How to do a sum of sums of the square of sum of sums?

Question

I have a sum of sums that I want to speed up. In one case it is:

S_{x,y,k,l} Fu_{ku} Fv_{lv} Fx_{kx} Fy_{ly}

In the other case it is:

S_{x,y} ( S_{k,l} F

Answer 1

I'll start a new answer since the problem has changed.

Try this:

E = np.einsum('uk, vl, xk, yl, xy, kl->uvxy', Fu, Fv, Fx, Fy, P, B)
E1 = np.einsum('uvxy->uv', E)
E2 = np.einsum('uvxy->uv', np.square(E))

I've found it runs just as fast as the time for I1_.

Here is my test code: http://pastebin.com/ufwy7cLy

Answer 2

(Update: Jump to the end to see the result expressed as a couple of matrix multiplications.)

I think you can greatly simplify the computation by using the identity:

For instance,

S_{k,l} Fu_{ku} Fv_{lv} Fx_{kx} Fy_{ly}
  = S_{k,l} Fu_{ku} Fx_{kx} Fv_{lv} Fy_{ly}            -- rearrange the factors
            \___ A ____/    \___ B ____/
  = ( S_k Fu_{ku} Fx_{kx} ) * ( S_l Fv_{lv} Fy_{ly} )  -- from the identity
  =   A_{ux}                * B_{vy}

where A_{ux} only depends on u and x and B_{vy} only depends on v and y.

For the square sum, we have:

S_k [ S_l Fu_{ku} Fv_{lv} Fx_{kx} Fy_{ly} ]^2
  = S_k Fu_{ku} Fx_{kx} * [ S_l Fv_{lv} Fy_{ly} ]^2
  = S_k Fu_{ku} Fx_{kx} * B_{vy}^2                 -- B is from the above calc.
  = B_{vy}^2 * S_k Fu_{ku} Fx_{kx}                 -- B_vy is free of k
  = B_{vy}^2 * A_{ux}                              -- A is from the above calc.

Similar reductions occur when continuing the sum over x and y:

S_{xy} A_{ux} * B_{vy}
  = S_x A_{ux} * S_y B_{vy}                        -- from the identity
  =  C_u       *    D_v

And then finally summing over u and v:

S_{uv} C_u D_v = (S_u C_u) * (S_v D_v)             -- from the identity

Hope this helps.

Update: I just realized that perhaps for the square sum you wanted to compute [ S_k S_l ... ]^2 in which case you can proceed like this:

[ S_k  S_l Fu_{ku} Fv_{lv} Fx_{kx} Fy_{ly} ]^2
  =  [ A_{ux}                * B_{vy} ]^2
  =  A_{ux}^2 * B_{vy}^2

So when we sum over the over variables we get:

S_{uvxy} A_{ux}^2 B_{vy}^2
  = S_{uv} ( S_{xy}  A_{ux}^2 B_{vy}^2 )
  = S_{uv} ( S_x A_{ux}^2 ) * ( S_y B_{vy}^2 )     -- from the identity
  = S_{uv}     C_u          *      D_v
  = (S_u C_u) * (S_v D_v)                          -- from the identity

Update 2: This does boil down to just a few matrix multiplications.

The definitions of A and B:

A_{uv} = S_k Fu_{ku} Fx_{kx}
B_{vy} = S_l Fv_{lv} Fy_{ly}

may also be written in matrix form as:

A = (transpose Fu) . Fx             -- . = matrix multiplication
B = (transpose Fv) . Fy

and the definition of C and D:

C_u = S_x A_{ux}
D_v = S_y B_{vy}

we see that the vector C is just the row sums of A and the vector D is just the row sums of B. Since the answer for the entire summation (not squared) is:

total = (S_u C_u) * (S_v D_v)

we see that the total is just the sum of all of the matrix elements of A times the sum of all of the matrix elements of B.

Here is the numpy code:

from numpy import *
# ... set up Fx, Fv, Fu, Fy as above...

A = Fx.dot(Fu.transpose())
B = Fv.dot(Fy.transpose())
sum1 = sum(A) * sum(B)

A2 = square(A)
B2 = square(B)
sum2 = sum(A2) * sum(B2)

print "sum of terms:", sum1
print "sum of squares of terms:", sum2