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 *** 作。因此，它相当快并且具有良好的内存占用量。

如果您选择购买该文章，我发现源代码中有错误。

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 *** 作等优化了该 *** 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 depre 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 enpre 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