Problem 323: Bitwise-OR operations on random integers

(see projecteuler.net/problem=323)

Let y_0, y_1, y_2,... be a sequence of random unsigned 32 bit integers (i.e. 0 <= y_i < 2^32, every value equally likely).

For the sequence x_i the following recursion is given:

x_0 = 0 and
x_i = x_{i-1} | y_{i-1}, for i > 0. ( | is the bitwise-OR operator)

It can be seen that eventually there will be an index N such that x_i = 2^{32}-1 (a bit-pattern of all ones) for all i >= N.

Find the expected value of N.
Give your answer rounded to 10 digits after the decimal point.

My Algorithm

I evaluate n rounds and find their likelihood of producing the final value.
The "chance of survival" of an unset bit decreases after each round so that the result eventually stabilizes.
I stop once the delta falls below my threshold 10^-11 (one digit more than needed to prevent rounding issues).

A single is zero after a certain number of rounds with probability:
(1) 0.5^{round}
And it became 1 after all those rounds:
(2) 1 - 0.5^{round}
The chance that all 32 bits are 1:
(3) (1 - 0.5^{round})^32
Therefore the chance that at least one out of 32 bits is still zero:
(4) 1 - (1 - 0.5^{round})^32

I keep adding all values of equation (4) until the value of (4) drops below 10^-11.
It takes a little more than 40 iterations to find the correct result.

Note

Actually, my first step was writing a simple Monte-Carlo simulation to get an approximation of the result.
Of course it's impossible to find all 10 digits after the decimal point (unless you invest a huge amount of time).
The code can still be found in montecarlo (and its helper function myrand).

It's worth mentioning that the results of my Monte-Carlo simulation differed significantly from their true values.
I spent more than an hour trying to find a bug in my "mathematical" approach because I trusted the Monte-Carlo result.
As it turned out, myrand was flawed because the simple LCG algorithm had bad parameters.
When I choose different parameters (from en.wikipedia.org/wiki/Linear_congruential_generator) then everything turned out to be fine.

My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#include <iostream>
#include <iomanip>
#include <cmath>
 
// 32 bits
unsigned int maxBits = 32;
 
// ---------- first approach: find a good estimation with Monte-Carlo simulation ----------
 
// a simple pseudo-random number generator
// (produces the same result no matter what compiler you have - unlike rand() from math.h)
unsigned int myrand()
{
static unsigned long long seed = 0;
seed = 6364136223846793005ULL * seed + 1;
return seed >> 30;
}
 
// randomized simulation
double montecarlo(unsigned int iterations)
{
// all 32 bits are set
const unsigned int allBits = (1ULL << maxBits) - 1;
 
unsigned int numSteps = 0;
for (unsigned int i = 0; i < iterations; i++)
{
// start with zero
unsigned int current = 0;
// and repeat until all bits are set
do
{
// and some random bits
current |= myrand() & allBits;
numSteps++;
} while (current != allBits);
}
 
// return average number of steps until "completion"
return numSteps / double(iterations);
}
 
// ---------- final approach: solve "smarter" ! ----------
 
int main()
{
std::cin >> maxBits;
 
// number of accurate digits
const unsigned int digits = 10;
// 0.00000000001 => one more zero ensures that the last digit is correct even after rounding
const double Epsilon = pow(10.0, -(double)(digits + 1));
 
// at least one round
double result = 0;
unsigned int round = 0;
 
// until change is smaller than Epsilon
while (true)
{
// chance of "survival" for a single zero after all those rounds
auto hasZero = pow(0.5, round);
// were all 32 bits flipped from 0 to 1 ?
auto isDone = pow(1 - hasZero, maxBits);
 
// remaining
auto delta = 1 - isDone;
// enough digits ?
if (delta < Epsilon)
break;
 
// next iteration
result += delta;
round++; // stop after 42 rounds
}
 
// display result
std::cout << std::fixed << std::setprecision(digits)
<< result << std::endl;
 
// run Monte-Carlo simulation
//while (true)
// std::cout << montecarlo(100000000) << std::endl;
 
return 0;
}

This solution contains 17 empty lines, 24 comments and 3 preprocessor commands.

Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Input data (separated by spaces or newlines):

This is equivalent to
echo 8 | ./323

Output:

(please click 'Go !')

Note: the original problem's input 32 cannot be entered
because just copying results is a soft skill reserved for idiots.

(this interactive test is still under development, computations will be aborted after one second)

Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=c++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

Changelog

July 31, 2017 submitted solution
July 31, 2017 added comments

Difficulty

20% Project Euler ranks this problem at 20% (out of 100%).

Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
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The 235 solved problems (level 9) had an average difficulty of 29.1% at Project Euler and
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