nested for loop
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I can't figure out why the S matrix is computed to have the same value for every element - 0.7654. All other matrices (A through E) are computed correctly.
format short
x=[-0.9239, -0.6533, 0, 0.6533, 0.9239, 0.6533, 0, -0.6533]; %yj
y=[0, 0.6533, 0.9239, 0.6533, 0, -0.6533, -0.9239, -0.6533]; %yi
X=[-0.9239, -0.9239, -0.3827, 0.3827, 0.9239, 0.9239, 0.3827, -0.3827]; %Xj
Xone=[-0.9239, -0.3827, 0.3827, 0.9239, 0.9239, 0.3827, -0.3827, -0.9239];%Xj+1
Y=[-0.3827, 0.3827, 0.9239, 0.9239, 0.3827, -0.3827, -0.9239, -0.9239]; %Yj
Yone=[0.3827, 0.9239, 0.9239, 0.3827, -0.3827, -0.9239, -0.9239, -0.3827]; %Yj+1
phi=pi/180*[90, 45, 0, 315, 270, 225, 180, 135]; %phi
for i=1:8
for j=1:8
A(i,j)=-(x(i)-X(j)).*cos(phi(j)) - (y(i)-Y(j)).*sin(phi(j));
B(i,j)=(x(i)-X(j)).^2 + (y(i)-Y(j)).^2;
C(i,j)=sin(phi(i) - phi(j));
D(i,j)=(y(i)-Y(j)).*cos(phi(i)) - (x(i)-X(j)).*sin(phi(i));
E(i,j)=(x(i)-X(j)).*sin(phi(j)) - (y(i)-Y(j)).*cos(phi(j));
S(i,j)=((Xone(j)-X(j))^2 + (Yone(j)-Y(j))^2)^0.5;
end
end
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Accepted Answer
Matt Fig
on 8 May 2011
The way you have defined S, it is independent of index i.
S(i,j) = ((Xone(j)-X(j))^2 + (Yone(j)-Y(j))^2)^0.5; % No i appears.
Therefore it will have the same value for every element of each column. Did you mean to have no dependence on index i?
4 Comments
Matt Fig
on 8 May 2011
You probably wouldn't notice the difference with such small arrays. But in general, this will make your code faster. This code can be vectorized with multiple calls to BSXFUN, but simply removing the inner FOR loop and dynamically pre-allocating the arrays would probably help speed-wise:
cosphi = cos(phi);
sinphi = sin(phi);
Sv = ((Xone-X).^2 + (Yone-Y).^2).^0.5;
for ii = 8:-1:1
A2(ii,:) = -(x(ii)-X).*cosphi - (y(ii)-Y).*sinphi;
B2(ii,:) = (x(ii)-X).^2 + (y(ii)-Y).^2;
C2(ii,:) = sin(phi(ii) - phi);
D2(ii,:) = (y(ii)-Y).*cos(phi(ii)) - (x(ii)-X).*sin(phi(ii));
E2(ii,:) = (x(ii)-X).*sinphi - (y(ii)-Y).*cosphi;
S2(ii,:) = Sv;
end
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