c row matrix corresponding to the minimum value of the distance of the matrix

m=3;
t=[1 0 0; 0 (1/sqrt(2)) (1/sqrt(2))];
v1=0.24;
v2=0.956;
a=[ 1 1 0];
b=[0 1 0];
p=b-a;
d1=m*a;
for x=1:1:m+1
    c=d1+p*(x-1)
    f=transpose(c);
    vndq=t*f;
    vnq=vndq(1,1);
    vnd=vndq(2,1);
    d(x)=abs(v2-vnq)+abs(v1-vnd)
    n=min(d)
end

In the given code c row matrix corresponding minimum value of d matrix must be output if the d changes corresponding c row matrix should be abtained to that minimum distance using for loop

[Walter Roberson]

Duplicates http://www.mathworks.com/matlabcentral/answers/323064-matrix-corresponing-to-the-minimum-value-of-the-difference

Answer 1

[Andrei Bobrov]

m  =  3;
t  =  [1 0 0; 0 (1/sqrt(2)) (1/sqrt(2))];
v1  =  0.24;
v2  =  0.956;
a  =  [ 1 1 0];
b = [0 1 0];
p = b - a;
d1 = m * a;
x = (0 : m)';

c = bsxfun(@plus, d1, x * p);
v = t * c';
d = sum(abs(bsxfun(@minus,[v2;v1], v)));
[~,n] = min(d);
out = c(n,:);

with loop for..end :(

m = 3;
t = [1 0 0; 0 (1/sqrt(2)) (1/sqrt(2))];
v1 = 0.24;
v2 = 0.956;
a = [ 1 1 0];
b = [0 1 0];
p = b - a;
d1 = m * a;
n0 = numel(d1);
out = zeros(1,n0);
ii = 0;
c = zeros(m+1,n0);
n = inf;
for x = 1:m + 1
    c(x,:) = d1 + p * (x - 1);
    vndq = t * c(x,:)';
    d(x) = sum(abs([v2;v1] - vndq));
    if d(x) < n
      out = c(x,:);
      ii = x;
      n = d(x);
    end
end