Let's say you have a 20-sided die. You start rolling that die and have to roll it a few dozen times before you finally roll all 20 values. You wonder, how many rolls do I need before I get a 50% chance of seeing all 20 values? And how many rolls of an n-sided die do I need to roll before I roll all n sides?

After some research, you find out that a formula exists for calculating the chance of rolling all n values after r rolls.

P(r, n) = n! * S(r, n) / n**r

where S(a, b) denotes Stirling numbers of the second kind, the number of ways to partition a set of n objects (each roll) into k non-empty subsets (each side).

You also find the OEIS sequence, which we'll call R(n), that corresponds to the smallest r where P(r, n) is at least 50%. The challenge is to calculate the nth term of this sequence as fast as possible.

The challenge

  • Given an n, find the smallest r where P(r, n) is greater than or equal to 0.5 or 50%.
  • Your code should theoretically handle any non-negative integer n as input, but we will only be testing your code in the range of 1 <= n <= 1000000.
  • For scoring, we will be take the total time required to run R(n) on inputs 1 through 10000.
  • We will check if your solutions are correct by running our version of R(n) on your output to see if P(your_output, n) >= 0.5 and P(your_output - 1, n) < 0.5, i.e. that your output is actually the smallest r for a given n.
  • You may use any definition for S(a, b) in your solution. Wikipedia has several definitions that may be helpful here.
  • You may use built-ins in your solutions, including those that calculate S(a, b), or even those that calculate P(r, n) directly.
  • You can hardcode up to 1000 values of R(n) and a million Stirling numbers, though neither of these are hard limits, and can be changed if you can make a convincing argument for raising or lowering them.
  • You don't need to check every possible r between n and the r we're looking for, but you do need to find the smallest r and not just any r where P(r, n) >= 0.5.
  • Your program must use a language that is freely runnable on Windows 10.

The specifications of the computer that will test your solutions are i7 4790k, 8 GB RAM. Thanks to @DJMcMayhem for providing his computer for the testing. Feel free to add your own unofficial timing for reference, but the official timing will be provided later once DJ can test it.

Test cases

n       R(n)
1       1
2       2
3       5
4       7
5       10
6       13
20      67       # our 20-sided die
52      225      # how many cards from a huge uniformly random pile until we get a full deck
100     497
366     2294     # number of people for to get 366 distinct birthdays
1000    7274
2000    15934
5000    44418
10000   95768
100000  1187943
1000000 14182022

Let me know if you have any questions or suggestions. Good luck and good optimizing!

  • 1
    \$\begingroup\$ @JonathanAllan Knew I should have chosen a different wording. Thanks for the heads up. \$\endgroup\$
    – Sherlock9
    Jun 5, 2017 at 9:03

1 Answer 1


Python + NumPy, 3.95 Seconds

from __future__ import division
import numpy as np

def rolls(n):
    if n == 1:
        return 1
    r = n * (np.log(n) - np.log(np.log(2)))
    x = np.log1p(np.arange(n) / -n)
    cx = x.cumsum()
    y = cx[:-1] + cx[-2::-1] - cx[-1]
    while True:
        r0 = np.round(r)
        z = np.exp(y + r0 * x[1:])
        z[::2] *= -1
        r = r0 - (z.sum() + 0.5) / z.dot(x[1:])
        if abs(r - r0) < 0.75:
            return np.ceil(r).astype(int)

for n in [1, 2, 3, 4, 5, 6, 20, 52, 100, 366, 1000, 2000, 5000, 10000, 100000, 1000000]:
    print('R({}) = {}'.format(n, rolls(n)))

import timeit
print('Benchmark: {:.2f}s'.format(timeit.timeit(lambda: sum(map(rolls, range(1, 10001))), number=1)))

Try it online!

How it works

This uses the closed-form series for P(r, n), and its derivative with respect to r, rearranged for numerical stability and vectorization, to do a Newton’s method search for r such that P(r, n) = 0.5, rounding r to an integer before each step, until the step moves r by less than 3/4. With a good initial guess, this usually takes just one or two iterations.

xi = log(1 − i/n) = log((ni)/n)
cxi = log(n!/((ni − 1)! ⋅ ni + 1)
yi = cxi + cxni − 2cxn − 1 = log binom(n, i + 1)
zi = (-1)i + 1 ⋅ binom(n, i + 1) ⋅ ((ni − 1)/n)r
1 + ∑zi = n! ⋅ S(r, n)/nr = P(r, n)
zixi + 1 = (-1)i + 1 ⋅ binom(n, i + 1) ⋅ ((ni − 1)/n)r log((ni − 1)/n)
zixi + 1 = d/dr P(r, n)

  • 1
    \$\begingroup\$ Excellent work on the whole answer! First, I should have realized that 0.366512 was the log of something ages ago. Will use -log(log(2) in my next iteration. Second, the idea to use Newton's method is also very clever and I'm glad to see that this works so well. Third, I'm almost definitely going to steal exp(log(binom(n, i+1)) + r * log((n-i-1)/n)) :P Kudos on a great answer! :D \$\endgroup\$
    – Sherlock9
    Jun 6, 2017 at 4:10
  • 1
    \$\begingroup\$ I've added the official timing! Nice answer BTW :) \$\endgroup\$
    – DJMcMayhem
    Jun 12, 2017 at 23:24
  • 2
    \$\begingroup\$ I'm really confused. I changed the numpy import to from numpy import * and for some reason the time dropped to basically 0... Try it online? \$\endgroup\$
    – notjagan
    Jul 31, 2017 at 4:08
  • \$\begingroup\$ @notjagan cache hit maybe? \$\endgroup\$ Jul 31, 2017 at 4:42
  • 1
    \$\begingroup\$ I'd like to apologize for several things: 1) my plagiarism of your answer when I tried to find improvements; 2) not owning up to it properly and just trying to fix my answer; 3) that this apology has taken so long. I was so mortified that at first I just abandoned this challenge. In a small attempt at reparation, I suppose it fair to tell you that my main improvement on this answer was changing from Newton's method to incrementing r, as your initial approximation is already quite good. Hope to see you in PPCG once more, and sorry for everything. \$\endgroup\$
    – Sherlock9
    Oct 9, 2017 at 17:07

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