Python 向不定筛添加轮因子分解

我正在修改一个不确定的埃拉托什尼筛，所以它使用轮分解来跳过更多的合成物，而不是目前只检查所有可能性的形式

我已经计算出了如何生成到达车轮上所有间隙所需的步骤。从那以后，我想我可以用+2来代替这些车轮踏板，但这会导致筛子跳过素数。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

---

当我将控制盘大小设置为0时，我得到了前100个素数的正确和24133，但当我使用任何其他控制盘大小时，我最终会丢失素数、错误和合成。即使是2-3轮（使用2和4的交替步骤）也会使筛子漏掉素数。我做错了什么？

赔率，即2个共线，是通过“滚动轮子”

[2]

产生的，即从初始值3开始重复相加2（类似于从5、7、9、…）

2-3-coprimes由2、4和2、4的重复相加生成，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，我们确实需要知道从哪里开始添加差值，2或4，这取决于初始值。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（即第一个元素）

我们怎么知道从哪里开始？车轮优化的要点是，我们只处理这一系列的共轮辋（在本例中为2-3-共轮辋）。因此，在我们得到递归生成的素数的代码部分，我们还将维护滚动轮流，并将其推进，直到我们看到其中的下一个素数。滚动顺序将需要产生两个结果-值和车轮位置。因此，当我们看到素数时，我们也得到了相应的轮子位置，我们可以从轮子上的那个位置开始生成它的倍数。我们用

p

乘以一切当然，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、基本素数和当前车轮位置（如果需要，环绕为0表示圆形）

为了产生结果素数，我们滚动另一个coprimes序列，只保留dict中没有的元素，就像参考代码一样

---

更新：经过几次迭代（非常感谢那里的贡献者！），为了提高速度，我尽可能多地使用itertools编写了这段代码：

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()

从itertools导入累积、链、循环、计数
def wsieve（）：#轮筛，根据意愿。ideone.com/mqO25A
wh11=[2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,6,8,4,2,4，
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=累积（链条（[11]，循环（wh11））#从11开始滚动车轮
收益率（下一步（cs））#参见ideone.com/WFv4f，
ps=wsieve（）#codereview.stackexchange.com/q/92365/9064
p=下一个（ps）#11
psq=p**2#121
D=dict（zip（累积（链（[0]，wh11）），计数（0））#车轮滚动查找dict
mults={}
对于cs中的c：#候选，与210的互质，从11
如果在mults中使用c：
车轮=多个弹跳（c）
elif c

---

与上面的描述不同，此代码直接计算每个素数从何处开始滚动，以生成其倍数。这是我提出的版本。它不像“我的”那么干净，但很管用。我在发布它，所以有另一个关于如何使用轮子分解的例子，以防有人来。我留下了选择轮子大小的能力，但是很容易确定一个更持久的轮子-只需生成你想要的大小并粘贴到代码中

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)

下面是当设置为生成一千万个素数时，它如何在我的计算机上使用PyPy（与Python 2.7兼容）运行：

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，但是你可以看到它们设置起来变得相当长。当车轮变大时，也有收益递减的规律-通过2-13车轮没有多大意义，因为它们实际上不会使速度变快。我最后还遇到了一个内存错误，超过了2-23轮（它的

间隙
列表中有大约3600万个数字）。
请注意，wsieve从11开始。如果您正在使用此命令生成素数列表，请将
[2,3,5,7]
添加到它的前面。我允许自己编辑您的代码并使其符合要求。此外，我将
p*p
更改为
p**2
-我保证没有其他更改（好的，添加了
primes
生成器）！请随意后退！并不是说如果你有这么好的素数生成器，你就需要一个ErasToSchenes筛，而是如果你感兴趣的话：这是我的一个古老的答案，可以实现一个内存效率高（而且相当快）的实现在
轮隙
中，您可以尝试测试
gcd（i，d）==1
，其中
d=whlCirm/2
；也许这会使安装速度更快？@WillNess我不太担心车轮间隙的速度，因为它们的时间对于长时间运行的黄金一代来说是微不足道的。但我会看看的。下一步是看看发电机是否可以并联，这样它就可以同时在多个磁芯上运行，从而获得更快的速度！使用primesFrom（int start）生成器，您只需将区域分成4（8）个部分，然后启动4（8）个进程，
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)

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)

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