finding index-positions of a composed-matrix

Question

I need help with my code. The code is used to find the minumin of a square-distance problem. I am providing my code through an example, I believe this will be the easiest way to explain what I need.

clear all
clc
x=10.8; % is a fixed value
y=34; % is a fixed value
z=12; % is a fixed value
A = [11 14 1; 5 8 18; 10 8 19; 13 20 16]; % a (4x3) matrix
B = [2 3 10; 6 15 16; 7 3 15; 14 14 19]; % a (4x3) matrix

I create a new matrix C which is composed in this following way:

C1 = bsxfun(@minus, A(:,1)',B(:,1)); 
C1=C1(:); % this is the first column of the new matrix C
C2 = bsxfun(@minus, A(:,2)',B(:,2));
C2=C2(:); % this is the second column of the new matrix C
C3 = bsxfun(@minus, A(:,3)',B(:,3));
C3=C3(:); % this is the third column of the new matrix C
C = [C1 C2 C3]; % the new matrix C of size (16x3)

C has to be formed in this way! And this is what I meant when I wrote in my title a composed-matrix

Then:

[d,p] = min((C(:,1)-x).^2 + (C(:,2)-y).^2 + (C(:,3)-z).^2);
d = sqrt(d);
outputs:
d = 18.0289;
p = 13;

Gives me the distance (d) and position (p) which satisfies this min problem.

MY PROBLEM: I need to find which combinations of A and B has given my this p value, in other words I need the index from ´A,B´ which gives me this optimal C1,C2,C3:

C1 = bsxfun(@minus, A(?,1)',B(?,1));
C2 = bsxfun(@minus, A(?,2)',B(?,2));
C3 = bsxfun(@minus, A(?,3)',B(?,3));

The ? is the index position I need, in this case the index position of the matrix A and the index position of B.

Calculated by hand I have the following illustration:

I know that:

C = [9    11    -9
 5    -1   -15
 4    11   -14
-3     0   -18
 3     5     8
-1    -7     2
-2     5     3
-9    -6    -1
 8     5     9
 4    -7     3
 3     5     4
-4    -6     0
11    17     6
 7     5     0
 6    17     1
-1     6    -3]

And I know that my optimal index is given in the position 13th. This index positions goes back to:

[13-2 20-3 16-10]

Which is A(4,:) - B(1,:)

I need a code which can help me to find this indexes from A and B

Thanks in advance!

PS. I am using the code in parameter estimation problems of ODEs.

Answer 1

First case: vector-matrix case

subvals = bsxfun(@minus,A,[x y z])
[distance,index] = min(sqrt(sum(subvals.^2,2)))

Second case: Two matrices case

subvals = bsxfun(@minus,A,permute(B,[3 2 1]));
[distances,indices] = min(sqrt(sum(subvals.^2,2)),[],3)

Testing for second case:

%%// Get some random data into A and B
A = randi(20,8,3)
B = randi(20,4,3)

%%// Just to test out out code for correctness, 
%%// let us make any one one row of B, say 3rd row equal to 
%%// any one row of A, say the 6th row -
B(3,:) = A(6,:)

%%// Use the earlier code
subvals = bsxfun(@minus,A,permute(B,[3 2 1]));
[distances,indices] = min(sqrt(sum(subvals.^2,2)),[],3)

%%// Get the minimum row index for A and B
[~,min_rowA] = min(distances)
min_rowB = indices(min_rowA)

Verification

min_rowA =
     6

min_rowB =
     3

Edit 1 [Response to simple example posted in question]:

The title says you are interested in finding the difference of two matrices and then find the shortest distance between it to a vector [x y z]. So I am hoping this is what you need -

x=10.8;
y=34;
z=12;
A = [11 14 1; 5 8 18; 10 8 19; 13 20 16];
B = [2 3 10; 6 15 16; 7 3 15; 14 14 19];

C = A -B; %%// Distance of two vectors as posted in title
subvals = bsxfun(@minus,C,[x y z])
[distance,index] = min(sqrt(sum(subvals.^2,2)))

Output

distance =
   31.0780

index =
     3

Edit 2: After you have done this -

[d,p] = min((C(:,1)-x).^2 + (C(:,2)-y).^2 + (C(:,3)-z).^2);

If you are looking to find the corresponding indices of A and B , you may do this -

[minindex_alongB,minindex_alongA] = ind2sub(size(A),p)