Question

我使用以下代码

在一个区域内分发了100个随机点（节点）

n=100; xm=100;ym=100;
    for i = 1 : 1 :n
    S(i).xd = rand(1,1)*xm ;
    S(i).yd = rand (1,1)*ym ;
    xd(i) = S(i).xd ;
    yd(i) = S(i).yd ;
    S(i).id = i;
    S(i).type = 0;
    S(i).g = 0 ;
    S(i).E = Eo ;
    S(i).type = 0 ;

end

接下来，我使用以下代码

修复了区域边缘的10个网关节点

 for i=1:1:10
    Sg(i).xd= 2+100*rand(1,1);
      Sg(i).yd=100;
    xd(i)=Sg(i).xd;
    yd(i)=Sg(i).yd;
    Sg(i).id=i;
     plot(Sg(i).xd,Sg(i).yd,'*k')
grid on;
hold on;   
end

现在我已经使用100个节点的LEACH协议形成了簇头。

我必须从CH找到最小距离网关节点。由于有10个网关节点，我必须找到哪个更靠近该区域的CH。

确定CH我使用了以下代码

for all nodes in the area


    temp_rand1=rand;
                if(temp_rand1<= (p/(1-p*mod(r,round(1/p)))))
                    countCHs1=countCHs1+1;

                    S3(i).type=1;
                    S3(i).g=round(1/p)-1;
                    C1(cluster1).xd=S3(i).xd;
                    C1(cluster1).yd=S3(i).yd;

         **distance1=sqrt((S3(i).xd-(gw_node.x) )^2 + (S3(i).yd-(gw_node.y) )^2 );**
                    C1(cluster1).distance1=distance1;
                    C1(cluster1).id=i;
                    X(cluster1)=S3(i).xd;
                    Y(cluster1)=S3(i).yd;
                    cluster1=cluster1+1; 
end

我知道如何确定CH和一个网关节点之间的距离，但我不知道如何从一组网关节点中找到closet网关节点。提前致谢

请回复

Answer 1

不太确定你要求的是什么，但我想你想找到两个最接近的点。

这就是你能做到的。

%// Generate some points
n = 3;    xm = 1:n;    ym =1:n;
m = [xm' ym']; %// This is the line
xd = [    2    8   3]; 
yd = [    1    4   3];
d = [xd' yd']; %// These are the random points
drep = repmat(d,n,1); %// This will repeat the matrix d n times
B= ones(n,1); 
mrep = kron(m,B); %// This will repeat each value in m n times        
distances = bsxfun(@minus,mrep,drep);  %// The distance from each point to every other point       
C_squared = sum(distances.^2,2)
[min_dist,idx] = min(sum(distances.^2,2)) %// Find the minimum distance

value_in_m = mrep(idx,:)

ans =

   3   3

value_in_d = drep(idx,:)
ans =

   3   3

解释

% Generate some points
n = 3;
xm = 1:n;
ym =1:n;
m = [xm' ym']; %// This is the line. Each row is a point
xd = [    2    8   3]; 
yd = [    1    4   3];
d = [xd' yd']; %// These are the random points

我们想要计算从一个点到每个点的距离其他一点，所以我们重复点nxn次

以下只是获取重复元素的一种方法。更多建议如下：https://stackoverflow.com/a/16269185

drep = repmat(d,n,1); %// This will repeat the matrix d n times
B= ones(n,1); 
mrep = kron(m,B); %// This will repeat each value in m n times

我们基于毕达哥拉斯定理，(m_x-d_x)^2 + (m_y-d_y)^2 = C^2

distances = bsxfun(@minus,mrep,drep);  %// The distance from each point to every other point       
C_squared = sum(distances.^2,2)

找到最小距离

[min_dist,idx] = min(sum(distances.^2,2));

找到距离最小的点

value_in_m = mrep(idx,:)
ans =

   3   3

value_in_d = drep(idx,:)
ans =

   3   3

Answer 2

不确定你在问什么。我假设你有一组固定点在一条直线上（标记为蓝色）和一组随机点（标记为红色）。你想要一个固定点和一个随机点之间的最小距离。

试试这段代码：

si = randi([0,100],10,2);    %// Random points (red)
sg(:,1) = randi([0,100],10,1);  %// fixed points x param (blue)
sg(:,2) = ones(10,1)*50;    %// fixed points y param (blue) r const, to make them a st. line

sgc = mat2cell(sg,ones(1,size(sg,1)),size(sg,2)); %// converting to cell array
sic = mat2cell(si,ones(1,size(si,1)),size(si,2)); %// converting to cell array

%// distMat gives n*n matrix of all the dist between each fixed point with every other point.
distMat = cellfun(@(x,y) pdist2(x,y),repmat(sic,1,size(sgc,1)),repmat(sgc',size(sic,1),1));

figure; axis('image');
plot(si(:,1),si(:,2),'r.');
text(si(:,1),si(:,2),[repmat('  ',size(si,1),1), num2str((1:size(si,1))')]);
hold on
plot(sg(:,1),sg(:,2),'b.');
text(sg(:,1),sg(:,2),[repmat('  ',size(sg,1),1), num2str((1:size(sg,1))')]);
hold off

[mdist, index] = min(distMat(:));
[randomP, fixedP] = ind2sub(size(distMat),index)

简介：

enter image description here

输出：

randomP =

 2


fixedP =

 2

>> mdist

mdist =

6.7082

如何在matlab中找到从一组点到另一组点的最小距离？

2 个答案:

解释