存储特定索引位置的矩阵向量乘法的迭代 [英] Iteration of matrix-vector multiplication which stores specific index-positions

问题描述

我需要解决最小距离问题，以查看正在尝试的一些工作:

I need to solve a min distance problem, to see some of the work which has being tried take a look at:

链接:点击这里

我有四个元素:两个列向量:昏暗(px1)的alpha和昏暗(qx1)的beta.在这种情况下，p = q = 50分别给出两个昏暗的(50x1)列向量.它们的定义如下:

I have four elements: two column vectors: alpha of dim (px1) and beta of dim (qx1). In this case p = q = 50 giving two column vectors of dim (50x1) each. They are defined as follows:

alpha = alpha = 0:0.05:2;
beta = beta = 0:0.05:2;

我有两个矩阵:L1和L2.

L1由每个尺寸为(kx1)的三个列向量组成.

L1 is composed of three column-vectors of dimension (kx1) each.

L2由每个维度为(mx1)的三个列向量组成.

L2 is composed of three column-vectors of dimension (mx1) each.

在这种情况下，它们的大小相等，这意味着k = m = 1000给出:L1和L2分别为昏暗的(1000x3).这些矩阵的值是预定义的.

In this case, they have equal size, meaning that k = m = 1000 giving: L1 and L2 of dim (1000x3) each. The values of these matrices are predefined.

尽管如此，它们仍具有以下结构:

They have, nevertheless, the following structure:

L1(kx3) = [t1(kx1) t2(kx1) t3(kx1)];
L2(mx3) = [t1(mx1) t2(mx1) t3(mx1)];

分钟(数学上)给出了我需要解决的距离问题:

The min. distance problem I need to solve is given (mathematically) as follows:

 d = min( (x-(alpha_p*t1_k - beta_q*t1_m)).^2 + (y-(alpha_p*t2_k - beta_q*t2_m)).^2 +
 (z-(alpha_p*t3_k - beta_q*t3_m)).^2 )

值x,y,z是三个固定常数.

我的问题

我需要开发一个迭代，该迭代可以使我从alpha, beta, L1和L2的组合返回索引位置，从而可以从上方满足最小距离问题.

I need to develop an iteration which can give me back the index positions from the combination of: alpha, beta, L1 and L2 which fulfills the min-distance problem from above.

我希望问题的表述清楚，我对索引符号非常小心.但是，如果仍然不清楚...的步长:

I hope the formulation for the problem is clear, I have been very careful with the index notations. But if it is still not so clear... the step size for:

alpha是p = 1，... 50

alpha is p = 1,...50

beta是q = 1，... 50

beta is q = 1,...50

对于L1; t1, t2, t3是k = 1，...，1000

for L1; t1, t2, t3 is k = 1,...,1000

对于L2; t1, t2, t3是m = 1，...，1000

for L2; t1, t2, t3 is m = 1,...,1000

并且我需要找到index of p，index of q，index of k和index of m，这些都为我提供了最小值.到点x,y,z的距离.

And I need to find the index of p, index of q, index of k and index of m which gives me the min. distance to the point x,y,z.

提前感谢您的帮助！

推荐答案

我不知道您的值，所以我无法检查我的代码.我正在使用循环，因为它是最明显的解决方案.可以肯定的是，来自bsxfun -brigarde(;-D)的人会找到更短/更有效的解决方案.

I don't know your values so i wasn't able to check my code. I am using loops because it is the most obvious solution. Pretty sure that someone from the bsxfun-brigarde ( ;-D ) will find a shorter/more effective solution.

alpha = 0:0.05:2;
beta = 0:0.05:2;

L1(kx3) = [t1(kx1) t2(kx1) t3(kx1)];
L2(mx3) = [t1(mx1) t2(mx1) t3(mx1)];
idx_smallest_d =[1,1,1,1];
smallest_d = min((x-(alpha(1)*t1(1) - beta(1)*t1(1))).^2 + (y-(alpha(1)*t2(1) - beta(1)*t2(1))).^2+...
                    (z-(alpha(1)*t3(1) - beta(1)*t3(1))).^2);

%The min. distance problem I need to solve is given (mathematically) as follows:
for p=1:1:50
    for q=1:1:50
        for k=1:1:1000
            for m=1:1:1000
                d = min((x-(alpha(p)*t1(k) - beta(q)*t1(m))).^2 + (y-(alpha(p)*t2(k) - beta(q)*t2(m))).^2+...
                    (z-(alpha(p)*t3(k) - beta(q)*t3(m))).^2);
                if d < smallest_d
                    smallest_d=d;
                    idx_smallest_d= [p,q,k,m];
                end
            end
        end
    end
end

我正在做的是将最小距离预定义为第一个组合的距离，然后检查每个组合，而该距离小于以前的最短距离.

What I am doing is predefining the smallest distance as the distance of the first combination and then checking for each combination rather the distance is smaller than the previous shortest distance.

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