问题描述
p = []
for x in range(1, 50000000):
count = 0
for y in range(1, x // 2 + 1):
if (x % y == 0):
count += y
if (count == x):
p.append(x)
这是我的代码,用于尝试查找源自 1 到 50000000 之间的所有完美数字.它适用于前 3 个数字,它们在 1 到 10000 之间.但随着它的进展,它变得非常缓慢.就像可能每 10 秒通过 1000 个数字.然后最终每 5 秒通过 10 个数字.
This is my code to try and find all the perfect numbers that originate between 1 and 50000000. It works fine for the first 3 numbers, they are between 1 and 10000. But as it progresses it becomes extremely slow. Like maybe going through 1000 numbers every 10 seconds. Then eventually going through 10 numbers every 5 seconds.
现在有什么办法可以让这更快吗?我尝试包含一些较小的东西,例如将 x 减 2 以确保我们不会超过一半(不会是 x 的倍数)
Now is there anyway I could make this faster? I tried including some smaller things, like diving x by 2 to make sure we don't go over half (not going to be a multiple of x)
推荐答案
正如已经提到的,没有找到奇完美数.根据 维基百科关于完美数的文章,如果存在任何奇数完全数,它们必须大于 101500.显然,寻找奇完美数需要复杂的方法和/或很多时间.:)
As has already been mentioned, no odd perfect numbers have been found. And according to the Wikipedia article on perfect numbers, if any odd perfect numbers exist they must be greater than 101500. Clearly, looking for odd perfect numbers takes sophisticated methods &/or a lot of time. :)
正如 Wikipedia 中所述,欧几里得证明了即使是完美的数字也可以从梅森素数和欧拉证明了,相反,所有偶数都具有这种形式.
As stated on Wikipedia, Euclid proved that even perfect numbers can be produced from Mersenne primes, and Euler proved that, conversely, all even perfect numbers have that form.
因此,我们可以通过生成梅森素数来生成一个偶数列表.我们可以(相对)通过 Lucas 快速测试一个数字是否是梅森素数-Lehmer 测试.
So we can produce a list of even perfect numbers by generating Mersenne primes. And we can (relatively) quickly test if a number is a Mersenne prime via the Lucas-Lehmer test.
这是一个可以做到这一点的简短程序.这里使用的 primes
函数是由 Robert William Hanks 编写的,其余代码是我几分钟前刚刚编写的.:)
Here's a short program that does just that. The primes
function used here was written by Robert William Hanks, the rest of the code was just written by me a few minutes ago. :)
''' Mersenne primes and perfect numbers '''
def primes(n):
""" Return a list of primes < n """
# From http://stackoverflow.com/a/3035188/4014959
sieve = [True] * (n//2)
for i in range(3, int(n**0.5) + 1, 2):
if sieve[i//2]:
sieve[i*i//2::i] = [False] * ((n - i*i - 1) // (2*i) + 1)
return [2] + [2*i + 1 for i in range(1, n//2) if sieve[i]]
def lucas_lehmer(p):
''' The Lucas-Lehmer primality test for Mersenne primes.
See https://en.wikipedia.org/wiki/Mersenne_prime#Searching_for_Mersenne_primes
'''
m = (1 << p) - 1
s = 4
for i in range(p - 2):
s = (s * s - 2) % m
return s == 0 and m or 0
#Lucas-Lehmer doesn't work on 2 because it's even, so we just print it manually :)
print('1 2
3
6
')
count = 1
for p in primes(1280):
m = lucas_lehmer(p)
if m:
count += 1
n = m << (p - 1)
print(count, p)
print(m)
print(n, '
')
输出
1 2
3
6
2 3
7
28
3 5
31
496
4 7
127
8128
5 13
8191
33550336
6 17
131071
8589869056
7 19
524287
137438691328
8 31
2147483647
2305843008139952128
9 61
2305843009213693951
2658455991569831744654692615953842176
10 89
618970019642690137449562111
191561942608236107294793378084303638130997321548169216
11 107
162259276829213363391578010288127
13164036458569648337239753460458722910223472318386943117783728128
12 127
170141183460469231731687303715884105727
14474011154664524427946373126085988481573677491474835889066354349131199152128
13 521
6864797660130609714981900799081393217269435300143305409394463459185543183397656052122559640661454554977296311391480858037121987999716643812574028291115057151
23562723457267347065789548996709904988477547858392600710143027597506337283178622239730365539602600561360255566462503270175052892578043215543382498428777152427010394496918664028644534128033831439790236838624033171435922356643219703101720713163527487298747400647801939587165936401087419375649057918549492160555646976
14 607
531137992816767098689588206552468627329593117727031923199444138200403559860852242739162502265229285668889329486246501015346579337652707239409519978766587351943831270835393219031728127
141053783706712069063207958086063189881486743514715667838838675999954867742652380114104193329037690251561950568709829327164087724366370087116731268159313652487450652439805877296207297446723295166658228846926807786652870188920867879451478364569313922060370695064736073572378695176473055266826253284886383715072974324463835300053138429460296575143368065570759537328128
15 1279
10407932194664399081925240327364085538615262247266704805319112350403608059673360298012239441732324184842421613954281007791383566248323464908139906605677320762924129509389220345773183349661583550472959420547689811211693677147548478866962501384438260291732348885311160828538416585028255604666224831890918801847068222203140521026698435488732958028878050869736186900714720710555703168729087
54162526284365847412654465374391316140856490539031695784603920818387206994158534859198999921056719921919057390080263646159280013827605439746262788903057303445505827028395139475207769044924431494861729435113126280837904930462740681717960465867348720992572190569465545299629919823431031092624244463547789635441481391719816441605586788092147886677321398756661624714551726964302217554281784254817319611951659855553573937788923405146222324506715979193757372820860878214322052227584537552897476256179395176624426314480313446935085203657584798247536021172880403783048602873621259313789994900336673941503747224966984028240806042108690077670395259231894666273615212775603535764707952250173858305171028603021234896647851363949928904973292145107505979911456221519899345764984291328
该输出是在 2GHz 32 位机器上在 4.5 秒内产生的.您可以轻松生成更大的梅森素数和完美数,但不要指望它会很快.
That output was produced in 4.5 seconds on a 2GHz 32 bit machine. You can easily produce larger Mersenne primes and perfect numbers, but don't expect it to be fast.
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