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📄 conjugate gradient minimization -minimize.m

📁 Conjugate Gradient Minimization在梯度下降算法中有着重要应用。可以解决一些一般方法不容易解决的问题
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function [X, fX, i] = minimize(X, f, length, varargin)% Minimize a differentiable multivariate function. %% Usage: [X, fX, i] = minimize(X, f, length, P1, P2, P3, ... )%% where the starting point is given by "X" (D by 1), and the function named in% the string "f", must return a function value and a vector of partial% derivatives of f wrt X, the "length" gives the length of the run: if it is% positive, it gives the maximum number of line searches, if negative its% absolute gives the maximum allowed number of function evaluations. You can% (optionally) give "length" a second component, which will indicate the% reduction in function value to be expected in the first line-search (defaults% to 1.0). The parameters P1, P2, P3, ... are passed on to the function f.%% The function returns when either its length is up, or if no further progress% can be made (ie, we are at a (local) minimum, or so close that due to% numerical problems, we cannot get any closer). NOTE: If the function% terminates within a few iterations, it could be an indication that the% function values and derivatives are not consistent (ie, there may be a bug in% the implementation of your "f" function). The function returns the found% solution "X", a vector of function values "fX" indicating the progress made% and "i" the number of iterations (line searches or function evaluations,% depending on the sign of "length") used.%% The Polack-Ribiere flavour of conjugate gradients is used to compute search% directions, and a line search using quadratic and cubic polynomial% approximations and the Wolfe-Powell stopping criteria is used together with% the slope ratio method for guessing initial step sizes. Additionally a bunch% of checks are made to make sure that exploration is taking place and that% extrapolation will not be unboundedly large.%% See also: checkgrad %% Copyright (C) 2001 - 2006 by Carl Edward Rasmussen (2006-09-08).INT = 0.1;    % don't reevaluate within 0.1 of the limit of the current bracketEXT = 3.0;                  % extrapolate maximum 3 times the current step-sizeMAX = 20;                         % max 20 function evaluations per line searchRATIO = 10;                                       % maximum allowed slope ratioSIG = 0.1; RHO = SIG/2; % SIG and RHO are the constants controlling the Wolfe-% Powell conditions. SIG is the maximum allowed absolute ratio between% previous and new slopes (derivatives in the search direction), thus setting% SIG to low (positive) values forces higher precision in the line-searches.% RHO is the minimum allowed fraction of the expected (from the slope at the% initial point in the linesearch). Constants must satisfy 0 < RHO < SIG < 1.% Tuning of SIG (depending on the nature of the function to be optimized) may% speed up the minimization; it is probably not worth playing much with RHO.% The code falls naturally into 3 parts, after the initial line search is% started in the direction of steepest descent. 1) we first enter a while loop% which uses point 1 (p1) and (p2) to compute an extrapolation (p3), until we% have extrapolated far enough (Wolfe-Powell conditions). 2) if necessary, we% enter the second loop which takes p2, p3 and p4 chooses the subinterval% containing a (local) minimum, and interpolates it, unil an acceptable point% is found (Wolfe-Powell conditions). Note, that points are always maintained% in order p0 <= p1 <= p2 < p3 < p4. 3) compute a new search direction using% conjugate gradients (Polack-Ribiere flavour), or revert to steepest if there% was a problem in the previous line-search. Return the best value so far, if% two consecutive line-searches fail, or whenever we run out of function% evaluations or line-searches. During extrapolation, the "f" function may fail% either with an error or returning Nan or Inf, and minimize should handle this% gracefully.if max(size(length)) == 2, red=length(2); length=length(1); else red=1; endif length>0, S='Linesearch'; else S='Function evaluation'; end i = 0;                                            % zero the run length counterls_failed = 0;                             % no previous line search has failed[f0 df0] = feval(f, X, varargin{:});          % get function value and gradientfX = f0;i = i + (length<0);                                            % count epochs?!s = -df0; d0 = -s'*s;           % initial search direction (steepest) and slopex3 = red/(1-d0);                                  % initial step is red/(|s|+1)while i < abs(length)                                      % while not finished  i = i + (length>0);                                      % count iterations?!  X0 = X; F0 = f0; dF0 = df0;                   % make a copy of current values  if length>0, M = MAX; else M = min(MAX, -length-i); end  while 1                             % keep extrapolating as long as necessary    x2 = 0; f2 = f0; d2 = d0; f3 = f0; df3 = df0;    success = 0;    while ~success && M > 0      try        M = M - 1; i = i + (length<0);                         % count epochs?!        [f3 df3] = feval(f, X+x3*s, varargin{:});        if isnan(f3) || isinf(f3) || any(isnan(df3)+isinf(df3)), error(''), end        success = 1;      catch                                % catch any error which occured in f        x3 = (x2+x3)/2;                                  % bisect and try again      end    end    if f3 < F0, X0 = X+x3*s; F0 = f3; dF0 = df3; end         % keep best values    d3 = df3'*s;                                                    % new slope    if d3 > SIG*d0 || f3 > f0+x3*RHO*d0 || M == 0  % are we done extrapolating?      break    end    x1 = x2; f1 = f2; d1 = d2;                        % move point 2 to point 1    x2 = x3; f2 = f3; d2 = d3;                        % move point 3 to point 2    A = 6*(f1-f2)+3*(d2+d1)*(x2-x1);                 % make cubic extrapolation    B = 3*(f2-f1)-(2*d1+d2)*(x2-x1);    x3 = x1-d1*(x2-x1)^2/(B+sqrt(B*B-A*d1*(x2-x1))); % num. error possible, ok!    if ~isreal(x3) || isnan(x3) || isinf(x3) || x3 < 0 % num prob | wrong sign?      x3 = x2*EXT;                                 % extrapolate maximum amount    elseif x3 > x2*EXT                  % new point beyond extrapolation limit?      x3 = x2*EXT;                                 % extrapolate maximum amount    elseif x3 < x2+INT*(x2-x1)         % new point too close to previous point?      x3 = x2+INT*(x2-x1);    end  end                                                       % end extrapolation  while (abs(d3) > -SIG*d0 || f3 > f0+x3*RHO*d0) && M > 0  % keep interpolating    if d3 > 0 || f3 > f0+x3*RHO*d0                         % choose subinterval      x4 = x3; f4 = f3; d4 = d3;                      % move point 3 to point 4    else      x2 = x3; f2 = f3; d2 = d3;                      % move point 3 to point 2    end    if f4 > f0                 x3 = x2-(0.5*d2*(x4-x2)^2)/(f4-f2-d2*(x4-x2));  % quadratic interpolation    else      A = 6*(f2-f4)/(x4-x2)+3*(d4+d2);                    % cubic interpolation      B = 3*(f4-f2)-(2*d2+d4)*(x4-x2);      x3 = x2+(sqrt(B*B-A*d2*(x4-x2)^2)-B)/A;        % num. error possible, ok!    end    if isnan(x3) || isinf(x3)      x3 = (x2+x4)/2;               % if we had a numerical problem then bisect    end    x3 = max(min(x3, x4-INT*(x4-x2)),x2+INT*(x4-x2));  % don't accept too close    [f3 df3] = feval(f, X+x3*s, varargin{:});    if f3 < F0, X0 = X+x3*s; F0 = f3; dF0 = df3; end         % keep best values    M = M - 1; i = i + (length<0);                             % count epochs?!    d3 = df3'*s;                                                    % new slope  end                                                       % end interpolation  if abs(d3) < -SIG*d0 && f3 < f0+x3*RHO*d0          % if line search succeeded    X = X+x3*s; f0 = f3; fX = [fX' f0]';                     % update variables    fprintf('%s %6i;  Value %4.6e\r', S, i, f0);    s = (df3'*df3-df0'*df3)/(df0'*df0)*s - df3;   % Polack-Ribiere CG direction    df0 = df3;                                               % swap derivatives    d3 = d0; d0 = df0'*s;    if d0 > 0                                      % new slope must be negative      s = -df0; d0 = -s'*s;                  % otherwise use steepest direction    end    x3 = x3 * min(RATIO, d3/(d0-realmin));          % slope ratio but max RATIO    ls_failed = 0;                              % this line search did not fail  else    X = X0; f0 = F0; df0 = dF0;                     % restore best point so far    if ls_failed || i > abs(length)         % line search failed twice in a row      break;                             % or we ran out of time, so we give up    end    s = -df0; d0 = -s'*s;                                        % try steepest    x3 = 1/(1-d0);                         ls_failed = 1;                                    % this line search failed  endendfprintf('\n');

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