I would like to use MATLAB function fmincon using a gradient vector only (without scalar function). But I have trouble with it. For instance, I tried the following, but it doesn't work. Any help please? Thanks!
fun = #rosenbrockwithgrad;
x0 = [-1,2];
A = [];
b = [];
Aeq = [];
beq = [];
lb = [-2,-2];
ub = [2,2];
x = fmincon(fun,x0,A,b,Aeq,beq,lb,ub)
function [grad] = rosenbrockwithgrad(x)
grad = [-400*(x(2)-x(1)^2)*x(1)-2*(1-x(1));
200*(x(2)-x(1)^2)];
end
You need to include both the function and its gradient. See below:
options = optimoptions('fmincon','SpecifyObjectiveGradient',true);
fun = #rosenbrockwithgrad;
x0 = [-1,2];
A = [];
b = [];
Aeq = [];
beq = [];
lb = [-2,-2];
ub = [2,2];
nonlcon = [];
x = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon,options)
function [f,g] = rosenbrockwithgrad(x)
% Calculate objective f
f = 100*(x(2) - x(1)^2)^2 + (1-x(1))^2;
if nargout > 1 % gradient required
g = [-400*(x(2)-x(1)^2)*x(1)-2*(1-x(1));
200*(x(2)-x(1)^2)];
end
end
This code should run properly.
Related
Trying to find the optimal hyperparameters for my svm model using a grid search, but it simply returns 1 for the hyperparameters.
function evaluations = inner_kfold_trainer(C,q,k,features_xy,labels)
features_xy_flds = kdivide(features_xy, k);
labels_flds = kdivide(labels, k);
evaluations = zeros(k,3);
for i = 1:k
fprintf('Fold %i of %i\n',i,k);
train_data = cell2mat(features_xy_flds(1:end ~= i));
train_labels = cell2mat(labels_flds(1:end ~= i));
test_data = cell2mat(features_xy_flds(i));
test_labels = cell2mat(labels_flds(i));
%AU1
train_labels = train_labels(:,1);
test_labels = test_labels(:,1);
[k,~] = size(test_labels);
%train
sv = fitcsvm(train_data,train_labels, 'KernelFunction','polynomial', 'PolynomialOrder',q,'BoxConstraint',C);
sv.predict(test_data);
%Calculate evaluative measures
%svm_outputs = zeros(k,1);
sv_predictions = sv.predict(test_data);
[precision,recall,F1] = evaluation(sv_predictions,test_labels);
evaluations(i,1) = precision;
evaluations(i,2) = recall;
evaluations(i,3) = F1;
end
save('eval.mat', 'evaluations');
end
an inner-fold cross validation function
and below the grid function where something seems to be going wrong
function [q,C] = grid_search(features_xy,labels,k)
% n x n grid
n = 3;
q_grid = linspace(1,19,n);
C_grid = linspace(1,59,n);
tic
evals = zeros(n,n,3);
for i = 1:n
for j = 1:n
fprintf('## i=%i, j=%i ##\n', i, j);
svm_results = inner_kfold_trainer(C_grid(i), q_grid(j),k,features_xy,labels);
evals(i,j,:) = mean(svm_results(:,:));
% precision only
%evals(i,j,:) = max(svm_results(:,1));
toc
end
end
f = evals;
% retrieving the best value of the hyper parameters, to use in the outer
% fold
[M1,I1] = max(f);
[~,I2] = max(M1(1,1,:));
index = I1(:,:,I2);
C = C_grid(index(1))
q = q_grid(index(2))
end
When I run grid_search(features_xy,labels,8) for example, I get C=1 and q=1, for any k(the no. of folds) value. Also features_xy is a 500*98 matrix.
I have a little problem, fseminf used to work just fine, but now as I reached to it, it happend to produce an error, I have no idea what happend. Here is the code with an example which does not work:
clc;
clear;
close;
% Zdefiniuj funkcje
global g f S
f = #(x) (x(1)-2).^2 + (x(2)-0.2).^2;
g = #(x,s) 5.*x(1).^2.*sin(pi.*sqrt(s))./(1+s.^2) - x(2);
lb = [-1,0];
ub = [1 ,0.2];
S = [0,1];
A = [];
b = [];
Aeq = [];
beq = [];
x0 = [0,0]';
[x,fval,exitflag,output,lambda] = fseminf(f,x0,1,#seminfcon2,A,b,Aeq,beq,lb,ub);
disp('x= ')
x
And seminfcon is defined:
function [c, ceq, K1, s] = seminfcon2(x,s)
global S g
% Dodatkowe parametry
c = [];
ceq = [];
if isnan(s(1,1))
s(1,1)=0.001;
s(1,2)=0;
end
t = S(1):s(1):S(2);
K1 = g(x,t);
The error is when calculating x, meaning in fseminf
I am trying to run fseminf function
f = #(x) (x(1)-2).^2 + (x(2)-0.2).^2;
g = #(x,s) 5*x(1).^2*sin(pi*sqrt(s))/(1+s.^2) - x(2);
lb = [-1,0];
ub = [1 ,0.2];
S = [0,1];
A = [];
b = [];
Aeq = [];
beq = [];
x0 = [0,0];
x = fseminf(f,x0,1,#seminfcon1,A,b,Aeq,beq,lb,ub);
Where:
function [c, ceq, K1, s] = seminfcon1(x,s)
global S g
% No finite nonlinear inequality and equality constraints
c = [];
ceq = [];
% Sample set
if isnan(s)
% Initial sampling interval
s = [0.0001 0];
end
t = S(1):s(1):S(2);
% Evaluate the semi-infinite constraint
K1 = g(x,t);
This is the code I use for creating a finite set from S, which is an interval. I do not understand how it works, why do we use isnan function instead of just declaring s=[0.0001] (which by the way does not work). It is used for fseminf alghoritm, I took the code from an example.
My question this time concerns the obtention of the degree distribution of a LDPC matrix through linear programming, under the following statement:
My code is the following:
function [v] = LP_Irr_LDPC(k,Ebn0)
options = optimoptions('fmincon','Display','iter','Algorithm','interior-point','MaxIter', 4000, 'MaxFunEvals', 70000);
fun = #(v) -sum(v(1:k)./(1:k));
A = [];
b = [];
Aeq = [0, ones(1,k-1)];
beq = 1;
lb = zeros(1,k);
ub = [0, ones(1,k-1)];
nonlcon = #(v)DensEv_SP(v,Ebn0);
l0 = [0 rand(1,k-1)];
l0 = l0./sum(l0);
v = fmincon(fun,l0,A,b,Aeq,beq,lb,ub,nonlcon,options)
end
Definition of nonlinear constraints:
function [c, ceq] = DensEv_SP(v,Ebn0)
% It is also needed to modify this function, as you cannot pass parameters from others to it.
h = [0 rand(1,19)];
h = h./sum(h); % This is where h comes from
syms x;
X = x.^(0:(length(h)-1));
R = h*transpose(X);
ebn0 = 10^(Ebn0/10);
Rm = 1;
LLR = (-50:50);
p03 = 0.3;
LLR03 = log((1-p03)/p03);
r03 = 1 - p03;
noise03 = (2*r03*Rm*ebn0)^-1;
pf03 = normpdf(LLR, LLR03, noise03);
sumpf03 = sum(pf03(1:length(pf03)/2));
divisions = 100;
Aj = zeros(1, divisions);
rho = zeros(1, divisions);
xj = zeros(1, divisions);
k = 10; % Length(v) -> Same value as in 'Complete.m'
for j=1:1:divisions
xj(j) = sumpf03*j/divisions;
rho(j) = subs(R,x,1-xj(j));
Aj(j) = 1 - rho(j);
end
c = zeros(1, length(xj));
lambda = zeros(1, length(Aj));
for j = 1:1:length(xj)
lambda(j) = sum(v(2:k).*(Aj(j).^(1:(k-1))));
c(j) = sumpf03*lambda(j) - xj(j);
end
save Almacen
ceq = [];
%ceq = sum(v)-1;
end
This question is linked to the one posted here. My problem is that I need that each element from vectors v and h resulting from this optimization problem is a fraction of x/N and x/(N(1-r) respectively.
How could I ensure that condition without losing convergence capability?
I came up with one possible solution, at least for vector h, within the function DensEv_SP:
function [c, ceq] = DensEv_SP(v,Ebn0)
% It is also needed to modify this function, as you cannot pass parameters from others to it.
k = 10; % Same as in Complete.m, desired sum of h
M = 19; % Number of integers
h = [0 diff([0,sort(randperm(k+M-1,M-1)),k+M])-ones(1,M)];
h = h./sum(h);
syms x;
X = x.^(0:(length(h)-1));
R = h*transpose(X);
ebn0 = 10^(Ebn0/10);
Rm = 1;
LLR = (-50:50);
p03 = 0.3;
LLR03 = log((1-p03)/p03);
r03 = 1 - p03;
noise03 = (2*r03*Rm*ebn0)^-1;
pf03 = normpdf(LLR, LLR03, noise03);
sumpf03 = sum(pf03(1:length(pf03)/2));
divisions = 100;
Aj = zeros(1, divisions);
rho = zeros(1, divisions);
xj = zeros(1, divisions);
N = 20; % Length(v) -> Same value as in 'Complete.m'
for j=1:1:divisions
xj(j) = sumpf03*j/divisions;
rho(j) = subs(R,x,1-xj(j));
Aj(j) = 1 - rho(j);
end
c = zeros(1, length(xj));
lambda = zeros(1, length(Aj));
for j = 1:1:length(xj)
lambda(j) = sum(v(2:k).*(Aj(j).^(1:(k-1))));
c(j) = sumpf03*lambda(j) - xj(j);
end
save Almacen
ceq = (N*v)-floor(N*v);
%ceq = sum(v)-1;
end
As above stated, there is no longer any problem with vector h; nevertheless the way I defined ceq value seemed to be insufficient to make the optimization work out (the problems with v have not diminished at all). Does anybody know how to find the solution?
I am using contourf function with binary image. I am trouble how i can get the area and centroid of the different surface in the image, need this task to classify the objects.
You need to use the the Contour Matrix output
Here is an example:
function data = ContourInfo(C)
data = [];
if isempty(C)
return
end
k = 1;
j = 1;
while j < size(C,2);
data(k).numxy = C(2,j);
data(k).x = C(1,j+1:j+data(k).numxy);
data(k).y = C(2,j+1:j+data(k).numxy);
data(k).level = C(1,j);
[data(k).centroid(1) data(k).centroid(2) data(k).area] = ...
polycentroid(data(k).x, data(k).y);
data(k).area = polyarea(data(k).x, data(k).y);
data(k).centroid = polycentroid(data(k).x, data(k).y);
j = j + data(k).numxy + 1;
k = k+1;
end
function [x0,y0,a] = polycentroid(x,y)
[m1,n1] = size(x); [m2,n2] = size(y);
n = max(m1,n1);
x = x(:); y = y(:);
x2 = [x(2:n);x(1)];
y2 = [y(2:n);y(1)];
a = 1/2*sum (x.*y2-x2.*y);
x0 = 1/6*sum((x.*y2-x2.*y).*(x+x2))/a;
y0 = 1/6*sum((x.*y2-x2.*y).*(y+y2))/a;
Call as follow:
Z = peaks(20);
[C, h] = contourf(Z,10);
contourData = ContourInfo(C)
disp('Area of contour 1:');
disp(contourData(1).area
disp('Centroid of contour 1:');
disp(contourData(1).centroid);