Numerical answer to Partial derivative of Hopfield-style Energy in symbolic expression

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Hello,
I'm implementing a gradient descent optimizer that uses Hopfield-style energy as function,
the hopfield-style energy is a function of p:
The hopfield energy equation can also be expressed as:
The gradient vector is defined as:
I'm attempting to calculate the gradient vector using the Symbolic Toolbox as follows:
r_mat = sym('r',[4,4]);
r_mat = r_mat(:);
p_mat = sym('p',[4*4,1]);
c_mat = sym('c',[4*4,1]);
Energy = -0.5*p_mat'*r_mat*p_mat' - c_mat'*p_mat;
%the first partial derivative:
q1 = -1*diff(testi1,p_mat(1,1));
which yields as the first term:
conj(c1) + r1_1*conj(p1) + (r1_2*conj(p5))/2 + (r1_3*conj(p9))/2 + (r2_1*conj(p2))/2 + (r1_4*conj(p13))/2 + (r2_2*conj(p6))/2 + (r2_3*conj(p10))/2 + (r3_1*conj(p3))/2 + (r2_4*conj(p14))/2 + (r3_2*conj(p7))/2 + (r3_3*conj(p11))/2 + (r4_1*conj(p4))/2 + (r3_4*conj(p15))/2 + (r4_2*conj(p8))/2 + (r4_3*conj(p12))/2 + (r4_4*conj(p16))/2
How can I replace the symbols with numbers to yield the q vector? I need the q-vector for the gradient descent
the algorithm I'm implementing: https://doi.org/10.1117/1.JEI.23.1.013007 (relevant part starts at Equation 5)

Risposta accettata

Torsten
Torsten il 28 Lug 2022
Modificato: Torsten il 28 Lug 2022
p is (4x1), not (16x1).
And keep r_mat as matrix, then
E = -0.5*p.'*r_mat*p - c.'*p
( r_mat can be assumed symmetric ).
By the way:
dE/dp = -r_mat*p - c
assuming r_mat is symmetric.
To replace symbols by numerical values, use "subs".
  5 Commenti
Pseudoscientist
Pseudoscientist il 2 Ago 2022
It turned out symbolic expressions was extremely slow, I used symbolic to calculate the partial derivative and then used that to think how to write the equation in vector form and was evetually succesful and the optimizer is now very fast !
Torsten
Torsten il 2 Ago 2022
I used symbolic to calculate the partial derivative and then used that to think how to write the equation in vector form
That's the best you can do, I guess.

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