我希望能够为32位、64位和128位生产一个莫顿码，同时也需要优化代码！有什么解决方案吗？
1个回答
12
这是一个使用Python脚本的解决方案：
我从他的评论中得到了提示：Fabian “ryg” Giesen
请阅读下面的长评论！我们需要跟踪哪些位需要移动多远！
然后在每个步骤中，我们选择这些位并将它们移动，并应用位掩码（见评论的最后几行）来屏蔽它们！
针对10位数字和2个交错位（32位），Python脚本的位掩码生成器输出如下：
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
因此，对于10位数字和2个交错位（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
对于一个21位的数字和2个交错位（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
对于一个42位数字和2个交错位（128位），您需要执行以下操作（如果需要）：
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
Python脚本用于生成和检查交错模式！！！
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!!!!")
- Gabriel
5
1好的，看起来像是通常的解决方案，但位数有点不同，我猜。也许你也对直接添加两个莫顿键感兴趣。 - harold
啊，好的，谢谢:-)。为什么我应该添加两个莫顿键？你是说，在交错部分直接构建一个莫顿键可以更快地完成吗？ - Gabriel
1这使得你可以例如，将莫顿键按照任意数量的偏移量在两个方向上进行偏移，而无需采用昂贵的“解绑 -> 添加 -> 重新绑定”路线，你只需要对偏移进行重新绑定（特别适用于偏移量是一个常数的情况），并将其添加到键上。 - harold
我们需要等多久才能等到解码器？ - Toby Speight
嗯，我认为有人应该写下来，并将其添加到这篇文章中。我认为我们应该将其制作成一个要点。 - Gabriel
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