# MATLAB: Matric “c” and “d” out of for loop and c row matrix corresponding to minimum distance

matric "c" and "d" out of for loop

m=2T=[1 0 0; 0 (1/sqrt(2)) (1/sqrt(2))];V1=0.956V2=2.4A=[ 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

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