解，他妈的怎么回事儿啊！"); target[0] = new CGPoint(pMiddle.x, pMiddle.y - radius); target[1] = new CGPoint(pMiddle.x, pMiddle.y + radius); } else { double x1 = (-B + sqrt(delta)) / (2 * A); double y1 = newLine.k * x1 + newLine.b; target[0] = new CGPoint((float)x1, (float)y1); double x2 = (-B - sqrt(delta)) / (2 * A); double y2 = newLine.k * x2 + newLine.b; target[1] = new CGPoint((float)x2, (float)y2); } } else { target[0] = new CGPoint(pMiddle.x, pMiddle.y - radius); target[1] = new CGPoint(pMiddle.x, pMiddle.y + radius); } } else { target[0] = new CGPoint(pMiddle.x - radius, pMiddle.y); target[1] = new CGPoint(pMiddle.x + radius, pMiddle.y); } System.out.println("target[0] 距离中点的距离为：" + getDistanceBetween2Points(target[0], pMiddle)); System.out.println("target[1] 距离中点的距离为：" + getDistanceBetween2Points(target[1], pMiddle)); return target; } /** * @param args */ public static void main(String[] args) { double currentRadian = 0; double deltaRadian = Math.PI / 180; float bigRadius = 50; float smallRadius = 20; CGPoint origin = new CGPoint(0, 0); // 原点~ CGPoint tail = null; // tail 是尾巴、末梢的意思~ for(int i = 0; i < 360; ++ i) { System.out.println(" -- 第 "+ i + "度!"); tail = new CGPoint(bigRadius*(float)cos(currentRadian), bigRadius*(float)sin(currentRadian)); currentRadian += deltaRadian; getWhatIWanted(origin, tail, smallRadius); } }}
[解决办法]
public CGLine(float k, CGPoint p) {
? /**
? * (y-y') = k*(x-x')
? * 变形成斜截式为：
? * y = k*x + y' - k*x'
? * k = k, b = y'-k*x'
? */
? this.kExists = true;
? this.k = k;
? this.b = p.y - k * p.y;
? }
? ?
this.b = p.y - k*p.x;
这里错了，查了好久，先查delta发现没问题。在查k也没问题。中点也没问题
查到这里出问题了
float kOfNewLine = -1 / l1.k;
? CGLine newLine = new CGLine(kOfNewLine, pMiddle);
发现那个b有错误。。
[解决办法]
写了一个求交点的方法，看看怎么样
- Java code
public class Test { public static void main(String[] args) { test(-3, 7, 5, -3, 3); } static void test(double x1,double y1,double x2,double y2,double r){ double k = (y2 -y1) / (x2 - x1);//斜率 double k2 = -1.0 / k;//中垂线斜率 double midX = (x1 + x2) / 2;//中点ｘ double midY = (y1 + y2) / 2;//中点ｙ double a = Math.atan(k2);//中垂线与ｘ轴夹角 double y = r * Math.sin(a);//ｙ轴偏移量 double x = r * Math.cos(a);//ｘ轴偏移量 double resultX1 = k > 0 ? midX - x : midX + x;//上方交点ｘ坐标 double resultY1 = midY + y;//上方交点ｙ坐标 double resultX2 = k > 0 ? midX + x : midX - x ;//下方交点ｘ坐标 double resultY2 = midY - y;//下方交点ｙ坐标 System.out.println("result: "); System.out.println(resultX1 + "," + resultY1); System.out.println(resultX2 + "," + resultY2); }}
[解决办法]
- Java code
public class NiDongDe{ /** * @param args */ public static void main(String[] args) { Point one = new Point(0, 0); Point two = new Point(); double smallR = 20; double bigR = 50; for (int i = 0; i < 360; ++i) { two.x = bigR * Math.cos(i * Math.PI / 180); two.y = bigR * Math.sin(i * Math.PI / 180); System.out.println(i + "度 " + two.toString() + " " + countCrossPoint(one, two, smallR)); } } private static String countCrossPoint(Point one, Point two, double smallR) { // 转90度后use ax + by + c = 0 double a = 1; double b = 1; double k = 0; if (one.x == two.x || (Math.abs(one.x - two.x))<Math.pow(1, -10)) { a = 0; } else if (one.y == two.y|| (Math.abs(one.y - two.y))<Math.pow(1, -10)) { b = 0; } else { k = (two.x - one.x) / (two.y - one.y); } if (k == 0) { // 两种特殊情况 if (a == 0) { double y = (one.y + two.y) / 2; double x1 = smallR + (one.x + two.x) / 2; double x2 = (one.x + two.x) / 2 - smallR; return "(" + x1 + ", " + y + ") (" + x2 + ", " + y + ")"; } else if (b == 0) { double x = (one.x + two.x) / 2; double y1 = smallR + (one.y + two.y) / 2; double y2 = (one.y + two.y) / 2 - smallR; return "(" + x + ", " + y1 + ") (" + x + ", " + y2 + ")"; } } else { // 一般情况了啦 // 公式整理你懂的： // [(y2-y1)/(x2-x1)*x + y2 -x2*(y2-y1)/(x2-x1) -(y1+y2)/2]^2 + [x - (x1+x2)/2]^2 = smallR^2 // |<- p ->| |<- w ->| |<- q ->| double p = (two.y -one.y)/(two.x - one.x); double w = two.y - two.x * (two.y - one.y)/(two.x - one.x) - (one.y + two.y)/2; double q = (one.x + two.x)/2; // reserve = y2 -x2*(y2-y1)/(x2-x1) we'll use it later double reserve = two.y - two.x*(two.y - one.y)/(two.x - one.x); // now we have [px + w]^2 + [x - q]^2 = smallR^2 //再整理你懂的： // (p^2 + 1)x^2 + (2pw -2q)x + (q^2-smallR^2)=0 // a1 b1 c1 double a1 = Math.pow(p, 2) + 1; double b1 = 2*p*w - 2*q; double c1 = Math.pow(q, 2) - Math.pow(smallR, 2); // finally~ double x1 = (-1*b1 + Math.sqrt((Math.pow(b1, 2) - 4*a1*c1))) / (2*a1); double x2 = (-1*b1 - Math.sqrt((Math.pow(b1, 2) - 4*a1*c1))) / (2*a1); // y = px + reserve double y1 = p * x1 + reserve; double y2 = p * x1 + reserve; return "(" + x1 + ", " + y1 + ") (" + x2 + ", " + y2 + ")"; } return ""; }}class Point{ public double x; public double y; Point() { } Point(double x, double y) { this.x = x; this.y = y; } public String toString(){ return "(" + this.x + ", " + this.y + ")"; }}
[解决办法]
如果要求那两个点，倒是很简单的。
中点 x0，y0，源直线的角度d。中垂线角度是d- 90 小圆半径 r
一个点是?
x1 = x0 + r*cos(d- 90)?
y1 = y0 + r*sin(d- 90)
另外一个点
x2 = x0 - r*cos(d- 90)?
y2 = y0 - r*sin(d- 90)