115 lines
3.1 KiB
Mathematica
115 lines
3.1 KiB
Mathematica
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function sdpvarExpr = parseLMI(X)
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%PARSELMI Internal function for dirty parsing of lmi string
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% Assume Error
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% TypeofConstraint = 1;
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% Check for obsolete notation .<, .>, =
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single_equality = findstr(X,'=');
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if isempty(findstr(X,'>=')) & isempty(findstr(X,'<=')) & (rem(length(single_equality),2) == 1) % There is a single =
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error('Obsolete constraint =, use == in equalities.');
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%X = strrep(X,'=','==');
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%X = strrep(X,'====','=='); % Whoops, == replaced with =====!
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end
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if ~isempty(findstr(X,'.<'))
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disp(' ');
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disp('Warning: Obsolete constraint .<, use < in equalities.');
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disp('If you have a Hermitian matrix which you want to')
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disp('constrain to have negative values, use ,e.g., P(:)<0')
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disp('The constraint has been changed to <')
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disp(' ');
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X = strrep(X,'.<','<');
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end
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if ~isempty(findstr(X,'.>'))
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disp(' ');
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disp('Warning: Obsolete constraint .>, use > in equalities.');
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disp('If you have a Hermitian matrix which you want to')
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disp('constrain to have negative values, use e.g. P(:)>0')
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disp('The constraint has been changed to >')
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disp(' ');
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X = strrep(X,'.>','>');
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end
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% Any norm? If not, we're done!
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if isempty(findstr(X,'||'))
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sdpvarExpr = X;
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return
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end
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% The only actual parsing needed is for the ||Axplusb||<cx+d notation
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delimiters = {'.<','.>','<.','>.','<','>','==',''};
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delindex = 0;indtodel = [];
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while isempty(indtodel) & (delindex<=7)
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delindex = delindex+1;
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indtodel = findstr(X,delimiters{delindex});
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end
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switch delindex
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case 5 %<
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TypeofConstraint = 1;
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ind_start = indtodel-1;
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ind_end = indtodel+1;
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REVERSE = 1;
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case 6 %>
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TypeofConstraint = 1;
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ind_start = indtodel-1;
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ind_end = indtodel+1;
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REVERSE = 0;
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case 7%=
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TypeofConstraint = 3;
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ind_start = indtodel-1;
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ind_end = indtodel+2;
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REVERSE = 0;
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case {3,4}
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error('Incorrect argument. Perhaps you mean .> or .<');
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otherwise
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error('Incorrect argument. Could not find <=>.>.<')
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end
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LeftHand = X(1:ind_start);
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RightHand = X(ind_end:end);
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if REVERSE
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temp = LeftHand;
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LeftHand = RightHand;
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RightHand = temp;
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end
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% Search for a norm expression
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ind_norm_Right = findstr(RightHand,'||');
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ind_norm_Left = findstr(LeftHand,'||');
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% Any norm at all, if not, we're done!
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if isempty(ind_norm_Right) & isempty(ind_norm_Left)
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sdpvarExpr = [LeftHand '-(' RightHand ')'];
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return
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end
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% Equality constrained norm?
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if (TypeofConstraint == 3) & (ind_norm_Right | ind_norm_Left)
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error('Equality constraints cannot be used with ||..|| operator');
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end
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% Convex normconstraint?
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if ~isempty(ind_norm_Left)
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error('Norm ||..|| must look like || column vector ||< scalar, or scalar>|| vector ||')
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end
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% Everthing seem ok in norm
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TypeofConstraint = 4;
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Axplusb = RightHand(2+ind_norm_Right(1):ind_norm_Right(2)-1);
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WithoutNorm = strrep(RightHand,RightHand(ind_norm_Right(1):ind_norm_Right(2)+1),'0');
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if length(WithoutNorm)==1
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cxplusd = LeftHand;
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else
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cxplusd = [LeftHand '-(' WithoutNorm ')'];
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end
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sdpvarExpr = ['cone( ' Axplusb ',' cxplusd ')'];
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