I cant seem to figure this out on my own. I need to minimize the following function where xi and yi are given data sets (points in 2D),
f(x,y) = minimize{maximize[sqrt((xi-x)^2+(yi-y)^2)]-minimize[sqrt((xi-x)^2+(yi-y)^2)]}
I will need to find x,y.

 Réponse acceptée

infinity
infinity le 27 Juin 2019
Modifié(e) : infinity le 27 Juin 2019

0 votes

Hello,
Here is an example for your problem, which you can refer
clear
xiyi = [0,1;
1,3;
-1 2];
x0 = zeros(size(xiyi));
fun = @(x) max(sqrt((x(:,1)-xiyi(:,1)).^2 + (x(:,2)-xiyi(:,2)).^2))...
- min(sqrt((x(:,1)-xiyi(:,1)).^2 + (x(:,2)-xiyi(:,2)).^2));
xsol = fminsearch(fun,x0)
It is assumed that xi and yi are the first and second column of vector xiyi (in the code).
The solution (x,y) will be stored in xsol.

3 commentaires

Joshua Woodard
Joshua Woodard le 28 Juin 2019
This answers my question, but leads me to one more question. Are the colons acting as 1,2,3,4,5...,n ? Or, are they allowing the search to be independent? Said in another way, are the colons allowing the search to pick any index independently (not all : acting as 1,2,3... together). I would like it to analyze any row in my data set in any combination.
Joshua Woodard
Joshua Woodard le 28 Juin 2019
I am looking for an x and y that is the center of two concentric circles where the zone of the two circles contain the data. One circle is bigger and contains all the data whereas the small circle does not. I am maximizing the small circle and minimizing the big circle.
Hello,
In the case of searching only one index, for example,
max(sqrt((x(1,1)-xiyi(1,1)).^2 + (x(1,2)-xiyi(1,2)).^2))...
- min(sqrt((x(1,1)-xiyi(1,1)).^2 + (x(1,2)-xiyi(1,2)).^2))
the usage of max and min functions are not necessary since only one value in these functions. I am still unclear with your description of the problem. You may illustrate by picture, which will be more easy.

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Automotive dans Centre d'aide et File Exchange

Produits

Community Treasure Hunt

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

Start Hunting!

Translated by