i want to find distances between coordinate points as i explained below?? any help??

i have coordinates as

A = (0,0); B = (1,0); C = (2,0)

D = (0,1); E = (1,1); F = (2,1)

G = (0,2); H = (1,2); I = (2,2)

i want to distance between A to D plus D to B and so on......

matrix 1 (2*3)

[(AtoD+DtoB) (AtoE+EtoB) (AtoF+FtoB)

(AtoG+GtoB) (AtoH+HtoB) (AtoI+ItoB)

matrix 2 (2*3)

[(AtoD+DtoC) (AtoE+EtoC) (AtoF+FtoC)

(AtoG+GtoC) (AtoH+HtoC) (AtoI+ItoC)

matrix 3 (2*3)

[(BtoD+DtoA) (BtoE+EtoA) (BtoF+FtoA)

(BtoG+GtoA) (BtoH+HtoA) (BtoI+ItoA) ]

matrix 4 (2*3)

[(BtoD+DtoC) (BtoE+EtoC) (BtoF+FtoC)

(BtoG+GtoC) (BtoH+HtoC) (BtoI+ItoC) ]

matrix 5 (2*3)

[(CtoD+DtoA) (CtoE+EtoA) (CtoF+FtoA)

(CtoG+GtoA) (CtoH+HtoA) (CtoI+ItoA) ]

matrix 6 (2*3)

[(CtoD+DtoB) (CtoE+EtoB) (CtoF+FtoB)

(CtoG+GtoB) (CtoH+HtoB) (CtoI+ItoB) ]

finally i want six matrices as i explained above

thank you

[KSSV]

Read about pdist and pdist2

[darova]

Did you try something? Any ideas?

[M.Prasanna kumar]

should i use 4D matrix???

[M.Prasanna kumar]

no_gridsx = 3;

no_gridsy = 3;

no_receivers = no_gridsy-1;

step_x = 1; %spacing between grids 0.5 cm in x-direction

step_y = 1; %spacing between grids 0.5 cm in y-direction

[X,Y] = meshgrid(0:step_x:no_gridsy-1, 0:step_y:no_gridsx-1);

for kk = 1:no_receivers

for ii = 1:no_gridsx-1

for jj =1:no_gridsy

dist(ii,jj,kk) = sqrt((X(1,1)-X(ii+1,jj))^2+(Y(1,1)-Y(ii+1,jj))^2)+sqrt((X(ii+1,jj)-X(1,kk+1))^2+(Y(ii+1,jj)-Y(1,kk+1))^2)

end

end

end

I am able to get first two matrices ..

how to get next four matrices any help would be greatly appreciated

[darova]

matrix 1 == matrix 3

matrix 2 == matrix 5

matrix 4 == matrix 6

What do you think? Does it make your task easier?

[M.Prasanna kumar]

no they are not equal..they are different sir

Answer 1

[darova]

With use of pdist2()

A = [0 0];    B = [1 0];    C = [2 0];
[x,y] = meshgrid(0:2,1:2);        % D G E H F I
D_I = [x(:) y(:)];
AtoXY = reshape(pdist2(A, D_I),size(x));    % matrix distance from A to (D G E H F I)
BtoXY = reshape(pdist2(B, D_I),size(x));    % distance from B
CtoXY = reshape(pdist2(C, D_I),size(x));    % distance from C
matrix1 = AtoXY + BtoXY;
matrix2 = AtoXY + CtoXY;
matrix4 = BtoXY + CtoXY;

[M.Prasanna kumar]

sorry sir what u said is correct i did not notice it

thanks for ur answer

[M.Prasanna kumar]

if i have 90*90 grid then i have to write so many lines right??

can we put that in to loop??

[darova]

The scheme is the same? If a grid 90x90 then each matrix is 89x90? How many matrix will be then?

3x3 grid: 3 matrices

4x4 grid: 6 matrices

5x5 grid: 10 matrices

NxN grid: matrices

[M.Prasanna kumar]

the y grid value of y is 90. so no. of matrices are 89+88+87+86+.......+1 = 4005 matrices

[darova]

Can you please show how to calculate matrix for 4x4 grid?

[M.Prasanna kumar]

i will explain with your code

A = [0 0]; B = [1 0]; C = [2 0];

[x,y] = meshgrid(0:2,1:2); % D G E H F I

D_I = [x(:) y(:)];

AtoXY = reshape(pdist2(A, D_I),size(x)); % matrix distance from A to (D G E H F I)

BtoXY = reshape(pdist2(B, D_I),size(x)); % distance from B

CtoXY = reshape(pdist2(C, D_I),size(x)); % distance from C

matrix1 = AtoXY + BtoXY;

matrix2 = AtoXY + CtoXY;

matrix4 = BtoXY + CtoXY;

1. lets fix A (A to B via (x y))

from distance from A to every point in (x y) plus from every point in (x y) to B = matrix 1

2. fix A (A to C via (x y))

from distance from A to every point in (x y) plus from every point in (x y) to C = matrix 2

3. fix B (B to A via (x y))

from distance from B to every point in (x y) plus from every point in (x y) to A = matrix 3

3. fix B and C(B to C via (x y))

from distance from B to every point in (x y) plus from every point in (x y) to A = matrix 4

3. fix C (C to A via (x y))

from distance from B to every point in (x y) plus from every point in (x y) to A = matrix 5

3. fix C (C to B via (x y))

from distance from B to every point in (x y) plus from every point in (x y) to A = matrix 6

as u said

matrix 1 == matrix 3

matrix 2 == matrix 5

matrix 4 == matrix 6

when we fix A remaing points are B and C (SO WE GET TWO MATRICES)

when we fix B remaing points are A and C (SO WE GET TWO MATRICES)

when we fix C remaing points are A and B (SO WE GET TWO MATRICES)

[darova]

How many matrices we'll get for 4x4 grid?

[M.Prasanna kumar]

we will get total 20 matrices

(as per your logic we get 10 matrices)

[M.Prasanna kumar]

4*4 grid means we will get 5 cordinates in x and y direction

[M.Prasanna kumar]

IN a grid of 4*4

A = (0 0); B = (1 0) ; C=(2 0) ;D = (3 0); E = (4 0)

A1 = (0 1); B1 = (1 1) ; C1=(2 1) ;D1 = (3 1); E1 = (4 1)

A2 = (0 2); B2 = (1 2) ; C2=(2 2) ;D2 = (3 2); E2 = (4 2)

A3= (0 3); B3= (1 3) ; C3=(2 3) ;D3 = (3 3); E3= (4 3)

A4 = (0 4); B4 = (1 4) ; C4=(2 4) ;D4 = (3 4); E4 = (4 4)

here A1 to E4 are x and y coordinates as u defined in ur code

from A to all x,y coordinates and from x,y coordinates to B we get matrix 1

from A to all x,y coordinates and from x,y coordinates to C we get matrix 2

from A to all x,y coordinates and from x,y coordinates to C we get matrix 3

from A to all x,y coordinates and from x,y coordinates to D we get matrix 4

from A to all x,y coordinates and from x,y coordinates to E we get matrix 5

I explained for A to

similarly for B,C,D,E

[darova]

It should work

clc,clear

n = 4;

AA = [0:(n-1); zeros(1,n)]';              % top row: A B C ...
[x,y] = meshgrid(0:n-1,1:n-1);            % XY grid
XY = [x(:) y(:)];                         % make vector
D = pdist2(AA,XY);                        

% AtoXY = reshape(D(1,:),size(x))
% BtoXY = reshape(D(2,:),size(x))
% CtoXY = reshape(D(3,:),size(x))
%     ...

k = 0;
matrix = cell((n-1)*n/2,1);
    % get every possible combination
    % [1 2], [1 3], [1 4] ...
    % [2 3], [2 4] ...
    % ...
for i = 1:n-1
    for j = i+1:n
        k = k + 1;
        d = D(i,:)+D(j,:);
        matrix{k} = reshape(d,size(x));
    end
end

[M.Prasanna kumar]

sir,

thank your very much for your kind reply. you were replying very patiently

code is working perfectly

[darova]

You are welcome