<< problem 266 - Pseudo Square Root | Counting numbers with at least four distinct ... - problem 268 >> |
Problem 267: Billionaire
(see projecteuler.net/problem=267)
You are given a unique investment opportunity.
Starting with £1 of capital, you can choose a fixed proportion, f, of your capital to bet on a fair coin toss repeatedly for 1000 tosses.
Your return is double your bet for heads and you lose your bet for tails.
For example, if f=1/4, for the first toss you bet £0.25, and if heads comes up you win £0.5 and so then have £1.5.
You then bet £0.375 and if the second toss is tails, you have £1.125.
Choosing f to maximize your chances of having at least £1,000,000,000 after 1,000 flips, what is the chance that you become a billionaire?
All computations are assumed to be exact (no rounding), but give your answer rounded to 12 digits behind the decimal point in the form 0.abcdefghijkl.
My Algorithm
My function getMinHeads
performs a simple linear search for the minimum number of heads required to become a billionaire.
ratio
(which is f in the problem description) is slowly increased from 0 to 1 with a step = 0.0001
(works with larger values, such as 0.01
, too).
The expected amount of money is (1 + 2f)^{heads} * (1 - f)^{tails}.
After 1000 tosses, this is equivalent to (1 + 2f)^{heads} * (1 - f)^{1000 - heads}.
The inner-most for
-loop tries to find the lowest value of heads where (1 + 2f)^{heads} * (1 - f)^{1000 - heads} >= 10^9.
My function probability
is simple Dynamic Programming code that returns the probability of at least minHeads
times observing a head after maxTosses
fair coin tosses.
Alternative Approaches
There are mathematical ways to figure out the results of getMinHeads
with pen and paper but I like the simple and obvious code of my solution.
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This is equivalent toecho "100 1000000" | ./267
Output:
Note: the original problem's input 1000 1000000000
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)
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 <vector>
#include <cmath>
// find minimum number of heads to become a billionaire
unsigned int getMinHeads(unsigned int tosses, double billion, double step = 0.0001)
{
unsigned int result = tosses;
// simple linear search
for (double ratio = 0; ratio < 1; ratio += step)
{
// won: gain twice the invested money
auto won = 1 + 2 * ratio;
// lost: invested money is lost
auto lost = 1 - ratio;
// try to have a billion with less heads than the total best
unsigned int heads = result;
for (; heads > 0; heads--)
{
// total amount of money
auto total = pow(won, heads) * pow(lost, tosses - heads);
// below one billion ? one step too far
if (total < billion)
{
heads++;
break;
}
}
// better than before ?
if (result > heads)
result = heads;
}
return result;
}
// return probability of tossing at least "minHeads" times head with a fair coin
// note: parameters "tosses" and "heads" are using during recursion only
double probability(unsigned int minHeads, unsigned int maxTosses, unsigned int tosses = 0, unsigned int heads = 0)
{
// enough heads ?
if (heads >= minHeads)
return 1;
// too few heads: can't achieve minHeads even if a series of heads only appears ?
if (maxTosses - tosses < minHeads - heads)
return 0;
// memoize
const double Unknown = -1;
static std::vector<double> cache(minHeads*maxTosses, Unknown);
auto id = heads * maxTosses + tosses;
if (cache[id] != Unknown)
return cache[id];
// fifty-fifty ...
auto result = 0.5 * probability(minHeads, maxTosses, tosses + 1, heads + 1) +
0.5 * probability(minHeads, maxTosses, tosses + 1, heads );
cache[id] = result;
return result;
}
int main()
{
// I wanna be a billionaire after 1000 coin tosses !
unsigned int tosses = 1000;
double money = 1000000000;
std::cin >> tosses >> money;
// determine minimum number of heads
auto minHeads = getMinHeads(tosses, money);
// compute probability for that number of heads (or more) after 1000 coin tosses
std::cout << std::fixed << std::setprecision(12)
<< probability(minHeads, tosses) << std::endl;
return 0;
}
This solution contains 13 empty lines, 17 comments and 4 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 6 MByte.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++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 11, 2017 submitted solution
July 11, 2017 added comments
Difficulty
Project Euler ranks this problem at 50% (out of 100%).
Links
projecteuler.net/thread=267 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
C# www.mathblog.dk/project-euler-267-billionaire/ (written by Kristian Edlund)
C# github.com/HaochenLiu/My-Project-Euler/blob/master/267.cs (written by Haochen Liu)
Python github.com/LaurentMazare/ProjectEuler/blob/master/e267.py (written by Laurent Mazare)
Python github.com/Meng-Gen/ProjectEuler/blob/master/267.py (written by Meng-Gen Tsai)
Python github.com/nayuki/Project-Euler-solutions/blob/master/python/p267.py (written by Nayuki)
Python github.com/smacke/project-euler/blob/master/python/267.py (written by Stephen Macke)
C++ github.com/roosephu/project-euler/blob/master/267.cpp (written by Yuping Luo)
Java github.com/nayuki/Project-Euler-solutions/blob/master/java/p267.java (written by Nayuki)
Java github.com/thrap/project-euler/blob/master/src/Java/Problem267.java (written by Magnus Solheim Thrap)
Go github.com/frrad/project-euler/blob/master/golang/Problem267.go (written by Frederick Robinson)
Mathematica github.com/steve98654/ProjectEuler/blob/master/267.nb
Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own. Maybe not all linked resources produce the correct result and/or exceed time/memory limits.
Heatmap
Please click on a problem's number to open my solution to that problem:
green | solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too | |
yellow | solutions score less than 100% at Hackerrank (but still solve the original problem easily) | |
gray | problems are already solved but I haven't published my solution yet | |
blue | solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much | |
orange | problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte | |
red | problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too | |
black | problems are solved but access to the solution is blocked for a few days until the next problem is published | |
[new] | the flashing problem is the one I solved most recently |
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I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.
My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.
Look at my progress and performance pages to get more details.
Copyright
I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.
All of my solutions can be used for any purpose and I am in no way liable for any damages caused.
You can even remove my name and claim it's yours. But then you shall burn in hell.
The problems and most of the problems' images were created by Project Euler.
Thanks for all their endless effort !!!
<< problem 266 - Pseudo Square Root | Counting numbers with at least four distinct ... - problem 268 >> |