如何解决将N维值映射到希尔伯特曲线上的一个点
我终于崩溃了,掏出一些钱。AIP(美国物理研究所)在C上有一篇很好的简短文章,带有源代码。John Skilling的“ Programming Hilbert curve”(来自AIP Conf。Proc。707,381(2004))有一个附录,其中包含用于双向映射。它适用于任何大于1的维数,不是递归的,不使用状态转换查找表,该表占用大量内存,并且主要使用位操作。因此,它相当快并且具有良好的内存占用量。
下面的代码行(在TransposetoAxes函数中找到)具有错误:
for(i = n-1; i> = 0; i–)X [i] ^ = X [i-1];
纠正方法是将大于或等于(> =)更改为大于(>)。如果不进行此更正,则当变量“ i”变为零时,将使用负索引访问X数组,从而导致程序失败。
我建议阅读该文章(包括代码在内,长达七页),因为它解释了算法的工作原理,这显然并不明显。
我将他的代码翻译成C#供我自己使用。代码如下。Skilling进行适当的转换,覆盖您传入的向量。我选择对输入向量进行克隆,然后返回一个新副本。另外,我将这些方法实现为扩展方法。
Skilling的代码将希尔伯特索引表示为转置,存储为数组。我发现交错这些位并形成一个BigInteger更方便(在Dictionary中更有用,更容易在循环中进行迭代等),但是我使用魔术数字,位操作等优化了该操作及其逆运算,并且代码很长,所以我省略了。
namespace HilbertExtensions
{
/// <summary>
/// Convert between Hilbert index and N-dimensional points.
///
/// The Hilbert index is expressed as an array of transposed bits.
///
/// Example: 5 bits for each of n=3 coordinates.
/// 15-bit Hilbert integer = A B C D E F G H I J K L M N O is stored
/// as its Transpose ^
/// X[0] = A D G J M X[2]| 7
/// X[1] = B E H K N <-------> | /X[1]
/// X[2] = C F I L O axes |/
/// high low 0------> X[0]
///
/// NOTE: This algorithm is derived from work done by John Skilling and published in "Programming the Hilbert curve".
/// (c) 2004 American Institute of Physics.
///
/// </summary>
public static class HilbertCurveTransform
{
/// <summary>
/// Convert the Hilbert index into an N-dimensional point expressed as a vector of uints.
///
/// Note: In Skilling's paper, this function is named TransposetoAxes.
/// </summary>
/// <param name="transposedindex">The Hilbert index stored in transposed form.</param>
/// <param name="bits">Number of bits per coordinate.</param>
/// <returns>Coordinate vector.</returns>
public static uint[] HilbertAxes(this uint[] transposedindex, int bits)
{
var X = (uint[])transposedindex.Clone();
int n = X.Length; // n: Number of dimensions
uint N = 2U << (bits - 1), P, Q, t;
int i;
// Gray decode by H ^ (H/2)
t = X[n - 1] >> 1;
// Corrected error in Skilling's paper on the following line. The appendix had i >= 0 leading to negative array index.
for (i = n - 1; i > 0; i--)
X[i] ^= X[i - 1];
X[0] ^= t;
// Undo excess work
for (Q = 2; Q != N; Q <<= 1)
{
P = Q - 1;
for (i = n - 1; i >= 0; i--)
if ((X[i] & Q) != 0U)
X[0] ^= P; // invert
else
{
t = (X[0] ^ X[i]) & P;
X[0] ^= t;
X[i] ^= t;
}
} // exchange
return X;
}
/// <summary>
/// Given the axes (coordinates) of a point in N-Dimensional space, find the distance to that point along the Hilbert curve.
/// That distance will be transposed; broken into pieces and distributed into an array.
///
/// The number of dimensions is the length of the hilbertAxes array.
///
/// Note: In Skilling's paper, this function is called AxestoTranspose.
/// </summary>
/// <param name="hilbertAxes">Point in N-space.</param>
/// <param name="bits">Depth of the Hilbert curve. If bits is one, this is the top-level Hilbert curve.</param>
/// <returns>The Hilbert distance (or index) as a transposed Hilbert index.</returns>
public static uint[] HilbertIndexTransposed(this uint[] hilbertAxes, int bits)
{
var X = (uint[])hilbertAxes.Clone();
var n = hilbertAxes.Length; // n: Number of dimensions
uint M = 1U << (bits - 1), P, Q, t;
int i;
// Inverse undo
for (Q = M; Q > 1; Q >>= 1)
{
P = Q - 1;
for (i = 0; i < n; i++)
if ((X[i] & Q) != 0)
X[0] ^= P; // invert
else
{
t = (X[0] ^ X[i]) & P;
X[0] ^= t;
X[i] ^= t;
}
} // exchange
// Gray encode
for (i = 1; i < n; i++)
X[i] ^= X[i - 1];
t = 0;
for (Q = M; Q > 1; Q >>= 1)
if ((X[n - 1] & Q)!=0)
t ^= Q - 1;
for (i = 0; i < n; i++)
X[i] ^= t;
return X;
}
}
}
我已经在C#中将工作代码发布到了github。
参见https://github.com/paulchernoch/HilbertTransformation
更新:我刚刚(2019年秋季)在crates.io上发布了一个名为“ hilbert”的Rust板条箱。它还使用了斯基林算法。参见https://crates.io/crates/hilbert
解决方法
我有大量的N维点(数千万; N接近100)。
我需要将这些点映射到一个维度,同时保留空间局部性。我想使用希尔伯特空间填充曲线来做到这一点。
对于每个点,我想选择曲线上最接近的点。该点的希尔伯特值(从曲线起点到拾取点的曲线长度)是我要寻找的单个尺寸值。
计算不一定必须是即时的,但我希望它在体面的现代家用PC硬件上不会超过几个小时。
对实施有什么建议吗?有没有什么图书馆对我有帮助?(语言无关紧要。)
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