我正在从here修改Eratosthenes的无限筛子，所以它使用轮子分解来跳过比现在只检查所有几率的复合物更多的复合物。
我已经研究出如何生成步骤以达到车轮上的所有间隙。从那里我想我可以用+ 2代替这些轮子步骤，但它导致筛子跳过素数。这是代码：
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
当我将轮尺寸设置为0时，我得到前100个素数的正确总和为24133，但是当我使用任何其他轮尺寸时，我最终会丢失素数，不正确的总和和复合材料。即使是2-3轮（使用2和4的交替步骤）也会使筛子错过质量。我做错了什么？
答案 0 :(得分：9)
赔率，即2-coprimes，由“滚动轮” [2]生成，即重复添加2，从初始值3开始（类似于5） ，7,9，......），
n=3; n+=2; n+=2; n+=2; ... # wheel = [2]
3 5 7 9
2-3次互质是通过重复添加2，然后是4，再次增加2，然后是4等产生的，依此类推：
n=5; n+=2; n+=4; n+=2; n+=4; ... # wheel = [2,4]
5 7 11 13 17
在这里，我们需要知道从哪里开始添加2或4的差异，具体取决于初始值。对于5,11,17，......，它是2（即车轮的第0个元素）;对于7,13,19，......，它是4（即1元素）。
我们怎么知道从哪里开始？轮优化的要点是我们只对这个互质序列（在这个例子中，2-3个互质）起作用。因此，在我们得到递归生成素数的代码部分中，我们还将保持滚动轮流，然后推进它直到我们看到其中的下一个素数。滚动序列需要产生两个结果 - 值和车轮位置。因此，当我们看到素数时，我们也得到相应的车轮位置，我们可以从车轮上的那个位置开始生成其倍数。当然，我们将所有内容乘以p，从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（在需要时环绕为0以获得圆度）。
为了产生最终的素数，我们滚动另一个互质序列，并且只保留那些不在dict中的元素，就像在参考代码中一样。
几次迭代之后更新： on codereview（非常感谢那里的贡献者！）我已经到了这个代码，尽可能使用itertools来提高速度：
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，此代码直接计算从每个素数开始滚动轮子的位置，以生成其倍数
答案 1 :(得分：1)
这是我提出的版本。它不像Ness＆＃39;那样干净。但它的确有效。我发布了它，所以还有另一个关于如何使用车轮分解的例子，以防任何人过来。我能够选择使用哪种轮尺寸，但很容易确定一个更永久的尺寸 - 只需生成您想要的尺寸并将其粘贴到代码中。
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)
留下车轮尺寸选项还可以让您了解更大的车轮不会做多快。下面是测试生成所选尺寸的轮子所需的时间以及筛子与该轮子的速度有关的代码。
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)
在设置为生成一千万个素数时，使用PyPy（兼容Python 2.7）在计算机上运行的方式：
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，但你可以看到它们设置得相当长。随着车轮越来越大，还有收益递减的规律 - 没有多少要超过2-13轮，因为它们并没有真正让它快得多。我最后还遇到了2-23轮（在gaps列表中有大约3600万个数字）的内存错误。