matric "c" and "d" out of for loop and c row matrix corresponding to minimum distance

m=2
T=[1 0 0; 0 (1/sqrt(2)) (1/sqrt(2))];
V1=0.956
V2=2.4
A=[ 1 0 0];
    B=[1 1 0];
    D=B-A;
    d1=m*A;
    for X=1:1:m+1
    c(x,:)=d1+D*(x-1)
    Vndq=T*c(x,:)';
    Vnq=Vndq(1,1)
    Vnd=Vndq(2,1)
    d(x)=abs(V2-Vnq)+abs(V1-Vnd)
    end

in the code given minimum distance is

 d =[1.3560    0.6489    0.8582]

and

 C =
     2     0     0
     2     1     0
     2     2     0

c row matrix 2 1 0 of the minimum distance 0.6489 should be obtained

Answer 1

[Stephen23]

Here is a simpler version of your code:

m  = 2;
T  = [1,0,0;0,(1/sqrt(2)),(1/sqrt(2))];
V1 = 0.956;
V2 = 2.4;
c = zeros(m+1,3);
c(:,1) = m;
c(:,2) = 0:m;
Vndq = T*c.'
d = sum(abs(bsxfun(@minus,[V2;V1],Vndq)),1)

Use min to get the closest value:

>> [val,idx] = min(d)
val =
     0.64889
idx =
     2
>> c(idx,:)
ans =
     2     1     0