MATLAB: Error with optimisation

error

Hi, I was looking to do optimization for this http://4.bp.blogspot.com/-WSb8BrAmc9w/T5Hy2QaXJqI/AAAAAAAAAac/ZwDB5y19jAs/s1600/toric_optimization.jpg shape. I was wondering if someone could kindly point out my errors in this code:
    % objective function for optimization
  function F = toric_opt_obj(XX)
  global alpha11 alpha12 alpha21 alpha22 alpha0 
  global icount
alpha11 = XX(1);
alpha12 = XX(2);
alpha21 = XX(3);
alpha22 = XX(4);
alpha0 = XX(5);
icount = icount + 1;
toric_obstacle;
% (x_1^2+x_2^2)^2+alpha12*x_1^2+alpha22*x_2^2+alpha11*x_1+alpha21*x_2+alpha0=0
toric =(z1.^2+z2.^2).^2+alpha12.*z1.^2+alpha22.*z2.^2+alpha11.*z1+alpha21.*z2+alpha0;
L1 = length(z1);
L2 = length(z2);
if z2 == -1*ones(1,L2)
    Err_1 = toric - z1;
end
if (z2 == .8*ones(1,L2) && (-1.2*ones(1,L1) <z1< -.5*ones(1,L1)))
    Err_2 = toric - z2;
end
if z2 == .5*ones(1,L2)
    Err_3 = toric - z1;
end
if (z2 == .8*ones(1,L2) && (.6*ones(1,L1) <z1< 1.2*ones(1,L1)))
    Err_4 = toric - z2;
end
Err = Err_1^2 + Err_2^2 + Err_3^2 + Err_4^2; 
% Err = (z1.^2+z2.^2).^2+alpha12*z1.^2+alpha22*z2.^2+alpha11*z1+alpha21*z2+alpha0;
F = norm(Err); %error square root of the sum of the error squared
[icount XX F]
this part of code is right which is
% Minimum error - toric section
clear all; clc;
global alpha11 alpha12 alpha21 alpha22 alpha0
alpha11 = 0;
alpha12 = -2;
alpha21 = .2;
alpha22 = -1.5;
alpha0 = .5;
% initial limity cycle shape
ii0 = 0;
for x10=-1.5:.01:1.5
   x20 = roots([1 0 (2*x10^2+alpha22) alpha21 (alpha0+alpha11*x10+alpha12*x10^2+x10^4)]);
   for ii=1:4
       if isreal(x20(ii))
           ii0 = ii0 + 1;
           x1lci(ii0) = x10;
           x2lci(ii0) = x20(ii);
       end
   end
end
XX0 = [alpha11 alpha12 alpha21 alpha22 alpha0];
lb = [-.1 -3 .01 -2 .3];
ub = [ .1 -1 .40 -1 .7];
optionso = optimset('LargeScale','off','MaxFunEvals',150);
%norm_tol = .01;
%icount = 0;
% [XX,fval,exitflag,output] = fmincon(@objfun,XX0,[],[],[],[],lb,ub,@confun,optionso)
[XX,fval,exitflag,output] = fminsearch(@toric_opt_obj,XX0)
alpha11 = XX(1);
alpha12 = XX(2);
alpha21 = XX(3);
alpha22 = XX(4);
alpha0 = XX(5);
%alpha22=-2.124;
% optimal limity cycle shape
ii0 = 0;
for x10=-1.5:.01:1.5
   x20 = roots([1 0 (2*x10^2+alpha22) alpha21 (alpha0+alpha11*x10+alpha12*x10^2+x10^4)]);
   for ii=1:4
       if isreal(x20(ii))
           ii0 = ii0 + 1;
           x1lco(ii0) = x10;
           x2lco(ii0) = x20(ii);
       end
   end
end
toric_obstacle;
figure (1)
plot(z1,z2,'k-',x1lci,x2lci,'b*',x1lco,x2lco,'ro','LineWidth',2);
legend('obstacle','initial','optimal');
and this is my obstacle m-file:
    % obstacle definition
    x0 = -1.2;
    y0 = -1;
    x1 = -0.6;
    y1 = -0.5;
    x2 = 0.6;
    x3 = 1.2;
    y3 = 0.8;
    delta = 0.01;
    %line 1
    i = 0;
    for x = x0:delta:x3
        i = i + 1;
        z1(i) = x;
        z2(i) = y0;
    end
    %line 2
    i = i - 1;
    for y = y0:delta:y3
        i = i + 1;
        z1(i) = x3;
        z2(i) = y;
    end
    %line 3
    i = i - 1;
    for x = x3:-delta:x2
        i = i + 1;
        z1(i) = x;
        z2(i) = y3;
    end
    %line 4
    i = i - 1;
    for y = y3:-delta:y1
        i = i + 1;
        z1(i) = x2;
        z2(i) = y;
    end
    %line 5
    i = i - 1;
    for x = x2:-delta:x1
        i = i + 1;
        z1(i) = x;
        z2(i) = y1;
    end
    %line 6
    i = i - 1;
    for y = y1:delta:y3
        i = i + 1;
        z1(i) = x1;
        z2(i) = y;
    end
    %line 7
    i = i - 1;
    for x = x1:-delta:x0
        i = i + 1;
        z1(i) = x;
        z2(i) = y3;
    end
    %line 8
    i = i - 1;
    for y = y3:-delta:y0
        i = i + 1;
        z1(i) = x0;
        z2(i) = y;
    end
    i = i - 1;
    z1 = z1(1:i);
    z2 = z2(1:i);

Best Answer

Use & instead of && .
Read http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F
Read the documentation for all() and any()
a < b < c is interpreted as (a < b) < c . The first part produces a logical value, 0 or 1, and it is that logical value that is compared to c. Consider using (a < b & b < c)
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