我有一个问题（在JAVA中）由于浮点精度错误而无法解决。
我有一个轴对齐的方形类，它通过指定2个点来定义。在构造函数中，确定最大距离是什么（在x方向或y方向上），并且这用于创建正方形，其中点的距离等于中心的距离。这意味着这些点位于边界上。
例如，如果我定义一个带有点（0,2）和（3,3）的正方形，则最大距离是x距离（3），正方形将定义如下：
可以看出，这些点位于边缘，点的中点正好是方形的中心（1.5,2.5）。
我创建了一个方法来检查方块是否包含某个点，如下所示：
see added code sample below
这意味着如果一个点在边界上，它就会被“包含”。
现在我想将方块细分为4个'同等'大小的部分（NorthEast，NorthWest，SouthWest和SouthEast），逻辑上定义原始正方形的初始点必须至少包含1个部分。但是当单元测试时，随机指向它失败并且它看起来像是因为双精度浮点错误。
我尝试了不同的解决方法，最后一次迭代如下：
这可以保证包含原始点，并且我可以轻松地创建细分，并且我思想如果我确保没有执行数学运算，它保证原始点至少包含1个部分关于定义正方形的点的特征（因为它们在边界上）。我的细分程序如下：
see added code sample below
但是当使用随机点运行大量迭代的单元测试时，差不多16个仍然失败，我不知道为什么边缘点会发生变化。在所有这些测试中，初始收容检查（无论父方是否包含点，尽管它们位于边缘）通过100％。
修改 一些实际的代码显示了我的实现：
public class Point implements IPoint {
double x, y;
public Point(double x, double y) {
this.x = x;
this.y = y;
}
@Override
public double x() {
return x;
}
@Override
public double y() {
return y;
}
@Override
public IPoint midPoint(IPoint other) {
return new Point((x() + other.x()) / 2, (y() + other.y()) / 2);
}
}
public class Rectangle implements IRectangle {
IPoint p0, p1, p2, p3, p4;
public Rectangle(IPoint v1, IPoint v2) {
double dx, dy, dl;
IPoint v0;
// calculate dominant length
dx = Math.abs(v1.x() - v2.x());
dy = Math.abs(v1.y() - v2.y());
dl = dx >= dy ? dx : dy;
if (dx >= dy) {
// make sure v0 = left-most
if (v1.x() <= v2.x()) {
v0 = v1;
v1 = v2;
} else {
v0 = v2;
}
} else {
// make sure v0 = bottom-most
if (v1.y() <= v2.y()) {
v0 = v1;
v1 = v2;
} else {
v0 = v2;
}
}
this.p0 = v0.midPoint(v1);
if (dx >= dy) {
// this way v0 and v1 are always on the vertical boundaries
this.p1 = new Point(v0.x(), this.p0.y() - dl / 2);
this.p2 = new Point(v0.x(), this.p0.y() + dl / 2);
this.p3 = new Point(v1.x(), this.p0.y() + dl / 2);
this.p4 = new Point(v1.x(), this.p0.y() - dl / 2);
} else {
// this way v0 and v1 are always on the horizontal boundaries
this.p1 = new Point(this.p0.x() - dl / 2, v0.y());
this.p2 = new Point(this.p0.x() - dl / 2, v1.y());
this.p3 = new Point(this.p0.x() + dl / 2, v1.y());
this.p4 = new Point(this.p0.x() + dl / 2, v0.y());
}
}
@Override
public boolean contains(IPoint p) {
if (p.x() < p1.x() || p.x() > p4.x()) return false;
if (p.y() < p1.y() || p.y() > p2.y()) return false;
return true;
}
@Override
public IRectangle[] subdivide() {
return new Rectangle[] {
new Rectangle(p0, p2),
new Rectangle(p0, p3),
new Rectangle(p0, p4),
new Rectangle(p0, p1)
};
}
}
以下是测试用例：
@Test
public void testMassedSubdivide() throws Exception {
Random r = new Random();
IPoint p1, p2;
IRectangle[] rects;
boolean check1, check2;
for (int i = 0; i < 100000; i++) {
p1 = new Point(r.nextDouble(), r.nextDouble());
p2 = new Point(r.nextDouble(), r.nextDouble());
q = new Rectangle(p1, p2);
assertTrue(q.contains(p1));
assertTrue(q.contains(p2));
rects = q.subdivide();
check1 = rects[0].contains(p1) || rects[1].contains(p1) || rects[2].contains(p1) || rects[3].contains(p1);
check2 = rects[0].contains(p2) || rects[1].contains(p2) || rects[2].contains(p2) || rects[3].contains(p2);
assertTrue(check1);
assertTrue(check2);
}
}
我的随机测试导致的一个失败案例：
p1 = (0.31587198758298796, 0.12796964677511913)
p2 = (0.04837609765424089, 0.6711236142940149)
这个失败了，因为p1应该在SouthEast扇区，但那个被定义为：
p0=(0.31791253449833834, 0.2637581386548431),
p1=(0.18212404261861442, 0.12796964677511916), <- wrong, last 6 should be 3
p2=(0.18212404261861442, 0.39954663053456707),
p3=(0.4537010263780623, 0.39954663053456707),
p4=(0.4537010263780623, 0.12796964677511916) <- wrong, last 6 should be 3
答案 0 :(得分：4)
在查看您的代码后，它没有任何意义，它会失败，因为据我所知，您没有对该失败案例中的Y值应用任何操作 - 你只是在没有任何操作的情况下通过它，所以浮点精度损失是无关紧要的。虽然p1 - p4可以代表任何一个角落，但我有点难以理解，所以我重写了Rectangle类，希望有点清楚：
public class Rectangle implements IRectangle {
IPoint centroid, bottomLeft, topLeft, bottomRight, topRight;
public Rectangle(IPoint v0, IPoint v1) {
IPoint bottomLeft = new Point(Math.min(v0.x, v1.x), Math.min(v0.y, v1.y));
IPoint topRight = new Point(Math.max(v0.x, v1.x), Math.max(v0.y, v1.y));
// calculate dominant length
double dx = topRight.x - bottomLeft.x;
double dy = topRight.y - bottomLeft.y; // Assumes (0, 0) is in the bottom-left.
double dl = dx >= dy ? dx : dy;
this.centroid = bottomLeft.midPoint(topRight);
if (dx >= dy) {
// this way bottomLeft and topRight are always on the vertical boundaries
this.bottomLeft = new Point(bottomLeft.x(), this.centroid.y() - dl / 2);
this.topLeft = new Point(bottomLeft.x(), this.centroid.y() + dl / 2);
this.bottomRight = new Point(topRight.x(), this.centroid.y() - dl / 2);
this.topRight = new Point(topRight.x(), this.centroid.y() + dl / 2);
} else {
// this way bottomLeft and topRight are always on the horizontal boundaries
this.bottomLeft = new Point(this.centroid.x() - dl / 2, bottomLeft.y());
this.topLeft = new Point(this.centroid.x() - dl / 2, topLeft.y());
this.bottomRight = new Point(this.centroid.x() + dl / 2, bottomLeft.y());
this.topRight = new Point(this.centroid.x() + dl / 2, topLeft.y());
}
}
@Override
public boolean contains(IPoint p) {
if (p.x() < bottomLeft.x() || p.x() > topRight.x()) return false;
if (p.y() < bottomLeft.y() || p.y() > topRight.y()) return false;
return true;
}
@Override
public IRectangle[] subdivide() {
return new Rectangle[] {
new Rectangle(centroid, bottomLeft),
new Rectangle(centroid, topLeft),
new Rectangle(centroid, bottomRight),
new Rectangle(centroid, topRight)
};
}
}
如果这还没有解决问题，可能会有一些日志记录会在0.12796964677511913更改为0.12796964677511916
答案 1 :(得分：1)
我修好了！感谢所有的帮助0x24a537r9，因为它使它更清晰。我添加了你的代码（并修正了一个拼写错误）但我们仍然错过了一个特殊情况，即它是一个完美的正方形，因此，dx == dy。
如果dx == dy，我们知道正方形的所有点，并且可以在不使用dl的情况下添加它们。在我的失败案例中，它是一个完美的正方形，因此它将在第一个 if 子句中结束，因此使用 dl 来计算2个新的角点（这将导致在浮点错误）。
逻辑上它最终将处于一个正方形状态，因为我强迫它变成方形......
我的最终构造函数代码如下：
IPoint centroid, bottomLeft, topLeft, topRight, bottomRight;
public Rectangle(IPoint v0, IPoint v1) {
IPoint bottomLeft = new Point(Math.min(v0.x(), v1.x()), Math.min(v0.y(), v1.y()));
IPoint topRight = new Point(Math.max(v0.x(), v1.x()), Math.max(v0.y(), v1.y()));
// calculate dominant length
double dx = topRight.x() - bottomLeft.x();
double dy = topRight.y() - bottomLeft.y(); // Assumes (0, 0) is in the bottom-left.
double dl = dx >= dy ? dx : dy;
this.centroid = bottomLeft.midPoint(topRight);
if (dx == dy) // special case where it is square <- important, because this one fixes the errors
{
this.bottomLeft = bottomLeft;
this.topLeft = new Point(bottomLeft.x(), topRight.y());
this.topRight = topRight;
this.bottomRight = new Point(topRight.x(), bottomLeft.y());
}
else if (dx >= dy) {
// this way bottomLeft and topRight are always on the vertical boundaries
this.bottomLeft = new Point(bottomLeft.x(), this.centroid.y() - dl / 2);
this.topLeft = new Point(bottomLeft.x(), this.centroid.y() + dl / 2);
this.bottomRight = new Point(topRight.x(), this.centroid.y() - dl / 2);
this.topRight = new Point(topRight.x(), this.centroid.y() + dl / 2);
} else {
// this way bottomLeft and topRight are always on the horizontal boundaries
this.bottomLeft = new Point(this.centroid.x() - dl / 2, bottomLeft.y());
this.topLeft = new Point(this.centroid.x() - dl / 2, topRight.y());
this.bottomRight = new Point(this.centroid.x() + dl / 2, bottomLeft.y());
this.topRight = new Point(this.centroid.x() + dl / 2, topRight.y());
}
}