81 lines
2.0 KiB
Mathematica
81 lines
2.0 KiB
Mathematica
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function varargout = boundingbox(F,ops,x)
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%BOUNDINGBOX Computes bounding box of a constraint
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%
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% If only one output is requested, only the symbolic model is returned
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% B = boundingbox(F)
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%
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% If three outputs are requested, numerical values are returned too
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% [B,L,U] = boundingbox(F)
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%
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% A second argument can be used to specificy solver settings
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% B = boundingbox(F,sdpsettings('solver','cplex'))
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%
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% A third argument can (should!) be used to obtain a bounding box in a
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% particular set of variables, i.e. the bounding box of a projection.
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% Unless you specify which variables you are computing the bounding box
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% w.r.t, it will be very hard for you to understand which variables the
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% values in L and U relate to
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% [B,L,U] = boundingbox(F,[],x)
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% B will now be the box [L <= x <= U] (infinite bounds not included)
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if nargin < 3
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x = recover(depends(F));
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else
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x = x(:);
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if ~isa(x,'sdpvar')
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error('The third argument should be an SDPVAR obeject');
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end
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end
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if nargin < 2
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ops = sdpsettings('verbose',0);
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end
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if isempty(ops)
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ops = sdpsettings('verbose',0);
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end
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if ~isa(ops,'struct')
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error('The second argument should be an SDPSETTINGS struct (or empty)');
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end
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sol = solvesdp(F,[x;-x],ops);
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n = length(x);
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for i = 1:n
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xi = x(i);
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if isa(xi,'sdpvar')
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if sol.problem(i)==0
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% selectsolution(i);
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L(i,1) = double(xi,i);
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else
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L(i,1) = -inf;
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end
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if sol.problem(n+i)==0
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% selectsolution(n+i);
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U(i,1) = double(xi,n+i);
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else
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U(i,1) = inf;
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end
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else
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L(i,1) = xi;
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U(i,1) = xi;
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end
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end
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% Only add finite bounds
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Lf = find(~isinf(L));
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Uf = find(~isinf(U));
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B = [];
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if ~isempty(Lf)
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xLf = x(Lf);
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if isa(xLf,'sdpvar')
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B = [B, (xLf >= L(Lf)):'Finite lower bounds'];
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end
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end
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if ~isempty(Uf)
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xUf = x(Uf);
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if isa(xUf,'sdpvar')
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B = [B, (xUf <= U(Uf)):'Finite upper bounds'];
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end
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end
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varargout{1} = B;
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varargout{2} = L;
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varargout{3} = U;
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