问题
This question: How to de-interleave bits (UnMortonizing?) has a good answer for extracting one of the two halves of a Morton number (just the odd bits), but I need a solution which extracts both parts (the odd bits and the even bits) in as few operations as possible.
For my use I would need to take a 32 bit int and extract two 16 bit ints, where one is the even bits and the other is the odd bits shifted right by 1 bit, e.g.
input, z: 11101101 01010111 11011011 01101110
output, x: 11100001 10110111 // odd bits shifted right by 1
y: 10111111 11011010 // even bits
There seem to be plenty of solutions using shifts and masks with magic numbers for generating Morton numbers (i.e. interleaving bits), e.g. Interleave bits by Binary Magic Numbers, but I haven't yet found anything for doing the reverse (i.e. de-interleaving).
UPDATE
After re-reading the section from Hacker's Delight on perfect shuffles/unshuffles I found some useful examples which I adapted as follows:
// morton1 - extract even bits
uint32_t morton1(uint32_t x)
{
x = x & 0x55555555;
x = (x | (x >> 1)) & 0x33333333;
x = (x | (x >> 2)) & 0x0F0F0F0F;
x = (x | (x >> 4)) & 0x00FF00FF;
x = (x | (x >> 8)) & 0x0000FFFF;
return x;
}
// morton2 - extract odd and even bits
void morton2(uint32_t *x, uint32_t *y, uint32_t z)
{
*x = morton1(z);
*y = morton1(z >> 1);
}
I think this can still be improved on, both in its current scalar form and also by taking advantage of SIMD, so I'm still interested in better solutions (either scalar or SIMD).
回答1:
If your processor handles 64 bit ints efficiently, you could combine the operations...
int64 w = (z &0xAAAAAAAA)<<31 | (z &0x55555555 )
w = (w | (w >> 1)) & 0x3333333333333333;
w = (w | (w >> 2)) & 0x0F0F0F0F0F0F0F0F;
...
回答2:
Code for the Intel Haswell and later CPUs. You can use the BMI2 instruction set which contains the pext and pdep instructions. These can (among other great things) be used to build your functions.
#include <immintrin.h>
#include <stdint.h>
// on GCC, compile with option -mbmi2, requires Haswell or better.
uint64_t xy_to_morton (uint32_t x, uint32_t y)
{
return _pdep_u32(x, 0x55555555) | _pdep_u32(y,0xaaaaaaaa);
}
uint64_t morton_to_xy (uint64_t m, uint32_t *x, uint32_t *y)
{
*x = _pext_u64(m, 0x5555555555555555);
*y = _pext_u64(m, 0xaaaaaaaaaaaaaaaa);
}
回答3:
In case someone is using morton codes in 3d, so he needs to read one bit every 3, and 64 bits here is the function I used:
uint64_t morton3(uint64_t x) {
x = x & 0x9249249249249249;
x = (x | (x >> 2)) & 0x30c30c30c30c30c3;
x = (x | (x >> 4)) & 0xf00f00f00f00f00f;
x = (x | (x >> 8)) & 0x00ff0000ff0000ff;
x = (x | (x >> 16)) & 0xffff00000000ffff;
x = (x | (x >> 32)) & 0x00000000ffffffff;
return x;
}
uint64_t bits;
uint64_t x = morton3(bits)
uint64_t y = morton3(bits>>1)
uint64_t z = morton3(bits>>2)
回答4:
If you need speed than you can use table-lookup for one byte conversion at once (two bytes table is faster but to big). Procedure is made under Delphi IDE but the assembler/algorithem is the same.
const
MortonTableLookup : array[byte] of byte = ($00, $01, $10, $11, $12, ... ;
procedure DeinterleaveBits(Input: cardinal);
//In: eax
//Out: dx = EvenBits; ax = OddBits;
asm
movzx ecx, al //Use 0th byte
mov dl, byte ptr[MortonTableLookup + ecx]
//
shr eax, 8
movzx ecx, ah //Use 2th byte
mov dh, byte ptr[MortonTableLookup + ecx]
//
shl edx, 16
movzx ecx, al //Use 1th byte
mov dl, byte ptr[MortonTableLookup + ecx]
//
shr eax, 8
movzx ecx, ah //Use 3th byte
mov dh, byte ptr[MortonTableLookup + ecx]
//
mov ecx, edx
and ecx, $F0F0F0F0
mov eax, ecx
rol eax, 12
or eax, ecx
rol edx, 4
and edx, $F0F0F0F0
mov ecx, edx
rol ecx, 12
or edx, ecx
end;
回答5:
I didn't want to be limited to a fixed size integer and making lists of similar commands with hardcoded constants, so I developed a C++11 solution which makes use of template metaprogramming to generate the functions and the constants. The assembly code generated with -O3 seems as tight as it can get without using BMI:
andl $0x55555555, %eax
movl %eax, %ecx
shrl %ecx
orl %eax, %ecx
andl $0x33333333, %ecx
movl %ecx, %eax
shrl $2, %eax
orl %ecx, %eax
andl $0xF0F0F0F, %eax
movl %eax, %ecx
shrl $4, %ecx
orl %eax, %ecx
movzbl %cl, %esi
shrl $8, %ecx
andl $0xFF00, %ecx
orl %ecx, %esi
TL;DR source repo and live demo.
Implementation
Basically every step in the morton1 function works by shifting and adding to a sequence of constants which look like this:
0b0101010101010101(alternate 1 and 0)0b0011001100110011(alternate 2x 1 and 0)0b0000111100001111(alternate 4x 1 and 0)0b0000000011111111(alternate 8x 1 and 0)
If we were to use D dimensions, we would have a pattern with D-1 zeros and 1 one. So to generate these it's enough to generate consecutive ones and apply some bitwise or:
/// @brief Generates 0b1...1 with @tparam n ones
template <class T, unsigned n>
using n_ones = std::integral_constant<T, (~static_cast<T>(0) >> (sizeof(T) * 8 - n))>;
/// @brief Performs `@tparam input | (@tparam input << @tparam width` @tparam repeat times.
template <class T, T input, unsigned width, unsigned repeat>
struct lshift_add :
public lshift_add<T, lshift_add<T, input, width, 1>::value, width, repeat - 1> {
};
/// @brief Specialization for 1 repetition, just does the shift-and-add operation.
template <class T, T input, unsigned width>
struct lshift_add<T, input, width, 1> : public std::integral_constant<T,
(input & n_ones<T, width>::value) | (input << (width < sizeof(T) * 8 ? width : 0))> {
};
Now that we can generate the constants at compile time for arbitrary dimensions with the following:
template <class T, unsigned step, unsigned dimensions = 2u>
using mask = lshift_add<T, n_ones<T, 1 << step>::value, dimensions * (1 << step), sizeof(T) * 8 / (2 << step)>;
With the same type of recursion, we can generate functions for each of the steps of the algorithm x = (x | (x >> K)) & M:
template <class T, unsigned step, unsigned dimensions>
struct deinterleave {
static T work(T input) {
input = deinterleave<T, step - 1, dimensions>::work(input);
return (input | (input >> ((dimensions - 1) * (1 << (step - 1))))) & mask<T, step, dimensions>::value;
}
};
// Omitted specialization for step 0, where there is just a bitwise and
It remains to answer the question "how many steps do we need?". This depends also on the number of dimensions. In general, k steps compute 2^k - 1 output bits; the maximum number of meaningful bits for each dimension is given by z = sizeof(T) * 8 / dimensions, therefore it is enough to take 1 + log_2 z steps. The problem is now that we need this as constexpr in order to use it as a template parameter. The best way I found to work around this is to define log2 via metaprogramming:
template <unsigned arg>
struct log2 : public std::integral_constant<unsigned, log2<(arg >> 1)>::value + 1> {};
template <>
struct log2<1u> : public std::integral_constant<unsigned, 0u> {};
/// @brief Helper constexpr which returns the number of steps needed to fully interleave a type @tparam T.
template <class T, unsigned dimensions>
using num_steps = std::integral_constant<unsigned, log2<sizeof(T) * 8 / dimensions>::value + 1>;
And finally, we can perform one single call:
/// @brief Helper function which combines @see deinterleave and @see num_steps into a single call.
template <class T, unsigned dimensions>
T deinterleave_first(T n) {
return deinterleave<T, num_steps<T, dimensions>::value - 1, dimensions>::work(n);
}
来源：https://stackoverflow.com/questions/4909263/how-to-efficiently-de-interleave-bits-inverse-morton