seed = 0;
randn('state',seed);
rand('state',seed);
delta = 1e-8;
m = 100;
n = 50;
A = randn(m,n);
x0 = randn(n,1);
b = A*x0 + rand(m,1);
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
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);
NUM_RUNS = 15;
nnzs = [];
W = ones(n,1);
disp([char(10) 'Log-based heuristic:']);
for k = 1:NUM_RUNS
cvx_begin quiet
variable x_log(n)
minimize( sum( W.*abs(x_log) ) )
subject to
A*x_log <= b;
cvx_end
nnz = length(find( abs(x_log) > delta ));
nnzs = [nnzs nnz];
fprintf(1,' found a solution with %d nonzeros...\n', nnz);
W = 1./(delta + abs(x_log));
end
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(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, 100 equality constraints
For improved efficiency, sedumi is solving the dual problem.
------------------------------------------------------------
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 = 100, order n = 201, dim = 201, blocks = 1
nnz(A) = 5200 + 0, nnz(ADA) = 2650, nnz(L) = 1375
it : b*y gap delta rate t/tP* t/tD* feas cg cg prec
0 : 8.50E-01 0.000
1 : 3.98E+01 3.79E-01 0.000 0.4456 0.9000 0.9000 6.27 1 1 6.6E+00
2 : -5.85E+00 1.47E-01 0.000 0.3882 0.9000 0.9000 1.78 1 1 1.9E+00
3 : -2.47E+01 6.78E-02 0.000 0.4609 0.9000 0.9000 1.49 1 1 7.5E-01
4 : -3.26E+01 2.55E-02 0.000 0.3769 0.9000 0.9000 1.32 1 1 2.5E-01
5 : -3.46E+01 1.01E-02 0.000 0.3936 0.9000 0.9000 1.13 1 1 9.5E-02
6 : -3.55E+01 3.53E-03 0.000 0.3513 0.9000 0.9000 1.05 1 1 3.3E-02
7 : -3.58E+01 1.30E-03 0.000 0.3695 0.9000 0.9000 1.02 1 1 1.2E-02
8 : -3.59E+01 2.58E-05 0.000 0.0198 0.9258 0.9000 1.01 1 1 1.9E-03
9 : -3.59E+01 6.19E-06 0.000 0.2397 0.9000 0.0000 1.00 1 1 8.5E-04
10 : -3.59E+01 1.39E-06 0.000 0.2251 0.9000 0.8569 1.00 1 1 2.0E-04
11 : -3.59E+01 1.50E-07 0.000 0.1077 0.9138 0.9000 1.00 1 1 2.9E-05
12 : -3.59E+01 2.89E-08 0.000 0.1924 0.9180 0.9000 1.00 1 1 6.1E-06
13 : -3.59E+01 3.29E-09 0.000 0.1139 0.9000 0.9112 1.00 1 1 6.8E-07
14 : -3.59E+01 4.33E-12 0.000 0.0013 0.9990 0.9990 1.00 1 1
iter seconds digits c*x b*y
14 0.1 Inf -3.5940776942e+01 -3.5940776942e+01
|Ax-b| = 2.7e-15, [Ay-c]_+ = 8.9E-15, |x|= 7.1e+00, |y|= 1.0e+01
Detailed timing (sec)
Pre IPM Post
1.000E-02 6.000E-02 0.000E+00
Max-norms: ||b||=1, ||c|| = 2.494989e+01,
Cholesky |add|=0, |skip| = 0, ||L.L|| = 1.58293.
------------------------------------------------------------
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 43 nonzeros...
found a solution with 43 nonzeros...
found a solution with 41 nonzeros...
found a solution with 40 nonzeros...
found a solution with 39 nonzeros...
found a solution with 39 nonzeros...
found a solution with 37 nonzeros...
found a solution with 37 nonzeros...
found a solution with 37 nonzeros...
found a solution with 37 nonzeros...
found a solution with 37 nonzeros...
found a solution with 37 nonzeros...
found a solution with 37 nonzeros...
Found a feasible x in R^50 that has 37 nonzeros using the log heuristic.