source: distools/ksvo_nu.m @ 89

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1%KSVO_NU Kernel Support Vector Optimizer: NU algorithm
2%
3%   [V,J,C,REG] = KSVO_NU(K,Y,NU,PD)
4%
5% INPUT
6%   K   Kernel matrix
7%   Y   Label list consisting of -1/+1
8%   NU  Regularization parameter (0 < NU < 1):
9%       expected fraction of support vectors (optional; default: 0.25)
10%   R   Parameter: -1,0,1,2
11%       -1 or 'flip',   Changes the kernel by flipping negative eigenvalues to
12%          positive
13%        0 or '',       Uses the kernel as it is
14%        1 or 'reg',    Checks positive definiteness and regularizes the kernel by
15%          adding the minimal constant in the form of 10^i to the diagonal
16%        2 or 'lamreg', Checks positive definiteness and regularizes the kernel by
17%          adding the minimal constant to the diagonal (equal to the magnitude of
18%          the smallest negative eigenvalue)
19%       (optional; default: 0, do not change the kernel)
20%
21% OUTPUT
22%   V   Vector of weights for the support vectors
23%   J   Index vector pointing to the support vectors
24%   T   Transformation matrix for the test kernel; to be used in testing
25%   REG Regularization parameter added to the diagonal, if used (R=1,2); a vector
26%       of eigenvalues of K (R=-1), or -1 if not checked (R=0)
27%
28% DESCRIPTION
29% A low level routine that optimizes the set of support vectors for a two-class
30% classification problem based on the kernel or similarity matrix K.
31% KSVO_NU is called directly from KSVC_NU. The labels Y should indicate the
32% two classes by +1 and -1. Optimization is done by a quadratic programming.
33% If available, the QLD function is used for a positive definite kernel,
34% otherwise an appropriate Matlab routine.
35%
36% NU is bounded from above by NU_MAX = (1 - ABS(Lp-Lm)/(Lp+Lm)), where Lp (Lm)
37% is the number of positive (negative) samples. If NU > NU_MAX is supplied,
38% it will be changed to the NU_MAX.
39%
40% If NU is less than some NU_MIN which depends on the overlap between the
41% classes, the algorithm will typically take long time to converge (if at all).
42% So, it is advisable to set NU larger than the expected overlap.
43% Weights V are rescaled in such a way as if returned by KSVO with the
44% parameter C.
45%
46% SEE ALSO
47% KSCV_NU, KSVC, KSVO, SVO
48
49% Copyright: Elzbieta Pekalska, Robert P.W. Duin, ela.pekalska@googlemail.com
50% Based on SVC.M by D.M.J. Tax, D. de Ridder and R.P.W. Duin and SVC_NU.M by S.Verzakov
51% Faculty EWI, Delft University of Technology and
52% School of Computer Science, University of Manchester
53
54
55function [v,J,T,reg,err] = ksvo_nu(K,y,nu,r)
56
57prtrace(mfilename);
58
59if (nargin < 4),
60  r = 0;      % Default: the kernel is used as provided
61end
62if (nargin < 3)
63  prwarning(3,'Third parameter (nu) not specified, assuming 0.25.');
64  nu = 0.25;
65end
66
67if all(r ~= [-1,0,1,2]),
68  error('Wrong parameter R.');
69end
70
71nu_max = 1 - abs(nnz(y == 1) - nnz(y == -1))/length(y);
72if nu > nu_max
73  prwarning(3,['nu==' num2str(nu)  ' is not feasible; set to ' num2str(nu_max)]);
74  nu = nu_max;
75end
76
77err  = 0;
78v    = [];     % Weights of the SVM
79J    = [];     % Index of support vectors
80reg = -1;      % Regularization on the diagonal or a list of eigenvalues, if used; -1, if not checked
81vmin = 1e-9;   % Accuracy to determine when an object becomes the support object
82iacc = -14;    % Accuracy of 10^i to determine whether the kernel is pd or not
83T    = [];     % Matrix to transform the test kernel if r = -1
84
85
86% Set up the variables for the optimization.
87n  = size(K,1);
88Ky = (y*y').*K;
89f  = zeros(1,n);
90A  = [ones(1,n); y'];
91b  = [nu*n; 0];
92lb = zeros(n,1);
93ub = ones(n,1);
94p  = rand(n,1);
95
96
97if r ~= 0,
98  % Find whether the kernel is positive definite.
99  i = -20;
100  while (pd_check (Ky + (10.0^i) * eye(n)) == 0)
101    i = i + 1;
102  end
103
104  if r == 1,
105    if i > iacc,  % if i is smaller, then the regularization is within numerical accuracy
106      prwarning(1,'K is not positive definite. The diagonal is regularized by 10.0^(%d)',i);
107    end
108
109    % Make the kernel positive definite by adding a constant to the diagonal.
110    reg = 10.0^i;
111    Ky  = Ky + (10.0^(i)) * eye(n);
112
113  elseif r == 2,
114    L = preig(Ky);
115    L = diag(real(L));
116    I = find(L < 0);
117    if ~isempty(I),
118      % Make the kernel positive definite by adding a constant to the diagonal.
119      reg = -1.001*min(L(I));
120      Ky  = Ky + reg * eye(n);
121      pow = round(log10(reg));
122      if pow >= 0
123        prwarning(1,['K is not positive definite. The diagonal is regularized by %' num2str(pow) '.2f'],reg);
124      elseif pow > iacc,
125        prwarning(1,['K is not positive definite. The diagonal is regularized by %0.3g'], reg);
126      else
127        ;
128      end
129    else
130      reg = 0;
131    end
132
133  else   % r = -1,
134    % reverse the negative eigenvalues of the kernel K
135    if i > iacc,
136      [Q,L] = preig(K);
137      Q     = real(Q);
138      L     = diag(real(L));
139      if all(L >= 0),
140        T = [];
141      else
142        K   = Q*diag(abs(L))*Q';
143        Ky  = (y*y').*K;
144        T   = Q* diag(sign(L)) *Q';   % transformation matrix for the test kernel
145      end
146    else
147      % if i < iacc, then the regularization is within numerical accuracy, so apply it
148      Ky    = Ky + (10.0^(i)) * eye(n);
149    end
150  end
151end
152
153
154
155% Minimization procedure
156% QP  minimizes:   0.5 x'*Ky*x + f'x
157% subject to:      Ax <= b
158%
159
160
161done = 0;
162if r ~= 0 & (exist('qld') == 3)
163  prwarning(5,'QLD routine is used.')
164  v = qld (Ky, f, -A, b, lb, ub, p, 2);
165  done = 1;
166end
167
168if r == 0 | ~done,
169  if (exist('quadprog') == 2)
170    prwarning(5,'Matlab QUADPROG is used.')
171    opt = optimset;
172    opt.Diagnostics = 'off';
173    opt.LargeScale  = 'off';
174    opt.Display     = 'off';
175    opt.MaxIter     = 500;
176    v = quadprog(Ky, f, [], [], A, b, lb, ub,[],opt);
177  else
178    error(5,'KSVC requires Matlab 6.x')
179  end
180end
181
182
183% Compute the square pseudo-norm of the weights in the corresponding
184% Hilbert/Krein space. If it is negative (which may happen for a highly
185% indefinite kernel), then the SVM is not proper.
186
187vnorm = v'*Ky*v;
188if vnorm < 0 | abs(vnorm) < 10^iacc,
189  prwarning(1,'SVM: ||w||^2_PE < 0. The SVM is not properly defined. Pseudo-Fisher is computed instead.');
190  err = 1;
191  v   = prpinv([K ones(n,1)])*y;
192  J   = [1:n]';
193  return;
194end
195
196
197% Find all the support vectors.
198J = find(v > vmin);
199
200% First find the SV on the boundary
201I  = J(v(J) < 1-vmin);
202Ip = find(y(I) ==  1);
203Im = find(y(I) == -1);
204if (isempty(v) | isempty(Ip) | isempty(Im))
205  prwarning(1,'Quadratic Optimization failed. Pseudo-Fisher is computed instead.');
206  v = prpinv([K ones(n,1)])*y;
207  J = [1:n]';
208  C = nan;
209  err = 1;
210  return;
211end
212
213
214
215v    = y.*v;
216wxI  = K(I,J)*v(J);
217wxIp = mean(wxI(Ip),1);
218wxIm = mean(wxI(Im),1);
219
220rho  = 0.5*(wxIp-wxIm);
221b    = - (0.5/rho)*(wxIp+wxIm);
222%b = mean(y(I) - wxI/rho);
223
224v = [v(J)/rho; b];
225C = 1/rho;
226
227return;
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