Contents

function [x,out] = lbreg_fixedstep(A,b,alpha,opts)
% lbreg_fixedstep: linearized Bregman iteration with fixed stepsize
%   minimize |x|_1 + 1/(2*alpha) |x|_2^2
%   subject to Ax = b
%
% input:
%       A: constraint matrix
%       b: constraint vector
%       alpha: smoothing parameter, typical value: 1 to 10 times estimated norm(x,inf)
%       opts.
%           stepsize: dual ascent stepsize, see below
%           tol: stop iteration once norm(Ax-b)<tol*norm(b), default: 1e-4
%           maxit: max number of iterations, default: 3000
%           maxT:  max running time in second, default: 1000
%           x_ref: if provided, norm(x^k - x_ref) is computed and saved in out.hist_err, default: []
%
% output:
%       x: solution vector
%       out.
%           iter: number of iterations
%           hist_obj: objective value at each iteration
%           hist_res: |Ax-b|_2 at each iteration
%           hist_err: if opts.x_ref is provided, contains norm(x^k - x_ref); otherwise, will be set to []
%
% Algorithm:
%   Linearized Bregman is the dual gradient ascent iteration.
%   The dual problem is: b'y - alpha/2 |shrink(A'y,1)|^2,  where shrink(z,1) = z - proj_[-1,1](z) = sign(z).*max(abs(z)-1,0)
%   Let y be the dual variable. The iteration is
%     y^{k+1} = y^k + stepsize (b - alpha A shrink(A'y^k,1))
%   Promal variable x is obtained as x^k = alpha shrink(A'y^k,1)
%
% How to set alpha:
%   There exists alpha0 so that any alpha >= alpha0 gives the solution to minimize |x|_1 subject to Ax = b.
%   The alpha depends on the data, but a typical value is 1 to 10 times the estimate of norm(solution_x,inf)
%
% How to set opts.stepsize:
%   Too large will cause divergence; too small will cause slow convergence.
%   A conservative value is 1.99/alpha/norm(A)^2, which guarantees convergence.
%   If norm(A)^2 is expensive to compute, one can compute norm(A*A') or norm(A'*A) (choose the easier one!),
%   or use the method in arXiv:1104.2076.
%   If opts.stepsize is not set, then the code uses 2/alpha/normest(A*A',1e-2)
%
% How is the algorithm stopped: see "% stopping" below
%
% More information can be found at
% http://www.caam.rice.edu/~optimization/linearized_bregman

Parameters and defaults

if isfield(opts,'stepsize'),  stepsize = opts.stepsize;  else stepsize = 1.99/alpha/normest(A*A',1e-2); end
if isfield(opts,'tol'),    tol = opts.tol;     else tol = 1e-4;   end
if isfield(opts,'maxit'),  maxit = opts.maxit; else maxit = 500;  end
if isfield(opts,'maxT'),   maxT = opts.maxT;   else maxT = 1e3;   end
if isfield(opts,'x_ref'),  x_ref = opts.x_ref; else x_ref = []; out.out.hist_err = [];  end

Data preprocessing and initialization

m = size(A,1);

y = zeros(m,1);
res = b; % residual (b - Ax)
norm_b = norm(b);

shrink = @(z) sign(z).*max(0,abs(z)-1);

Main iterations

start_time = tic;

for k = 1:maxit

    % --- y-update ---
    y = y + stepsize * res; % for k=1, (stepsize + 1/max(abs(b.'*A))) can be used as stepsize

    % --- x-update ---
    x = alpha * shrink(y.'*A).';
    Ax = A*x;
    res = b - Ax; % res will be used in next y-update

    % --- diagnostics, reporting, stopping checks ---
    % reporting
    out.hist_obj(k) = b.'*y - norm(x)^2/alpha/2; % dual objective
    out.hist_res(k) = norm(res); % residual size |b - Ax|_2
    if ~isempty(x_ref); out.hist_err(k) = norm(x - x_ref); end

    % stopping
    if toc(start_time) > maxT; break; end; % running time check
    if k > 1
        % dual objective check
        obj_diff = out.hist_obj(k) - out.hist_obj(k-1);
        if obj_diff < -1e-10; error('dual objective decreases; opts.stepsize is too large!'); end
        % primal residual check
        if norm(res) < tol*norm_b; break; end
    end
end

out.iter = k;

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