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[英]How to avoid floating point precision errors with floats or doubles in Java?
[英]Subdividing square problems due to floating point precision errors
我有一個問題(在JAVA中)由於浮點精度錯誤我無法解決。
我有一個軸對齊的方形類,通過指定2個點來定義。 在構造函數中,確定最大距離是什么(在x方向或y方向上),並且這用於創建正方形,其中點的距離等於中心的距離。 這意味着點位於邊界上。
例如,如果我定義一個帶有點(0,2)和(3,3)的正方形,則最大距離是x距離(3),正方形將定義如下:
可以看出,點位於邊緣,點的中點正好是正方形的中心(1.5,2.5)。
我創建了一個方法來檢查方塊是否包含某個點,看起來像這樣:
see added code sample below
這意味着如果一個點位於邊界上,則它被“包含”。
現在我想將方塊細分為4個“同等”大小的部分(NorthEast,NorthWest,SouthWest和SouthEast),邏輯上定義原始方塊的初始點必須包含在至少1個部分中。 但是當單元測試時,隨機指向它失敗並且它看起來像是因為雙精度浮點錯誤。
我嘗試了不同的解決方法,我的最后一次迭代如下:
這保證了原始點被包含,我可以輕松地創建細分,我認為如果我確保沒有對定義正方形的點的特征進行數學運算,我保證原始點至少包含1個部分(因為他們在邊界上)。 我的細分程序如下:
see added code sample below
但是當使用隨機點運行大量迭代的單元測試時,差不多16個仍然失敗,我不知道為什么邊緣點會發生變化。 在所有這些測試中,初始收容檢查(無論父方是否包含點,盡管它們位於邊緣)都通過了100%。
編輯顯示我的實現的一些實際代碼:
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)
};
}
}
以下是測試用例:
@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);
}
}
我的隨機測試導致的失敗案例之一:
p1 = (0.31587198758298796, 0.12796964677511913)
p2 = (0.04837609765424089, 0.6711236142940149)
這個失敗了,因為p1應該在SouthEast扇區,但是那個被定義為:
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
在看了你的代碼后,它沒有任何意義,它會失敗,因為據我所知,你沒有在那個失敗的情況下對Y
值應用任何操作 - 你只是通過它沒有任何操作,所以浮點精度損失是無關緊要的。 雖然p1
- p4
可以代表任何一個角落,但我有點難以理解,所以我重寫了Rectangle類如下,希望有點清楚:
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)
};
}
}
如果這不能解決問題,也許一些日志記錄會在0.12796964677511913
變為0.12796964677511916
我修好了它! 感謝所有的幫助0x24a537r9,因為它使它更清晰。 我添加了你的代碼(並修正了一個錯字),但我們仍然錯過了一個特殊情況,即它是一個完美的正方形,因此,dx == dy。
如果dx == dy,我們知道正方形的所有點,並且可以在不使用dl的情況下添加它們。 在我的失敗情況下,它是一個完美的正方形,因此它最終將在第一個if子句中,因此使用dl計算2個新的角點(這將導致浮點錯誤)。
從邏輯上講,它最終將處於一個正方形的情況,因為我強迫它成為正方形......
我的最終構造函數代碼如下:
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());
}
}
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