MATLAB 中 R^5 超立方体的蒙特卡罗积分

发布于 2024-12-06 03:51:37 字数 727 浏览 2 评论 0原文

我需要编写 MATLAB 代码,使用 Monte Carlo 在 R^5 超立方体上进行积分。当我有通用函数时,我有一个基本算法。但我需要积分的函数是:

∫dA

A 是 R^5 的元素。

如果我有 ∫f(x)dA 那么我认为我的算法会起作用。

这是算法:

% Writen by Jerome W Lindsey III

clear;
n = 10000;

% Make a matrix of the same dimension
% as the problem.  Each row is a dimension

A = rand(5,n);

% Vector to contain the solution

B = zeros(1,n);


    for k = 1:n
        % insert the integrand here
        % I don't know how to enter a function {f(1,n), f(2,n), … f(5n)} that
        % will give me the proper solution
        % I threw in a function that will spit out 5!
        % because that is the correct solution.
        B(k) = 1 / (2 * 3 * 4 * 5);

    end

mean(B) 

I need to write MATLAB code that will integrate over a R^5 hypercube using Monte Carlo. I have a basic algorithm that works when I have a generic function. But the function I need to integrate is:

∫dA

A is an element of R^5.

If I had ∫f(x)dA then I think my algorithm would work.

Here is the algorithm:

% Writen by Jerome W Lindsey III

clear;
n = 10000;

% Make a matrix of the same dimension
% as the problem.  Each row is a dimension

A = rand(5,n);

% Vector to contain the solution

B = zeros(1,n);


    for k = 1:n
        % insert the integrand here
        % I don't know how to enter a function {f(1,n), f(2,n), … f(5n)} that
        % will give me the proper solution
        % I threw in a function that will spit out 5!
        % because that is the correct solution.
        B(k) = 1 / (2 * 3 * 4 * 5);

    end

mean(B) 

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评论(3

明明#如月 2024-12-13 03:51:37

无论如何,我想我明白这里的意图是什么,尽管它看起来确实有点做作。考虑一下尝试通过 MC 求圆面积的问题,如此处所述。这里的样本是从一个单位正方形中抽取的,函数在圆内取值 1,在圆外取值 0。为了求出 R^5 中立方体的体积,我们可以从包含该立方体的其他物体中进行采样,并使用类似的过程来计算所需的体积。希望这足以让其余的实现变得简单。

In any case, I think I understand what the intent here is, although it does seem like somewhat of a contrived exercise. Consider the problem of trying to find the area of a circle via MC, as discussed here. Here samples are being drawn from a unit square, and the function takes on the value 1 inside the circle and 0 outside. To find the volume of a cube in R^5, we could sample from something else that contains the cube and use an analogous procedure to compute the desired volume. Hopefully this is enough of a hint to make the rest of the implementation straightforward.

和我恋爱吧 2024-12-13 03:51:37

我在这里猜测一下,因为您作为“正确”答案给出的数字与您陈述练习的方式不匹配(单位超立方体的体积为 1)。

鉴于结果应该是 1/120 - 是否您应该集成 超立方体中的标准单纯形?

你的功能会很清楚。如果 sum(x) <,则 f(x) = 1 1;否则为 0

I'm guessing here a bit since the numbers you give as "correct" answer don't match to how you state the exercise (volume of unit hypercube is 1).

Given the result should be 1/120 - could it be that you are supposed to integrate the standard simplex in the hypercube?

The your function would be clear. f(x) = 1 if sum(x) < 1; 0 otherwise

愿得七秒忆 2024-12-13 03:51:37
%Question 2, problem set 1
% Writen by Jerome W Lindsey III

clear;
n = 10000;

% Make a matrix of the same dimension
% as the problem.  Each row is a dimension
A = rand(5,n);

% Vector to contain the solution
B = zeros(1,n);


    for k = 1:n
        % insert the integrand here
        % this bit of code works as the integrand
        if sum(A(:,k)) < 1
            B(k) = 1;
        end

    end
    clear k;

clear A;

    % Begin error estimation calculations
    std_mc = std(B);
    clear n;
    clear B;

    % using the error I calculate a new random
    % vector of corect length
    N_new = round(std_mc ^ 2 * 3.291 ^ 2 * 1000000);
    A_new = rand(5, N_new);
    B_new = zeros(1,N_new);
    clear std_mc;

        for k = 1:N_new
            if sum(A_new(:,k)) < 1
                B_new(k) = 1;
            end
        end
        clear k;

    clear A_new;

    % collect descriptive statisitics
    M_new = mean(B_new);
    std_new = std(B_new);
    MC_new_error_999 = std_new * 3.921 / sqrt(N_new);
    clear N_new; 
    clear B_new;
    clear std_new;

% Display Results
disp('Integral in question #2 is');
disp(M_new);
disp(' ');
disp('Monte Carlo Error');
disp(MC_new_error_999);
%Question 2, problem set 1
% Writen by Jerome W Lindsey III

clear;
n = 10000;

% Make a matrix of the same dimension
% as the problem.  Each row is a dimension
A = rand(5,n);

% Vector to contain the solution
B = zeros(1,n);


    for k = 1:n
        % insert the integrand here
        % this bit of code works as the integrand
        if sum(A(:,k)) < 1
            B(k) = 1;
        end

    end
    clear k;

clear A;

    % Begin error estimation calculations
    std_mc = std(B);
    clear n;
    clear B;

    % using the error I calculate a new random
    % vector of corect length
    N_new = round(std_mc ^ 2 * 3.291 ^ 2 * 1000000);
    A_new = rand(5, N_new);
    B_new = zeros(1,N_new);
    clear std_mc;

        for k = 1:N_new
            if sum(A_new(:,k)) < 1
                B_new(k) = 1;
            end
        end
        clear k;

    clear A_new;

    % collect descriptive statisitics
    M_new = mean(B_new);
    std_new = std(B_new);
    MC_new_error_999 = std_new * 3.921 / sqrt(N_new);
    clear N_new; 
    clear B_new;
    clear std_new;

% Display Results
disp('Integral in question #2 is');
disp(M_new);
disp(' ');
disp('Monte Carlo Error');
disp(MC_new_error_999);
~没有更多了~
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