在任何地方都有Java库可以对IEEE 754 half-precision数字进行计算或将它们转换为双精度数据吗？
这两种方法都适用：
修改：转换需要100％准确 - 输入文件中存在大量NaN，无穷大和次正规。
相关问题但适用于JavaScript：Decompressing Half Precision Floats in Javascript
答案 0 :(得分：51)
您可以使用Float.intBitsToFloat()和Float.floatToIntBits()将它们与原始浮点值进行转换。如果您可以使用截断精度（而不是舍入），则只需几位移位即可实现转换。
我现在已经付出了更多的努力，结果并不像我在开始时预期的那么简单。这个版本现在在我能想象的每个方面进行测试和验证，我非常有信心它可以为所有可能的输入值生成精确的结果。它支持任意方向的精确舍入和次正规转换。
// ignores the higher 16 bits
public static float toFloat( int hbits )
{
int mant = hbits & 0x03ff; // 10 bits mantissa
int exp = hbits & 0x7c00; // 5 bits exponent
if( exp == 0x7c00 ) // NaN/Inf
exp = 0x3fc00; // -> NaN/Inf
else if( exp != 0 ) // normalized value
{
exp += 0x1c000; // exp - 15 + 127
if( mant == 0 && exp > 0x1c400 ) // smooth transition
return Float.intBitsToFloat( ( hbits & 0x8000 ) << 16
| exp << 13 | 0x3ff );
}
else if( mant != 0 ) // && exp==0 -> subnormal
{
exp = 0x1c400; // make it normal
do {
mant <<= 1; // mantissa * 2
exp -= 0x400; // decrease exp by 1
} while( ( mant & 0x400 ) == 0 ); // while not normal
mant &= 0x3ff; // discard subnormal bit
} // else +/-0 -> +/-0
return Float.intBitsToFloat( // combine all parts
( hbits & 0x8000 ) << 16 // sign << ( 31 - 15 )
| ( exp | mant ) << 13 ); // value << ( 23 - 10 )
}
// returns all higher 16 bits as 0 for all results
public static int fromFloat( float fval )
{
int fbits = Float.floatToIntBits( fval );
int sign = fbits >>> 16 & 0x8000; // sign only
int val = ( fbits & 0x7fffffff ) + 0x1000; // rounded value
if( val >= 0x47800000 ) // might be or become NaN/Inf
{ // avoid Inf due to rounding
if( ( fbits & 0x7fffffff ) >= 0x47800000 )
{ // is or must become NaN/Inf
if( val < 0x7f800000 ) // was value but too large
return sign | 0x7c00; // make it +/-Inf
return sign | 0x7c00 | // remains +/-Inf or NaN
( fbits & 0x007fffff ) >>> 13; // keep NaN (and Inf) bits
}
return sign | 0x7bff; // unrounded not quite Inf
}
if( val >= 0x38800000 ) // remains normalized value
return sign | val - 0x38000000 >>> 13; // exp - 127 + 15
if( val < 0x33000000 ) // too small for subnormal
return sign; // becomes +/-0
val = ( fbits & 0x7fffffff ) >>> 23; // tmp exp for subnormal calc
return sign | ( ( fbits & 0x7fffff | 0x800000 ) // add subnormal bit
+ ( 0x800000 >>> val - 102 ) // round depending on cut off
>>> 126 - val ); // div by 2^(1-(exp-127+15)) and >> 13 | exp=0
}
与 book 相比，我实现了两个小扩展，因为16位浮点数的一般精度相当低，这可能使浮点格式的固有异常在视觉上可感知，而较大的浮点类型由于精确度很高，通常不会被注意到。
第一个是toFloat()函数中的这两行：
if( mant == 0 && exp > 0x1c400 ) // smooth transition
return Float.intBitsToFloat( ( hbits & 0x8000 ) << 16 | exp << 13 | 0x3ff );
在类型大小的正常范围内的浮点数采用指数，因此精度为值的大小。但这并非顺利采用，而是按步骤进行：切换到下一个更高的指数会导致精度降低一半。对于尾数的所有值，精度现在保持不变，直到下一个跳到下一个更高的指数。上面的扩展代码通过返回该特定半浮点值的覆盖32位浮点范围的地理中心中的值，使这些转换更平滑。每个正常的半浮点值都精确映射到8192个32位浮点值。返回的值应该恰好位于这些值的中间。但是在半浮点指数的转换处，较低的4096值具有两倍于上4096值的精度，因此覆盖的数字空间仅为另一侧的一半。所有这些8192 32位浮点值映射到相同的半浮点值，因此将半浮点数转换为32位并返回导致相同的半浮点值，无论8192 中间 32位值中的哪一个是选择。现在，扩展现在会在转换时产生类似于sqrt（2）因子的更平滑的半步，如下面的右图片所示，而左图片应该是在没有抗锯齿的情况下将锐利步骤可视化为两倍。您可以安全地从代码中删除这两行以获得标准行为。
covered number space on either side of the returned value:
6.0E-8 ####### ##########
4.5E-8 | #
3.0E-8 ######### ########
第二个扩展名在fromFloat()函数中：
{ // avoid Inf due to rounding
if( ( fbits & 0x7fffffff ) >= 0x47800000 )
...
return sign | 0x7bff; // unrounded not quite Inf
}
此扩展稍微扩展了半浮点格式的数字范围，方法是将某些32位值保存为无限升级。受影响的值是那些在没有舍入的情况下小于无穷大的值，并且由于舍入而仅变为无穷大。如果您不想要此扩展名，可以安全地删除上面显示的行。
我尝试尽可能地优化fromFloat()函数中正常值的路径，这使得它由于使用预先计算和未移位的常量而使其可读性稍差。我没有在'toFloat（）'中投入太多精力，因为它无论如何都不会超过查找表的性能。因此，如果速度真的很重要，可以使用toFloat()函数仅填充具有0x10000元素的静态查找表，而不是使用此表进行实际转换。使用当前的x64服务器虚拟机大约快3倍，使用x86客户端虚拟机大约快5倍。
我将此代码置于公共领域。
答案 1 :(得分：1)
x4u的代码将值1正确编码为0x3c00（ref：https://en.wikipedia.org/wiki/Half-precision_floating-point_format）。但是具有平滑度改进的解码器将其解码为1.000122。维基百科条目表示可以精确表示整数值0..2048。不太好......从forFloat代码中删除"| 0x3ff"可确保toFloat(fromFloat(k)) == k表示-2048..2048范围内的整数k，可能代价不太平滑。< / p>
答案 2 :(得分：0)
在我看到这里发布的解决方案之前，我已经掀起了一些简单的事情：
public static float toFloat(int nHalf)
{
int S = (nHalf >>> 15) & 0x1;
int E = (nHalf >>> 10) & 0x1F;
int T = (nHalf ) & 0x3FF;
E = E == 0x1F
? 0xFF // it's 2^w-1; it's all 1's, so keep it all 1's for the 32-bit float
: E - 15 + 127; // adjust the exponent from the 16-bit bias to the 32-bit bias
// sign S is now bit 31
// exp E is from bit 30 to bit 23
// scale T by 13 binary digits (it grew from 10 to 23 bits)
return Float.intBitsToFloat(S << 31 | E << 23 | T << 13);
}
我确实喜欢其他发布解决方案中的方法。供参考：
// notes from the IEEE-754 specification:
// left to right bits of a binary floating point number:
// size bit ids name description
// ---------- ------------ ---- ---------------------------
// 1 bit S sign
// w bits E[0]..E[w-1] E biased exponent
// t=p-1 bits d[1]..d[p-1] T trailing significant field
// The range of the encoding’s biased exponent E shall include:
// ― every integer between 1 and 2^w − 2, inclusive, to encode normal numbers
// ― the reserved value 0 to encode ±0 and subnormal numbers
// ― the reserved value 2w − 1 to encode +/-infinity and NaN
// The representation r of the floating-point datum, and value v of the floating-point datum
// represented, are inferred from the constituent fields as follows:
// a) If E == 2^w−1 and T != 0, then r is qNaN or sNaN and v is NaN regardless of S
// b) If E == 2^w−1 and T == 0, then r=v=(−1)^S * (+infinity)
// c) If 1 <= E <= 2^w−2, then r is (S, (E−bias), (1 + 2^(1−p) * T))
// the value of the corresponding floating-point number is
// v = (−1)^S * 2^(E−bias) * (1 + 2^(1−p) * T)
// thus normal numbers have an implicit leading significand bit of 1
// d) If E == 0 and T != 0, then r is (S, emin, (0 + 2^(1−p) * T))
// the value of the corresponding floating-point number is
// v = (−1)^S * 2^emin * (0 + 2^(1−p) * T)
// thus subnormal numbers have an implicit leading significand bit of 0
// e) If E == 0 and T ==0, then r is (S, emin, 0) and v = (−1)^S * (+0)
// parameter bin16 bin32
// -------------------------------------------- ----- -----
// k, storage width in bits 16 32
// p, precision in bits 11 24
// emax, maxiumum exponent e 15 127
// bias, E-e 15 127
// sign bit 1 1
// w, exponent field width in bits 5 8
// t, trailing significant field width in bits 10 23
答案 3 :(得分：0)
我创建了一个名为 HalfPrecisionFloat 的java类，它使用x4u的解决方案。该类具有便捷方法和错误检查。它更进一步，有从2字节半精度值返回Double和Float的方法。
希望这会对某人有所帮助。
==＆GT;
import java.nio.ByteBuffer;
/**
* Accepts various forms of a floating point half-precision (2 byte) number
* and contains methods to convert to a
* full-precision floating point number Float and Double instance.
* <p>
* This implemention was inspired by x4u who is a user contributing
* to stackoverflow.com.
* (https://stackoverflow.com/users/237321/x4u).
*
* @author dougestep
*/
public class HalfPrecisionFloat {
private short halfPrecision;
private Float fullPrecision;
/**
* Creates an instance of the class from the supplied the supplied
* byte array. The byte array must be exactly two bytes in length.
*
* @param bytes the two-byte byte array.
*/
public HalfPrecisionFloat(byte[] bytes) {
if (bytes.length != 2) {
throw new IllegalArgumentException("The supplied byte array " +
"must be exactly two bytes in length");
}
final ByteBuffer buffer = ByteBuffer.wrap(bytes);
this.halfPrecision = buffer.getShort();
}
/**
* Creates an instance of this class from the supplied short number.
*
* @param number the number defined as a short.
*/
public HalfPrecisionFloat(final short number) {
this.halfPrecision = number;
this.fullPrecision = toFullPrecision();
}
/**
* Creates an instance of this class from the supplied
* full-precision floating point number.
*
* @param number the float number.
*/
public HalfPrecisionFloat(final float number) {
if (number > Short.MAX_VALUE) {
throw new IllegalArgumentException("The supplied float is too "
+ "large for a two byte representation");
}
if (number < Short.MIN_VALUE) {
throw new IllegalArgumentException("The supplied float is too "
+ "small for a two byte representation");
}
final int val = fromFullPrecision(number);
this.halfPrecision = (short) val;
this.fullPrecision = number;
}
/**
* Returns the half-precision float as a number defined as a short.
*
* @return the short.
*/
public short getHalfPrecisionAsShort() {
return halfPrecision;
}
/**
* Returns a full-precision floating pointing number from the
* half-precision value assigned on this instance.
*
* @return the full-precision floating pointing number.
*/
public float getFullFloat() {
if (fullPrecision == null) {
fullPrecision = toFullPrecision();
}
return fullPrecision;
}
/**
* Returns a full-precision double floating point number from the
* half-precision value assigned on this instance.
*
* @return the full-precision double floating pointing number.
*/
public double getFullDouble() {
return new Double(getFullFloat());
}
/**
* Returns the full-precision float number from the half-precision
* value assigned on this instance.
*
* @return the full-precision floating pointing number.
*/
private float toFullPrecision() {
int mantisa = halfPrecision & 0x03ff;
int exponent = halfPrecision & 0x7c00;
if (exponent == 0x7c00) {
exponent = 0x3fc00;
} else if (exponent != 0) {
exponent += 0x1c000;
if (mantisa == 0 && exponent > 0x1c400) {
return Float.intBitsToFloat(
(halfPrecision & 0x8000) << 16 | exponent << 13 | 0x3ff);
}
} else if (mantisa != 0) {
exponent = 0x1c400;
do {
mantisa <<= 1;
exponent -= 0x400;
} while ((mantisa & 0x400) == 0);
mantisa &= 0x3ff;
}
return Float.intBitsToFloat(
(halfPrecision & 0x8000) << 16 | (exponent | mantisa) << 13);
}
/**
* Returns the integer representation of the supplied
* full-precision floating pointing number.
*
* @param number the full-precision floating pointing number.
* @return the integer representation.
*/
private int fromFullPrecision(final float number) {
int fbits = Float.floatToIntBits(number);
int sign = fbits >>> 16 & 0x8000;
int val = (fbits & 0x7fffffff) + 0x1000;
if (val >= 0x47800000) {
if ((fbits & 0x7fffffff) >= 0x47800000) {
if (val < 0x7f800000) {
return sign | 0x7c00;
}
return sign | 0x7c00 | (fbits & 0x007fffff) >>> 13;
}
return sign | 0x7bff;
}
if (val >= 0x38800000) {
return sign | val - 0x38000000 >>> 13;
}
if (val < 0x33000000) {
return sign;
}
val = (fbits & 0x7fffffff) >>> 23;
return sign | ((fbits & 0x7fffff | 0x800000)
+ (0x800000 >>> val - 102) >>> 126 - val);
}
这是单元测试
import org.junit.Assert;
import org.junit.Test;
import java.nio.ByteBuffer;
public class TestHalfPrecision {
private byte[] simulateBytes(final float fullPrecision) {
HalfPrecisionFloat halfFloat = new HalfPrecisionFloat(fullPrecision);
short halfShort = halfFloat.getHalfPrecisionAsShort();
ByteBuffer buffer = ByteBuffer.allocate(2);
buffer.putShort(halfShort);
return buffer.array();
}
@Test
public void testHalfPrecisionToFloatApproach() {
final float startingValue = 1.2f;
final float closestValue = 1.2001953f;
final short shortRepresentation = (short) 15565;
byte[] bytes = simulateBytes(startingValue);
HalfPrecisionFloat halfFloat = new HalfPrecisionFloat(bytes);
final float retFloat = halfFloat.getFullFloat();
Assert.assertEquals(new Float(closestValue), new Float(retFloat));
HalfPrecisionFloat otherWay = new HalfPrecisionFloat(retFloat);
final short shrtValue = otherWay.getHalfPrecisionAsShort();
Assert.assertEquals(new Short(shortRepresentation), new Short(shrtValue));
HalfPrecisionFloat backAgain = new HalfPrecisionFloat(shrtValue);
final float backFlt = backAgain.getFullFloat();
Assert.assertEquals(new Float(closestValue), new Float(backFlt));
HalfPrecisionFloat dbl = new HalfPrecisionFloat(startingValue);
final double retDbl = dbl.getFullDouble();
Assert.assertEquals(new Double(startingValue), new Double(retDbl));
}
@Test(expected = IllegalArgumentException.class)
public void testInvalidByteArray() {
ByteBuffer buffer = ByteBuffer.allocate(4);
buffer.putFloat(Float.MAX_VALUE);
byte[] bytes = buffer.array();
new HalfPrecisionFloat(bytes);
}
@Test(expected = IllegalArgumentException.class)
public void testInvalidMaxFloat() {
new HalfPrecisionFloat(Float.MAX_VALUE);
}
@Test(expected = IllegalArgumentException.class)
public void testInvalidMinFloat() {
new HalfPrecisionFloat(-35000);
}
@Test
public void testCreateWithShort() {
HalfPrecisionFloat sut = new HalfPrecisionFloat(Short.MAX_VALUE);
Assert.assertEquals(Short.MAX_VALUE, sut.getHalfPrecisionAsShort());
}
}
答案 4 :(得分：0)
我对小的正浮点数感兴趣，因此我构建了具有 12位尾数，无符号位和4位指数（偏差为15）的变体，这样它可以表示介于0和1.00（不包括）完全可以。在尾数上有2位分辨率，但指数较低。
Text测试给出：
public static float toFloat(int hbits) {
int mant = hbits & 0x0fff; // 12 bits mantissa
int exp = (hbits & 0xf000) >>> 12; // 4 bits exponent
if (exp == 0xf) {
exp = 0xff;
} else {
if (exp != 0) { // normal value
exp += 127 - 15;
} else { // subnormal value
if (mant != 0) { // not zero
exp += 127 - 15;
// make it noral
exp++;
do {
mant <<= 1;
exp--;
} while ((mant & 0x1000) == 0);
mant &= 0x0fff;
}
}
}
return Float.intBitsToFloat(exp << 23 | mant << 11);
}
public static int fromFloat(float fval) {
int fbits = Float.floatToIntBits( fval );
int val = ( fbits & 0x7fffffff ) + 0x400; // rounded value
if( val < 0x32000000 ) // too small for subnormal or negative
return 0; // becomes 0
if( val >= 0x47800000 ) // might be or become NaN/Inf
{ // avoid Inf due to rounding
if( ( fbits & 0x7fffffff ) >= 0x47800000 )
{ // is or must become NaN/Inf
if( val < 0x7f800000 ) // was value but too large
return 0xf000; // make it +/-Inf
return 0xf000 | // remains +/-Inf or NaN
( fbits & 0x007fffff ) >>> 11; // keep NaN (and Inf) bits
}
return 0x7fff; // unrounded not quite Inf
}
if( val >= 0x38800000 ) // remains normalized value
return val - 0x38000000 >>> 11; // exp - 127 + 15
val = ( fbits & 0x7fffffff ) >>> 23; // tmp exp for subnormal calc
return ( ( fbits & 0x7f_ffff | 0x80_0000 ) // add subnormal bit
+ ( 0x800000 >>> val - 100 ) // round depending on cut off
>>> 124 - val ); // div by 2^(1-(exp-127+15)) and >> 11 | exp=0
}
法线：
Smallest subnormal float : 0.0000000149
Largest subnormal float : 0.0000610203
Smallest normal float : 0.0000610352
Smallest normal float + ups: 0.0000610501
E=1, M=fff (max) : 0.0001220554
Largest normal float : 0.0078115463
对于次正规检验：
0.9990000129 => 3f7fbe77 => eff8 => 0.9990234375 | error: 0.002%
0.8991000056 => 3f662b6b => ecc5 => 0.8990478516 | error: 0.006%
0.8091899753 => 3f4f2713 => e9e5 => 0.8092041016 | error: 0.002%
0.7282709479 => 3f3a6ff7 => e74e => 0.7282714844 | error: 0.000%
0.6554438472 => 3f27cb2b => e4f9 => 0.6553955078 | error: 0.007%
0.5898994207 => 3f1703a6 => e2e0 => 0.5898437500 | error: 0.009%
0.5309094787 => 3f07e9af => e0fd => 0.5308837891 | error: 0.005%
0.4778185189 => 3ef4a4a1 => de95 => 0.4778442383 | error: 0.005%
0.4300366640 => 3edc2dc4 => db86 => 0.4300537109 | error: 0.004%
0.3870329857 => 3ec62930 => d8c5 => 0.3870239258 | error: 0.002%
0.3483296633 => 3eb25844 => d64b => 0.3483276367 | error: 0.001%
0.3134966791 => 3ea082a3 => d410 => 0.3134765625 | error: 0.006%
0.2821469903 => 3e907592 => d20f => 0.2821655273 | error: 0.007%
0.2539322972 => 3e82036a => d040 => 0.2539062500 | error: 0.010%
0.2285390645 => 3e6a0625 => cd41 => 0.2285461426 | error: 0.003%
0.2056851536 => 3e529f21 => ca54 => 0.2056884766 | error: 0.002%
0.1851166338 => 3e3d8f37 => c7b2 => 0.1851196289 | error: 0.002%
0.1666049659 => 3e2a9a7e => c553 => 0.1665954590 | error: 0.006%
0.1499444693 => 3e198b0b => c331 => 0.1499328613 | error: 0.008%
0.1349500120 => 3e0a3056 => c146 => 0.1349487305 | error: 0.001%
0.1214550063 => 3df8bd67 => bf18 => 0.1214599609 | error: 0.004%
0.1093095019 => 3ddfdda9 => bbfc => 0.1093139648 | error: 0.004%
0.0983785465 => 3dc97ab1 => b92f => 0.0983734131 | error: 0.005%
0.0885406882 => 3db554d2 => b6ab => 0.0885467529 | error: 0.007%
0.0796866193 => 3da332bd => b466 => 0.0796813965 | error: 0.007%
0.0717179552 => 3d92e0dd => b25c => 0.0717163086 | error: 0.002%
0.0645461604 => 3d8430c7 => b086 => 0.0645446777 | error: 0.002%
0.0580915436 => 3d6df166 => adbe => 0.0580902100 | error: 0.002%
0.0522823893 => 3d56260f => aac5 => 0.0522842407 | error: 0.004%
0.0470541492 => 3d40bbda => a817 => 0.0470504761 | error: 0.008%
0.0423487313 => 3d2d75dd => a5af => 0.0423507690 | error: 0.005%
0.0381138586 => 3d1c1d47 => a384 => 0.0381164551 | error: 0.007%
0.0343024731 => 3d0c80c0 => a190 => 0.0343017578 | error: 0.002%
0.0308722258 => 3cfce7c0 => 9f9d => 0.0308723450 | error: 0.000%
0.0277850032 => 3ce39d60 => 9c74 => 0.0277862549 | error: 0.005%
0.0250065029 => 3cccda70 => 999b => 0.0250053406 | error: 0.005%
0.0225058515 => 3cb85e31 => 970c => 0.0225067139 | error: 0.004%
0.0202552658 => 3ca5ee5f => 94be => 0.0202560425 | error: 0.004%
0.0182297379 => 3c955688 => 92ab => 0.0182304382 | error: 0.004%
0.0164067633 => 3c86677a => 90cd => 0.0164070129 | error: 0.002%
0.0147660868 => 3c71ed75 => 8e3e => 0.0147666931 | error: 0.004%
0.0132894777 => 3c59bc1c => 8b38 => 0.0132904053 | error: 0.007%
0.0119605297 => 3c43f619 => 887f => 0.0119609833 | error: 0.004%
0.0107644768 => 3c305d7d => 860c => 0.0107650757 | error: 0.006%
0.0096880291 => 3c1eba8a => 83d7 => 0.0096874237 | error: 0.006%
0.0087192263 => 3c0edb16 => 81db => 0.0087184906 | error: 0.008%
0.0078473035 => 3c0091fa => 8012 => 0.0078468323 | error: 0.006%
0.0070625730 => 3be76d28 => 7cee => 0.0070629120 | error: 0.005%
0.0063563157 => 3bd048a4 => 7a09 => 0.0063562393 | error: 0.001%
0.0057206838 => 3bbb7493 => 776f => 0.0057210922 | error: 0.007%
0.0051486152 => 3ba8b5b7 => 7517 => 0.0051488876 | error: 0.005%
0.0046337536 => 3b97d6be => 72fb => 0.0046339035 | error: 0.003%
0.0041703782 => 3b88a7ab => 7115 => 0.0041704178 | error: 0.001%
0.0037533403 => 3b75fa9a => 6ebf => 0.0037531853 | error: 0.004%
0.0033780062 => 3b5d618a => 6bac => 0.0033779144 | error: 0.003%
0.0030402055 => 3b473e2f => 68e8 => 0.0030403137 | error: 0.004%
0.0027361847 => 3b335190 => 666a => 0.0027360916 | error: 0.003%
0.0024625661 => 3b216301 => 642c => 0.0024623871 | error: 0.007%
0.0022163095 => 3b113f81 => 6228 => 0.0022163391 | error: 0.001%
0.0019946785 => 3b02b927 => 6057 => 0.0019946098 | error: 0.003%
0.0017952106 => 3aeb4d46 => 5d6a => 0.0017952919 | error: 0.005%
0.0016156895 => 3ad3c58b => 5a79 => 0.0016157627 | error: 0.005%
0.0014541205 => 3abe9830 => 57d3 => 0.0014541149 | error: 0.000%
0.0013087085 => 3aab88f8 => 5571 => 0.0013086796 | error: 0.002%
0.0011778376 => 3a9a61ac => 534c => 0.0011777878 | error: 0.004%
0.0010600538 => 3a8af181 => 515e => 0.0010600090 | error: 0.004%
0.0009540484 => 3a7a191b => 4f43 => 0.0009540319 | error: 0.002%
0.0008586436 => 3a611698 => 4c23 => 0.0008586645 | error: 0.002%
0.0007727792 => 3a4a9455 => 4953 => 0.0007728338 | error: 0.007%
0.0006955012 => 3a36524c => 46ca => 0.0006954670 | error: 0.005%
0.0006259511 => 3a2416de => 4483 => 0.0006259680 | error: 0.003%
0.0005633560 => 3a13ae2e => 4276 => 0.0005633831 | error: 0.005%
0.0005070204 => 3a04e990 => 409d => 0.0005069971 | error: 0.005%
0.0004563183 => 39ef3e03 => 3de8 => 0.0004563332 | error: 0.003%
0.0004106865 => 39d75169 => 3aea => 0.0004106760 | error: 0.003%
0.0003696179 => 39c1c945 => 3839 => 0.0003696084 | error: 0.003%
0.0003326561 => 39ae6857 => 35cd => 0.0003326535 | error: 0.001%
0.0002993904 => 399cf781 => 339f => 0.0002993941 | error: 0.001%
0.0002694514 => 398d4527 => 31a9 => 0.0002694726 | error: 0.008%
0.0002425062 => 397e4946 => 2fc9 => 0.0002425015 | error: 0.002%
0.0002182556 => 3964db8b => 2c9b => 0.0002182424 | error: 0.006%
0.0001964300 => 394df8ca => 29bf => 0.0001964271 | error: 0.001%
0.0001767870 => 39395fe9 => 272c => 0.0001767874 | error: 0.000%
0.0001591083 => 3926d651 => 24db => 0.0001591146 | error: 0.004%
0.0001431975 => 39162749 => 22c5 => 0.0001432002 | error: 0.002%
0.0001288777 => 3907235b => 20e4 => 0.0001288652 | error: 0.010%
0.0001159900 => 38f33fa3 => 1e68 => 0.0001159906 | error: 0.001%
0.0001043910 => 38daec79 => 1b5e => 0.0001043975 | error: 0.006%
0.0000939519 => 38c50806 => 18a1 => 0.0000939518 | error: 0.000%
0.0000845567 => 38b15405 => 162b => 0.0000845641 | error: 0.009%
0.0000761010 => 389f986b => 13f3 => 0.0000761002 | error: 0.001%
0.0000684909 => 388fa2c6 => 11f4 => 0.0000684857 | error: 0.008%
0.0000616418 => 388145b2 => 1029 => 0.0000616461 | error: 0.007%