@freemo surely it should be doable, as I proved out the result using 'bc' on the command line.

My work friend tried it in his head while doing a run the other day, and came up with a result far larger than appropriate.

Another friend solved it as a function that is nested 5 deep, and at the bottom it takes 5x as a starter.

f(f(f(f(f(5x)))))
With the function being f(i) {5/4 * i + 1}... Something like that. I think there was more to it than that, but once you found the right value for x it should all balance out. In that case I think you still had to figure out the x such that 5/4 x was equally divisible by 5. I guess that would still require some brute force testing.

@Absinthe @freemo I did it like your second friend (i.e., the nested function, checking that the result is an integer).

@zingbretsen @freemo did you post it somewhere?

@Absinthe @freemo will try to post it today!

@zingbretsen @freemo

How about something like this:

def func(x):
if x % 5 == 1:
return (x - 1) / 5 * 4
return 0

x = 6
while True:
if func(func(func(func(func(x))))):
break
x += 5

print(x)

@Absinthe @freemo I forked the repo and added my solution here: https://git.qoto.org/zingbretsen/coconuts/blob/master/coco.py

It should solve the 5 person problem when run as-is, but it could be run for any number of people. I should probably have added that as a command line argument.

It works backwards from possible ending amounts, but is somewhat selective in the numbers it tries. e.g., for the 5 person case, we know that the ending pile must be divisible by both 4 and 5, so the final pile will be a multiple of 20. In addition to that, we know that 5/4 + 1 of the ending pile must be divisible by 5, so 60 is the lowest number that would work for the first round of stepping backwards.

There is a regular pattern that seems to work for all of the numbers of people I tested, not just 5, so we can reduce the search space pretty significantly. For the 5 person problem, I only have to test 13 numbers before finding the answer.

@zingbretsen

very cool, I like how you try to limit the problem space for efficiency. Reminds me of common examples of primality checkers that use a sort of smart-brute force to reduce the problem space :)

@Absinthe

@zingbretsen @freemo

I do like the solution, and to my eye the code looks great. I could not get it to compile on :

https://rextester.com/l/python3_online_compiler

I might be doing something wrong though.

@Absinthe @freemo ah, I used f-strings for formatting which isn't available in 3.5--they're only available in 3.6+

@zingbretsen @freemo that might do it :) Yep, works on 3.7 that will teach me to try a shortucut! :)

@zingbretsen @freemo Also, I think we need to include #toyprogrammingchallenge hashtag in the conversational posts of they are lost on other instances

@zingbretsen @freemo Things that sound dirty, but aren't...

Of course, everything is dirty if you just add "...if you know what I mean..." after you say it. :D

@freemo the numbers are good, but I can't follow the screen grabs on my phone. Maybe more readable on the computer screen. Will see later.

@Absinthe They are readable here, should be good. Let me know if you have any questions on the math

@freemo
Very nice 👍
@Absinthe