Tensor product of 3 arrays

Question

I have an equation to count, that looks like this:

F(a,b) = sum(c=1...n, a*b*c[i])

I can count it this way:

for a=
  for b=
    for c=
    F(a,b)=F(a,b)+a*b*c
    end
  end
end

But I've heard, that on big arrays matlab "freezes" on nested loops. So I can do it this way:

a=(1:n)';
b=1:n2;
Fs=a*b;
for c=
   F=F+Fs*c;
end

But I want to solve this problem without any visible cycle. So I can create three orthogonal arrays of a,b,c tensor product them to get 3-D array, and then use sum function by third dimension.

a(:,1,1)=1:3;
b(1,:,1)=4:6;
c(1,1,:)=7:9;
d=tensorprod (a,b,2,1)
e=tensorprod (d,c);

But I ran into tensor multiplication problem in matlab. e has to be 3x3x3 array, but it is 3x3x1x1x3 array. It is correct and all, and even

g=sum(e,5);

returns matrix, but I don't understand why the third dimension has moved to the fifth

But I've heard, that on big arrays matlab "freezes" on nested loops -> Where have you heard this? MATLAB does fine with big loops, but they're not always necessary. It's often easier to get something working with loops first, otherwise you're doing premature optimisation at the expense of having a minimal reproducible example. Do you have something which works with a loop? Is it too slow? You could show us that, presumably it's not too slow for some small input... — Wolfie
– Wolfie, Commented Sep 9, 2022 at 16:39

aitzkora · Accepted Answer · 2022-09-09 22:54:06Z

1

I'm not sure that a no loop version is better. Using a broadcast (Matlab took that from Fortran and numpy), you could do it without loop, but I think the java JIT compiler of matlab could give better results with simple loops, nevertheless, a no-loop version could be

n2=3;
n2=4;
n3=5;
a=1:n1;
b=1:n2;
Fs=a'.*b;
F=zeros(size(Fs))
for c=1:n3
   F=F+Fs*c;
end
# the broadcast : all dimensions equal to one will be duplicate
# for the other tensor : after that we contract the tensor along the 
# third dimension
F2 = sum(Fs(:,:,1) .* reshape(1:n3,1,1,5), 3)
norm(F2-F,2)

answered Sep 9, 2022 at 22:54

aitzkora

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AboAmmar · Accepted Answer · 2022-09-09 19:06:21Z

1

Based on your loop, this should give the desired sum.

a = 1:3;
b = 4:6;
c = 7:9;
F = a' * b * sum(c) 

F = 
    96   120   144
   192   240   288
   288   360   432

answered Sep 9, 2022 at 19:06

AboAmmar

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2 Comments

aitzkora Over a year ago

If I test this code, the resultat is wrong, but mathematically I do not see why il would be illegal to put a'*b outer of the sum.

lazba Over a year ago

I've used a bad example, my wrong. The real equation ( it counts radiation pattern) is more complicated, and has more parts, so I cannot do it this way, but as my problem was in tensor dimension, I've decided not to use the whole in this MWE

mhopeng · Accepted Answer · 2022-09-09 17:01:14Z

0

You can do this with vectors and a dot product (inner product):

a=1:3;
b=4:6;
c=7:9;
D=dot(a,b);
E=sum(D*c)

E =

   768

answered Sep 9, 2022 at 17:01

mhopeng

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1 Comment

lazba Over a year ago

Answer has to be matrix, so I need outer product, not inner

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