0042算法笔记——【随机化算法】计算π值和计算定积分

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1、计算π值

问题描述

设有一半径为r的圆及其外切四边形。向该正方形随机地投掷n个点。设落入圆内的点数为k。由于所投入的点在正方形上均匀分布，因而所投入的点落入圆内的概率为。所以当n足够大时，k与n之比就逼近这一概率。从而。

程序具体代码如下：

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//随机化算法用随机投点法计算π值
#include "stdafx.h"
#include "RandomNumber.h"
#include <iostream>
using namespace std;
double Darts(int n);
int main()
{
int n1 = 100,n2 = 1000,n3 = 1000,n4 = 10000,n5 = 10000000;
cout<<"n1="<<n1<<",π1="<<Darts(n1)<<endl;
cout<<"n2="<<n2<<",π2="<<Darts(n2)<<endl;
cout<<"n3="<<n3<<",π3="<<Darts(n3)<<endl;
cout<<"n4="<<n4<<",π4="<<Darts(n4)<<endl;
cout<<"n5="<<n5<<",π5="<<Darts(n5)<<endl;
return 0;
}
//用随机投点法计算π值
double Darts(int n)
{
static RandomNumber dart;
int k = 0;
for(int i=1; i<=n; i++)
{
double x = dart.fRandom();
double y = dart.fRandom();
if((x*x + y*y)<=1)
{
k++;
}
}
return 4*k/double(n);
}

//随机化算法 用随机投点法计算π值#include "stdafx.h"#include "RandomNumber.h"#include <iostream>using namespace std;double Darts(int n);int main(){int n1 = 100,n2 = 1000,n3 = 1000,n4 = 10000,n5 = 10000000;cout<<"n1="<<n1<<",π1="<<Darts(n1)<<endl;cout<<"n2="<<n2<<",π2="<<Darts(n2)<<endl;cout<<"n3="<<n3<<",π3="<<Darts(n3)<<endl;cout<<"n4="<<n4<<",π4="<<Darts(n4)<<endl;cout<<"n5="<<n5<<",π5="<<Darts(n5)<<endl;return 0;}//用随机投点法计算π值double Darts(int n){static RandomNumber dart;int k = 0;for(int i=1; i<=n; i++){double x = dart.fRandom();double y = dart.fRandom();if((x*x + y*y)<=1){k++;}}return 4*k/double(n);}

程序运行结果如图：

2、计算定积分

例：设f(x)=x^2,求

解：

1)随机投点法计算定积分

基本思想是在矩形区域上随机均匀的投点实现。本算法的基本思想是在积分区间上随机均匀的产生点, 即在[a,b]上随机均匀的取点, 求出由这些点产生的函数值的算术平均值, 再乘以区间宽度, 即可解出定积分得近似解。

算法具体代码如下：

[cpp] view plaincopyprint?

//随机化算法用随机投点法计算定积分
#include "stdafx.h"
#include "RandomNumber.h"
#include <iostream>
using namespace std;
double Darts(int n,double a,double b);
double f(double x);
int main()
{
int n1 = 100,n2 = 1000,n3 = 1000,n4 = 10000,n5 = 10000000;
double a = 2.0,b = 3.0;
cout<<"n1="<<n1<<",r1="<<Darts(n1,a,b)<<endl;
cout<<"n2="<<n2<<",r2="<<Darts(n2,a,b)<<endl;
cout<<"n3="<<n3<<",r3="<<Darts(n3,a,b)<<endl;
cout<<"n4="<<n4<<",r4="<<Darts(n4,a,b)<<endl;
cout<<"n5="<<n5<<",r5="<<Darts(n5,a,b)<<endl;
return 0;
}
/*
* 基本思想是在矩形区域内随机均匀投点，求出由这些点
* 产生的函数值的算术平均值，再乘以区间宽度，即可得
* 出定积分的近似解
*/
double Darts(int n,double a,double b)
{
static RandomNumber dart;
double sum = 0.0;
for(int i=0; i<n; i++)
{
double x = (b-a)*dart.fRandom() + a;//产生[a,b)之间的随机数
sum = sum + f(x);
}
return (b-a)*sum/n;
}
double f(double x)
{
return x*x;
}

//随机化算法 用随机投点法计算定积分#include "stdafx.h"#include "RandomNumber.h"#include <iostream>using namespace std;double Darts(int n,double a,double b);double f(double x);int main(){int n1 = 100,n2 = 1000,n3 = 1000,n4 = 10000,n5 = 10000000;double a = 2.0,b = 3.0;cout<<"n1="<<n1<<",r1="<<Darts(n1,a,b)<<endl;cout<<"n2="<<n2<<",r2="<<Darts(n2,a,b)<<endl;cout<<"n3="<<n3<<",r3="<<Darts(n3,a,b)<<endl;cout<<"n4="<<n4<<",r4="<<Darts(n4,a,b)<<endl;cout<<"n5="<<n5<<",r5="<<Darts(n5,a,b)<<endl;return 0;}/* * 基本思想是在矩形区域内随机均匀投点，求出由这些点 * 产生的函数值的算术平均值，再乘以区间宽度，即可得 * 出定积分的近似解 */double Darts(int n,double a,double b){static RandomNumber dart;double sum = 0.0;for(int i=0; i<n; i++){double x = (b-a)*dart.fRandom() + a;//产生[a,b)之间的随机数sum = sum + f(x);}return (b-a)*sum/n;}double f(double x){return x*x;}

程序运行结果如图：

2)概率法法计算定积分

设f:[a,b]→[c,d]连续函数(如图2 所示), 则由曲线y=f(x)以及x 轴和直线x=a,x=b 围成的面积由定积分给出。根据几何概型可知。假设向矩形区域随机均匀的投镖n 次, 落入阴影为K次, 又设M为x=a、x=b、y=c、y=d 所围成的矩形面积, s 为定积分面积,则, 所以s= k/n×M。

算法具体代码如下：

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//随机化算法用概率法计算定积分
#include "stdafx.h"
#include "RandomNumber.h"
#include <iostream>
using namespace std;
double Darts(int n,double a,double b,double d);
double f(double x);
int main()
{
int n1 = 100,n2 = 1000,n3 = 1000,n4 = 10000,n5 = 10000000;
double a = 2.0,b = 3.0;
double d = f(b);
cout<<"n1="<<n1<<",r1="<<Darts(n1,a,b,d)<<endl;
cout<<"n2="<<n2<<",r2="<<Darts(n2,a,b,d)<<endl;
cout<<"n3="<<n3<<",r3="<<Darts(n3,a,b,d)<<endl;
cout<<"n4="<<n4<<",r4="<<Darts(n4,a,b,d)<<endl;
cout<<"n5="<<n5<<",r5="<<Darts(n5,a,b,d)<<endl;
return 0;
}
/*
* f 为积分函数, n 为投镖
* 总数, a,b 为积分区间, c,d 为函
* 数f 的值域的端点值
*/
double Darts(int n,double a,double b,double d)
{
static RandomNumber dart;
int k = 0;
for(int i=0; i<n; i++)
{
double x = (b-a)*dart.fRandom() + a;//产生[a,b)之间的随机数
double y = d * dart.fRandom();
if(y<=f(x))
{
k++;
}
}
return d*(b-a)*k/n;
}
double f(double x)
{
return x*x;
}

//随机化算法 用概率法计算定积分#include "stdafx.h"#include "RandomNumber.h"#include <iostream>using namespace std;double Darts(int n,double a,double b,double d);double f(double x);int main(){int n1 = 100,n2 = 1000,n3 = 1000,n4 = 10000,n5 = 10000000;double a = 2.0,b = 3.0;double d = f(b);cout<<"n1="<<n1<<",r1="<<Darts(n1,a,b,d)<<endl;cout<<"n2="<<n2<<",r2="<<Darts(n2,a,b,d)<<endl;cout<<"n3="<<n3<<",r3="<<Darts(n3,a,b,d)<<endl;cout<<"n4="<<n4<<",r4="<<Darts(n4,a,b,d)<<endl;cout<<"n5="<<n5<<",r5="<<Darts(n5,a,b,d)<<endl;return 0;}/* * f 为积分函数, n 为投镖 * 总数, a,b 为积分区间, c,d 为函 * 数f 的值域的端点值 */double Darts(int n,double a,double b,double d){static RandomNumber dart;int k = 0;for(int i=0; i<n; i++){double x = (b-a)*dart.fRandom() + a;//产生[a,b)之间的随机数double y = d * dart.fRandom();if(y<=f(x)){k++;}}return d*(b-a)*k/n;}double f(double x){return x*x;}

程序运行结果如图：