[英]Algorithm to find ALL factorizations of an integer
是否有算法可以找到整數的所有因子分解,最好是在Python / Java中,但歡迎任何反饋。
我有一個算法來計算素因子。 例如[2,2,5]
是20
的主要因素。
def prime_factorization(n):
primfac = []
d = 2
while d*d <= n:
while (n % d) == 0:
primfac.append(d)
n /= d
d += 1
if n > 1:
primfac.append(n)
return primfac
我還有一個算法來計算算法的所有因子(素數和非素數)。 例如, 20
的因子是[1, 2, 4, 5, 10, 20]
。
def factors(n):
a, r = 1, [1]
while a * a < n:
a += 1
if n % a: continue
b, f = 1, []
while n % a == 0:
n //= a
b *= a
f += [i * b for i in r]
r += f
if n > 1: r += [i * n for i in r]
return sorted(r)
我正在尋找的是一種返回給定整數的所有因子分解(而不是因子)的算法。 對於整數20
,算法將產生以下結果:
[1,20]
[2,10]
[4,5]
[2,2,5]
謝謝!
這是一種非常低效的方法。 它會生成大量重復項,然后在返回之前將其過濾掉。
這個想法是你從n=1
到len(factors)
包含因子中選擇乘以,然后你重復使用未使用的因子。
import itertools
def mult(fs):
res = 1
for f in fs:
res *= f
return res
def _generate_all_factorizations(factors):
if len(factors) <= 1:
yield factors
return
for factors_in_mult in xrange(1, len(factors)+1):
for which_is in itertools.combinations(range(len(factors)), factors_in_mult):
this_mult = mult(factors[i] for i in which_is)
rest = [factors[i] for i in xrange(len(factors)) if i not in which_is]
for remaining in _generate_all_factorizations(rest):
yield [this_mult] + remaining
我添加了一個函數來刪除重復項並返回它們很好地排序:
def generate_all_factorizations(factors):
seen = set()
res = []
for f in _generate_all_factorizations(factors):
f = tuple(sorted(f))
if f in seen:
continue
seen.add(f)
yield f
現在只需將它歸結為您的主要因素:
for factorization in generate_all_factorizations([2, 2, 5]):
print factorization
print "----"
for factorization in generate_all_factorizations([2, 3, 5, 7]):
print factorization
結果:
(2, 2, 5)
(2, 10)
(4, 5)
(20,)
----
(2, 3, 5, 7)
(2, 3, 35)
(2, 5, 21)
(2, 7, 15)
(2, 105)
(3, 5, 14)
(3, 7, 10)
(3, 70)
(5, 6, 7)
(5, 42)
(7, 30)
(6, 35)
(10, 21)
(14, 15)
(210,)
純娛樂:
from itertools import combinations_with_replacement
from operator import mul
my_integer = 20
factorizations = {t for t in {list(t).remove(1) if 1 in t and len(t)>2 else t if len(t)>1 else None for combo in [combinations_with_replacement(factors(my_integer), n) for n in xrange(len(factors(my_integer)))] for t in combo if reduce(mul, t, 1) == my_integer} if t}
print factorizations
set([(4, 5), (2, 2, 5), (1, 20), (2, 10)])
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