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I am trying to write a script in python to find the 1000th prime number. I don't understand why this isn't working here. Basically while the mod is less than the square root of the number and still has a remainder, the mod goes up by one. This should continue until the mod equals the square root of the number. Then the check should remain at 0 and the number should be prime. Every time I try to run the script it tells me theres a system error.

import math
b=2
count=2
next_odd=3
next_prime=1
check = 0

while count<=10:
    while b<float(math.sqrt(next_odd)):
        if next_odd%b>0:
                b+=1
        if next_odd%b == 0:
                check+=1
    if check > 0:
        next_prime=next_odd
        next_odd+=2
        print(next_prime)
        b=2
        count+=1`
Jahanara Nares
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2 Answers2

2

I understand what you are trying to do, but unfortunately there were too many things wrong with your program. Here is a working version. I made minimal changes. Hopefully you can compare the below version with your own and see where you went wrong.

import math

count=2
next_odd=3
next_prime=1

while count<=1000:
    b=1
    check = 0
    while b<float(math.sqrt(next_odd)):
        b+=1
        if next_odd%b == 0:
            check+=1
    if check == 0:
        next_prime=next_odd
        print(next_prime)
        count+=1
    next_odd+=2

With the above program, 1000th prime can be successfully determined to be 7919.

wookie919
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2

(first, I assume the tick on the end of your code is a typo in your stack overflow post, not the code itself)

Consider what happens when next_odd is prime. This block:

while b<float(math.sqrt(next_odd)):
    if next_odd%b>0:
            b+=1
    if next_odd%b == 0:
            check+=1

will increment b up until the square root of next_odd without ever incrementing check. That means that if check > 0: won't pass, and thus count never increments, and you then you just spin around in the while count<=10:, skipping both if blocks because their conditions are false.

In other words, you don't actually say what to do when next_odd is prime. This is also an example of why while shouldn't really be used when all you want to do is increment through numbers (which is what you're using it for here). Try something like this:

max_num = 10000 # or whatever
for odd in range(3, max_num, 2):
    factor_count = 0
    for factor in range(2, math.floor(math.sqrt(max_num)) + 1):
        if odd % factor == 0:
            factor_count += 1
    if factor_count == 0:
        print(odd)

A couple points about this code:

  • There's no (non-constant) variables in the global scope. That makes it much easier to reason about how the script's state changes over time.
  • The use of for-loops over while-loops guarantees that our script won't get caught in an infinite loop due to an erroneous (or unaccounted for) condition.
  • The use of for-loops means that we don't have to worry about incrementing all of the variables ourselves, which dramatically reduces the amount of state that we have to manage.

Hope that helps!

Oh, note that there are much more efficient ways to compute primes also. See https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

Bryan Head
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