How can I implement cubic interpolation in Gradient Descent algorithm?

5 vues (au cours des 30 derniers jours)
OzzWal
OzzWal le 15 Jan 2019
Commenté : OzzWal le 18 Jan 2019
After computing the gradient of z along the entire computational grid, I want to calculate Z0=Z(X0,Y0) and gradient(Z0) using cubic interpolation. I'm unsure of the best way to do this but it is essential that X0 and Y0 are set outside of the function. They are coordinates of an arbitrary starting point to begin the Gradient Descent algorithm.
Below is the problem part of my code with some attempted statements:
%Function to implement the 2-Dimensional Gradient Descent method. The
%output of this function should be an [N x 3] matrix containing, on each
%row, the x, y and z co-ordinates of every position considered in the
%Gradient Descent algorithm (including start and end points)
function [xi, yi, Z] = gradient_descent2(Z,X0,Y0,gamma,tau)%Declare function
x = -9:0.2:9;% Set the input of comp_z function two be two vectors with a set stepsize
y = -8:0.2:8;
[Xg, Yg] = meshgrid(x,y); %make mesh grid
[px,py] = gradient(Z); %compute gradient of z along entire computational
%mesh grid (z computed outside).
%---------------------------------2--------------------------------------%
%Use cubic interpolation to obtain starting point z(X0,Y0) and gradient at
%this point
Z0 = griddata(Xg,Yg,Z,X0,Y0,'cubic');
t = (x == X0) & (y == Y0);
indt = find(t);
Z0_grad = [px0(indt) py0(indt)];
This is supposed to be a function that implements a 2D Gradient Descent method. x, y & z are related as follows:
and is computed outside the function.

Réponse acceptée

Matt J
Matt J le 15 Jan 2019
Modifié(e) : Matt J le 15 Jan 2019
Since you have an analytical formula for z(x,y), why don't you just use that to calculate Z and its gradient exactly? Why are you working with discrete approximations and interpolaton?
In other words, why don't you just do as follows?
Z = @(x,y) cos(x/2)*cos(y/2) - x/5 +y/10;
gradZ = @(x,y) [-sin(x/2)*cos(y/2)/2 -1/5 ; -cos(x/2)*sin(y/2)/2 + 1/10 ];
  3 commentaires
Matt J
Matt J le 15 Jan 2019
OK. Well, I think it should look like this,
xgrid = -9:0.2:9;% Set the input of comp_z function two be two vectors with a set stepsize
ygrid = -8:0.2:8;
[Gx,Gy] = gradient(Z);
Z0=interp2(xgrid,ygrid,Z,X0,Y0,'cubic');
gradZ0_X=interp2(xgrid,ygrid,Gx,X0,Y0,'cubic');
gradZ0_Y=interp2(xgrid,ygrid,Gy,X0,Y0,'cubic');
OzzWal
OzzWal le 18 Jan 2019
Thank you!

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur MATLAB dans Help Center et File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by