multiplicative persistance

multiplicative_persistance.py

# time everything precisely
from time import perf_counter_ns
from multiprocessing import Pool
from functools import partial

def persistance(n, minimum=10, sequence=False):
    if sequence:
        print(f"Started with {n}")
    depth = persinstence_rec(n, sequence=sequence)
    # only print numbers with a persistance greater than 10
    if depth >= minimum:
        print(f"Found {n} with persistence of {depth}")
    else:
        # uncomment if you want to print every depth
        # print(f"Found {n} with persistence of {depth}")
        pass


def persinstence_rec(n, depth=0, sequence=False):
    # convert number to int at the start, we generate it as a string
    if depth == 0:
        n = int(n)
    # number is single digit, we are done
    if n < 10:
        return depth
    # convert it to string to be able to loop over the digits
    number_as_string = str(n)
    result = 1
    for x in number_as_string:
        result *= int(x)
    # print current result if we want to see each step
    if sequence:
        print(result)
    return persinstence_rec(result, depth=depth + 1, sequence=sequence)


def create_testset(n, with_fives=False):
    # create the set of numbers we want to check
    # the numbers we need to check will start with (blank), 2,3, 34 or 4
    # if you had 24, you could just use an 8 to get a smaller number
    # same with 23
    # then followed by sixes, sevens, eights and nines
    # zeros would lead to a persistance of 1
    # ones dont make a difference in persistance but make the number bigger
    # so we dont use ones.
    # for the 'with_fives' setting we can remove all equal numbers because that would lead to a 0
    start_time = perf_counter_ns()
    if with_fives:
        # remove even numbers in testset with fives
        testset = {
            f"{start}" + f"{five}" + f"{seven}" + f"{nine}"
            for start in ["", "3"]
            for five in ["5" * x for x in range(1, n)] # <- always have one 5
            for seven in ["7" * x for x in range(n)]
            for nine in ["9" * x for x in range(n)]
        }
    else:
        testset = {
            f"{start}" + f"{six}" + f"{seven}" + f"{eight}" + f"{nine}"
            for start in ["", "2", "3", "34", "4"]
            for six in ["6" * x for x in range(n)]
            for seven in ["7" * x for x in range(n)]
            for eight in ["8" * x for x in range(n)]
            for nine in ["9" * x for x in range(n)]
        }
    # remove 'empty' number
    testset.discard("")
    total_time = (perf_counter_ns() - start_time) / 1000000000
    print(f"testset created in {total_time} sec")
    return testset


def find_numbers(n, minimum=0, with_fives=False):
    # get partial function for pool.map
    partial_persistance = partial(persistance, minimum=minimum)
    # create testset with up to n sevens, eights, and/or nines
    numbers = create_testset(n, with_fives)
    # check how many numbers we have in the testset
    amount = len(numbers)
    print("numbers: length", amount)
    # get the longest number
    maxlen = max(len(s) for s in numbers)
    print("numbers: longest string", maxlen)
    # start timer
    gen = (x for x in numbers)
    print("start search")
    start_time = perf_counter_ns()
    # calculate the persistance of each number
    # i have a quadcore, so i choose 4 processes
    with Pool(processes=4) as pool:
        pool.map(partial_persistance, gen)
    # for number in numbers:
    #     partial_persistance(number)
    total_time = (perf_counter_ns() - start_time) / 1000000000
    # print(f"complete in {round(total_time,3)} sec ({round(amount/total_time,3)} checked numbers per sec)")
    print(f"complete in {round(total_time,3)} sec")


if __name__ == "__main__":
    # check single numbers and print they persistance with sequence=True
    # persistance(277777788888899, sequence=True)
    # persistance(27777899, sequence=True)
    # check all numbers we create in testset with up to n fives, sixes, sevens, eights and/or nines
    # use 'with_fives=True' to create numbers with fives
    # print only if persistance higher than 'minimum'
    # find_numbers(100, minimum=5, with_fives=True)  # takes ~35 sec
    find_numbers(30, minimum=10)