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I am solving problem 3 in euler project to find the largest prime factor of a certain number.
def findFactors(num: int)->list:
factors = []
for i in range(1, num+1):
if num%i == 0:
factors.append(i)
return factors
prime_factors = (findFactors(600851475143))
max= prime_factors[0]
num = 600851475143
for i in range(0, len(prime_factors)):
if (prime_factors[i] > max):
max = prime_factors[i]
print(f"The largest prime factor of the {num} is {max}")
When I run the code for "13195", the code runs correctly but when I run the code for the actual number i.e 600851475143, the code is not giving any output, neither any errors
I think while loop will perform better:
n = 600851475143
x = 2
while x * x < n:
while n % x == 0:
n = n // x
x = x + 1
print(n)
#6857
You need two functions. One to determine the factors of a positive integer and another to determine if a number is prime.
from functools import cache
from math import sqrt, floor
from functools import reduce
from time import perf_counter
#cache
def isprime(n):
if n < 2:
return False
if n == 2 or n == 3:
return True
if n % 2 == 0 or n % 3 == 0:
return False
for i in range(5, floor(sqrt(n))+1, 6):
if n % i == 0 or n % (i + 2) == 0:
return False
return True
def factors(n):
if n > 0:
for f in sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, floor(sqrt(n)) + 1) if n % i == 0)))):
yield f
N = 600851475143
start_time = perf_counter()
f = [_f for _f in factors(N) if isprime(_f)]
end_time = perf_counter()
print(f[-1] if f else None, f'{end_time-start_time:.4f}s')
Output:
6857 0.0452s
Your code does work, but just takes too long. As someone pointed out, the number evaluated is indeed very big... Here it is required that you compute at least 10^12 division before it finishes.
Of course there are better ways to answer this problem. For instance, you need to look for factors only up to sqrt(n); or you could try to think about the sieve of eratosthenes.
You could use a recursive function:
from math import sqrt, floor
import time
def findFactors(n):
if n%2==0:
findFactors(int(n/i))
return
for i in range (3, floor(sqrt(n))+2, 2):
if n%i==0:
print (i)
findFactors(int(n/i))
return
print (n)
return
start=time.time()
findFactors(600851475143)
print (time.time()-start)
Output:
71
839
1471
6857
0.00026988983154296875
Two part question:
Trying to determine the largest prime factor of 600851475143, I found this program online that seems to work. The problem is, I'm having a hard time figuring out how it works exactly, though I understand the basics of what the program is doing. Also, I'd like if you could shed some light on any method you may know of finding prime factors, perhaps without testing every number, and how your method works.
Here's the code that I found online for prime factorization [NOTE: This code is incorrect. See Stefan's answer below for better code.]:
n = 600851475143
i = 2
while i * i < n:
while n % i == 0:
n = n / i
i = i + 1
print(n)
#takes about ~0.01secs
Why is that code so much faster than this code, which is just to test the speed and has no real purpose other than that?
i = 1
while i < 100:
i += 1
#takes about ~3secs
This question was the first link that popped up when I googled "python prime factorization".
As pointed out by #quangpn88, this algorithm is wrong (!) for perfect squares such as n = 4, 9, 16, ... However, #quangpn88's fix does not work either, since it will yield incorrect results if the largest prime factor occurs 3 or more times, e.g., n = 2*2*2 = 8 or n = 2*3*3*3 = 54.
I believe a correct, brute-force algorithm in Python is:
def largest_prime_factor(n):
i = 2
while i * i <= n:
if n % i:
i += 1
else:
n //= i
return n
Don't use this in performance code, but it's OK for quick tests with moderately large numbers:
In [1]: %timeit largest_prime_factor(600851475143)
1000 loops, best of 3: 388 µs per loop
If the complete prime factorization is sought, this is the brute-force algorithm:
def prime_factors(n):
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors
Ok. So you said you understand the basics, but you're not sure EXACTLY how it works. First of all, this is a great answer to the Project Euler question it stems from. I've done a lot of research into this problem and this is by far the simplest response.
For the purpose of explanation, I'll let n = 20. To run the real Project Euler problem, let n = 600851475143.
n = 20
i = 2
while i * i < n:
while n%i == 0:
n = n / i
i = i + 1
print (n)
This explanation uses two while loops. The biggest thing to remember about while loops is that they run until they are no longer true.
The outer loop states that while i * i isn't greater than n (because the largest prime factor will never be larger than the square root of n), add 1 to i after the inner loop runs.
The inner loop states that while i divides evenly into n, replace n with n divided by i. This loop runs continuously until it is no longer true. For n=20 and i=2, n is replaced by 10, then again by 5. Because 2 doesn't evenly divide into 5, the loop stops with n=5 and the outer loop finishes, producing i+1=3.
Finally, because 3 squared is greater than 5, the outer loop is no longer true and prints the result of n.
Thanks for posting this. I looked at the code forever before realizing how exactly it worked. Hopefully, this is what you're looking for in a response. If not, let me know and I can explain further.
It looks like people are doing the Project Euler thing where you code the solution yourself. For everyone else who wants to get work done, there's the primefac module which does very large numbers very quickly:
#!python
import primefac
import sys
n = int( sys.argv[1] )
factors = list( primefac.primefac(n) )
print '\n'.join(map(str, factors))
For prime number generation I always use the Sieve of Eratosthenes:
def primes(n):
if n<=2:
return []
sieve=[True]*(n+1)
for x in range(3,int(n**0.5)+1,2):
for y in range(3,(n//x)+1,2):
sieve[(x*y)]=False
return [2]+[i for i in range(3,n,2) if sieve[i]]
In [42]: %timeit primes(10**5)
10 loops, best of 3: 60.4 ms per loop
In [43]: %timeit primes(10**6)
1 loops, best of 3: 1.01 s per loop
You can use Miller-Rabin primality test to check whether a number is prime or not. You can find its Python implementations here.
Always use timeit module to time your code, the 2nd one takes just 15us:
def func():
n = 600851475143
i = 2
while i * i < n:
while n % i == 0:
n = n / i
i = i + 1
In [19]: %timeit func()
1000 loops, best of 3: 1.35 ms per loop
def func():
i=1
while i<100:i+=1
....:
In [21]: %timeit func()
10000 loops, best of 3: 15.3 us per loop
If you are looking for pre-written code that is well maintained, use the function sympy.ntheory.primefactors from SymPy.
It returns a sorted list of prime factors of n.
>>> from sympy.ntheory import primefactors
>>> primefactors(6008)
[2, 751]
Pass the list to max() to get the biggest prime factor: max(primefactors(6008))
In case you want the prime factors of n and also the multiplicities of each of them, use sympy.ntheory.factorint.
Given a positive integer n, factorint(n) returns a dict containing the
prime factors of n as keys and their respective multiplicities as
values.
>>> from sympy.ntheory import factorint
>>> factorint(6008) # 6008 = (2**3) * (751**1)
{2: 3, 751: 1}
The code is tested against Python 3.6.9 and SymPy 1.1.1.
"""
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
"""
from sympy import primefactors
print(primefactors(600851475143)[-1])
def find_prime_facs(n):
list_of_factors=[]
i=2
while n>1:
if n%i==0:
list_of_factors.append(i)
n=n/i
i=i-1
i+=1
return list_of_factors
Isn't largest prime factor of 27 is 3 ??
The above code might be fastest,but it fails on 27 right ?
27 = 3*3*3
The above code returns 1
As far as I know.....1 is neither prime nor composite
I think, this is the better code
def prime_factors(n):
factors=[]
d=2
while(d*d<=n):
while(n>1):
while n%d==0:
factors.append(d)
n=n/d
d+=1
return factors[-1]
Another way of doing this:
import sys
n = int(sys.argv[1])
result = []
for i in xrange(2,n):
while n % i == 0:
#print i,"|",n
n = n/i
result.append(i)
if n == 1:
break
if n > 1: result.append(n)
print result
sample output :
python test.py 68
[2, 2, 17]
The code is wrong with 100. It should check case i * i = n:
I think it should be:
while i * i <= n:
if i * i = n:
n = i
break
while n%i == 0:
n = n / i
i = i + 1
print (n)
My code:
# METHOD: PRIME FACTORS
def prime_factors(n):
'''PRIME FACTORS: generates a list of prime factors for the number given
RETURNS: number(being factored), list(prime factors), count(how many loops to find factors, for optimization)
'''
num = n #number at the end
count = 0 #optimization (to count iterations)
index = 0 #index (to test)
t = [2, 3, 5, 7] #list (to test)
f = [] #prime factors list
while t[index] ** 2 <= n:
count += 1 #increment (how many loops to find factors)
if len(t) == (index + 1):
t.append(t[-2] + 6) #extend test list (as much as needed) [2, 3, 5, 7, 11, 13...]
if n % t[index]: #if 0 does else (otherwise increments, or try next t[index])
index += 1 #increment index
else:
n = n // t[index] #drop max number we are testing... (this should drastically shorten the loops)
f.append(t[index]) #append factor to list
if n > 1:
f.append(n) #add last factor...
return num, f, f'count optimization: {count}'
Which I compared to the code with the most votes, which was very fast
def prime_factors2(n):
i = 2
factors = []
count = 0 #added to test optimization
while i * i <= n:
count += 1 #added to test optimization
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors, f'count: {count}' #print with (count added)
TESTING, (note, I added a COUNT in each loop to test the optimization)
# >>> prime_factors2(600851475143)
# ([71, 839, 1471, 6857], 'count: 1472')
# >>> prime_factors(600851475143)
# (600851475143, [71, 839, 1471, 6857], 'count optimization: 494')
I figure this code could be modified easily to get the (largest factor) or whatever else is needed. I'm open to any questions, my goal is to improve this much more as well for larger primes and factors.
In case you want to use numpy here's a way to create an array of all primes not greater than n:
[ i for i in np.arange(2,n+1) if 0 not in np.array([i] * (i-2) ) % np.arange(2,i)]
Check this out, it might help you a bit in your understanding.
#program to find the prime factors of a given number
import sympy as smp
try:
number = int(input('Enter a number : '))
except(ValueError) :
print('Please enter an integer !')
num = number
prime_factors = []
if smp.isprime(number) :
prime_factors.append(number)
else :
for i in range(2, int(number/2) + 1) :
"""while figuring out prime factors of a given number, n
keep in mind that a number can itself be prime or if not,
then all its prime factors will be less than or equal to its int(n/2 + 1)"""
if smp.isprime(i) and number % i == 0 :
while(number % i == 0) :
prime_factors.append(i)
number = number / i
print('prime factors of ' + str(num) + ' - ')
for i in prime_factors :
print(i, end = ' ')
This is my python code:
it has a fast check for primes and checks from highest to lowest the prime factors.
You have to stop if no new numbers came out. (Any ideas on this?)
import math
def is_prime_v3(n):
""" Return 'true' if n is a prime number, 'False' otherwise """
if n == 1:
return False
if n > 2 and n % 2 == 0:
return False
max_divisor = math.floor(math.sqrt(n))
for d in range(3, 1 + max_divisor, 2):
if n % d == 0:
return False
return True
number = <Number>
for i in range(1,math.floor(number/2)):
if is_prime_v3(i):
if number % i == 0:
print("Found: {} with factor {}".format(number / i, i))
The answer for the initial question arrives in a fraction of a second.
Below are two ways to generate prime factors of given number efficiently:
from math import sqrt
def prime_factors(num):
'''
This function collectes all prime factors of given number and prints them.
'''
prime_factors_list = []
while num % 2 == 0:
prime_factors_list.append(2)
num /= 2
for i in range(3, int(sqrt(num))+1, 2):
if num % i == 0:
prime_factors_list.append(i)
num /= i
if num > 2:
prime_factors_list.append(int(num))
print(sorted(prime_factors_list))
val = int(input('Enter number:'))
prime_factors(val)
def prime_factors_generator(num):
'''
This function creates a generator for prime factors of given number and generates the factors until user asks for them.
It handles StopIteration if generator exhausted.
'''
while num % 2 == 0:
yield 2
num /= 2
for i in range(3, int(sqrt(num))+1, 2):
if num % i == 0:
yield i
num /= i
if num > 2:
yield int(num)
val = int(input('Enter number:'))
prime_gen = prime_factors_generator(val)
while True:
try:
print(next(prime_gen))
except StopIteration:
print('Generator exhausted...')
break
else:
flag = input('Do you want next prime factor ? "y" or "n":')
if flag == 'y':
continue
elif flag == 'n':
break
else:
print('Please try again and enter a correct choice i.e. either y or n')
Since nobody has been trying to hack this with old nice reduce method, I'm going to take this occupation. This method isn't flexible for problems like this because it performs loop of repeated actions over array of arguments and there's no way how to interrupt this loop by default. The door open after we have implemented our own interupted reduce for interrupted loops like this:
from functools import reduce
def inner_func(func, cond, x, y):
res = func(x, y)
if not cond(res):
raise StopIteration(x, y)
return res
def ireducewhile(func, cond, iterable):
# generates intermediary results of args while reducing
iterable = iter(iterable)
x = next(iterable)
yield x
for y in iterable:
try:
x = inner_func(func, cond, x, y)
except StopIteration:
break
yield x
After that we are able to use some func that is the same as an input of standard Python reduce method. Let this func be defined in a following way:
def division(c):
num, start = c
for i in range(start, int(num**0.5)+1):
if num % i == 0:
return (num//i, i)
return None
Assuming we want to factor a number 600851475143, an expected output of this function after repeated use of this function should be this:
(600851475143, 2) -> (8462696833 -> 71), (10086647 -> 839), (6857, 1471) -> None
The first item of tuple is a number that division method takes and tries to divide by the smallest divisor starting from second item and finishing with square root of this number. If no divisor exists, None is returned.
Now we need to start with iterator defined like this:
def gener(prime):
# returns and infinite generator (600851475143, 2), 0, 0, 0...
yield (prime, 2)
while True:
yield 0
Finally, the result of looping is:
result = list(ireducewhile(lambda x,y: div(x), lambda x: x is not None, iterable=gen(600851475143)))
#result: [(600851475143, 2), (8462696833, 71), (10086647, 839), (6857, 1471)]
And outputting prime divisors can be captured by:
if len(result) == 1: output = result[0][0]
else: output = list(map(lambda x: x[1], result[1:]))+[result[-1][0]]
#output: [2, 71, 839, 1471]
Note:
In order to make it more efficient, you might like to use pregenerated primes that lies in specific range instead of all the values of this range.
You shouldn't loop till the square root of the number! It may be right some times, but not always!
Largest prime factor of 10 is 5, which is bigger than the sqrt(10) (3.16, aprox).
Largest prime factor of 33 is 11, which is bigger than the sqrt(33) (5.5,74, aprox).
You're confusing this with the propriety which states that, if a number has a prime factor bigger than its sqrt, it has to have at least another one other prime factor smaller than its sqrt. So, with you want to test if a number is prime, you only need to test till its sqrt.
def prime(n):
for i in range(2,n):
if n%i==0:
return False
return True
def primefactors():
m=int(input('enter the number:'))
for i in range(2,m):
if (prime(i)):
if m%i==0:
print(i)
return print('end of it')
primefactors()
Another way that skips even numbers after 2 is handled:
def prime_factors(n):
factors = []
d = 2
step = 1
while d*d <= n:
while n>1:
while n%d == 0:
factors.append(d)
n = n/d
d += step
step = 2
return factors
Two part question:
Trying to determine the largest prime factor of 600851475143, I found this program online that seems to work. The problem is, I'm having a hard time figuring out how it works exactly, though I understand the basics of what the program is doing. Also, I'd like if you could shed some light on any method you may know of finding prime factors, perhaps without testing every number, and how your method works.
Here's the code that I found online for prime factorization [NOTE: This code is incorrect. See Stefan's answer below for better code.]:
n = 600851475143
i = 2
while i * i < n:
while n % i == 0:
n = n / i
i = i + 1
print(n)
#takes about ~0.01secs
Why is that code so much faster than this code, which is just to test the speed and has no real purpose other than that?
i = 1
while i < 100:
i += 1
#takes about ~3secs
This question was the first link that popped up when I googled "python prime factorization".
As pointed out by #quangpn88, this algorithm is wrong (!) for perfect squares such as n = 4, 9, 16, ... However, #quangpn88's fix does not work either, since it will yield incorrect results if the largest prime factor occurs 3 or more times, e.g., n = 2*2*2 = 8 or n = 2*3*3*3 = 54.
I believe a correct, brute-force algorithm in Python is:
def largest_prime_factor(n):
i = 2
while i * i <= n:
if n % i:
i += 1
else:
n //= i
return n
Don't use this in performance code, but it's OK for quick tests with moderately large numbers:
In [1]: %timeit largest_prime_factor(600851475143)
1000 loops, best of 3: 388 µs per loop
If the complete prime factorization is sought, this is the brute-force algorithm:
def prime_factors(n):
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors
Ok. So you said you understand the basics, but you're not sure EXACTLY how it works. First of all, this is a great answer to the Project Euler question it stems from. I've done a lot of research into this problem and this is by far the simplest response.
For the purpose of explanation, I'll let n = 20. To run the real Project Euler problem, let n = 600851475143.
n = 20
i = 2
while i * i < n:
while n%i == 0:
n = n / i
i = i + 1
print (n)
This explanation uses two while loops. The biggest thing to remember about while loops is that they run until they are no longer true.
The outer loop states that while i * i isn't greater than n (because the largest prime factor will never be larger than the square root of n), add 1 to i after the inner loop runs.
The inner loop states that while i divides evenly into n, replace n with n divided by i. This loop runs continuously until it is no longer true. For n=20 and i=2, n is replaced by 10, then again by 5. Because 2 doesn't evenly divide into 5, the loop stops with n=5 and the outer loop finishes, producing i+1=3.
Finally, because 3 squared is greater than 5, the outer loop is no longer true and prints the result of n.
Thanks for posting this. I looked at the code forever before realizing how exactly it worked. Hopefully, this is what you're looking for in a response. If not, let me know and I can explain further.
It looks like people are doing the Project Euler thing where you code the solution yourself. For everyone else who wants to get work done, there's the primefac module which does very large numbers very quickly:
#!python
import primefac
import sys
n = int( sys.argv[1] )
factors = list( primefac.primefac(n) )
print '\n'.join(map(str, factors))
For prime number generation I always use the Sieve of Eratosthenes:
def primes(n):
if n<=2:
return []
sieve=[True]*(n+1)
for x in range(3,int(n**0.5)+1,2):
for y in range(3,(n//x)+1,2):
sieve[(x*y)]=False
return [2]+[i for i in range(3,n,2) if sieve[i]]
In [42]: %timeit primes(10**5)
10 loops, best of 3: 60.4 ms per loop
In [43]: %timeit primes(10**6)
1 loops, best of 3: 1.01 s per loop
You can use Miller-Rabin primality test to check whether a number is prime or not. You can find its Python implementations here.
Always use timeit module to time your code, the 2nd one takes just 15us:
def func():
n = 600851475143
i = 2
while i * i < n:
while n % i == 0:
n = n / i
i = i + 1
In [19]: %timeit func()
1000 loops, best of 3: 1.35 ms per loop
def func():
i=1
while i<100:i+=1
....:
In [21]: %timeit func()
10000 loops, best of 3: 15.3 us per loop
If you are looking for pre-written code that is well maintained, use the function sympy.ntheory.primefactors from SymPy.
It returns a sorted list of prime factors of n.
>>> from sympy.ntheory import primefactors
>>> primefactors(6008)
[2, 751]
Pass the list to max() to get the biggest prime factor: max(primefactors(6008))
In case you want the prime factors of n and also the multiplicities of each of them, use sympy.ntheory.factorint.
Given a positive integer n, factorint(n) returns a dict containing the
prime factors of n as keys and their respective multiplicities as
values.
>>> from sympy.ntheory import factorint
>>> factorint(6008) # 6008 = (2**3) * (751**1)
{2: 3, 751: 1}
The code is tested against Python 3.6.9 and SymPy 1.1.1.
"""
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
"""
from sympy import primefactors
print(primefactors(600851475143)[-1])
def find_prime_facs(n):
list_of_factors=[]
i=2
while n>1:
if n%i==0:
list_of_factors.append(i)
n=n/i
i=i-1
i+=1
return list_of_factors
Isn't largest prime factor of 27 is 3 ??
The above code might be fastest,but it fails on 27 right ?
27 = 3*3*3
The above code returns 1
As far as I know.....1 is neither prime nor composite
I think, this is the better code
def prime_factors(n):
factors=[]
d=2
while(d*d<=n):
while(n>1):
while n%d==0:
factors.append(d)
n=n/d
d+=1
return factors[-1]
Another way of doing this:
import sys
n = int(sys.argv[1])
result = []
for i in xrange(2,n):
while n % i == 0:
#print i,"|",n
n = n/i
result.append(i)
if n == 1:
break
if n > 1: result.append(n)
print result
sample output :
python test.py 68
[2, 2, 17]
The code is wrong with 100. It should check case i * i = n:
I think it should be:
while i * i <= n:
if i * i = n:
n = i
break
while n%i == 0:
n = n / i
i = i + 1
print (n)
My code:
# METHOD: PRIME FACTORS
def prime_factors(n):
'''PRIME FACTORS: generates a list of prime factors for the number given
RETURNS: number(being factored), list(prime factors), count(how many loops to find factors, for optimization)
'''
num = n #number at the end
count = 0 #optimization (to count iterations)
index = 0 #index (to test)
t = [2, 3, 5, 7] #list (to test)
f = [] #prime factors list
while t[index] ** 2 <= n:
count += 1 #increment (how many loops to find factors)
if len(t) == (index + 1):
t.append(t[-2] + 6) #extend test list (as much as needed) [2, 3, 5, 7, 11, 13...]
if n % t[index]: #if 0 does else (otherwise increments, or try next t[index])
index += 1 #increment index
else:
n = n // t[index] #drop max number we are testing... (this should drastically shorten the loops)
f.append(t[index]) #append factor to list
if n > 1:
f.append(n) #add last factor...
return num, f, f'count optimization: {count}'
Which I compared to the code with the most votes, which was very fast
def prime_factors2(n):
i = 2
factors = []
count = 0 #added to test optimization
while i * i <= n:
count += 1 #added to test optimization
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors, f'count: {count}' #print with (count added)
TESTING, (note, I added a COUNT in each loop to test the optimization)
# >>> prime_factors2(600851475143)
# ([71, 839, 1471, 6857], 'count: 1472')
# >>> prime_factors(600851475143)
# (600851475143, [71, 839, 1471, 6857], 'count optimization: 494')
I figure this code could be modified easily to get the (largest factor) or whatever else is needed. I'm open to any questions, my goal is to improve this much more as well for larger primes and factors.
In case you want to use numpy here's a way to create an array of all primes not greater than n:
[ i for i in np.arange(2,n+1) if 0 not in np.array([i] * (i-2) ) % np.arange(2,i)]
Check this out, it might help you a bit in your understanding.
#program to find the prime factors of a given number
import sympy as smp
try:
number = int(input('Enter a number : '))
except(ValueError) :
print('Please enter an integer !')
num = number
prime_factors = []
if smp.isprime(number) :
prime_factors.append(number)
else :
for i in range(2, int(number/2) + 1) :
"""while figuring out prime factors of a given number, n
keep in mind that a number can itself be prime or if not,
then all its prime factors will be less than or equal to its int(n/2 + 1)"""
if smp.isprime(i) and number % i == 0 :
while(number % i == 0) :
prime_factors.append(i)
number = number / i
print('prime factors of ' + str(num) + ' - ')
for i in prime_factors :
print(i, end = ' ')
This is my python code:
it has a fast check for primes and checks from highest to lowest the prime factors.
You have to stop if no new numbers came out. (Any ideas on this?)
import math
def is_prime_v3(n):
""" Return 'true' if n is a prime number, 'False' otherwise """
if n == 1:
return False
if n > 2 and n % 2 == 0:
return False
max_divisor = math.floor(math.sqrt(n))
for d in range(3, 1 + max_divisor, 2):
if n % d == 0:
return False
return True
number = <Number>
for i in range(1,math.floor(number/2)):
if is_prime_v3(i):
if number % i == 0:
print("Found: {} with factor {}".format(number / i, i))
The answer for the initial question arrives in a fraction of a second.
Below are two ways to generate prime factors of given number efficiently:
from math import sqrt
def prime_factors(num):
'''
This function collectes all prime factors of given number and prints them.
'''
prime_factors_list = []
while num % 2 == 0:
prime_factors_list.append(2)
num /= 2
for i in range(3, int(sqrt(num))+1, 2):
if num % i == 0:
prime_factors_list.append(i)
num /= i
if num > 2:
prime_factors_list.append(int(num))
print(sorted(prime_factors_list))
val = int(input('Enter number:'))
prime_factors(val)
def prime_factors_generator(num):
'''
This function creates a generator for prime factors of given number and generates the factors until user asks for them.
It handles StopIteration if generator exhausted.
'''
while num % 2 == 0:
yield 2
num /= 2
for i in range(3, int(sqrt(num))+1, 2):
if num % i == 0:
yield i
num /= i
if num > 2:
yield int(num)
val = int(input('Enter number:'))
prime_gen = prime_factors_generator(val)
while True:
try:
print(next(prime_gen))
except StopIteration:
print('Generator exhausted...')
break
else:
flag = input('Do you want next prime factor ? "y" or "n":')
if flag == 'y':
continue
elif flag == 'n':
break
else:
print('Please try again and enter a correct choice i.e. either y or n')
Since nobody has been trying to hack this with old nice reduce method, I'm going to take this occupation. This method isn't flexible for problems like this because it performs loop of repeated actions over array of arguments and there's no way how to interrupt this loop by default. The door open after we have implemented our own interupted reduce for interrupted loops like this:
from functools import reduce
def inner_func(func, cond, x, y):
res = func(x, y)
if not cond(res):
raise StopIteration(x, y)
return res
def ireducewhile(func, cond, iterable):
# generates intermediary results of args while reducing
iterable = iter(iterable)
x = next(iterable)
yield x
for y in iterable:
try:
x = inner_func(func, cond, x, y)
except StopIteration:
break
yield x
After that we are able to use some func that is the same as an input of standard Python reduce method. Let this func be defined in a following way:
def division(c):
num, start = c
for i in range(start, int(num**0.5)+1):
if num % i == 0:
return (num//i, i)
return None
Assuming we want to factor a number 600851475143, an expected output of this function after repeated use of this function should be this:
(600851475143, 2) -> (8462696833 -> 71), (10086647 -> 839), (6857, 1471) -> None
The first item of tuple is a number that division method takes and tries to divide by the smallest divisor starting from second item and finishing with square root of this number. If no divisor exists, None is returned.
Now we need to start with iterator defined like this:
def gener(prime):
# returns and infinite generator (600851475143, 2), 0, 0, 0...
yield (prime, 2)
while True:
yield 0
Finally, the result of looping is:
result = list(ireducewhile(lambda x,y: div(x), lambda x: x is not None, iterable=gen(600851475143)))
#result: [(600851475143, 2), (8462696833, 71), (10086647, 839), (6857, 1471)]
And outputting prime divisors can be captured by:
if len(result) == 1: output = result[0][0]
else: output = list(map(lambda x: x[1], result[1:]))+[result[-1][0]]
#output: [2, 71, 839, 1471]
Note:
In order to make it more efficient, you might like to use pregenerated primes that lies in specific range instead of all the values of this range.
You shouldn't loop till the square root of the number! It may be right some times, but not always!
Largest prime factor of 10 is 5, which is bigger than the sqrt(10) (3.16, aprox).
Largest prime factor of 33 is 11, which is bigger than the sqrt(33) (5.5,74, aprox).
You're confusing this with the propriety which states that, if a number has a prime factor bigger than its sqrt, it has to have at least another one other prime factor smaller than its sqrt. So, with you want to test if a number is prime, you only need to test till its sqrt.
def prime(n):
for i in range(2,n):
if n%i==0:
return False
return True
def primefactors():
m=int(input('enter the number:'))
for i in range(2,m):
if (prime(i)):
if m%i==0:
print(i)
return print('end of it')
primefactors()
Another way that skips even numbers after 2 is handled:
def prime_factors(n):
factors = []
d = 2
step = 1
while d*d <= n:
while n>1:
while n%d == 0:
factors.append(d)
n = n/d
d += step
step = 2
return factors
I am a beginner. My Solution below for Project Euler Problem No. 3 takes forever to return answer. Can anybody suggest improvement? what am i doing wrong? I have written pieces of codes to help me think of all the puzzle pieces of the problem.
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
#limiter identification for iteration
def limiter(x):
for i in range (2, x):
if x % i == 0:
return int (x/i)
#Prime checker
def is_prime(a):
for i in range (2, a):
if a % i == 0:
return False
return True
#Lists all factors of a given no.
def factorlist(b):
list = []
for i in range (2, limiter(b)+1):
if b % i == 0:
list.append(i)
return list
#Lists all prime factors of a given no.
def primefactor(p):
plist = []
for i in factorlist(p):
if is_prime(i)== True:
plist.append(i)
return plist
print (primefactor(600851475143)[-1])
I do not have the answer, but I will say that sometimes print() statements in the functions will prove to be useful to debug in functions. An example that your limiter() function hangs at 71. I found out running your program on my machine:
def limiter(x):
for i in range (2, x):
print(i) <------------------ I added this
if x % i == 0:
return int (x/i)
So, adding print() in your functions may help.
A recursive approach to find the largest prime factor is going to be simpler and probably much faster:
def lpf(N,p=2): # start from preceding prime factor
for f in range(p,int(N**0.5)+1): # factors up to square root
if N%f==0 : return max(f,lpf(N//f,f)) # 1st prime vs lpf of counterpart
return N # none found, N is prime itself
lpf(600851475143) # 6857
here is lot simpler and shorter code:
# Largest Prime Factor
def prime_number(b):
c1 = 0
for i in range(1,int(b/2)+1):
if b % i == 0:
c1 += 1
if c1 < 2:
return(True)
else:
return(False)
def prime_factor(a):
n = 0
count = 1
prime = 0
while n < a+1:
n += 1
if a % n == 0:
if prime_number(n) == True:
prime = n
print(prime)
prime_factor(600851475143)
This is my solution to Project Euler Problem 3. I have written this code for Project Euler but if I put in "49" i get "49". What seems to be the problem?
n = 600851475143
i = 2
while (i * i < n):
while (n % i == 0):
n = n / i
i = i + 1
print (n)
I'm assuming you meant set n = 49.
Your outer while loop checks the condition i * i < n, which is not true for i == 7, so the outer loop breaks as soon as it hits 7. Change the < to <=.
However, your code isn't correct in the first place-- perhaps something like this is what you meant?
n = 600851475143
i = 2
factors = []
while (i <= n):
while (n % i == 0):
n = n / i
factors.append(i)
i = i + 1
print factors
You're printing n you want to print i...
Probably the fastest way to solve it by finding all the prime factors and then return the max.
Brute force solution took me less then 1 sec
I'm assuming you mean n = 49.
Your code isn't right, but the error is small -- change the < to <= and it works for Project Euler #3.
The problem of the code not working for squares such as 49 still remains though. Here is a modified piece of code that should work for squares as well.
n = 49
i = 2
while i * i <= n:
while n % i == 0:
x = n
n = n / i
i = i + 1
if n == 1:
print x
else:
print n
finding factors of N only need to check upto √N.
First basic solution:-
value of N never change
find factors upto √N & store biggest factor.
from math import sqrt
ans = -1
n = input() #or directly put
for i in range(2,int(sqrt(n))+1):
while (n%i==0):
if i > ans:
ans = i
print(ans)
Little optimized solution:-
if we change value of N, it iterate less than previous method.
only need to check % (modulus) of N with primes.
if have prime numbers list, then check/iterate with that only
unless, ignoring even numbers check numbers like 9,15,21... is prime or not, is worthless so...
excluding 2 all prime is odd.
so after check N with 2, check N with only odd numbers.
find factors upto √N & store biggest factor.
when get factor, divide N until N no have that factor
find the next factor do same process, until N become 1 (no have any factors)
from math import sqrt
ans = 2
n = input() #or directly put
while (n%2 == 0):
n /= 2
i = 3
while n > 1:
while (n%i == 0):
ans = i
n /= i
i += 2
print(ans)
find prime factors and return largest obviously
from math import sqrt
def is_prime(n):
if n ==2:return True
if n<2:return False
if n%2==0:return False
for i in range(3,int(sqrt(n))+1,2):
if n%i == 0:
return False;
return True;
n = 600851475143
i = n
while(i>1):
if is_prime(i) and is_prime(i) and n%i==0:
print(i);
break
i = i-1;
Your code is written assuming there are more than one factor, but in the case of n=49, it turns out that it has only one factor that is 7. So you can add a line checking whether it has more than one factor, or if not then it should be printed.