Question:
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
And I tried to solve it by using the below snippet, its taking long time to run. Is there any best way to solve this problem?
def find_prime(num, ranger):
count = 0
prime = True
for i in range(2, num):
if num % i == 0:
count = count + 1
if count > 1:
prime = False
return prime
return prime
prime_list = []
target = 600851475143
for i in range(2,target ):
if target % i == 0:
print(i)
if find_prime(i,target) and i % 2 == 0:
prime_list.append(i)
print(prime_list)
print(max(prime_list))
Please suggest.
n=600851475143
d=2
while d < n/d:
if n%d==0:
n/=d
else:
d+=1
print n
Related
I have written a simple code to print the largest prime factor of a given number from Project Euler. It works just fine for numbers like 24, but there is no response from the python shell for large numbers!
a=600851475143
b=[]
for i in range(2,600851475143):
if a%i==0:
if i==2:
b.append(i)
continue
for j in range(2,i+1):
if j==i:
b.append(i)
if i%j==0:
break
print(max(b))
print(b)
You can use an algorithm like this to get large factors of prime numbers:
import math
def LLL(N):
p = 1<<N.bit_length()-1
if N == 2:
return 2
if N == 3:
return 3
s = 4
M = pow(p, 2) - 1
for x in range (1, 100000):
s = (((s * N ) - 2 )) % M
xx = [math.gcd(s, N)] + [math.gcd(s*p+x,N) for x in range(7)] + [math.gcd(s*p-x,N) for x in range(1,7)]
try:
prime = min(list(filter(lambda x: x not in set([1]),xx)))
except:
prime = 1
if prime == 1:
continue
else:
break
return N/prime
sum_ans=17
for i in range(11,2000000):
for j in range(2,int(i**0.5)):
if i%j==0:
break
else:
sum_ans+=i
print(sum_ans)
The code i have return gives answer 143064094781 and the correct answer is 142913828922 but i can not figure out where i have gone wrong. So can any one help me.
Range's stop parameter is exclusive. This means that your code is only calculating j from 2 to 1 less than i**0.5. To fix this you can add 1, meaning that your end code will look a little like this, providing the correct output of 142913828922:
sum_ans=17
for i in range(11,2000000):
for j in range(2,int(i**0.5+1)):
if i%j==0:
break
else:
sum_ans+=i
print(sum_ans)
What about about this:
def isPrime(x):
prime = True
for i in range(2, x):
if x % i == 0:
prime = False
break
else:
continue
return prime
primes = (a for a in range(2, 2000000) if isPrime(a))
print(sum(primes))
# output
142913828922
This code will take a few minutes to execute. Buckle up!!
def sumPrimes(n):
sum =0
for i in range(2 ,n):
for j in range(2, int(i / 2) +1):
if (i % j) == 0:
break
else:
sum += i
return sum
print(sumPrimes(2000000))
If you iterating over all numbers below 2 million, you might as well sieve them using the Sieve of Eratoshenes:
from math import ceil, sqrt
def sieve(n):
nums = list(range(2,n+1))
primes = []
p = 2
k = 1+ ceil(sqrt(n))
while p < k:
primes.append(p)
nums = [num for num in nums[1:] if num % p > 0]
p = nums[0]
return primes + nums
Then sum(sieve(2000000)) evaluates to 142913828922. It takes about 5 seconds to run on my machine.
I am writing a small program to find all prime number from a range from 0 to whatever number the user want. How should I fix my code?
I tried to swap the loop position and the algorithm but it didn't work
g=int(input("type"))
def check_prime(g):
if(g<2):
return False
for x in range(2, g+1):
v=int(math.sqrt(x))
if(x%v==0):
return False
return True
if check_prime(g):
print(x)
I expect when the user input g they will get all the prime numbers from 0 to g and also the program will tell them how many prime numbers found
It's pretty simple, this code below will return you both the count of primes and their values.
g=int(input("type"))
def check_prime(g):
count = 0
primes = []
for number in range(g+1): #considering 'g' is inclusive
primes.append(number)
count += 1
if(number < 2 or number%2 == 0):
primes = primes[:-1]
count -= 1
continue
for x in range(3, number, 2):
if (number % x) == 0:
primes = primes[:-1]
count -= 1
break
return count, primes
print (check_prime(g))
I tried to find the prime factor for a number by the following code:
def pf(n):
f=[]
while n != 1:
for i in range(2, n+1):
if n%i == 0:
f.append(i)
n //=i
return f
The output for pf(8) is [2, 4], rather than [2,2,2] as my expect.
The output for pf(16) is [2,4,2], rather than [2,2,2,2].
can anyone help me to figure out what's wrong with my code?
If you break inside the loop, you will get what you want :
def pf(n):
f=[]
while n != 1:
for i in range(2, n+1):
if n%i == 0:
f.append(i)
n //=i
break
return f
Why was it wrong ?
in you first for loop, 8 is can be divied by 2, so n is equal to 4. And when i is then equals to 4, then 4 is added as a prime factor
Other implementation:
I prefer this implementation :)
def pf(n):
f=[]
i = 2
while n != 1:
if n%i == 0:
f.append(i)
n //=i
else:
i += 1
return f
This implementation avoids you to recompute the numbers in range you already know is not dividable by.
Im a newbie at python and i have a task. Given a number as input, i have to print the prime that belongs in the number/position on a list of primes starting from position 1 and not 0, until the input is 'END'. For example, if the input is 1, output should be the first prime which is 2, if the input is 5, output should be the 5th prime which is 11 and so. It works fine but after 3/4-digit numbers the output has a delay until i get the Error: Time limit exceeded. How can i make it run faster? Here's the code:
def primes_function(n):
primes = []
num = 2
while len(primes) <= n:
x = num // 2
while x > 1:
if num % x == 0:
break
x -= 1
else:
primes.append(num)
num += 1
return primes[n - 1]
#main
while True:
n = input()
if n == 'END':
break
elif n > '0':
n = int(n)
value = primes_function(n)
print(value)
Sorry if i made any mistakes in the description
enter image description here
I combined this answer (1) and this answer (2) to speed up the function. The two key ideas are: When testing primality of a candidate ...
... do not divide by every number (2, 3, 4, 5, 6, ...) but only by the preceding prime numbers (2, 3, 5, ...). Every non-prime number > 2 is has to have some prime factor.
... divide only by numbers that are ≤ sqrt(candidate).
import math
def nth_prime(n):
prime_list = [2]
candidate = 3
while len(prime_list) < n:
max_factor = math.sqrt(candidate)
is_prime = True
for p in prime_list:
if p > max_factor:
break
elif candidate % p == 0:
is_prime = False
break
if is_prime:
prime_list.append(candidate)
candidate += 2
return prime_list[-1]
Benchmark of different solutions:
n=9000 n=15000 n=25000 n=75000
your solution 1m38.455s - - -
linked answer (1) 0m 2.954s 8.291s 22.482s -
linked answer (2) 0m 0.352s 0.776s 1.685s 9.567s
this answer 0m 0.120s 0.228s 0.410s 1.857s
Brij's answer 0m 0.315s 0.340s 0.317s 0.318s
For every n the programs where started from scratch.
As we can see, Brij's Sieve Of Eratosthenes takes a fairly low constant amount of time. If you want to find big prime numbers below a fixed limit then that's the best solution (here n < 78499, as the 78499-th prime number is 1 000 003 which is bigger than the sieve list).
If you also want to find a lot of smaller or medium sized prime numbers or cannot accept a fixed limit then go with this solution.
def SieveOfEratosthenes():
n = 1000000
prime = [True for i in range(n+1)]
p = 2
count = 0
while (p * p <= n):
if (prime[p] == True):
count = count + 1
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
seive = []
for p in range(2, n):
if prime[p]:
seive.append(p)
return seive
def primes_function(n , seive):
return seive[n - 1]
#main
seive = SieveOfEratosthenes()
while True:
n = input()
if n == 'END':
break
elif n > '0':
n = int(n)
value = primes_function(n,seive)
print(value)
Full working : https://ide.geeksforgeeks.org/QTSGQfhFV3
I have precomputed the primes below 10^6 and made a list of primes and accessed the nth prime number by the index.