Given a number n, determine what the nth prime is.
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
If your language provides methods in the standard library to deal with prime numbers, pretend they don't exist and implement them yourself.
I tried my best to have something as optimal as I could, any comment is welcome :).
I focused on speed and sacrificed memory (memoization of previously computed primes).
This solution works well even for incremental requests, for example both: [nth_prime(n) for n in range(1, 1001)]
and nth_prime(1000) have very similar executions as they will both call has_factor_in the same
number of times. Execution times for these two on my machine are the following:
def test():
nth_prime = prime_finder()
nth_prime(1000)
%timeit test()
# 9.72 ms ± 84.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)def test():
nth_prime = prime_finder()
[nth_prime(n) for n in range(1, 1001)]
%timeit test()
#10.5 ms ± 124 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)Any subsequent request for a smaller n would be served in constant time (reading in the primes dictionary).
The inner algorithm for finding one prime, is actually quite simple, it iterates through the odd numbers until enough primes are found.
For each odd number x it checks if x has any factor amongst the smaller primes, if it
doesn't have one, then by definition x prime number: every non-prime (composite) number
can be decomposed in prime factors.
I simply use the fact that one of these factors is necessarily smaller than the number itself.
More precisely, inside has_factor_in there is one optimization (lines 27-28) that uses the
fact that at least one prime factor of x is smaller than sqrt(x).
Proof Suppose that x is a composite number for which the smallest prime factor is p1 > sqrt(x).
Since x has several factors, it has at least one factor p2 > p1. Since both p1 and p2 are
greater than sqrt(x) the factorisation p1*p2*... is greater than x, a contradiction. QED.
As always the code is here.
Sometimes it is necessary to raise an exception. When you do this, you should include a meaningful error message to indicate what the source of the error is. This makes your code more readable and helps significantly with debugging. Not every exercise will require you to raise an exception, but for those that do, the tests will only pass if you include a message.
To raise a message with an exception, just write it as an argument to the exception type. For example, instead of
raise Exception, you should write:
raise Exception("Meaningful message indicating the source of the error")To run the tests, run the appropriate command below (why they are different):
- Python 2.7:
py.test nth_prime_test.py - Python 3.4+:
pytest nth_prime_test.py
Alternatively, you can tell Python to run the pytest module (allowing the same command to be used regardless of Python version):
python -m pytest nth_prime_test.py
-v: enable verbose output-x: stop running tests on first failure--ff: run failures from previous test before running other test cases
For other options, see python -m pytest -h
Note that, when trying to submit an exercise, make sure the solution is in the $EXERCISM_WORKSPACE/python/nth-prime directory.
You can find your Exercism workspace by running exercism debug and looking for the line that starts with Workspace.
For more detailed information about running tests, code style and linting, please see Running the Tests.
A variation on Problem 7 at Project Euler http://projecteuler.net/problem=7
It's possible to submit an incomplete solution so you can see how others have completed the exercise.