piecewise function with N conditions

4 次查看（过去 30 天）
Matthieu
Matthieu 2023-2-10
评论： Walter Roberson 2023-2-12
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 个评论
Matt J
Matt J 2023-2-11
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 2023-2-11
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.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Walter Roberson
Walter Roberson 2023-2-11
Ran in:
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 个评论
显示 1更早的评论
Walter Roberson
Walter Roberson 2023-2-12
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 2023-2-12
Ran in:
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 = 

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Code Performance 的更多信息

标签

产品


版本

R2022b

Community Treasure Hunt

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

Start Hunting!

Translated by