<< 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.

# 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.

# 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.

# Interactive test

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

This is equivalent to`echo "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)*

# 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

May 19, 2017 submitted solution

May 19, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler121

My code solved **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 never 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:

Python: 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)

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.

yellow problems score less than 100% at Hackerrank (but still solve the original problem).

gray problems are already solved but I haven't published my solution yet.

blue problems are already solved and there wasn't a Hackerrank version of it (at the time I solved it) or I didn't care about it because it differed too much.

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<< problem 120 - Square remainders | Efficient exponentiation - problem 122 >> |