<< problem 120 - Square remainders | Efficient exponentiation - problem 122 >> |
Problem 121: Disc game prize fund
(see projecteuler.net/problem=121)
A bag contains one red disc and one blue disc. In a game of chance a player takes a disc at random and its colour is noted.
After each turn the disc is returned to the bag, an extra red disc is added, and another disc is taken at random.
The player pays £1 to play and wins if they have taken more blue discs than red discs at the end of the game.
If the game is played for four turns, the probability of a player winning is exactly 11/120, and so the maximum prize fund the banker
should allocate for winning in this game would be £10 before they would expect to incur a loss.
Note that any payout will be a whole number of pounds and also includes the original £1 paid to play the game, so in the example given the player actually wins £9.
Find the maximum prize fund that should be allocated to a single game in which fifteen turns are played.
My Algorithm
I compute all possible states after each round.
Before the first round, the player had obviously no red and no blue disc. This State
was the only possibility
.
In each round, the number of possible combinations are possibilities_n = possibilities_{n-1} * (red + blue) because the player picks any disc at random.
Several choices lead to the same next state, e.g. it doesn't matter which red disc he/she picks.
Therefore the total number of states increases by one per round but the possibilities explode in a factorial way.
The final prize is \lfloor dfrac{sum{possibilities}}{sum{possibilities(blue>red)}} \rfloor, e.g. \lfloor dfrac{120}{11} \rfloor = 10 → £10 payout.
Modifications by HackerRank
Solving the last test case requires more than 40 rounds. That's beyond the reach of unsigned long long
and/or double
.
19 rounds are fine with my current implemenation. Switching to double
works surprisingly well up to 30 rounds.
Note
My states don't need to store red
and blue
because they can be derived from the index of the State
object in the open
container.
I kept those member variables for clarity, though.
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This is equivalent toecho "1 4" | ./121
Output:
Note: the original problem's input 15
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 <vector>
typedef unsigned long long Possibilities; // replace unsigned long long by double to compute up to 30 rounds
struct State
{
// picked red and blue disc
unsigned int red;
unsigned int blue;
// number of different choices leading to this state
Possibilities possibilities;
};
int main()
{
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
{
unsigned int maxRounds = 15;
std::cin >> maxRounds;
// number of red and blue discs in the bag
unsigned int availableRed = 1;
unsigned int availableBlue = 1;
Possibilities possibilities = 1;
// initially no disc taken yet
std::vector<State> open = { { 0, 0, 1 } };
// evaluateall 15 rounds
for (unsigned int round = 1; round <= maxRounds; round++)
{
// total number of possible choices
possibilities *= availableRed + availableBlue;
// set number of blue and red discs per state
std::vector<State> next;
for (unsigned int blue = 0; blue <= round; blue++)
{
State state;
state.red = round - blue;
state.blue = blue;
state.possibilities = 0;
next.push_back(state);
}
// update possibilities of each state
for (auto current : open)
{
// picking red
next[current.blue ].possibilities += current.possibilities * availableRed;
// picking blue
next[current.blue + 1].possibilities += current.possibilities * availableBlue;
}
// prepare next iteration
open = next;
availableRed++;
}
// compute chance that blue > red
Possibilities moreBlue = 0;
for (auto current : open)
if (current.blue > current.red)
moreBlue += current.possibilities;
// chance of winning is moreBlue / possibilities, therefore maximum prize is the inverse
unsigned int prize = possibilities / moreBlue; // integer division !
std::cout << prize << std::endl;
}
return 0;
}
This solution contains 13 empty lines, 13 comments and 2 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.
(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
May 19, 2017 submitted solution
May 19, 2017 added comments
Hackerrank
see https://www.hackerrank.com/contests/projecteuler/challenges/euler121
My code solves 2 out of 3 test cases (score: 50%)
I failed 1 test cases due to wrong answers and 0 because of timeouts
Difficulty
Project Euler ranks this problem at 35% (out of 100%).
Hackerrank describes this problem as easy.
Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.
Links
projecteuler.net/thread=121 - 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-121-coloured-discs/ (written by Kristian Edlund)
Java github.com/nayuki/Project-Euler-solutions/blob/master/java/p121.java (written by Nayuki)
Scala github.com/samskivert/euler-scala/blob/master/Euler121.scala (written by Michael Bayne)
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 | |
the flashing problem is the one I solved most recently |
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I scored 13,486 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 120 - Square remainders | Efficient exponentiation - problem 122 >> |