```% Boyd & Vandenberghe "Convex Optimization"
% Joelle Skaf - 10/09/05
% (a figure is generated)
%
% If the two polyhedra C = {x | A1*x <= b1} and D = {y | A2*y <= b2} can be
% separated by a hyperplane, it will be of the  form
%           z'*x - z'*y >= -lambda'*b1 - mu'*b2 > 0
% where z, lambda and mu are the optimal variables of the problem:
%           maximize    -b1'*lambda - b2'*mu
%               s.t.    A1'*lambda + z = 0
%                       A2'*mu  - z = 0
%                       norm*(z) <= 1
%                       lambda >=0 , mu >= 0
% Note: here x is in R^2

% Input data
randn('seed',0);
n  = 2;
m = 2*n;
A1 = [1 1; 1 -1; -1 1; -1 -1];
A2 = [1 0; -1 0; 0 1; 0 -1];
b1 = 2*ones(m,1);
b2 = [5; -3; 4; -2];

% Solving with CVX
fprintf(1,'Finding a separating hyperplane between the 2 polyhedra...');

cvx_begin
variables lam(m) muu(m) z(n)
maximize ( -b1'*lam - b2'*muu)
A1'*lam + z == 0;
A2'*muu - z == 0;
norm(z) <= 1;
-lam <=0;
-muu <=0;
cvx_end

fprintf(1,'Done! \n');

% Displaying results
disp('------------------------------------------------------------------');
disp('The distance between the 2 polyhedra C and D is: ' );
disp(['dist(C,D) = ' num2str(cvx_optval)]);

% Plotting
t = linspace(-3,6,100);
p = -z(1)*t/z(2) + (muu'*b2 - lam'*b1)/(2*z(2));
figure;
fill([-2; 0; 2; 0],[0;2;0;-2],'b', [3;5;5;3],[2;2;4;4],'r')
axis([-3 6 -3 6])
axis square
hold on;
plot(t,p)
title('Separating 2 polyhedra by a hyperplane');
```
```Finding a separating hyperplane between the 2 polyhedra...
Calling sedumi: 12 variables, 5 equality constraints
------------------------------------------------------------
SeDuMi 1.21 by AdvOL, 2005-2008 and Jos F. Sturm, 1998-2003.
Alg = 2: xz-corrector, Adaptive Step-Differentiation, theta = 0.250, beta = 0.500
eqs m = 5, order n = 12, dim = 13, blocks = 2
nnz(A) = 18 + 0, nnz(ADA) = 25, nnz(L) = 15
it :     b*y       gap    delta  rate   t/tP*  t/tD*   feas cg cg  prec
0 :            1.26E+01 0.000
1 :  -1.46E+00 4.21E+00 0.000 0.3333 0.9000 0.9000   2.47  1  1  2.1E+00
2 :  -2.14E+00 4.21E-01 0.000 0.1000 0.9900 0.9900   1.34  1  1  1.9E-01
3 :  -2.12E+00 2.81E-02 0.000 0.0668 0.9900 0.9900   1.03  1  1  1.2E-02
4 :  -2.12E+00 2.77E-03 0.019 0.0986 0.9900 0.9900   1.00  1  1  1.3E-03
5 :  -2.12E+00 8.19E-06 0.000 0.0030 0.9990 0.9990   1.00  1  1  3.8E-06
6 :  -2.12E+00 2.62E-07 0.000 0.0320 0.9900 0.9900   1.00  1  1  1.4E-07
7 :  -2.12E+00 1.11E-08 0.210 0.0422 0.9900 0.5465   1.00  1  2  1.6E-08
8 :  -2.12E+00 1.66E-10 0.000 0.0150 0.9902 0.9900   1.00  2  2  3.9E-10

iter seconds digits       c*x               b*y
8      0.1   Inf -2.1213203424e+00 -2.1213203423e+00
|Ax-b| =   2.0e-11, [Ay-c]_+ =   1.3E-09, |x|=  1.9e+00, |y|=  3.9e+00

Detailed timing (sec)
Pre          IPM          Post
0.000E+00    5.000E-02    0.000E+00
Max-norms: ||b||=1, ||c|| = 5,
Cholesky |add|=0, |skip| = 0, ||L.L|| = 2163.65.
------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): +2.12132
Done!
------------------------------------------------------------------
The distance between the 2 polyhedra C and D is:
dist(C,D) = 2.1213
```