问题
I'm doing this problem on a site that I found (project Euler), and there is a question that involves finding the largest prime factor of a number. My solution fails at really large numbers so I was wondering how this code could be streamlined?
""" Find the largest prime of a number """
def get_factors(number):
factors = []
for integer in range(1, number + 1):
if number%integer == 0:
factors.append(integer)
return factors
def test_prime(number):
prime = True
for i in range(1, number + 1):
if i!=1 and i!=2 and i!=number:
if number%i == 0:
prime = False
return prime
def test_for_primes(lst):
primes = []
for i in lst:
if test_prime(i):
primes.append(i)
return primes
################################################### program starts here
def find_largest_prime_factor(i):
factors = get_factors(i)
prime_factors = test_for_primes(factors)
print prime_factors
print find_largest_prime_factor(22)
#this jams my computer
print find_largest_prime_factor(600851475143)
it fails when using large numbers, which is the point of the question I guess. (computer jams, tells me I have run out of memory and asks me which programs I would like to stop).
************************************ thanks for the answer. there was actually a couple bugs in the code in any case. so the fixed version of this (inefficient code) is below.
""" Find the largest prime of a number """
def get_factors(number):
factors = []
for integer in xrange(1, number + 1):
if number%integer == 0:
factors.append(integer)
return factors
def test_prime(number):
prime = True
if number == 1 or number == 2:
return prime
else:
for i in xrange(2, number):
if number%i == 0:
prime = False
return prime
def test_for_primes(lst):
primes = []
for i in lst:
if test_prime(i):
primes.append(i)
return primes
################################################### program starts here
def find_largest_prime_factor(i):
factors = get_factors(i)
print factors
prime_factors = test_for_primes(factors)
return prime_factors
print find_largest_prime_factor(x)
回答1:
From your approach you are first generating all divisors of a number n in O(n) then you test which of these divisors is prime in another O(n) number of calls of test_prime (which is exponential anyway).
A better approach is to observe that once you found out a divisor of a number you can repeatedly divide by it to get rid of all of it's factors. Thus, to get the prime factors of, say 830297 you test all small primes (cached) and for each one which divides your number you keep dividing:
830297is divisible by13so now you'll test with830297 / 13 = 6386963869is still divisible by13, you are at49134913doesn't divide by 13, next prime is17which divides4913to get289289is still a multiple of17, you have17which is the divisor and stop.
For further speed increase, after testing the cached prime numbers below say 100, you'll have to test for prime divisors using your test_prime function (updated according to @Ben's answer) but go on reverse, starting from sqrt. Your number is divisible by 71, the next number will give an sqrt of 91992 which is somewhat close to 6857 which is the largest prime factor.
回答2:
Here is my favorite simple factoring program for Python:
def factors(n):
wheel = [1,2,2,4,2,4,2,4,6,2,6]
w, f, fs = 0, 2, []
while f*f <= n:
while n % f == 0:
fs.append(f)
n /= f
f, w = f + wheel[w], w+1
if w == 11: w = 3
if n > 1: fs.append(n)
return fs
The basic algorithm is trial division, using a prime wheel to generate the trial factors. It's not quite as fast as trial division by primes, but there's no need to calculate or store the prime numbers, so it's very convenient.
If you're interested in programming with prime numbers, you might enjoy this essay at my blog.
回答3:
My solution is in C#. I bet you can translate it into python. I've been test it with random long integer ranging from 1 to 1.000.000.000 and it's doing good. You can try to test the result with online prime calculator Happy coding :)
public static long biggestPrimeFactor(long num) {
for (int div = 2; div < num; div++) {
if (num % div == 0) {
num \= div
div--;
}
}
return num;
}
回答4:
The naive primality test can be improved upon in several ways:
- Test for divisibility by 2 separately, then start your loop at 3 and go by 2's
- End your loop at ceil(sqrt(num)). You're guaranteed to not find a prime factor above this number
- Generate primes using a sieve beforehand, and only move onto the naive way if you've exhausted the numbers in your sieve.
Beyond these easy fixes, you're going to have to look up more efficient factorization algorithms.
回答5:
Use a Sieve of Eratosthenes to calculate your primes.
from math import sqrt
def sieveOfEratosthenes(n):
primes = range(3, n + 1, 2) # primes above 2 must be odd so start at three and increase by 2
for base in xrange(len(primes)):
if primes[base] is None:
continue
if primes[base] >= sqrt(n): # stop at sqrt of n
break
for i in xrange(base + (base + 1) * primes[base], len(primes), primes[base]):
primes[i] = None
primes.insert(0,2)
return filter(None, primes)
回答6:
The point to prime factorization by trial division is, the most efficient solution for factorizing just one number doesn't need any prime testing.
You just enumerate your possible factors in ascending order, and keep dividing them out of the number in question - all thus found factors are guaranteed to be prime. Stop when the square of current factor exceeds the current number being factorized. See the code in user448810's answer.
Normally, prime factorization by trial division is faster on primes than on all numbers (or odds etc.), but when factorizing just one number, to find the primes first to test divide by them later, will might cost more than just going ahead with the increasing stream of possible factors. This enumeration is O(n), prime generation is O(n log log n), with the Sieve of Eratosthenes (SoE), where n = sqrt(N) for the top limit N. With trial division (TD) the complexity is O(n1.5/(log n)2).
Of course the asymptotics are to be taken just as a guide, actual code's constant factors might change the picture. Example, execution times for a Haskell code derived from here and here, factorizing 600851475149 (a prime):
2.. 0.57 sec
2,3,5,... 0.28 sec
2,3,5,7,11,13,17,19,... 0.21 sec
primes, segmented TD 0.65 sec first try
0.05 sec subsequent runs (primes are memoized)
primes, list-based SoE 0.44 sec first try
0.05 sec subsequent runs (primes are memoized)
primes, array-based SoE 0.15 sec first try
0.06 sec subsequent runs (primes are memoized)
so it depends. Of course factorizing the composite number in question, 600851475143, is near instantaneous, so it doesn't matter there.
回答7:
Here is an example in JavaScript
function largestPrimeFactor(val, divisor = 2) {
let square = (val) => Math.pow(val, 2);
while ((val % divisor) != 0 && square(divisor) <= val) {
divisor++;
}
return square(divisor) <= val
? largestPrimeFactor(val / divisor, divisor)
: val;
}
回答8:
I converted the solution from @under5hell to Python (2.7x). what an efficient way!
def largest_prime_factor(num, div=2):
while div < num:
if num % div == 0 and num/div > 1:
num = num /div
div = 2
else:
div = div + 1
return num
>> print largest_prime_factor(600851475143)
6857
>> print largest_prime_factor(13195)
29
回答9:
Try this piece of code:
from math import *
def largestprime(n):
i=2
while (n>1):
if (n % i == 0):
n = n/i
else:
i=i+1
print i
strinput = raw_input('Enter the number to be factorized : ')
a = int(strinput)
largestprime(a)
回答10:
Old one but might help
def isprime(num):
if num > 1:
# check for factors
for i in range(2,num):
if (num % i) == 0:
return False
return True
def largest_prime_factor(bignumber):
prime = 2
while bignumber != 1:
if bignumber % prime == 0:
bignumber = bignumber / prime
else:
prime = prime + 1
while isprime(prime) == False:
prime = prime+1
return prime
number = 600851475143
print largest_prime_factor(number)
回答11:
I Hope this would help and easy to understand.
A = int(input("Enter the number to find the largest prime factor:"))
B = 2
while (B <(A/2)):
if A%B != 0:
B = B+1
else:
A = A/B
C = B
B = 2
print (A)
回答12:
This code for getting the largest prime factor, with nums value of prime_factor(13195) when I run it, will return the result in less than a second. but when nums value gets up to 6digits it will return the result in 8seconds.
Any one has an idea of what is the best algorithm for the solution...
def prime_factor(nums):
if nums < 2:
return 0
primes = [2]
x = 3
while x <= nums:
for i in primes:
if x%i==0:
x += 2
break
else:
primes.append(x)
x += 2
largest_prime = primes[::-1]
# ^^^ code above to gets all prime numbers
intermediate_tag = []
factor = []
# this code divide nums by the largest prime no. and return if the
# result is an integer then append to primefactor.
for i in largest_prime:
x = nums/i
if x.is_integer():
intermediate_tag.append(x)
# this code gets the prime factors [29.0, 13.0, 7.0, 5.0]
for i in intermediate_tag:
y = nums/i
factor.append(y)
print(intermediate_tag)
print(f"prime factor of {nums}:==>",factor)
prime_factor(13195)
[455.0, 1015.0, 1885.0, 2639.0] prime factor of 13195:==> [29.0, 13.0, 7.0, 5.0]
来源：https://stackoverflow.com/questions/24166478/efficient-ways-of-finding-the-largest-prime-factor-of-a-number