我正在从这里修改埃拉托斯特尼的无限筛子，因此它使用车轮分解来跳过比当前仅检查所有赔率的形式更多的复合材料。
我已经想出了如何生成到达轮子上所有间隙的步骤。从那里，我想我可以用2来代替这些轮子步骤，但这会导致筛子跳过质数。代码如下：
from itertools import count, cycle
def dvprm(end):
"finds primes by trial division. returns a list"
primes=[2]
for i in range(3, end+1, 2):
if all(map(lambda x:i%x, primes)):
primes.append(i)
return primes
def prod(seq, factor=1):
"sequence -> product"
for i in seq:factor*=i
return factor
def wheelGaps(primes):
"""returns list of steps to each wheel gap
that start from the last value in primes"""
strtPt= primes.pop(-1)#where the wheel starts
whlCirm= prod(primes)# wheel's circumference
#spokes are every number that are divisible by primes (composites)
gaps=[]#locate where the non-spokes are (gaps)
for i in xrange(strtPt, strtPt+whlCirm+1, 2):
if not all(map(lambda x:i%x,primes)):continue#spoke
else: gaps.append(i)#non-spoke
#find the steps needed to jump to each gap (beginning from the start of the wheel)
steps=[]#last step returns to start of wheel
for i,j in enumerate(gaps):
if i==0:continue
steps.append(j - gaps[i-1])
return steps
def wheel_setup(num):
"builds initial data for sieve"
initPrms=dvprm(num)#initial primes from the "roughing" pump
gaps = wheelGaps(initPrms[:])#get the gaps
c= initPrms.pop(-1)#prime that starts the wheel
return initPrms, gaps, c
def wheel_psieve(lvl=0, initData=None):
'''postponed prime generator with wheels
Refs: http://stackoverflow.com/a/10733621
http://stackoverflow.com/a/19391111'''
whlSize=11#wheel size, 1 higher prime than
# 5 gives 2- 3 wheel 11 gives 2- 7 wheel
# 7 gives 2- 5 wheel 13 gives 2-11 wheel
#set to 0 for no wheel
if lvl:#no need to rebuild the gaps, just pass them down the levels
initPrms, gaps, c = initData
else:#but if its the top level then build the gaps
if whlSize>4:
initPrms, gaps, c = wheel_setup(whlSize)
else:
initPrms, gaps, c= dvprm(7), [2], 9
#toss out the initial primes
for p in initPrms:
yield p
cgaps=cycle(gaps)
compost = {}#found composites to skip
ps=wheel_psieve(lvl+1, (initPrms, gaps, c))
p=next(ps)#advance lower level to appropriate square
while p*p < c:
p=next(ps)
psq=p*p
while True:
step1 = next(cgaps)#step to next value
step2=compost.pop(c, 0)#step to next multiple
if not step2:
#see references for details
if c < psq:
yield c
c += step1
continue
else:
step2=2*p
p=next(ps)
psq=p*p
d = c + step2
while d in compost:
d+= step2
compost[d]= step2
c += step1
我正在用它来检查它：
def test(num=100):
found=[]
for i,p in enumerate(wheel_psieve(), 1):
if i>num:break
found.append(p)
print sum(found)
return found
当我将轮子大小设置为0时，我得到了前100个素数的正确总和24133，但当我使用任何其他轮子大小时，我最终会得到缺失的素数、错误的总和和合成。即使是2-3个轮子（使用2和4的交替步骤）也会使筛子漏掉底漆。我做错了什么？
赔率，即2-coprimes，是通过“滚动轮子”[2]产生的，即通过重复添加2，从初始值3开始（类似地从5,7,9，…），
n=3; n+=2; n+=2; n+=2; ... # wheel = [2]
3 5 7 9
2-3余数是通过重复相加2，然后4，再2，然后4等等而生成的:
n=5; n+=2; n+=4; n+=2; n+=4; ... # wheel = [2,4]
5 7 11 13 17
在这里，我们确实需要知道从哪里开始添加差异，2或4，具体取决于初始值。对于5，11，17，...，它是2（即轮子的第0个元素）;对于 7， 13， 19， ...， 它是 4 （即 1-st 元素）。
我们如何知道从哪里开始？轮子优化的要点是我们只处理这个共素序列(在这个例子中，2-3个共素)。因此，在我们获得递归生成的素数的代码部分，我们还将维护滚动的车轮流，并推进它，直到我们看到其中的下一个素数。滚动序列将需要产生两个结果-值和车轮位置。因此，当我们看到质数时，我们也得到相应的轮位置，并且我们可以从轮上的该位置开始产生它的倍数。我们用< code>p乘以所有的东西，当然，从< code>p*p开始:
for (i, p) # the (wheel position, summated value)
in enumerated roll of the wheel:
when p is the next prime:
multiples of p are m = p*p; # map (p*) (roll wheel-at-i from p)
m += p*wheel[i];
m += p*wheel[i+1]; ...
因此，dict中的每个条目都必须保持其当前值、基本素数和当前轮位置（如果需要，将其环绕为0表示圆度）。
为了生成得到的素数，我们滚动另一个共素数序列，并仅保留其中不在字典中的那些元素，就像在参考代码中一样。
更新：在codereview上进行了几次迭代之后（非常感谢那里的贡献者！为了速度，我尽可能多地使用迭代工具，得出了这段代码：
from itertools import accumulate, chain, cycle, count
def wsieve(): # wheel-sieve, by Will Ness. ideone.com/mqO25A
wh11 = [ 2,4,2,4,6,2,6,4,2,4,6, 6,2,6,4,2,6,4,6,8,4,2, 4,
2,4,8,6,4,6,2,4,6,2,6, 6,4,2,4,6,2,6,4,2,4,2, 10,2,10]
cs = accumulate(chain([11], cycle(wh11))) # roll the wheel from 11
yield(next(cs)) # cf. ideone.com/WFv4f,
ps = wsieve() # codereview.stackexchange.com/q/92365/9064
p = next(ps) # 11
psq = p**2 # 121
D = dict(zip(accumulate(chain([0], wh11)), count(0))) # wheel roll lookup dict
mults = {}
for c in cs: # candidates, coprime with 210, from 11
if c in mults:
wheel = mults.pop(c)
elif c < psq:
yield c
continue
else: # c==psq: map (p*) (roll wh from p) = roll (wh*p) from (p*p)
i = D[(p-11) % 210] # look up wheel roll starting point
wheel = accumulate( chain( [psq],
cycle( [p*d for d in wh11[i:] + wh11[:i]])))
next(wheel)
p = next(ps)
psq = p**2
for m in wheel: # pop, save in m, and advance
if m not in mults:
break
mults[m] = wheel # mults[143] = wheel@187
def primes():
yield from (2, 3, 5, 7)
yield from wsieve()
与上面的描述不同，这段代码直接计算每个质数从哪里开始滚动轮子，以生成它的倍数
这是我想到的版本。它不像Ness那样干净，但是很有效。我把它贴出来，这样就有了另一个如何使用轮子因式分解的例子，以防有人路过。我可以选择使用什么样的轮子尺寸，但是很容易确定一个更永久的尺寸——只需生成您想要的尺寸并粘贴到代码中。
from itertools import count
def wpsieve():
"""prime number generator
call this function instead of roughing or turbo"""
whlSize = 11
initPrms, gaps, c = wheel_setup(whlSize)
for p in initPrms:
yield p
primes = turbo(0, (gaps, c))
for p, x in primes:
yield p
def prod(seq, factor=1):
"sequence -> product"
for i in seq: factor *= i
return factor
def wheelGaps(primes):
"""returns list of steps to each wheel gap
that start from the last value in primes"""
strtPt = primes.pop(-1) # where the wheel starts
whlCirm = prod(primes) # wheel's circumference
# spokes are every number that are divisible by primes (composites)
gaps = [] # locate where the non-spokes are (gaps)
for i in xrange(strtPt, strtPt + whlCirm + 1, 2):
if not all(map(lambda x: i%x, primes)): continue # spoke
else: gaps.append(i) # non-spoke
# find the steps needed to jump to each gap (beginning from the start of the wheel)
steps = [] # last step returns to start of wheel
for i, j in enumerate(gaps):
if i == 0: continue
steps.append(int(j - gaps[i-1]))
return steps
def wheel_setup(num):
"builds initial data for sieve"
initPrms = roughing(num) # initial primes from the "roughing" pump
gaps = wheelGaps(initPrms[:]) # get the gaps
c = initPrms.pop(-1) # prime that starts the wheel
return initPrms, gaps, c
def roughing(end):
"finds primes by trial division (roughing pump)"
primes = [2]
for i in range(3, end + 1, 2):
if all(map(lambda x: i%x, primes)):
primes.append(i)
return primes
def turbo(lvl=0, initData=None):
"""postponed prime generator with wheels (turbo pump)
Refs: http://stackoverflow.com/a/10733621
http://stackoverflow.com/a/19391111"""
gaps, c = initData
yield (c, 0)
compost = {} # found composites to skip
# store as current value: (base prime, wheel index)
ps = turbo(lvl + 1, (gaps, c))
p, x = next(ps)
psq = p*p
gapS = len(gaps) - 1
ix = jx = kx = 0 # indices for cycling the wheel
def cyc(x): return 0 if x > gapS else x # wheel cycler
while True:
c += gaps[ix] # add next step on c's wheel
ix = cyc(ix + 1) # and advance c's index
bp, jx = compost.pop(c, (0,0)) # get base prime and its wheel index
if not bp:
if c < psq: # prime
yield c, ix # emit index for above recursive level
continue
else:
jx = kx # swap indices as a new prime comes up
bp = p
p, kx = next(ps)
psq = p*p
d = c + bp * gaps[jx] # calc new multiple
jx = cyc(jx + 1)
while d in compost:
step = bp * gaps[jx]
jx = cyc(jx + 1)
d += step
compost[d] = (bp, jx)
保留车轮尺寸的选项还可以让您看到较大的车轮不会起多大作用。下面是测试代码，说明生成选定尺寸的砂轮需要多长时间，以及筛子与该轮子的速度有多快。
import time
def speed_test(num, whlSize):
print('-'*50)
t1 = time.time()
initPrms, gaps, c = wheel_setup(whlSize)
t2 = time.time()
print('2-{} wheel'.format(initPrms[-1]))
print('setup time: {} sec.'.format(round(t2 - t1, 5)))
t3 = time.time()
prm = initPrms[:]
primes = turbo(0, (gaps, c))
for p, x in primes:
prm.append(p)
if len(prm) > num:
break
t4 = time.time()
print('run time : {} sec.'.format(len(prm), round(t4 - t3, 5)))
print('prime sum : {}'.format(sum(prm)))
for w in [5, 7, 11, 13, 17, 19, 23, 29]:
speed_test(1e7-1, w)
当设置为生成一千万个素数时，它是如何使用PyPy（兼容Python 2.7）在我的计算机上运行的：
2- 3 wheel
setup time: 0.0 sec.
run time : 18.349 sec.
prime sum : 870530414842019
--------------------------------------------------
2- 5 wheel
setup time: 0.001 sec.
run time : 13.993 sec.
prime sum : 870530414842019
--------------------------------------------------
2- 7 wheel
setup time: 0.001 sec.
run time : 7.821 sec.
prime sum : 870530414842019
--------------------------------------------------
2- 11 wheel
setup time: 0.03 sec.
run time : 6.224 sec.
prime sum : 870530414842019
--------------------------------------------------
2- 13 wheel
setup time: 0.011 sec.
run time : 5.624 sec.
prime sum : 870530414842019
--------------------------------------------------
2- 17 wheel
setup time: 0.047 sec.
run time : 5.262 sec.
prime sum : 870530414842019
--------------------------------------------------
2- 19 wheel
setup time: 1.043 sec.
run time : 5.119 sec.
prime sum : 870530414842019
--------------------------------------------------
2- 23 wheel
setup time: 22.685 sec.
run time : 4.634 sec.
prime sum : 870530414842019
更大的轮子是可能的，但你可以看到他们变得相当长的设置。随着轮子变大，回报率也在递减——没有太多的点可以超过2-13轮子，因为它们并没有使它变得更快。我还遇到了一个2-23轮的内存错误（它的gaps列表中有大约3600万个数字）。
