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2 | function [y,info] = lmirank(At,c,K,pars,y0)
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3 | % [y,info] = lmirank(At,c,K,pars,y0);
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4 | %
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5 | % LMIRANK can be used to try to solve rank constrained LMI problems such as
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6 | %
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7 | % F(y)>=0, (1)
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8 | % G(y)>=0, rank G(y)<=r. (2)
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9 | %
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10 | % More precisely it can be used to try to solve feasibility problems
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11 | % involving any number of LMI constraints and one or more rank constraints.
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12 | %
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13 | % LMI data is entered in standard SeDuMi format. Rank constraints are entered
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14 | % using K.rank. For example,
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15 | %
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16 | % K.s=[4 7 6]; K.rank=[4 4 6];
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17 | %
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18 | % specifies 3 LMI constraints with the 2nd LMI constrained to have rank <= 4.
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19 | %
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20 | % LP inequality constraints can also be included.
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21 | %
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22 | % LMIRANK can be called using any of following
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23 | % [y,info] = lmirank(At,c,K,pars,y0);
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24 | % [y,info] = lmirank(At,c,K);
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25 | % [y,info] = lmirank(At,c,K,pars);
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26 | % [y,info] = lmirank(At,c,K,y0);
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27 | %
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28 | % Inputs:
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29 | % At,c,K.l,K.s : data in SeDuMi format
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30 | % K.rank : rank constraints
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31 | % Optional Inputs:
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32 | % pars.maxiter : max. no. of iterations, default is 1000
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33 | % pars.eps : constraint tolerance, default is 1e-9
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34 | % pars.fid : set to 0 to suppress on-screen output. The default
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35 | % is 1 which displays on-screen output.
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36 | % pars.itermod : output results to screen every pars.itermod
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37 | % iterations, default is 1
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38 | % y0 : initial condition
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39 | % (If y0 is not given, the trace heuristic is
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40 | % used to initialize the algorithm. SeDuMi is
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41 | % used to do this calculation and hence must
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42 | % be installed if y0 is not given.)
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43 | % Outputs:
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44 | % y : solution
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45 | % info.solved : 1 if a solution was found, 0 otherwise
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46 | % info.cpusec : solution time
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47 | % info.iters : no. of iterations required to find a solution
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48 | % info.gap : constraint gap
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49 | % info.rank : ranks (with respect to tolerance pars.eps)
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50 | %
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51 | % The algorithm is described in
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52 | % R. Orsi, U. Helmke, and J. B. Moore. A Newton-like method for solving
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53 | % rank constrained linear matrix inequalities. In Proceedings of the
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54 | % 43rd IEEE Conference on Decision and Control, pages 3138-3144,
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55 | % Paradise Island, Bahamas, 2004.
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56 | %
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57 | % Algorithm notes
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58 | % 1. Projection onto the set B (see Section V.C of the paper) is now done
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59 | % in a much simpler manner. (It still results in a linearly constrained
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60 | % least squares problem.)
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61 | % 2. The above paper describes only the basic case given by (1) and (2)
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62 | % above, i.e., it does not deal with multiple non-rank constrained LMIs,
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63 | % multiple rank constrained LMIs, or LP inequality constraints.
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64 | %
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65 | %Feedback should be sent to robert.orsi@anu.edu.au
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66 |
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67 | % Author Robert Orsi
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68 | % Feb 2005
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69 |
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70 |
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71 | t=cputime;
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72 |
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73 | %%%% If no LP ineq. constraints are present, set K.l=0
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74 | if ~isfield(K,'l')
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75 | K.l=0;
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76 | end
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77 |
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78 | %%%% Set unspecified pars, calculate an initial condition if none given
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79 | if nargin==3
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80 | pars.fid=1;
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81 | y = trheuristic(At,c,K);
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82 | end
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83 | if nargin==4
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84 | if isstruct(pars)
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85 | y = trheuristic(At,c,K);
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86 | else
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87 | y=pars;
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88 | pars.fid=1;
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89 | end
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90 | end
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91 | if nargin==5
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92 | y=y0;
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93 | end
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94 | if ~isfield(pars,'maxiter') pars.maxiter=1000; end
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95 | if ~isfield(pars,'eps') pars.eps=1e-9; end
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96 | if ~isfield(pars,'fid') pars.fid=1; end
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97 | if ~isfield(pars,'itermod') pars.itermod=1; end
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98 |
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99 | %%%% Initialize X1
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100 | X1=c-At*y;
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101 |
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102 | %%%% Calculate m
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103 | m=size(At,2);
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104 |
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105 | %%%% Print output header
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106 | fprintf(pars.fid,'\nLMIRank by Robert Orsi, 2005.\n');
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107 | fprintf(pars.fid,'maxiter = %5d | rank bounds\n',pars.maxiter);
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108 | fprintf(pars.fid,'eps = %0.2e | ',pars.eps);
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109 | fprintf(pars.fid,'%5d',K.rank);
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110 | fprintf(pars.fid,'\n iter : gap ranks\n');
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111 |
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112 | %%%% MAIN LOOP
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113 | for iters=1:pars.maxiter,
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114 |
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115 | %% Calculate DX1 : eigs of X1
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116 | %%
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117 | %% X2LP : proj. of LP ineq. component of X1
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118 | %% X2LPindex : index of zeroes of X2LP
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119 | %% DX2 : eigs of X2.
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120 | %% : Sorted, non-neg., rank constained version of DX1
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121 | %% rankX2 : ranks of X2
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122 | %%
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123 | %% Attrans : At after transformation
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124 | %% ctrans : c after transformation
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125 | %% Attrans2 :
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126 | %% ctrans2 :
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127 | DX1=zeros(sum(K.s),1);
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128 | %
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129 | X2LP=max(X1(1:K.l),0);
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130 | X2LPindex=find(X2LP==0);
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131 | DX2=zeros(sum(K.s),1);
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132 | rankX2=zeros(length(K.s),1);
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133 | %
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134 | Attrans=zeros(size(At));
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135 | Attrans(1:K.l,:)=At(1:K.l,:);
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136 | ctrans=zeros(size(c));
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137 | ctrans(1:K.l)=c(1:K.l);
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138 | Attrans2=zeros(size(At)); % incorrect size, truncated later
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139 | Attrans2(1:length(X2LPindex),:)=Attrans(X2LPindex,:);
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140 | ctrans2=zeros(size(c)); % incorrect size, truncated later
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141 | ctrans2(1:length(X2LPindex))=ctrans(X2LPindex);
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142 | %
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143 | index=0;
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144 | index2=0;
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145 | index3=length(X2LPindex);
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146 | for j=1:length(K.s),
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147 | [V,D]=eig(reshape(X1(K.l+index2+1:K.l+index2+K.s(j)^2),K.s(j),K.s(j)));
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148 | DX1(index+1:index+K.s(j))=diag(D);
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149 | [Dsort,I]=sort(-diag(D));
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150 | Dsort=-Dsort;
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151 | V=V(:,I);
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152 | for i=1:K.s(j),
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153 | if (rankX2(j)<K.rank(j)) & (Dsort(i)>0)
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154 | rankX2(j)=rankX2(j)+1;
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155 | DX2(index+i)=Dsort(i);
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156 | end
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157 | end
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158 | for k=1:m,
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159 | Q=V'*reshape(At(K.l+index2+1:K.l+index2+K.s(j)^2,k),K.s(j),K.s(j))*V;
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160 | Q=(Q+Q')/2;
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161 | Attrans(K.l+index2+1:K.l+index2+K.s(j)^2,k)=Q(:);
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162 | Attrans2(index3+1:index3+(K.s(j)-rankX2(j))^2,k)=...
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163 | reshape(Q(rankX2(j)+1:end,rankX2(j)+1:end),(K.s(j)-rankX2(j))^2,1);
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164 | end
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165 | Q=V'*reshape(c(K.l+index2+1:K.l+index2+K.s(j)^2),K.s(j),K.s(j))*V;
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166 | Q=(Q+Q')/2;
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167 | ctrans(K.l+index2+1:K.l+index2+K.s(j)^2)=Q(:);
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168 | ctrans2(index3+1:index3+(K.s(j)-rankX2(j))^2)=...
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169 | reshape(Q(rankX2(j)+1:end,rankX2(j)+1:end),(K.s(j)-rankX2(j))^2,1);
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170 | %
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171 | index=index+K.s(j);
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172 | index2=index2+K.s(j)^2;
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173 | index3=index3+(K.s(j)-rankX2(j))^2;
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174 | end
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175 | %% Truncate Attrans2 and ctrans2 to correct sizes
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176 | q=length(X2LPindex);
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177 | for j=1:length(K.s),
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178 | q=q+(K.s(j)-rankX2(j))^2;
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179 | end
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180 | Attrans2=Attrans2(1:q,1:m);
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181 | ctrans2=ctrans2(1:q,1);
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182 |
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183 | %% BREAK if a solution has been found
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184 | %% Decision is based on X1, not X2
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185 | bound= (K.l+length(DX1)) * max([abs(X1(1:K.l)); abs(DX1)]) * eps * 10;
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186 | bound= max(bound,pars.eps);
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187 | gap=min(DX1);
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188 | if K.l~=0
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189 | gap=min([gap; X1(1:K.l)]);
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190 | end
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191 | breakflag=(gap >= -bound);
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192 | rankX1=zeros(1,length(K.s));
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193 | index=0;
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194 | for j=1:length(K.s),
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195 | rankX1(j)=sum(abs(DX1(index+1:index+K.s(j)))>bound);
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196 | breakflag=breakflag & (rankX1(j) <= K.rank(j));
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197 | index=index+K.s(j);
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198 | end
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199 | if (pars.fid)&(mod(iters,pars.itermod)==0) %% Output iters, gap & ranks
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200 | fprintf(pars.fid,'%5d : %0.2e ',iters,gap);
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201 | fprintf(pars.fid,'%5d',rankX1);
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202 | fprintf(pars.fid,'\n');
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203 | end
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204 | if breakflag
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205 | break
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206 | end
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207 |
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208 | %% Calculate SVD and rank of Attrans2
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209 | [U,S,V]=svd(Attrans2);
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210 | s=diag(S);
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211 | tol=max(size(Attrans2))*s(1)*eps;
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212 | rankAttrans2=sum(s>tol);
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213 |
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214 | %% Calculate y
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215 | if rankAttrans2 == size(Attrans2,2),
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216 | y=Attrans2\ctrans2;
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217 | else
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218 | if size(Attrans2,1)~=size(Attrans2,2)
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219 | warning off;
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220 | y0=Attrans2\ctrans2;
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221 | warning on;
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222 | else
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223 | y0=pinv(Attrans2)*ctrans2;
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224 | end
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225 | W=V(:,rankAttrans2+1:end);
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226 | e=zeros(size(c));
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227 | e(1:K.l)=X2LP;
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228 | index=0;
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229 | index2=K.l;
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230 | for j=1:length(K.s),
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231 | P=diag(DX2(index+1:index+K.s(j)));
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232 | e(index2+1:index2+K.s(j)^2)=P(:);
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233 | index=index+K.s(j);
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234 | index2=index2+K.s(j)^2;
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235 | end
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236 | q=(Attrans*W)\( ctrans - Attrans*y0 - e );
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237 | y=y0+W*q;
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238 | end
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239 |
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240 | %% Calculate new X1
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241 | X1=c-At*y;
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242 |
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243 | end
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244 |
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245 | info.solved=breakflag;
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246 | info.cpusec=cputime-t;
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247 | info.iters=iters;
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248 | info.gap=gap;
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249 | info.rank=rankX1;
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250 |
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251 | fprintf(pars.fid,'iters solved seconds\n');
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252 | fprintf(pars.fid,'%5d %4d %11.1e\n',info.iters,info.solved,info.cpusec);
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