Persistent Bugger

Hello there. I was digging in Codewars and I found a kata that piqued my curiosity:

Write a function, persistence, that takes in a positive parameter num and returns its multiplicative persistence, which is the number of times you must multiply the digits in num until you reach a single digit. For example:

persistence(39) # returns 3, because 3*9=27, 2*7=14, 1*4=4
                # and 4 has only one digit
                                            
persistence(999) # returns 4, because 9*9*9=729, 7*2*9=126,
                 # 1*2*6=12, and finally 1*2=2`

persistence(4)   # returns 0, because 4 is already a one-digit number

If it is not fully understandable, perhaps it is better to clarify first what multiplicative persistence is.

Persistence of a number

From Wikipedia:

In mathematics, the persistence of a number is the number of times one must apply a given operation to an integer before reaching a fixed point at which the operation no longer alters the number. For example: the additive persistence of 2718 is 2: first we find that 2 + 7 + 1 + 8 = 18, and then that 1 + 8 = 9. The multiplicative persistence of 39 is 3, because it takes three steps to reduce 39 to a single digit: 39 → 27 → 14 → 4. Also, 39 is the smallest number of multiplicative persistence 3.

In our case, we have to consider not the sum, but the product of the digits.

My solution

Here’s how I solved the problem, in Python:

import math

def persistence(n):
                    
    mp = 0

    while(n >= 10):
        prod_of_digits = math.prod(int(digit) for digit in str(n))
        n = prod_of_digits
        mp += 1
    
    return mp

I declared a function that takes the number n as input. About the while loop: as long as the number n always has more than two digits, these are multiplied by each other. The counter mp (Multiplicative Persistence) increases and the loop renews n with the number obtained. When the fixed point is reached, then the function will return mp.

Tested with numerous scenarios and it works!

Hope you found it interesting. Thanks for reading and SYL with the next solution.