Calculate Pi using Monte CarloHamiltonian Monte Carlo in ScalaMonte Carlo pi calculationPi by Monte-CarloPoker Odds Calculation with Monte CarloMonte Carlo Pi (MASM)Approximating π via Monte Carlo simulationApproximating Pi, Monte Carlo integrationMonte Carlo estimation of πMonte Carlo simulation to approximate the value of PIMultithreaded Monte-Carlo Integration

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Calculate Pi using Monte Carlo


Hamiltonian Monte Carlo in ScalaMonte Carlo pi calculationPi by Monte-CarloPoker Odds Calculation with Monte CarloMonte Carlo Pi (MASM)Approximating π via Monte Carlo simulationApproximating Pi, Monte Carlo integrationMonte Carlo estimation of πMonte Carlo simulation to approximate the value of PIMultithreaded Monte-Carlo Integration













2












$begingroup$


#include <iostream>
#include <iomanip>


#ifdef USE_OLD_RAND

#include <stdlib.h>
inline double getRandDart() return rand() * 1.0 / RAND_MAX;

#else

#include <random>
std::default_random_engine generator;
std::uniform_real_distribution<double> distribution(0,1);
inline double getRandDart() return distribution(generator);

#endif


// Monte Carlo Simulator to estimate the value of PI.
//
// If we have a circle with a radius of 1.
// Then the smallest square that encloses the circle as sides of length 2.
//
// Area of circle pi r^2 = pi
// Area of square 2.r.2.r = 4
//
// Ratio of overlapping area: pi/4
//
// If we throw darts randomly at a dart board (with an even distribution) and always hit the square.
// Then the ratio of darts falling into the circle should be pi/4 of the total number of darts thrown.
//
// pi/4 * countInSquare = countInCircle
//
// pi = 4 . countInCircle / countInSquare
//
// To simplify the maths.
// We will set the center point as 0,0 and only use the top right quadrant of the circle.
// We have 1/4 the size of the square and circle but the same maths still apply.
//
// A dart thrown has a random x/y value in the range 0->1 (top right quadrant).
// A dart is outside the circle if x^2 + y^2 > 1 (note 1^2 is 1)
//

int main()


long countInSquare = 0;
long countInCircle = 0;

for(long iteration = 0; iteration <= 10'000'000'000; ++iteration)
double x = getRandDart();
double y = getRandDart();

double d = (x * x) + (y * y);

countInSquare += 1;
countInCircle += (d >= 1.0) ? 0 : 1;

if (iteration % 10'000'000 == 0)
std::cout << iteration << " " << (4.0 * countInCircle / countInSquare) << "n";


std::cout << "nn" << std::setprecision(9) << (4.0 * countInCircle / countInSquare) << "n";




Output:



> ./a.out
9990000000 3.14158
10000000000 3.14158


3.14158355









share|improve this question











$endgroup$











  • $begingroup$
    @Juho The question is: Please review the code. Any comments about the code would be useful.
    $endgroup$
    – Martin York
    1 hour ago











  • $begingroup$
    Your comment says that xx+yy==1 is not outside the circle (hence it is inside), but your code considers it outside. Which way is the correct interpretation?
    $endgroup$
    – 1201ProgramAlarm
    1 hour ago










  • $begingroup$
    @1201ProgramAlarm: Good catch. Stupid comments; why don't they compile so we can check the code matches the comments.
    $endgroup$
    – Martin York
    1 hour ago











  • $begingroup$
    There are languages where that's kind of true (but also means that you can get syntax errors with your comments)
    $endgroup$
    – Foon
    45 mins ago















2












$begingroup$


#include <iostream>
#include <iomanip>


#ifdef USE_OLD_RAND

#include <stdlib.h>
inline double getRandDart() return rand() * 1.0 / RAND_MAX;

#else

#include <random>
std::default_random_engine generator;
std::uniform_real_distribution<double> distribution(0,1);
inline double getRandDart() return distribution(generator);

#endif


// Monte Carlo Simulator to estimate the value of PI.
//
// If we have a circle with a radius of 1.
// Then the smallest square that encloses the circle as sides of length 2.
//
// Area of circle pi r^2 = pi
// Area of square 2.r.2.r = 4
//
// Ratio of overlapping area: pi/4
//
// If we throw darts randomly at a dart board (with an even distribution) and always hit the square.
// Then the ratio of darts falling into the circle should be pi/4 of the total number of darts thrown.
//
// pi/4 * countInSquare = countInCircle
//
// pi = 4 . countInCircle / countInSquare
//
// To simplify the maths.
// We will set the center point as 0,0 and only use the top right quadrant of the circle.
// We have 1/4 the size of the square and circle but the same maths still apply.
//
// A dart thrown has a random x/y value in the range 0->1 (top right quadrant).
// A dart is outside the circle if x^2 + y^2 > 1 (note 1^2 is 1)
//

int main()


long countInSquare = 0;
long countInCircle = 0;

for(long iteration = 0; iteration <= 10'000'000'000; ++iteration)
double x = getRandDart();
double y = getRandDart();

double d = (x * x) + (y * y);

countInSquare += 1;
countInCircle += (d >= 1.0) ? 0 : 1;

if (iteration % 10'000'000 == 0)
std::cout << iteration << " " << (4.0 * countInCircle / countInSquare) << "n";


std::cout << "nn" << std::setprecision(9) << (4.0 * countInCircle / countInSquare) << "n";




Output:



> ./a.out
9990000000 3.14158
10000000000 3.14158


3.14158355









share|improve this question











$endgroup$











  • $begingroup$
    @Juho The question is: Please review the code. Any comments about the code would be useful.
    $endgroup$
    – Martin York
    1 hour ago











  • $begingroup$
    Your comment says that xx+yy==1 is not outside the circle (hence it is inside), but your code considers it outside. Which way is the correct interpretation?
    $endgroup$
    – 1201ProgramAlarm
    1 hour ago










  • $begingroup$
    @1201ProgramAlarm: Good catch. Stupid comments; why don't they compile so we can check the code matches the comments.
    $endgroup$
    – Martin York
    1 hour ago











  • $begingroup$
    There are languages where that's kind of true (but also means that you can get syntax errors with your comments)
    $endgroup$
    – Foon
    45 mins ago













2












2








2





$begingroup$


#include <iostream>
#include <iomanip>


#ifdef USE_OLD_RAND

#include <stdlib.h>
inline double getRandDart() return rand() * 1.0 / RAND_MAX;

#else

#include <random>
std::default_random_engine generator;
std::uniform_real_distribution<double> distribution(0,1);
inline double getRandDart() return distribution(generator);

#endif


// Monte Carlo Simulator to estimate the value of PI.
//
// If we have a circle with a radius of 1.
// Then the smallest square that encloses the circle as sides of length 2.
//
// Area of circle pi r^2 = pi
// Area of square 2.r.2.r = 4
//
// Ratio of overlapping area: pi/4
//
// If we throw darts randomly at a dart board (with an even distribution) and always hit the square.
// Then the ratio of darts falling into the circle should be pi/4 of the total number of darts thrown.
//
// pi/4 * countInSquare = countInCircle
//
// pi = 4 . countInCircle / countInSquare
//
// To simplify the maths.
// We will set the center point as 0,0 and only use the top right quadrant of the circle.
// We have 1/4 the size of the square and circle but the same maths still apply.
//
// A dart thrown has a random x/y value in the range 0->1 (top right quadrant).
// A dart is outside the circle if x^2 + y^2 > 1 (note 1^2 is 1)
//

int main()


long countInSquare = 0;
long countInCircle = 0;

for(long iteration = 0; iteration <= 10'000'000'000; ++iteration)
double x = getRandDart();
double y = getRandDart();

double d = (x * x) + (y * y);

countInSquare += 1;
countInCircle += (d >= 1.0) ? 0 : 1;

if (iteration % 10'000'000 == 0)
std::cout << iteration << " " << (4.0 * countInCircle / countInSquare) << "n";


std::cout << "nn" << std::setprecision(9) << (4.0 * countInCircle / countInSquare) << "n";




Output:



> ./a.out
9990000000 3.14158
10000000000 3.14158


3.14158355









share|improve this question











$endgroup$




#include <iostream>
#include <iomanip>


#ifdef USE_OLD_RAND

#include <stdlib.h>
inline double getRandDart() return rand() * 1.0 / RAND_MAX;

#else

#include <random>
std::default_random_engine generator;
std::uniform_real_distribution<double> distribution(0,1);
inline double getRandDart() return distribution(generator);

#endif


// Monte Carlo Simulator to estimate the value of PI.
//
// If we have a circle with a radius of 1.
// Then the smallest square that encloses the circle as sides of length 2.
//
// Area of circle pi r^2 = pi
// Area of square 2.r.2.r = 4
//
// Ratio of overlapping area: pi/4
//
// If we throw darts randomly at a dart board (with an even distribution) and always hit the square.
// Then the ratio of darts falling into the circle should be pi/4 of the total number of darts thrown.
//
// pi/4 * countInSquare = countInCircle
//
// pi = 4 . countInCircle / countInSquare
//
// To simplify the maths.
// We will set the center point as 0,0 and only use the top right quadrant of the circle.
// We have 1/4 the size of the square and circle but the same maths still apply.
//
// A dart thrown has a random x/y value in the range 0->1 (top right quadrant).
// A dart is outside the circle if x^2 + y^2 > 1 (note 1^2 is 1)
//

int main()


long countInSquare = 0;
long countInCircle = 0;

for(long iteration = 0; iteration <= 10'000'000'000; ++iteration)
double x = getRandDart();
double y = getRandDart();

double d = (x * x) + (y * y);

countInSquare += 1;
countInCircle += (d >= 1.0) ? 0 : 1;

if (iteration % 10'000'000 == 0)
std::cout << iteration << " " << (4.0 * countInCircle / countInSquare) << "n";


std::cout << "nn" << std::setprecision(9) << (4.0 * countInCircle / countInSquare) << "n";




Output:



> ./a.out
9990000000 3.14158
10000000000 3.14158


3.14158355






c++ numerical-methods






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago









200_success

130k17154419




130k17154419










asked 1 hour ago









Martin YorkMartin York

73.7k488270




73.7k488270











  • $begingroup$
    @Juho The question is: Please review the code. Any comments about the code would be useful.
    $endgroup$
    – Martin York
    1 hour ago











  • $begingroup$
    Your comment says that xx+yy==1 is not outside the circle (hence it is inside), but your code considers it outside. Which way is the correct interpretation?
    $endgroup$
    – 1201ProgramAlarm
    1 hour ago










  • $begingroup$
    @1201ProgramAlarm: Good catch. Stupid comments; why don't they compile so we can check the code matches the comments.
    $endgroup$
    – Martin York
    1 hour ago











  • $begingroup$
    There are languages where that's kind of true (but also means that you can get syntax errors with your comments)
    $endgroup$
    – Foon
    45 mins ago
















  • $begingroup$
    @Juho The question is: Please review the code. Any comments about the code would be useful.
    $endgroup$
    – Martin York
    1 hour ago











  • $begingroup$
    Your comment says that xx+yy==1 is not outside the circle (hence it is inside), but your code considers it outside. Which way is the correct interpretation?
    $endgroup$
    – 1201ProgramAlarm
    1 hour ago










  • $begingroup$
    @1201ProgramAlarm: Good catch. Stupid comments; why don't they compile so we can check the code matches the comments.
    $endgroup$
    – Martin York
    1 hour ago











  • $begingroup$
    There are languages where that's kind of true (but also means that you can get syntax errors with your comments)
    $endgroup$
    – Foon
    45 mins ago















$begingroup$
@Juho The question is: Please review the code. Any comments about the code would be useful.
$endgroup$
– Martin York
1 hour ago





$begingroup$
@Juho The question is: Please review the code. Any comments about the code would be useful.
$endgroup$
– Martin York
1 hour ago













$begingroup$
Your comment says that xx+yy==1 is not outside the circle (hence it is inside), but your code considers it outside. Which way is the correct interpretation?
$endgroup$
– 1201ProgramAlarm
1 hour ago




$begingroup$
Your comment says that xx+yy==1 is not outside the circle (hence it is inside), but your code considers it outside. Which way is the correct interpretation?
$endgroup$
– 1201ProgramAlarm
1 hour ago












$begingroup$
@1201ProgramAlarm: Good catch. Stupid comments; why don't they compile so we can check the code matches the comments.
$endgroup$
– Martin York
1 hour ago





$begingroup$
@1201ProgramAlarm: Good catch. Stupid comments; why don't they compile so we can check the code matches the comments.
$endgroup$
– Martin York
1 hour ago













$begingroup$
There are languages where that's kind of true (but also means that you can get syntax errors with your comments)
$endgroup$
– Foon
45 mins ago




$begingroup$
There are languages where that's kind of true (but also means that you can get syntax errors with your comments)
$endgroup$
– Foon
45 mins ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

One could consider at least the following points:



  • Instead of including <stdlib.h>, I'd include <cstdlib>.


  • In getRandDart(), it might in this case be more readable to do static_cast<double>(rand()) / RAND_MAX; instead of multiplying by 1.0.


  • In the for loop, all of x, y and d can be const, so I'd make them const. This has the potential to protect the programmer from unintended mistakes, and can sometimes allow the compiler to optimize better.


  • When you increment by one (in countInSquare += 1;), it makes more sense to use the ++ operator, i.e., to just write ++countInSquare. This is more idiomatic and protects us from unintended mistakes: ++ conveys the meaning of increment (by one), whereas with += we might accidentally write += 2; and that would be perfectly valid (but not what we wanted).


  • Regardless of the above point, notice that during the for-loop, it holds that iteration == countInSquare. So strictly speaking, the variable countInSquare is unnecessary and could be replaced by just iteration when needed.


  • You could consider making the number of iterations and the second operand of the % operand constants to allow for easier modification and perhaps to slightly improve readability.


  • Instead of typing (4.0 * countInCircle / countInSquare) twice, we could make a function that takes the two variables as parameters. This would allow us to save some typing, and again to protect us from unintended mistakes.






share|improve this answer











$endgroup$












  • $begingroup$
    Thanks. I like all those points.
    $endgroup$
    – Martin York
    1 hour ago










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

One could consider at least the following points:



  • Instead of including <stdlib.h>, I'd include <cstdlib>.


  • In getRandDart(), it might in this case be more readable to do static_cast<double>(rand()) / RAND_MAX; instead of multiplying by 1.0.


  • In the for loop, all of x, y and d can be const, so I'd make them const. This has the potential to protect the programmer from unintended mistakes, and can sometimes allow the compiler to optimize better.


  • When you increment by one (in countInSquare += 1;), it makes more sense to use the ++ operator, i.e., to just write ++countInSquare. This is more idiomatic and protects us from unintended mistakes: ++ conveys the meaning of increment (by one), whereas with += we might accidentally write += 2; and that would be perfectly valid (but not what we wanted).


  • Regardless of the above point, notice that during the for-loop, it holds that iteration == countInSquare. So strictly speaking, the variable countInSquare is unnecessary and could be replaced by just iteration when needed.


  • You could consider making the number of iterations and the second operand of the % operand constants to allow for easier modification and perhaps to slightly improve readability.


  • Instead of typing (4.0 * countInCircle / countInSquare) twice, we could make a function that takes the two variables as parameters. This would allow us to save some typing, and again to protect us from unintended mistakes.






share|improve this answer











$endgroup$












  • $begingroup$
    Thanks. I like all those points.
    $endgroup$
    – Martin York
    1 hour ago















4












$begingroup$

One could consider at least the following points:



  • Instead of including <stdlib.h>, I'd include <cstdlib>.


  • In getRandDart(), it might in this case be more readable to do static_cast<double>(rand()) / RAND_MAX; instead of multiplying by 1.0.


  • In the for loop, all of x, y and d can be const, so I'd make them const. This has the potential to protect the programmer from unintended mistakes, and can sometimes allow the compiler to optimize better.


  • When you increment by one (in countInSquare += 1;), it makes more sense to use the ++ operator, i.e., to just write ++countInSquare. This is more idiomatic and protects us from unintended mistakes: ++ conveys the meaning of increment (by one), whereas with += we might accidentally write += 2; and that would be perfectly valid (but not what we wanted).


  • Regardless of the above point, notice that during the for-loop, it holds that iteration == countInSquare. So strictly speaking, the variable countInSquare is unnecessary and could be replaced by just iteration when needed.


  • You could consider making the number of iterations and the second operand of the % operand constants to allow for easier modification and perhaps to slightly improve readability.


  • Instead of typing (4.0 * countInCircle / countInSquare) twice, we could make a function that takes the two variables as parameters. This would allow us to save some typing, and again to protect us from unintended mistakes.






share|improve this answer











$endgroup$












  • $begingroup$
    Thanks. I like all those points.
    $endgroup$
    – Martin York
    1 hour ago













4












4








4





$begingroup$

One could consider at least the following points:



  • Instead of including <stdlib.h>, I'd include <cstdlib>.


  • In getRandDart(), it might in this case be more readable to do static_cast<double>(rand()) / RAND_MAX; instead of multiplying by 1.0.


  • In the for loop, all of x, y and d can be const, so I'd make them const. This has the potential to protect the programmer from unintended mistakes, and can sometimes allow the compiler to optimize better.


  • When you increment by one (in countInSquare += 1;), it makes more sense to use the ++ operator, i.e., to just write ++countInSquare. This is more idiomatic and protects us from unintended mistakes: ++ conveys the meaning of increment (by one), whereas with += we might accidentally write += 2; and that would be perfectly valid (but not what we wanted).


  • Regardless of the above point, notice that during the for-loop, it holds that iteration == countInSquare. So strictly speaking, the variable countInSquare is unnecessary and could be replaced by just iteration when needed.


  • You could consider making the number of iterations and the second operand of the % operand constants to allow for easier modification and perhaps to slightly improve readability.


  • Instead of typing (4.0 * countInCircle / countInSquare) twice, we could make a function that takes the two variables as parameters. This would allow us to save some typing, and again to protect us from unintended mistakes.






share|improve this answer











$endgroup$



One could consider at least the following points:



  • Instead of including <stdlib.h>, I'd include <cstdlib>.


  • In getRandDart(), it might in this case be more readable to do static_cast<double>(rand()) / RAND_MAX; instead of multiplying by 1.0.


  • In the for loop, all of x, y and d can be const, so I'd make them const. This has the potential to protect the programmer from unintended mistakes, and can sometimes allow the compiler to optimize better.


  • When you increment by one (in countInSquare += 1;), it makes more sense to use the ++ operator, i.e., to just write ++countInSquare. This is more idiomatic and protects us from unintended mistakes: ++ conveys the meaning of increment (by one), whereas with += we might accidentally write += 2; and that would be perfectly valid (but not what we wanted).


  • Regardless of the above point, notice that during the for-loop, it holds that iteration == countInSquare. So strictly speaking, the variable countInSquare is unnecessary and could be replaced by just iteration when needed.


  • You could consider making the number of iterations and the second operand of the % operand constants to allow for easier modification and perhaps to slightly improve readability.


  • Instead of typing (4.0 * countInCircle / countInSquare) twice, we could make a function that takes the two variables as parameters. This would allow us to save some typing, and again to protect us from unintended mistakes.







share|improve this answer














share|improve this answer



share|improve this answer








edited 1 hour ago

























answered 1 hour ago









JuhoJuho

1,201410




1,201410











  • $begingroup$
    Thanks. I like all those points.
    $endgroup$
    – Martin York
    1 hour ago
















  • $begingroup$
    Thanks. I like all those points.
    $endgroup$
    – Martin York
    1 hour ago















$begingroup$
Thanks. I like all those points.
$endgroup$
– Martin York
1 hour ago




$begingroup$
Thanks. I like all those points.
$endgroup$
– Martin York
1 hour ago

















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