HackerRank: Project Euler 03

euler03.py

"""
HR: https://www.hackerrank.com/contests/projecteuler/challenges/euler003/problem
PE: https://projecteuler.net/problem=3
Discussion: https://www.hackerrank.com/contests/projecteuler/challenges/euler003/forum/comments/1195942

Some hints:

* Search backwards from what you think the largest prime factor might be.
* Start with the base case:
 * If `n` is even, then you know the largest prime factor will be `<= n/2`.
 * If it's odd, check against some other smaller primes that divide `n`.
 * If nothing is found, subtract 2 from `n` and try again.

Furthermore, you can keep dividing by 2 (for example) until you get an odd
number. This means you can *reduce* `n` successively by 2 and its other smallest
prime factors, greatly reducing the search space.

Using 30 as an example, whose prime factors are 2, 3, and 5:

* 30 / 2 = 15
 * 15 / 3 = 5 <-- largest prime factor
* 30 / 3 = 10
 * 10 / 2 = 5 <--

The link below is a (Python) solution that does this while also retrying the
prime factor reduction using newly found prime numbers.

[Click here for the solution code.](https://gist.github.com/sitbon/833c0a523b8839ff8a4576f64df239f3)

It's very fast for numbers in the test case range (`10^12`), in line with
or even slightly faster than `max(sympy.primefactors(n))`.
Performance starts to degrade as the prime factors get very big, around `n = 10**50-1`.
I split the prime search which helps from there, but there are better ways to check for primality
that can be dropped in.

Here are some test cases to try for yourself before checking out the code.

```
--> Test Case 1
8
45182
28
70
359
71
77
977
77777777

--> Output:
41
7
7
359
71
11
977
137


--> Test Case 2
10
999999999997
99999999993
9999999991
999999999993
999999999991
99999999997
999999999745
9999999997
999999999989
999999999983

--> Output:
181587071
108577633
100003
211371803
1000003
5882352941
199999999949
769230769
999999999989
4366812227
```
"""
import math
from functools import cache

known_primes = {2, 3, 5, 7, 11, 13, 17}


@cache
def isprime(p):
    if p & 1 == 0 or p < 2:
        return False

    if p in known_primes:  # Needed for initial (non-cached) calls
        return True

    for kp in filter(lambda kpf: kpf <= p, known_primes):
        if p % kp == 0:
            return False

    du = d_min = 3
    d_max = int(math.sqrt(p))

    if d_max & 1 == 0:
        d_max -= 1

    d_mid = d_max // 2
    dd = d_max

    while du <= d_mid:
        if p % du == 0:
            if du not in known_primes:
                isprime(du)
            return False
        du += 2

        if p % dd == 0:
            if dd not in known_primes:
                isprime(dd)
            return False
        dd -= 2

    known_primes.add(p)
    return True


def reduce(nn):
    while 1:
        nn0 = nn

        if nn in known_primes:
            return nn

        for pp in filter(lambda ppf: ppf <= nn, set(known_primes)):
            nni = nn

            while nn != pp and nn % pp == 0:
                nn //= pp

            if nn == pp or isprime(nn):
                return nn

            if nn != nni:
                break

        if nn == nn0:
            break

    return nn


def largest_prime_factor(n):
    nrr = nr = n

    while 1:
        nrn = reduce(nr)

        if nrn != nr:
            nr = nrn

            if nrn < nrr:
                nrr = nrn

        while not isprime(nrr):
            nrr -= 2

        if n % nrr == 0:
            return nrr

        nrr -= 2


for _ in range(int(input().strip())):
    ni = int(input().strip())
    print(largest_prime_factor(ni))

# # Interactive version for testing:
#
# try:
#     while 1:
#         print(largest_prime_factor(int(input().strip())))
# except KeyboardInterrupt:
#     pass