Effacer les filtres
Effacer les filtres

piecewise function with N conditions

3 vues (au cours des 30 derniers jours)
Matthieu
Matthieu le 10 Fév 2023
Commenté : Walter Roberson le 12 Fév 2023
I'm trying to achieve the construction of a second maximum logic function f using symbolic expressions/functions : I want f(X) to yield the 2nd highest input Xi with X = [X1, ... , XN] .
my problem resides in the N-dependant number of conditions (I figured i couldn't use "piecewise()" ?) and the fact I'm having trouble using "maxk()" in symbolic expressions/functions.
Also, I'm using symbolic functions because I need the partiate derivatives of f.
Even if you don't have the answer, ideas and leads are very welcome !
  2 commentaires
Matt J
Matt J le 11 Fév 2023
It's a bit hard to imagine why this would be useful, since you already know the result it would lead to is going to have a highly complex symbolic form at best. The point of using symbolic math is to try to arrive at simple expressions for things, because if it is not simple, non-symbolic processing is always going to be more efficient.
Walter Roberson
Walter Roberson le 11 Fév 2023
The derivative of a min() or max() function is not continuous. Indeed, on discrete inputs, the derivative is not defined. You would need to have the inputs be expressions in free variables for a derivative to have any meaning.

Connectez-vous pour commenter.

Réponse acceptée

Walter Roberson
Walter Roberson le 11 Fév 2023
For whatever good it will do you (probably not much)
N = 5;
syms f(x) [N 1] %f will be a function
F = formula(f); %F will be an array not a function
P = perms(1:N);
FP = F(P);
pieces = [arrayfun(@(IDX) fold(@le, FP(IDX,:)), 1:size(FP,1), 'uniform', 0);
num2cell(FP(:,end-1)).'];
second_largest = piecewise(pieces{:});
dsecond = diff(second_largest, x)
dsecond = 
  3 commentaires
Walter Roberson
Walter Roberson le 12 Fév 2023
It would be possible to combine the cases together so that their was only N cases for N different variables. This way is easier though.
Walter Roberson
Walter Roberson le 12 Fév 2023
I realized that we do not need to impose a total ordering, so the expression can be much smaller.
I think you might need an "otherwise" clause. As long as you are comparing functions or expressions with free variables, then a lot of the time f1(x) <= f2(x) might not be known, or might be difficult to prove.
N = 5;
syms f(x) [N 1] %f will be a function
F = formula(f); %F will be an array not a function
parts = cell(2,N-1,N);
for NMidx = 1 : N
nominal_max = F(NMidx);
second_candidates = setdiff(1 : N, NMidx);
for M2idx = 1 : N - 1
second_best_idx = second_candidates(M2idx);
second_best = F(second_best_idx);
smaller_candidates = F(setdiff(second_candidates, second_best_idx));
parts{1, M2idx, NMidx} = second_best <= nominal_max & fold(@and, second_best >= smaller_candidates);
parts{2, M2idx, NMidx} = second_best;
end
end
sb_piecewise = piecewise(parts{:});
dsecond = diff(sb_piecewise, x)
dsecond = 

Connectez-vous pour commenter.

Plus de réponses (0)

Produits


Version

R2022b

Community Treasure Hunt

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

Start Hunting!

Translated by