# Computing a sparse solution of a set of linear inequalities

    % Section 6.2, Boyd & Vandenberghe "Convex Optimization"
    % "Just relax: Convex programming methods for subset selection
    %  and sparse approximation" by J. A. Tropp
    % Written for CVX by Almir Mutapcic - 02/28/06
    %
    % We consider a set of linear inequalities A*x <= b which are
    % feasible. We apply two heuristics to find a sparse point x that
    % satisfies these inequalities.
    %
    % The (standard) l1-norm heuristic for finding a sparse solution is:
    %
    %   minimize   ||x||_1
    %       s.t.   Ax <= b
    %
    % The log-based heuristic is an iterative method for finding
    % a sparse solution, by finding a local optimal point for the problem:
    %
    %   minimize   sum(log( delta + |x_i| ))
    %       s.t.   Ax <= b
    %
    % where delta is a small threshold value (determines what is close to zero).
    % We cannot solve this problem since it is a minimization of a concave
    % function and thus it is not a convex problem. However, we can apply
    % a heuristic in which we linearize the objective, solve, and re-iterate.
    % This becomes a weighted l1-norm heuristic:
    %
    %   minimize sum( W_i * |x_i| )
    %       s.t. Ax <= b
    %
    % which in each iteration re-adjusts the weights W_i based on the rule:
    %
    %   W_i = 1/(delta + |x_i|), where delta is a small threshold value
    %
    % This algorithm is described in papers:
    % "An Affine Scaling Methodology for Best Basis Selection"
    %  by B. D. Rao and K. Kreutz-Delgado
    % "Portfolio optimization with linear and ï¬?xed transaction costs"
    %  by M. S. Lobo, M. Fazel, and S. Boyd

    % fix random number generator so we can repeat the experiment
    seed = 0;
    randn('state',seed);
    rand('state',seed);

    % the threshold value below which we consider an element to be zero
    delta = 1e-8;

    % problem dimensions (m inequalities in n-dimensional space)
    m = 100;
    n = 50;

    % construct a feasible set of inequalities
    % (this system is feasible for the x0 point)
    A  = randn(m,n);
    x0 = randn(n,1);
    b  = A*x0 + rand(m,1);

    % l1-norm heuristic for finding a sparse solution
    fprintf(1, 'Finding a sparse feasible point using l1-norm heuristic ...')
    cvx_begin
      variable x_l1(n)
      minimize( norm( x_l1, 1 ) )
      subject to
        A*x_l1 <= b;
    cvx_end

    % number of nonzero elements in the solution (its cardinality or diversity)
    nnz = length(find( abs(x_l1) > delta ));
    fprintf(1,['\nFound a feasible x in R^%d that has %d nonzeros ' ...
               'using the l1-norm heuristic.\n'],n,nnz);

    % iterative log heuristic
    NUM_RUNS = 15;
    nnzs = [];
    W = ones(n,1); % initial weights

    disp([char(10) 'Log-based heuristic:']);
    cvx_quiet(true);
    for k = 1:NUM_RUNS
      cvx_begin
        variable x_log(n)
        minimize( sum( W.*abs(x_log) ) )
        subject to
          A*x_log <= b;
      cvx_end

      % display new number of nonzeros in the solution vector
      nnz = length(find( abs(x_log) > delta ));
      nnzs = [nnzs nnz];
      fprintf(1,'   found a solution with %d nonzeros...\n', nnz);

      % adjust the weights and re-iterate
      W = 1./(delta + abs(x_log));
    end
    cvx_quiet(false);

    % number of nonzero elements in the solution (its cardinality or diversity)
    nnz = length(find( abs(x_log) > delta ));
    fprintf(1,['\nFound a feasible x in R^%d that has %d nonzeros ' ...
               'using the log heuristic.\n'],n,nnz);

    % plot number of nonzeros versus iteration
    plot(1:NUM_RUNS, nnzs, [1 NUM_RUNS],[nnzs(1) nnzs(1)],'--');
    axis([1 NUM_RUNS 0 n])
    xlabel('iteration'), ylabel('number of nonzeros (cardinality)');
    legend('log heuristic','l1-norm heuristic','Location','SouthEast')

    Finding a sparse feasible point using l1-norm heuristic ...
    Calling SeDuMi: 200 variables (0 free), 100 equality constraints
    ------------------------------------------------------------------------
    SeDuMi 1.1 by AdvOL, 2005 and Jos F. Sturm, 1998, 2001-2003.
    Alg = 2: xz-corrector, Adaptive Step-Differentiation, theta = 0.250, beta = 0.500
    eqs m = 100, order n = 201, dim = 201, blocks = 1
    nnz(A) = 10100 + 0, nnz(ADA) = 10000, nnz(L) = 5050
     it :     b*y       gap    delta  rate   t/tP*  t/tD*   feas cg cg  prec
      0 :            6.74E+002 0.000
      1 : -1.73E+001 2.29E+002 0.000 0.3397 0.9000 0.9000   2.99  1  1  1.7E+000
      2 : -8.94E+000 6.85E+001 0.000 0.2991 0.9000 0.9000   1.57  1  1  7.5E-001
      3 :  1.43E+001 3.66E+001 0.000 0.5336 0.9000 0.9000   1.55  1  1  4.1E-001
      4 :  2.71E+001 1.70E+001 0.000 0.4653 0.9000 0.9000   1.40  1  1  1.9E-001
      5 :  3.25E+001 7.34E+000 0.000 0.4313 0.9000 0.9000   1.20  1  1  8.4E-002
      6 :  3.43E+001 3.21E+000 0.000 0.4376 0.9000 0.9000   1.08  1  1  3.8E-002
      7 :  3.54E+001 1.13E+000 0.000 0.3514 0.9000 0.9000   1.04  1  1  1.3E-002
      8 :  3.57E+001 3.91E-001 0.000 0.3460 0.9045 0.9000   1.02  1  1  4.6E-003
      9 :  3.59E+001 1.20E-001 0.000 0.3083 0.9000 0.9241   1.01  1  1  1.5E-003
     10 :  3.59E+001 2.91E-002 0.000 0.2420 0.9106 0.9000   1.00  1  1  3.5E-004
     11 :  3.59E+001 6.58E-003 0.000 0.2259 0.9291 0.9000   1.00  1  1  7.4E-005
     12 :  3.59E+001 9.89E-004 0.000 0.1502 0.9000 0.9165   1.00  1  1  1.1E-005
     13 :  3.59E+001 1.90E-004 0.000 0.1918 0.9127 0.9000   1.00  2  2  2.2E-006
     14 :  3.59E+001 6.42E-007 0.000 0.0034 0.9990 0.9990   1.00  2  2
    iter seconds digits       c*x               b*y
     14      0.2   Inf  3.5940776942e+001  3.5940776942e+001
    |Ax-b| =  2.6e-014, [Ay-c]_+ =  1.7E-015, |x|= 1.5e+001, |y|= 1.6e+000

    Detailed timing (sec)
       Pre          IPM          Post
    6.009E-002    2.003E-001    0.000E+000
    Max-norms: ||b||=2.494989e+001, ||c|| = 1,
    Cholesky |add|=0, |skip| = 0, ||L.L|| = 4.38669.
    ------------------------------------------------------------------------
    Status: Solved
    Optimal value (cvx_optval): +35.9408

    Found a feasible x in R^50 that has 44 nonzeros using the l1-norm heuristic.

    Log-based heuristic:
       found a solution with 44 nonzeros...
       found a solution with 44 nonzeros...
       found a solution with 37 nonzeros...
       found a solution with 37 nonzeros...
       found a solution with 36 nonzeros...
       found a solution with 36 nonzeros...
       found a solution with 36 nonzeros...
       found a solution with 35 nonzeros...
       found a solution with 35 nonzeros...
       found a solution with 35 nonzeros...
       found a solution with 35 nonzeros...
       found a solution with 35 nonzeros...
       found a solution with 35 nonzeros...
       found a solution with 35 nonzeros...
       found a solution with 35 nonzeros...

    Found a feasible x in R^50 that has 35 nonzeros using the log heuristic.