HOW Calculate the distance of points form one center in 2-D space and display output in a distance matrix?

Hello all, I am new to Matlab and I am having quite a bit of difficulty understanding how to calculate the distance between points with equal distances in a 2-D space. I need to develop a function to create an array of 5 points and find the distances between them. The output of the function will be a distance matrix wich size is 5*5.

I know about "pdist2" but I canot develop this for my problem. I need this matrix(look at the picture). The out put matrix should be a 5*5 Symmetric matrix and the elemet of diagonal should be zero( because the distance of every point with each self will be zero)

[Rik]

What you tried so far? Have you figured out how to do it with pen and paper but struggle to implement it?

[sona nil]

I want to obtain this matrix.yes I can do it with pen. but I can not develop it in matlab.

[Rik]

And why can't you use pdist2?

[Rik]

Regarding the now deleted answer by Bruno: that code used implicit expansion which was introduced in R2016b. On prior releases you had to use bsxfun. You didn't specify your release in your question, which makes the process a bit more difficult.

[Rik]

I don't really see what is different about your edited question. Can you explain in a comment what your remaining issue is? Why don't the answers by Andrei and myself work for you?

[sona nil]

Did you see my other comment?

Because one of the assumptions of my question is that the distance is equal. look at this picture.

[Rik]

That is a totally different question. You never said you needed to generate the coordinates as well. Our functions work for any set of points. If you want to randomly generate a point cloud with exactly the shape you describe, you can do this:

x_center=rand;
y_center=rand;
h=randi([2 5]);
X=x_center+[0 0 h  0 -h];
Y=y_center+[0 h 0 -h  0];

Or this:

x_center=rand;
y_center=rand;
h=randi([2 5]);
phi_offset=2*pi*rand;

phi=[0.5 0 -0.5 -1]*pi;
X=x_center+[0 h*cos(phi_offset+phi)];
Y=y_center+[0 h*sin(phi_offset+phi)];

[sona nil]

fine, I think this is what I need. Thank you Rik.

Answer 1

[Rik]

This should do the trick:

X=2*(rand(5,5)-0.5);
Y=2*(rand(5,5)-0.5);
x=0.3;y=0.3;
d=point2pointcloud(X,Y,x,y)

function d=point2pointcloud(X,Y,x,y)
%X and Y are the coordinates of the point cloud
%x and y are the coordinates of the point
if ~isequal(size(X),size(Y)) || ~strcmp(class(X),class(Y))
    error('X and Y must be same size and type')
end
if numel(x)~=1 || numel(y)~=1 || ~strcmp(class(x),class(y))
    error('x and y must be scalar and same type')
end
if ~strcmp(class(x),class(X))
    error('all inputs must be same type')
end
d=sqrt((X-x).^2+(Y-y).^2);
end

Edit:

To make the function easier to find, I'll paste it here. This function seems to return the required matrix.

function d=pointcloud_dist(X,Y)
%X and Y are the coordinates of the point cloud
if ~isequal(size(X),size(Y)) || ~strcmp(class(X),class(Y))
    error('X and Y must be same size and type')
end
[ind1,ind2]=ndgrid(1:numel(X));
X1=X(ind1);X2=X(ind2);
Y1=Y(ind1);Y2=Y(ind2);
d=sqrt((X1-X2).^2+(Y1-Y2).^2);
end

[sona nil]

Thanks Rik for your answer .I tried your code .But the out put matrix should be a Symmetric matrix and the elemet of diagonal should be zero( because the distance of every point with each self will be zero)

[Rik]

Note that this ignores the center point you were talking about earlier.

X=2*(rand(5,5)-0.5);
Y=2*(rand(5,5)-0.5);
d=pointcloud_dist(X,Y);

function d=pointcloud_dist(X,Y)
%X and Y are the coordinates of the point cloud
if ~isequal(size(X),size(Y)) || ~strcmp(class(X),class(Y))
    error('X and Y must be same size and type')
end
[ind1,ind2]=ndgrid(1:numel(X));
X1=X(ind1);X2=X(ind2);
Y1=Y(ind1);Y2=Y(ind2);
d=sqrt((X1-X2).^2+(Y1-Y2).^2);
end

[sona nil]

yes you are right,you suppose that we dont consider center. your code is worked . your code obtains a Symmetric matrix and the elemet of diagonal is zero(my target). but the matrix size is 25*25.

but I need 5*5 matrix(the first row shows the distance from point 1 from others(other four points), the second row shows the distance from point 2 to others,.....) so we will have 5 rows and five columns.

[Rik]

My input was 25 points, so if you don't provide my 25 randomly generated points but put in your 5 points instead, you will get a 5*5 matrix.

[sona nil]

Thanks rik. I learned applicable information from you.

I did your suggestion about the release of matlab.

you answered all of my question, but one thing is wrong.

If you saw the picture of the points. They have equal distances(distance is contsant) but when we use "rand " it generates some random number with unequal distances.

Is there a code in matlab that generates random number with equal distance in (0 1)?

for example (0 , 0.1) , (0.1 , 0.1) , (0.2 , 0.1),(0.1 , 0),(0.1 , 0.2)

x=[0 0.1 0.2]

y=[0 0.1 0.2]

Answer 2

[Andrei Bobrov]

X = rand(5,1)
Y = rand(5,1)
D = squareform(pdist([X,Y]))

or

XY = [X, Y];
D = sqrt(squeeze(sum((XY - permute(XY,[3,2,1])).^2,2)))
%For old  MATLAB
D = sqrt(squeeze(sum(bsxfun(@minus,XY,permute(XY,[3,2,1])).^2,2)))

[sona nil]

Thanks andrei for your useful and brief answer.

Is there a code in matlab that generates random number with equal distance in (0 1)?

for example (0 , 0.1) , (0.1 , 0.1) , (0.2 , 0.1),(0.1 , 0),(0.1 , 0.2)

x=[0 0.1 0.2]

y=[0 0.1 0.2]