Hi, I got a homework and I have to ask for some help.
My task is:
Listing the first six prime numbers, we get 2, 3, 5, 7, 11, and 13, where 13 is prime number 6. Write a program that defines the prime number N. Create an effective implementation that can calculate 10,000 prime numbers in real time.
Please somebody help me.
Thanks!
Please show what you have tried so far.
We're here to help, but not do the work for you.
My recent program can only list the prime numbers in an interval.
number = int(input("How many prime numbers do you want ? "))
cnt = 0
for i in range (1, number+1):
for j in range(1, i+1):
if(i%j == 0):
cnt += 1
if(cnt == 2):
print(i)
cnt = 0
Write a function, which checks a number, if it's a prime.
In another function you collect primes, until you have the number of primes you want to have.
from itertools import count
def is_prime(number):
"""Your implementation"""
return True
def get_primes(n):
primes = []
while len(primes) < n:
for number in count():
if is_prime(number):
primes.append(number)
return primesThe function to check the prime can also taken from other libraries.
If you make the easiest implementation, it takes more time to check primes.
I wrote this code, but i don't know why is it exit with exit code 0.
from itertools import count
def is_prime(number):
if number <= 1:
return False
for i in range(2, number):
if number % i == 0:
return False;
return True
def get_primes(n):
primes = []
while len(primes) < n:
for number in count():
if is_prime(number):
primes.append(number)
return primes
Okay, I realized I didn't printed it last time, but if I print it like this, it is just running and running and never stops.
Can you help me what did I wrong?
from itertools import count
def is_prime(number):
if number <= 1:
return False
for i in range(2, number):
if number % i == 0:
return False;
return True
def get_primes(n):
primes = []
while len(primes) < n:
for number in count():
if is_prime(number):
primes.append(number)
return primes
print(get_primes(1))
from itertools import count will keep returning values forever.
for number in count(): has no way of ending
while len(primes) < n: only happens once
The following code Removes the while loop and gives the for loop a way out.
from itertools import count
def is_prime(number):
if number <= 1:
return False
for i in range(2, number):
if number % i == 0:
return False;
return True
def get_primes(n):
primes = []
for number in count():
if is_prime(number):
primes.append(number)
if len(primes) == n:
return primes
print(get_primes(1))Output:
[2]
Thank you so much, this is exactly what i was whinking for!
Can be improved as it only needs to check len of found primes after appending a new prime, no need to check otherwise.
from itertools import count
def is_prime(number):
if number <= 1:
return False
for i in range(2, number):
if number % i == 0:
return False;
return True
def get_primes(n):
primes = []
for number in count():
if is_prime(number):
primes.append(number)
if len(primes) == n:
return primes
print(get_primes(6))Output:
[2, 3, 5, 7, 11, 13]
More efficient a second count to keep count
from itertools import count
def is_prime(number):
if number <= 1:
return False
for i in range(2, number):
if number % i == 0:
return False;
return True
def get_primes(n):
primes = []
qty_found = count(1)
for number in count():
if is_prime(number):
primes.append(number)
if next(qty_found) == n:
return primes
print(get_primes(6))Output:
[2, 3, 5, 7, 11, 13]
Yes it is better. My task is to list 10,000 prime numbers in real time, this code can list it in 30 seconds,
but i think it's real time and enough for me.