# Reduce length of sum of products of symbolic variables

3 views (last 30 days)
Moritz Schappler on 21 Nov 2022
Commented: Paul on 22 Nov 2022
I am using Matlab symbolics toolbox to simplify an expression which is a sum of many single-variable products, where variables re-occur:
G1*P3 + G1*P5 + G1*P6 + G2*P3 + G2*P6 + G3*P2 + G3*P5 + G4*P2 + G4*P3 + G4*P5 + G4*P6 + G5*P2 + G5*P3 + G5*P5 + G5*P6 + G6*P3 + G6*P6 + G7*P2 + G7*P5 + G8*P2 + G8*P3 + G8*P5 + G8*P6
(created by using a loop with sym(sprintf('G%d', i))
Using simplify does not change the expression. I tried a custom command using collect(tmp,sym(sprintf('P%d', i))) in a loop. However, this simplifies up to
(G3 + G4 + G5 + G7 + G8)*P2 + (G1 + G2 + G4 + G5 + G6 + G8)*P3 + (G1 + G2 + G4 + G5 + G6 + G8)*P6 + (G1 + G3 + G4 + G5 + G7 + G8)*P5
How do I get
(G1 + G2 + G4 + G5 + G6 + G8)*(P3+P6) + (G1 + G3 + G4 + G5 + G7 + G8)*P5 + (G3 + G4 + G5 + G7 + G8)*P2
which is shorter and which you can get from directly looking at the expression? I started creating my own logic for simplification, but that seems not feasible.
I already tried this in Maple (with help by the Maple forum), but I have the expression in Matlab (and hundreds others, which are similiar). I want to omit calling Maple from Matlab for this.
A similar solution in Matlab does not work for me (tmp is expression):
matlabFunction(tmp, 'File', filename_test, 'Optimize', true)
The generated code file is unchanged relative to the initial expression.

Paul on 21 Nov 2022
Edited: Paul on 21 Nov 2022
Hi Moritz,
Would you mind explaining why it's necessary to get the expression into a particular form?
Recreating the expression
G=sym('G',[1 8]);
syms(G)
P=sym('P',[1 8]);
syms(P)
E = G1*P3 + G1*P5 + G1*P6 + G2*P3 + G2*P6 + G3*P2 + G3*P5 + G4*P2 + G4*P3 + G4*P5 + G4*P6 + G5*P2 + G5*P3 + G5*P5 + G5*P6 + G6*P3 + G6*P6 + G7*P2 + G7*P5 + G8*P2 + G8*P3 + G8*P5 + G8*P6
E =
Surprised that simplify can't do better.
simplify(collect(E,P),100)
ans =
One option might be to use coeffs.
[co,terms] = coeffs(E,P)
co =
terms =
Use unique to find unique coeffiecients.
[uniqueco,iuniqueco,ico] = unique(co)
uniqueco =
iuniqueco = 3×1
1 2 3
ico = 4×1
1 2 3 2
Loop through the unique coefficients and combine with sum of corresponding terms:
Enew = 0;
for ii = 1:numel(unique(co))
Enew = Enew + uniqueco(ii)*sum(terms(ico == ico(ii)));
end
Enew
Enew =
I don't know how robust unique is for vectors of sym expressions. It may be better to loop through manually and find matches using isAlways.
##### 2 CommentsShowHide 1 older comment
Paul on 22 Nov 2022
Hi Moritz,
Thank you for pointing out that error. The problem isn't with unique(), rather it was a bug in my use of unique. Here is an update that works for both of the examples. It uses unique to find the unique coefficients and then loops over those to find the correct mapping to the P terms
G=sym('G',[1 9]);
syms(G)
P=sym('P',[1 9]);
syms(P)
% first example
E = G1*P3 + G1*P5 + G1*P6 + G2*P3 + G2*P6 + G3*P2 + G3*P5 + G4*P2 + G4*P3 + G4*P5 + G4*P6 + G5*P2 + G5*P3 + G5*P5 + G5*P6 + G6*P3 + G6*P6 + G7*P2 + G7*P5 + G8*P2 + G8*P3 + G8*P5 + G8*P6;
Enew = func(E,P);
E,Enew,simplify(Enew - E)
E =
Enew =
ans =
0
% second example
E = G1*P2 + G2*P1 + G1*P3 + G3*P1 + G2*P3 + G3*P2 + G1*P8 + G9*P1 + G2*P9 + G9*P8;
Enew = func(E,P);
E,Enew,simplify(Enew - E)
E =
Enew =
ans =
0
function Enew = func(E,P)
[co,terms] = coeffs(E,P);
uniqueco = unique(co);
Enew = 0;
index = 1:numel(co);
for ii = 1:numel(uniqueco)
Enew = Enew + uniqueco(ii)*sum(terms(index(isAlways(uniqueco(ii) == co,'unknown','false'))));
end
end

### Categories

Find more on Number Theory in Help Center and File Exchange

R2022b

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by