Finding the nth prime using Python

Question

When I run this code even to just count to the 10th prime number (instead of 1000), I get a skewed / convex result - all the "not simple" headers for my is_composite variable, my test_num gives me prime and compound numbers , and my main_component is disabled

Some of the answers that developers use shared functions and mathematical import are something that we have not yet considered. I am not trying to get the most effective answer; I'm just trying to write workable python code to understand the basics of a loop.

---

  # test a prime by diving number by previous sequence of number(s) (% == 0).  Do this by
  # counting up from 1 all the way to 1000.

test_num = 2 #these are the numbers that are being tested for primality
is_composite = 'not prime' # will be counted by prime_count
prime_count = 0 #count the number of primes


while (prime_count<10): #counts number primes and make sures that loop stops after the 1000th prime (here: I am just running it to the tenth for quick testing)


 test_num = test_num + 1   # starts with two, tested for primality and counted if so
 x = test_num - 1  #denominator for prime equation

 while (x>2):   
  if test_num%(x) == 0:
   is_composite = 'not prime'
  else: 
   prime_count = prime_count + 1 
  x = x - 1 


  print is_composite
  print test_num
  print prime_count 

Answer 1

. MIT . :

• () a > 1

• ,

  3,1. - , > 1 0 . , , a% b a b.

  3,2. , - , , , , ?

• , , , ,

• , . , (2).

:

def primes(n):
    # http://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
    """ Returns  a list of primes < n """
    sieve = [True] * n
    for i in xrange(3,int(n**0.5)+1,2):
        if sieve[i]:
            sieve[i*i::2*i]=[False]*((n-i*i-1)/(2*i)+1)
    return [2] + [i for i in xrange(3,n,2) if sieve[i]]

Answer 2

, , , , . . , , ( , 2 ), :

def isprime(test):
    if test == 2: return True
    if test < 2 or test % 2 == 0: return False
    return not any(test % i == 0 for i in range(3, int(sqrt(test)) + 1, 2))

, , 2 , , , . 1000 , isprime.

, , . , .

: , " / " - , , . , , , , , , , .

from math import sqrt

def isprime(test):
    if test == 2: return True
    if test < 2 or test % 2 == 0: return False
    return not any(test % i == 0 for i in range(3, int(sqrt(test)) + 1, 2))

test_num = 2
prime_count = 1

while (prime_count< 1000): 

 test_num = test_num + 1  

 if (isprime(test_num)):
     prime_count += 1

print test_num

Answer 3

, ++. .

// This code was written by Mert Ener
#include <time.h>
#include <vector>
#include <iostream>

private: System::Void button1_Click_1(System::Object^  sender, 
                                          System::EventArgs^  e) { 
    using namespace std;
    UInt64 cloc = clock();
    long m = 1;
    long k = long::Parse(textBox1->Text)-2;   // The text you entered 
    std::vector<long> p(1,2);                 //   for the nth prime
    for( long n = 0; n <= k; n++ ) {
        m += 2;
        for( long l = 1; l <= n; l++ ) {
            if (m % p[l] == 0) {
                m += 2;
                l=0;}}
        p.push_back(m);}
    textBox2->Text = p[k+1].ToString(); // The textbox for the result.
    MessageBox::Show("It took me " + (clock() - cloc).ToString() 
                     + " milliseconds to find your prime.");}

Answer 4

1 . , primes < 1 . 10-12 .

import math
from itertools import islice
# number of primes below the square root of x
# works well when x is large (x > 20 and much larger)
numchecks = lambda x: int((math.sqrt(x))/(math.log(math.sqrt(x)) - 1.084)) + 1

primes = [2,3,5]
primes = primes + [x for x in range(7, 48, 2) if all((x%y for y in islice( primes, 1, int(math.sqrt(x)) )))]
primes = primes + [x for x in range(49, 2400, 2) if all((x%y for y in islice( primes, 1, numchecks(x) )))]
primes = primes + [x for x in range(2401, 1000000, 2) if all((x%y for y in islice( primes, 1, numchecks(x) )))]

, , ( ).

In your code, you can check if "test_num" is simple using the following ...

test_num = 23527631
if test_num<100:
    checks = int(math.sqrt(test_num))
else:
    checks = numchecks(test_num)

isPrime = all(test_num%x for x in islice(primes, 0, checks))
print 'The number is', 'prime' if isPrime else 'not prime'
print 'Tested in', checks, 'divisions'