# Combining 3 for loops into 1 in Matlab

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Aftab Ahmed Khan on 25 Feb 2016
Commented: Star Strider on 26 Feb 2016
I have this section of code which is working fine but i don't like the way i have implemented those 3 separate for loops. Can anyone suggest to me that how can i merge them together to make it even more efficient and compact ? Thank you
clear variables;
close all;
clc;
ilambda=5;
mu=2;
n=2;
jmax=n+1;
P= sym('P',[jmax,jmax]);
for j1 = 1:jmax
for j2 = 2:jmax-1
[c1,c2,c3,c4,c5]=coefficients(ilambda,mu,j1,j2);
if j1<jmax
E(j1, j2) = c1*P(j1, j2) - c2 * P(j1+1, j2) - c3 * P(j1, j2+1) - c4 * P(j1, j2-1);
else
E(j1, j2) = c1 * P(j1, j2) - c3 * P(j1, j2+1) - c4 * P(j1, j2-1);
end
end
end
j2=1;
for j1=1:jmax;
[c1,c2,c3,c4,c5]=coefficients(ilambda,mu,j1,j2);
if (j1<jmax)
E(j1, j2) = c1*P(j1, j2) - c2 * P(j1+1, j2) - c3 * P(j1, j2+1);
else
E(j1, j2) = c1*P(j1, j2) - c3 * P(j1, j2+1);
end
end
j2=jmax;
for j1=1:jmax;
[c1,c2,c3,c4,c5]=coefficients(ilambda,mu,j1,j2);
if (j1==1)
E(j1, j2) = c1*P(j1, j2) - c2 * P(j1+1, j2) - c4 * P(j1, j2-1);
elseif (j1==jmax)
c1=((j1-1)*mu)+((j2-1)*mu);
E(j1, j2) = c1 * P(j1, j2) - c4 * P(j1, j2-1) - c5 * P(j1-1, j2);
else
E(j1, j2) = c1*P(j1, j2) - c2 * P(j1+1, j2)- c4 * P(j1, j2-1)- c5 * P(j1-1, j2);
end
end

Star Strider on 25 Feb 2016
Unfortunately, I doubt it, because coeffs cannot be vectorised (to the best of my knowledge), since the Symbolic Math Toolbox isn’t intended for such calculations.
It would be nice for you to be able to get away from the Symbolic Math Toolbox for your calculations, but I can’t figure out how to numerically get the coefficients you want. The best I can come up with is that for a statement such as:
E(j1, j2) = c1*P(j1, j2) - c2 * P(j1+1, j2) - c4 * P(j1, j2-1);
creating a coefficient matrix ‘C’ (for want of originality this morning), you could define it as:
C(j1, j2) = c1;
C(j1+1, j2) = -c2;
C(j1, j2-1) = -c4;
and so for the others. That would probably work providing you don’t overwrite them with other values later. That could eliminate the later coeffs call and the computational overhead of the Symbolic Math Toolbox, since your ‘E’ matrix just seems to be a way to store the coefficients anyway.
This just a guess. You might code it as an experiment to see if it gives you the results you want.
Aftab Ahmed Khan on 26 Feb 2016
Hi, I think your guess is right. The way i am approaching this problem is like this but i can't map those coefficients in a right order. I want to map them like the table given below.
clear variables;
close all;
clc;
jmax=3;
mu=2;
ilambda=5;
C=zeros(jmax^2);
for j1=1:jmax
for j2=1:jmax
switch j2
case 1
c1=ilambda+((j1-1)*mu)+((j2-1)*mu);
c2=(((j1-1)+1)*mu);
c3=(((j2-1)+1)*mu);
c4=ilambda;
c5=ilambda;
if (j1<jmax)
C(j1,j2)=c1;
C(j1+1,j2)=c2;
C(j1,j2+1)=c3;
else
C(j1,j2)=c1;
C(j1,j2+1)=c3;
end
case jmax
c1=ilambda+((j1-1)*mu)+((j2-1)*mu);
c2=(((j1-1)+1)*mu);
c3=(((j2-1)+1)*mu);
c4=ilambda;
c5=ilambda;
if (j1==1)
C(j1,j2)=c1;
C(j1+1,j2)=c2;
C(j1,j2-1)=c4;
elseif (j1==jmax)
c1=((j1-1)*mu)+((j2-1)*mu);
C(j1,j2)=c1;
C(j1,j2-1)=c4;
C(j1-1,j2)=c5;
else
C(j1,j2)=c1;
C(j1+1,j2)=c2;
C(j1,j2-1)=c4;
C(j1-1,j2)=c5;
end
otherwise
c1=ilambda+((j1-1)*mu)+((j2-1)*mu);
c2=(((j1-1)+1)*mu);
c3=(((j2-1)+1)*mu);
c4=ilambda;
c5=ilambda;
if j1<jmax
C(j1,j2)=c1;
C(j1+1,j2)=c2;
C(j1,j2+1)=c3;
C(j1,j2-1)=c4;
else
C(j1,j2)=c1;
C(j1,j2+1)=c3;
C(j1,j2-1)=c4;
end
end
end
end
Star Strider on 26 Feb 2016
The matrix you created looks as though it will work. It may seem awkward code, but if it maps your coefficients correctly, and is faster and more efficient than using the Symbolic Math Toolbox, that’s likely the best you can hope for. It solves the problem of getting the coefficients in the correct order, because you’re mapping them directly to your matrix. Conditional statements are difficult to make into efficient, vectorised code.
If it works, go with it!