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

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JAGADEESH JAGA on 3 Feb 2017
Commented: Walter Roberson on 3 Feb 2017
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

Andrei Bobrov on 3 Feb 2017
Edited: Andrei Bobrov on 3 Feb 2017
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