为 32 位、64 位和 128 位生成交错位模式（morton 密钥）

Posted 2023-02-19

【中文标题】为 32 位、64 位和 128 位生成交错位模式（morton 密钥）

【英文标题】：Produce interleaving bit patterns (morton keys) for 32 bit , 64 bit and 128bit

【发布时间】：2013-09-02 22:56:39

【问题描述】：

我想为 32 位、64 位和 128 位生成一个 morton 密钥，并使用最佳代码！ 解决办法是什么？

【参考方案1】：

这是我使用 python 脚本的解决方案：

我从他的评论中得到了暗示：F*** “ryg” Giesen 阅读下面的长评论！我们需要跟踪哪些位需要走多远！ 然后在每一步中，我们选择这些位并移动它们并应用位掩码（参见最后几行的注释）来屏蔽它们！

python 脚本的位掩码生成器输出（见下文），用于 10 位数字和 2 个交错位（用于 32 位）：

Bit Distances: [0, 2, 4, 6, 8, 10, 12, 14, 16, 18]
Shifting bits by 1   for bits idx: []
Shifting bits by 2   for bits idx: [1, 3, 5, 7, 9]
Shifting bits by 4   for bits idx: [2, 3, 6, 7]
Shifting bits by 8   for bits idx: [4, 5, 6, 7]
Shifting bits by 16  for bits idx: [8, 9]
BitPositions: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Current Mask:           0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1111 1111
Which bits to shift:    0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0000 0000  hex: 0x300
Shifted part (<< 16):   0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0000 0000 0000 0000 0000 0000  hex: 0x3000000
NonShifted Part:        0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1111 1111  hex: 0xff
Bitmask is now :        0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0000 0000 0000 0000 1111 1111  hex: 0x30000ff
 (this is : bitMask = shifted | nonshifted) 


Current Mask:           0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0000 0000 0000 0000 1111 1111
Which bits to shift:    0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1111 0000  hex: 0xf0
Shifted part (<< 8):    0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1111 0000 0000 0000  hex: 0xf000
NonShifted Part:        0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0000 0000 0000 0000 0000 1111  hex: 0x300000f
Bitmask is now :        0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0000 0000 1111 0000 0000 1111  hex: 0x300f00f
 (this is : bitMask = shifted | nonshifted) 


Current Mask:           0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0000 0000 1111 0000 0000 1111
Which bits to shift:    0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 0000 0000 1100  hex: 0xc00c
Shifted part (<< 4):    0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 0000 0000 1100 0000  hex: 0xc00c0
NonShifted Part:        0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0000 0000 0011 0000 0000 0011  hex: 0x3003003
Bitmask is now :        0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0000 1100 0011 0000 1100 0011  hex: 0x30c30c3
 (this is : bitMask = shifted | nonshifted) 


Current Mask:           0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0000 1100 0011 0000 1100 0011
Which bits to shift:    0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0000 1000 0010 0000 1000 0010  hex: 0x2082082
Shifted part (<< 2):    0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 0010 0000 1000 0010 0000 1000  hex: 0x8208208
NonShifted Part:        0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0000 0100 0001 0000 0100 0001  hex: 0x1041041
Bitmask is now :        0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 0010 0100 1001 0010 0100 1001  hex: 0x9249249
 (this is : bitMask = shifted | nonshifted) 

x &= 0x3ff
x = (x | (x << 16)) & 0x30000ff
x = (x | (x << 8)) & 0x300f00f
x = (x | (x << 4)) & 0x30c30c3
x = (x | (x << 2)) & 0x9249249

所以对于一个 10 位数字和 2 个交错位（对于 32 位），您需要执行以下操作！：

x &= 0x3ff
x = (x | x << 16) & 0x30000ff   #<<< THIS IS THE MASK for shifting 16 (for bit 8 and 9)
x = (x | x << 8) & 0x300f00f
x = (x | x << 4) & 0x30c30c3
x = (x | x << 2) & 0x9249249

对于 21 位数字和 2 个交织位（对于 64 位），您需要执行以下操作！：

x &= 0x1fffff
x = (x | x << 32) & 0x1f00000000ffff
x = (x | x << 16) & 0x1f0000ff0000ff
x = (x | x << 8) & 0x100f00f00f00f00f
x = (x | x << 4) & 0x10c30c30c30c30c3
x = (x | x << 2) & 0x1249249249249249

对于 42 位数字和 2 个交织位（对于 128 位），您需要执行以下操作（以防万一；-)）：

x &= 0x3ffffffffff
x = (x | x << 64) & 0x3ff0000000000000000ffffffffL
x = (x | x << 32) & 0x3ff00000000ffff00000000ffffL
x = (x | x << 16) & 0x30000ff0000ff0000ff0000ff0000ffL
x = (x | x << 8) & 0x300f00f00f00f00f00f00f00f00f00fL
x = (x | x << 4) & 0x30c30c30c30c30c30c30c30c30c30c3L
x = (x | x << 2) & 0x9249249249249249249249249249249L

生成和检查交错模式的 Python 脚本！！！

import random;

def prettyBinString(x,d=32,steps=4,sep=".",emptyChar="0"):
    b = bin(x)[2:]
    zeros = d - len(b)


    if zeros <= 0: 
        zeros = 0
        k = steps - (len(b) % steps)
    else:
        k = steps - (d % steps)

    s = ""
    #print("zeros" , zeros)
    #print("k" , k)
    for i in range(zeros): 
        #print("k:",k)
        if(k%steps==0 and i!= 0):
            s+=sep
        s += emptyChar
        k+=1

    for i in range(len(b)):
        if( (k%steps==0 and i!=0 and zeros == 0) or  (k%steps==0 and zeros != 0) ):
            s+=sep
        s += b[i]
        k+=1
    return s    

def binStr(x): return prettyBinString(x,64,4," ","0")


def computeBitMaskPatternAndCode(numberOfBits, numberOfEmptyBits):
    bitDistances=[ i*numberOfEmptyBits for i in range(numberOfBits) ]
    print("Bit Distances: " + str(bitDistances))
    bitDistancesB = [bin(dist)[2:] for dist in  bitDistances]
    #print("Bit Distances (binary): " + str(bitDistancesB))
    moveBits=[] #Liste mit allen Bits welche aufsteigend um 2, 4,8,16,32,64,128 stellen geschoben werden müssen

    maxLength = len(max(bitDistancesB, key=len))
    abort = False
    for i in range(maxLength):
        moveBits.append([])
        for idx,bits in enumerate(bitDistancesB):
            if not len(bits) - 1 < i:
                if(bits[len(bits)-i-1] == "1"):
                    moveBits[i].append(idx)

    for i in range(len(moveBits)):
        print("Shifting bits by " + str(2**i) + "\t for bits idx: " + str(moveBits[i]))

    bitPositions = list(range(numberOfBits));
    print("BitPositions: " + str(bitPositions))
    maskOld = (1 << numberOfBits) -1

    codeString = "x &= " + hex(maskOld) + "\n"
    for idx in range(len(moveBits)-1, -1, -1):
        if len(moveBits[idx]):


           shifted = 0
           for bitIdxToMove in moveBits[idx]:
                shifted |= 1<<bitPositions[bitIdxToMove];
                bitPositions[bitIdxToMove] += 2**idx; # keep track where the actual bit stands! might get moved several times

           # Get the non shifted part!     
           nonshifted = ~shifted & maskOld
           print("\nCurrent Mask:\t\t" + binStr(maskOld))
           print("Which bits to shift:\t" + binStr(shifted) + "\t hex: " + hex(shifted))
           shifted = shifted << 2**idx
           print("Shifted part (<< " + str(2**idx) + "):\t" + binStr(shifted)+ "\t hex: " + hex(shifted))

           print("NonShifted Part:\t" + binStr(nonshifted) + "\t hex: " + hex(nonshifted))
           maskNew =  shifted | nonshifted
           print("Bitmask is now :\t" + binStr(maskNew) + "\t hex: " + hex(maskNew) +"\n (this is : bitMask = shifted | nonshifted) \n")
           #print("Code: " + "x = x | x << " +str(2**idx)+ " & " +hex(maskNew))

           codeString += "x = (x | (x << " +str(2**idx)+")) & " + hex(maskNew) + "\n"
           maskOld = maskNew
    return codeString


numberOfBits = 10;
numberOfEmptyBits = 2;
codeString = computeBitMaskPatternAndCode(numberOfBits,numberOfEmptyBits);
print(codeString)

def partitionBy2(x):
    l=locals();
    exec(codeString,None,l)
    return l['x']

def checkPartition(x):
    print("Check partition for: \t" + binStr(x))
    part = partitionBy2(x);
    print("Partition is : \t\t" + binStr(part))
    #make the pattern manualy
    partC = int(0);
    for bitIdx in range(numberOfBits):
        partC  = partC | (x & (1<<bitIdx)) << numberOfEmptyBits*bitIdx
    print("Partition check is :\t" + binStr(partC))
    if(partC == part):
        return True
    else:
        return False

checkError = False        
for i in range(20):
    x = random.getrandbits(numberOfBits);
    if(checkPartition(x) == False):
        checkError = True
        break
if not checkError:
    print("CHECK PARTITION SUCCESSFUL!!!!!!!!!!!!!!!!...")
else:
    print("checkPartition has ERROR!!!!")

【讨论】：

好的，看起来像the usual solution，但我猜位数有点不同。或许你也对adding two morton keys directly感兴趣

啊，好的，谢谢 :-)，我为什么要添加两个 morton 键？，你的意思是，通过在交错部分直接执行来更快地构造一个 morton 键？

这可以让你，例如，使用一个 morton 键并在两个方向上以任意数量偏移它，而无需采用昂贵的 de-interleave -> add -> interleave 路线，你只需交错偏移量（如果偏移量特别好是一个常数）并将其添加到键中。

我们需要等待解码器多长时间？

嗯，我认为应该有人写这个，并将其添加到这篇文章中，我认为我们应该从中总结一下:-)