Figure 8.17: Fourth-order placement problem

% Section 8.7.3, Boyd & Vandenberghe "Convex Optimization"
% Original by Lieven Vandenberghe
% Adapted for CVX by Joelle Skaf - 10/24/05
% (a figure is generated)
%
% Placement problem with 6 free points, 8 fixed points and 27 links.
% The coordinates of the free points minimize the sum of the squares of
% Euclidean lengths of the links, i.e.
%           minimize    sum_{i<j) h(||x_i - x_j||)
% where h(z) = z^4.

linewidth = 1;      % in points;  width of dotted lines
markersize = 5;    % in points;  marker size

fixed = [ 1   1  -1 -1    1   -1  -0.2  0.1; % coordinates of fixed points
          1  -1  -1  1 -0.5 -0.2    -1    1]';
M = size(fixed,1);  % number of fixed points
N = 6;              % number of free points

% first N columns of A correspond to free points,
% last M columns correspond to fixed points

A = [ 1  0  0 -1  0  0    0  0  0  0  0  0  0  0
      1  0 -1  0  0  0    0  0  0  0  0  0  0  0
      1  0  0  0 -1  0    0  0  0  0  0  0  0  0
      1  0  0  0  0  0   -1  0  0  0  0  0  0  0
      1  0  0  0  0  0    0 -1  0  0  0  0  0  0
      1  0  0  0  0  0    0  0  0  0 -1  0  0  0
      1  0  0  0  0  0    0  0  0  0  0  0  0 -1
      0  1 -1  0  0  0    0  0  0  0  0  0  0  0
      0  1  0 -1  0  0    0  0  0  0  0  0  0  0
      0  1  0  0  0 -1    0  0  0  0  0  0  0  0
      0  1  0  0  0  0    0 -1  0  0  0  0  0  0
      0  1  0  0  0  0    0  0 -1  0  0  0  0  0
      0  1  0  0  0  0    0  0  0  0  0  0 -1  0
      0  0  1 -1  0  0    0  0  0  0  0  0  0  0
      0  0  1  0  0  0    0 -1  0  0  0  0  0  0
      0  0  1  0  0  0    0  0  0  0 -1  0  0  0
      0  0  0  1 -1  0    0  0  0  0  0  0  0  0
      0  0  0  1  0  0    0  0 -1  0  0  0  0  0
      0  0  0  1  0  0    0  0  0 -1  0  0  0  0
      0  0  0  1  0  0    0  0  0  0  0 -1  0  0
      0  0  0  1  0 -1    0  0  0  0  0 -1  0  0        % error in data!!!
      0  0  0  0  1 -1    0  0  0  0  0  0  0  0
      0  0  0  0  1  0   -1  0  0  0  0  0  0  0
      0  0  0  0  1  0    0  0  0 -1  0  0  0  0
      0  0  0  0  1  0    0  0  0  0  0  0  0 -1
      0  0  0  0  0  1    0  0 -1  0  0  0  0  0
      0  0  0  0  0  1    0  0  0  0 -1  0  0  0 ];
nolinks = size(A,1);    % number of links

fprintf(1,'Computing the optimal locations of the 6 free points...');

cvx_begin
    variable x(N+M,2)
    minimize ( sum(square_pos(square_pos(norms( A*x,2,2 )))))
    x(N+[1:M],:) == fixed;
cvx_end

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

% Plots
free_sum = x(1:N,:);
figure(1);
dots = plot(free_sum(:,1), free_sum(:,2), 'or', fixed(:,1), fixed(:,2), 'bs');
set(dots(1),'MarkerFaceColor','red');
hold on
legend('Free points','Fixed points','Location','Best');
for i=1:nolinks
  ind = find(A(i,:));
  line2 = plot(x(ind,1), x(ind,2), ':k');
  hold on
  set(line2,'LineWidth',linewidth);
end
axis([-1.1 1.1 -1.1 1.1]) ;
axis equal;
title('Fourth-order placement problem');
% print -deps placement-quartic.eps

figure(2)
all = [free_sum; fixed];
bins = 0.05:0.1:1.95;
lengths = sqrt(sum((A*all).^2')');
[N2,hist2] = hist(lengths,bins);
bar(hist2,N2);
hold on;
xx = linspace(0,2,1000);  yy = (6/1.5^4)*xx.^4;
plot(xx,yy,'--');
axis([0 1.5 0 4.5]);
hold on
plot([0 2], [0 0 ], 'k-');
title('Distribution of the 27 link lengths');
% print -deps placement-quartic-hist.eps
Computing the optimal locations of the 6 free points... 
Calling SDPT3: 297 variables, 147 equality constraints
   For improved efficiency, SDPT3 is solving the dual problem.
------------------------------------------------------------

 num. of constraints = 147
 dim. of sdp    var  = 108,   num. of sdp  blk  = 54
 dim. of socp   var  = 81,   num. of socp blk  = 27
 dim. of linear var  = 54
*******************************************************************
   SDPT3: Infeasible path-following algorithms
*******************************************************************
 version  predcorr  gam  expon  scale_data
   HKM      1      0.000   1        0    
it pstep dstep pinfeas dinfeas  gap      mean(obj)   cputime
-------------------------------------------------------------------
 0|0.000|0.000|1.4e+01|6.6e+00|7.5e+03| 1.080000e+02| 0:0:00| chol  1  1 
 1|0.664|0.670|4.7e+00|2.2e+00|3.3e+03| 1.294054e+02| 0:0:00| chol  1  1 
 2|0.754|0.748|1.2e+00|5.7e-01|1.5e+03| 2.859040e+01| 0:0:00| chol  1  1 
 3|0.816|0.927|2.1e-01|4.3e-02|2.4e+02|-3.240148e+01| 0:0:00| chol  1  1 
 4|1.000|0.311|2.5e-06|2.9e-02|2.1e+02|-5.084695e+00| 0:0:00| chol  1  1 
 5|0.767|1.000|5.8e-07|1.0e-05|9.8e+01|-2.477078e+01| 0:0:00| chol  1  1 
 6|1.000|0.846|2.4e-08|2.6e-06|3.3e+01|-2.405122e+01| 0:0:00| chol  1  1 
 7|0.821|0.907|9.3e-09|3.4e-07|5.2e+00|-2.069173e+01| 0:0:00| chol  1  1 
 8|0.951|1.000|2.5e-09|1.2e-08|1.8e+00|-2.081896e+01| 0:0:00| chol  1  1 
 9|0.972|0.930|1.0e-09|2.3e-09|1.8e-01|-2.068494e+01| 0:0:01| chol  1  1 
10|1.000|1.000|7.7e-15|3.0e-10|2.8e-02|-2.065029e+01| 0:0:01| chol  1  1 
11|0.964|0.991|7.0e-15|1.4e-11|1.1e-03|-2.064631e+01| 0:0:01| chol  1  1 
12|0.982|0.985|1.9e-14|1.2e-12|1.9e-05|-2.064632e+01| 0:0:01| chol  1  1 
13|0.999|1.000|2.2e-12|1.0e-12|9.0e-07|-2.064632e+01| 0:0:01| chol  1  1 
14|0.994|1.000|9.7e-12|1.0e-12|2.9e-08|-2.064632e+01| 0:0:01|
  stop: max(relative gap, infeasibilities) < 1.49e-08
-------------------------------------------------------------------
 number of iterations   = 14
 primal objective value = -2.06463236e+01
 dual   objective value = -2.06463236e+01
 gap := trace(XZ)       = 2.85e-08
 relative gap           = 6.75e-10
 actual relative gap    = 6.64e-10
 rel. primal infeas     = 9.69e-12
 rel. dual   infeas     = 1.00e-12
 norm(X), norm(y), norm(Z) = 4.1e+01, 8.4e+00, 1.5e+01
 norm(A), norm(b), norm(C) = 2.0e+01, 6.2e+00, 1.0e+01
 Total CPU time (secs)  = 0.8  
 CPU time per iteration = 0.1  
 termination code       =  0
 DIMACS: 3.0e-11  0.0e+00  5.1e-12  0.0e+00  6.6e-10  6.7e-10
-------------------------------------------------------------------
------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): +20.6463
Done!