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[交流] 通過不等式和等式約束得到封閉的多邊體

以下的代碼來自Matt，我拿出來與大家進(jìn)行交流，希望各位能夠參與并分析這個程序的算法。

function [V,nr,nre]=lcon2vert(A,b,Aeq,beq,TOL,checkbounds)
%An extension of Michael Kleder's con2vert function, used for finding the
%vertices of a bounded polyhedron in R^n, given its representation as a set
%of linear constraints. This wrapper extends the capabilities of con2vert to
%also handle cases where the  polyhedron is not solid in R^n, i.e., where the
%polyhedron is defined by both equality and inequality constraints.
%
%SYNTAX:
%
%  [V,nr,nre]=lcon2vert(A,b,Aeq,beq,TOL)
%
%The rows of the N x n matrix V are a series of N vertices of the polyhedron
%in R^n, defined by the linear constraints
%
% A*x  <= b
% Aeq*x = beq
%
%By default, Aeq=beq=[], implying no equality constraints. The output "nr"
%lists non-redundant inequality constraints, and "nre" lists non-redundant
%equality constraints.
%
%The optional TOL argument is a tolerance used for both rank-estimation and
%for testing feasibility of the equality constraints. Default=1e-10.
%The default can also be obtained by passing TOL=[];
%
%
%EXAMPLE:
%
%The 3D region defined by x+y+z=1, x>=0, y>=0, z>=0
%is described by the following constraint data.
%
%
%    A =
%
%       0.4082 -0.8165 0.4082
%       0.4082 0.4082 -0.8165
%       -0.8165 0.4082 0.4082
%
%
%    b =
%
%       0.4082
%       0.4082
%       0.4082
%
%
%    Aeq =
%
%       0.5774 0.5774 0.5774
%
%
%    beq =
%
%       0.5774
%
%
%  >> V=lcon2vert(A,b,Aeq,beq)
%
%       V =
%
%          1.0000 0.0000 0.0000
%          -0.0000 1.0000       0
%                0 0.0000 1.0000
%
%

  %%initial argument parsing

  nre=[];
  nr=[];
  if nargin<5 || isempty(TOL), TOL=1e-10; end
  if nargin<6, checkbounds=true; end

  switch nargin

   case 0

         error 'At least 1 input argument required'


   case 1

      b=[]; Aeq=[]; beq=[];


   case 2

      Aeq=[]; beq=[];

   case 3

      beq=[];
      error 'Since argument Aeq specified, beq must also be specified'

  end


  b=b(:); beq=beq(:);

  if xor(isempty(A), isempty(b))
   error 'Since argument A specified, b must also be specified'
  end

  if xor(isempty(Aeq), isempty(beq))
      error 'Since argument Aeq specified, beq must also be specified'
  end


  nn=max(size(A,2)*~isempty(A),size(Aeq,2)*~isempty(Aeq));

  if ~isempty(A) && ~isempty(Aeq) && ( size(A,2)~=nn || size(Aeq,2)~=nn)

   error 'A and Aeq must have the same number of columns if both non-empty'

  end


  inequalityConstrained=~isempty(A);
  equalityConstrained=~isempty(Aeq);

[A,b]=rownormalize(A,b);
[Aeq,beq]=rownormalize(Aeq,beq);

  if equalityConstrained && nargout>2


      [discard,nre]=lindep([Aeq,beq].',TOL);

      if ~isempty(nre) %reduce the equality constraints

         Aeq=Aeq(nre,:);
         beq=beq(nre);

      else
         equalityConstrained=false;
      end

end



%%Find 1 solution to equality constraints within tolerance


if equalityConstrained


   Neq=null(Aeq);

   x0=pinv(Aeq)*beq;

   if norm(Aeq*x0-beq)>TOL*norm(beq),  %infeasible

      nre=[]; nr=[]; %All constraints redundant for empty polytopes
      V=[];
      return;

   elseif isempty(Neq)

         V=x0(:).';
         nre=(1:nn).'; %Equality constraints determine everything.
         nr=[];%All inequality constraints are therefore redundant.
         return

   end

   rkAeq= nn - size(Neq,2);


  end

%%
  if inequalityConstrained && equalityConstrained

AAA=A*Neq;
bbb=b-A*x0;

  elseif inequalityConstrained

AAA=A;
bbb=b;

  elseif equalityConstrained && ~inequalityConstrained

   error('Non-bounding constraints detected. (Consider box constraints on variables.)')


  end

  nnn=size(AAA,2);


  if nnn==1 %Special case

   idxu=sign(AAA)==1;
   idxl=sign(AAA)==-1;
   idx0=sign(AAA)==0;

   Q=bbb./AAA;
   U=Q;
   U(~idxu)=inf;
   L=Q;
   L(~idxl)=-inf;


   [ub,uloc]=min(U);
   [lb,lloc]=max(L);

   if ~all(bbb(idx0)>=0) || ub
      V=[]; nr=[]; nre=[];
      return

   elseif ~isfinite(ub) || ~isfinite(lb)

      error('Non-bounding constraints detected. (Consider box constraints on variables.)')

   end

   Zt=[lb;ub];

   if nargout>1
      nr=unique([lloc,uloc]); nr=nr(:);
   end


  else

      if nargout>1
         [Zt,nr]=con2vert(AAA,bbb,TOL,checkbounds);
      else
         Zt=con2vert(AAA,bbb,TOL,checkbounds);
      end

  end


  if equalityConstrained

   V=bsxfun(@plus,Zt*Neq.',x0(:).');

  else

   V=Zt;

  end

function [V,nr] = con2vert(A,b,TOL,checkbounds)
% CON2VERT - convert a convex set of constraint inequalities into the set
%          of vertices at the intersections of those inequalities;i.e.,
%          solve the "vertex enumeration" problem. Additionally,
%          identify redundant entries in the list of inequalities.
%
% V = con2vert(A,b)
% [V,nr] = con2vert(A,b)
%
% Converts the polytope (convex polygon, polyhedron, etc.) defined by the
% system of inequalities A*x <= b into a list of vertices V. Each ROW
% of V is a vertex. For n variables:
% A = m x n matrix, where m >= n (m constraints, n variables)
% b = m x 1 vector (m constraints)
% V = p x n matrix (p vertices, n variables)
% nr = list of the rows in A which are NOT redundant constraints
%
% NOTES: (1) This program employs a primal-dual polytope method.
%       (2) In dimensions higher than 2, redundant vertices can
%          appear using this method. This program detects redundancies
%          at up to 6 digits of precision, then returns the
%          unique vertices.
%       (3) Non-bounding constraints give erroneous results; therefore,
%          the program detects non-bounding constraints and returns
%          an error. You may wish to implement large "box" constraints
%          on your variables if you need to induce bounding. For example,
%          if x is a person's height in feet, the box constraint
%          -1 <= x <= 1000 would be a reasonable choice to induce
%          boundedness, since no possible solution for x would be
%          prohibited by the bounding box.
%       (4) This program requires that the feasible region have some
%          finite extent in all dimensions. For example, the feasible
%          region cannot be a line segment in 2-D space, or a plane
%          in 3-D space.
%       (5) At least two dimensions are required.
%       (6) See companion function VERT2CON.
%       (7) ver 1.0: initial version, June 2005
%       (8) ver 1.1: enhanced redundancy checks, July 2005
%       (9) Written by Michael Kleder
%
%Modified by Matt Jacobson - March 30, 2011
%

%%%3/4/2012 Improved boundedness test - unfortunately slower than Michael Kleder's
if checkbounds

[aa,bb,aaeq,bbeq]=vert2lcon(A,TOL);

if any(bb<=0) || ~isempty(bbeq)
      error('Non-bounding constraints detected. (Consider box constraints on variables.)')
end

clear aa bb aaeq bbeq

end

%%%Matt J initialization
dim=size(A,2);
slackfun=@(c)b-A*c;

%Initializer0
c = pinv(A)*b; %02/17/2012 -replaced with pinv()
s=slackfun(c);

if ~approxinpoly(s,TOL) %Initializer1

      c=Initializer1(TOL,A,b,c);
      s=slackfun(c);

end

if  ~approxinpoly(s,TOL)  %Attempt refinement

      %disp 'It is unusually difficult to find an interior point of your polytope. This may take some time... '
      %disp ' '

      c=Initializer2(TOL,A,b,c);
      %[c,fval]=Initializer1(TOL,A,b,c,10000);
      s=slackfun(c);


end


if ~approxinpoly(s,TOL)
         error('Unable to locate a point near the interior of the feasible region.')
end

if ~strictinpoly(s,TOL) %Added 02/17/2012 to handle initializers too close to polytope surface

      %disp 'Recursing...'


      idx=(  abs(s)<=max(s)*TOL );

      Amod=A; bmod=b;
      Amod(idx,:)=[];
      bmod(idx)=[];

      Aeq=A(idx,:); %pick the nearest face to c
      beq=b(idx);


      faceVertices=lcon2vert(Amod,bmod,Aeq,beq,TOL,1);
      if isempty(faceVertices)
         disp 'Something''s wrong. Couldn''t find face vertices. Possibly polyhedron is unbounded.'
         keyboard
      end

      c=faceVertices(1,:).';  %Take any vertex - find local recession cone vector
      s=slackfun(c);

      idx=(  abs(s)<=max(s)*TOL );

      Asub=A(idx,:); bsub=b(idx,:);

      [aa,bb,aaeq,bbeq]=vert2lcon(Asub);
      aa=[aa;aaeq;-aaeq];
      bb=[bb;bbeq;-bbeq];

      clear aaeq bbeq


      [bmin,idx]=min(bb);

      if bmin>=0
         disp 'Something''s wrong. We should have found a recession vector (bb<0).'
         keyboard
      end



      Aeq2=null(aa(idx,:)).';
      beq2=Aeq2*c;  %find intersection of polytope with line through facet centroid.

      lineCentroid=mean(lcon2vert(A,b,Aeq2,beq2,TOL,1));%Relies on boundedness

      clear aa bb

      c=lineCentroid(:);
      s=slackfun(c);


end


b = s;

%%%end Matt J initialization

D=bsxfun(@rdivide,A,b);

k = convhulln(D);
nr = unique(k(:));

G  = zeros(size(k,1),dim);
ee=ones(size(k,2),1);
discard=false( 1, size(k,1) );

for ix = 1:size(k,1) %02/17/2012 - modified

      F = D(k(ix,:),:);
      if lindep(F,TOL)          discard(ix)=1;
         continue;
      end

      G(ix,:)=F\ee;

end

G(discard,:)=[];

V = bsxfun(@plus, G, c.');

[discard,I]=unique( round(V*1e6),'rows');
V=V(I,:);

return

function [c,fval]=Initializer1(TOL, A,b,c,maxIter)


thresh=-10*max(eps(b));

if nargin>4
   [c,fval]=fminsearch(@(x) max([thresh;A*x-b]), c,optimset('MaxIter',maxIter));
else
   [c,fval]=fminsearch(@(x) max([thresh;A*x-b]), c);
end

return

function c=Initializer2(TOL,A,b,c)
%norm(  (I-A*pinv(A))*(s-b) )  subj. to s>=0


maxIter=100000;

[mm,nn]=size(A);

   Ap=pinv(A);
   Aaug=speye(mm)-A*Ap;
   Aaugt=Aaug.';

M=Aaugt*Aaug;
C=sum(abs(M),2);
   C(C<=0)=min(C(C>0));

slack=b-A*c;
slack(slack<0)=0;


      %    relto=norm(b);
      %    relto =relto + (relto==0);
      %
      %    relres=norm(A*c-b)/relto;


IterThresh=maxIter;
s=slack;
ii=0;
%for ii=1:maxIter
while ~approxinpoly(b-A*c,TOL)

   ii=ii+1;
   if ii>IterThresh,
         warning 'This is taking a lot of iterations'
         IterThresh=IterThresh+maxIter;
   end

   s=s-Aaugt*(Aaug*(s-b))./C;
   s(s<0)=0;


   c=Ap*(b-s);
   %slack=b-A*c;
   %relres=norm(slack)/relto;
   %if all(0

end

return

function [r,idx,Xsub]=lindep(X,tol)
%Extract a linearly independent set of columns of a given matrix X
%
% [r,idx,Xsub]=lindep(X)
%
%in:
%
%  X: The given input matrix
%  tol: A rank estimation tolerance. Default=1e-10
%
%out:
%
% r: rank estimate
% idx:  Indices (into X) of linearly independent columns
% Xsub: Extracted linearly independent columns of X

if ~nnz(X) %X has no non-zeros and hence no independent columns

   Xsub=[]; idx=[];
   return
end

if nargin<2, tol=1e-10; end


   [Q, R, E] = qr(X,0);

   diagr = abs(diag(R));

   %Rank estimation
   r = find(diagr >= tol*diagr(1), 1, 'last'); %rank estimation

   if nargout>1
   idx=sort(E(1:r));
      idx=idx(:);
   end


   if nargout>2
   Xsub=X(:,idx);
   end


function [A,b]=rownormalize(A,b)
%Modifies A,b data pair so that norm of rows of A is either 0 or 1

  if isempty(A), return; end

  normsA=sqrt(sum(A.^2,2));
  idx=normsA>0;
  A(idx,:)=bsxfun(@rdivide,A(idx,:),normsA(idx));
  b(idx)=b(idx)./normsA(idx);

function tf=approxinpoly(s,TOL)


smax=max(s);

if smax<=0
   tf=false; return
end

tf=all(s>=-smax*TOL);

  function tf=strictinpoly(s,TOL)

smax=max(s);

if smax<=0
   tf=false; return
end

tf=all(s>=smax*TOL);

function [A,b,Aeq,beq]=vert2lcon(V,tol)
%An extension of Michael Kleder's vert2con function, used for finding the
%linear constraints defining a polyhedron in R^n given its vertices. This
%wrapper extends the capabilities of vert2con to also handle cases where the
%polyhedron is not solid in R^n, i.e., where the polyhedron is defined by
%both equality and inequality constraints.
%
%SYNTAX:
%
%  [A,b,Aeq,beq]=vert2lcon(V,TOL)
%
%The rows of the N x n matrix V are a series of N vertices of a polyhedron
%in R^n. TOL is a rank-estimation tolerance (Default = 1e-10).
%
%Any point x inside the polyhedron will/must satisfy
%
% A*x  <= b
% Aeq*x = beq
%
%up to machine precision issues.
%
%
%EXAMPLE:
%
%Consider V=eye(3) corresponding to the 3D region defined
%by x+y+z=1, x>=0, y>=0, z>=0.
%
%
% >>[A,b,Aeq,beq]=vert2lcon(eye(3))
%
%
%    A =
%
%       0.4082 -0.8165 0.4082
%       0.4082 0.4082 -0.8165
%       -0.8165 0.4082 0.4082
%
%
%    b =
%
%       0.4082
%       0.4082
%       0.4082
%
%
%    Aeq =
%
%       0.5774 0.5774 0.5774
%
%
%    beq =
%
%       0.5774

  %%initial stuff

if nargin<2, tol=1e-10; end

[M,N]=size(V);

if M==1
   A=[];b=[];
   Aeq=eye(N); beq=V(:);
   return
end

p=V(1,:).';
X=bsxfun(@minus,V.',p);

%In the following, we need Q to be full column rank
%and we prefer E compact.

if M>N  %X is wide

   [Q, R, E] = qr(X,0);  %economy-QR ensures that E is compact.
                        %Q automatically full column rank since X wide

else%X is tall, hence non-solid polytope

   [Q, R, P]=qr(X);  %non-economy-QR so that Q is full-column rank.

   [~,E]=max(P);  %No way to get E compact. This is the alternative.
      clear P
end

diagr = abs(diag(R));

if nnz(diagr)

      %Rank estimation
      r = find(diagr >= tol*diagr(1), 1, 'last'); %rank estimation

      iE=1:length(E);
      iE(E)=iE;



      Rsub=R(1:r,iE).';

      if r>1

      [A,b]=vert2con(Rsub,tol);

      elseif r==1

         A=[1;-1];
         b=[max(Rsub);-min(Rsub)];

      end

      A=A*Q(:,1:r).';
      b=bsxfun(@plus,b,A*p);

      if r       Aeq=Q(:,r+1:end).';
      beq=Aeq*p;
      else
         Aeq=[];
         beq=[];
      end

else %Rank=0. All points are identical

   A=[]; b=[];
   Aeq=eye(N);
   beq=p;


end

%          ibeq=abs(beq);
%          ibeq(~beq)=1;
%
%          Aeq=bsxfun(@rdivide,Aeq,ibeq);
%          beq=beq./ibeq;



function [A,b] = vert2con(V,tol)
% VERT2CON - convert a set of points to the set of inequality constraints
%          which most tightly contain the points; i.e., create
%          constraints to bound the convex hull of the given points
%
% [A,b] = vert2con(V)
%
% V = a set of points, each ROW of which is one point
% A,b = a set of constraints such that A*x <= b defines
%    the region of space enclosing the convex hull of
%    the given points
%
% For n dimensions:
% V = p x n matrix (p vertices, n dimensions)
% A = m x n matrix (m constraints, n dimensions)
% b = m x 1 vector (m constraints)
%
% NOTES: (1) In higher dimensions, duplicate constraints can
%          appear. This program detects duplicates at up to 6
%          digits of precision, then returns the unique constraints.
%       (2) See companion function CON2VERT.
%       (3) ver 1.0: initial version, June 2005.
%       (4) ver 1.1: enhanced redundancy checks, July 2005
%       (5) Written by Michael Kleder,
%
%Modified by Matt Jacobson - March 29,2011
%

k = convhulln(V);
c = mean(V(unique(k),:));

V = bsxfun(@minus,V,c);
A  = nan(size(k,1),size(V,2));

dim=size(V,2);
ee=ones(size(k,2),1);
rc=0;

for ix = 1:size(k,1)
F = V(k(ix,:),:);
if lindep(F,tol) == dim
      rc=rc+1;
      A(rc,:)=F\ee;
end
end

A=A(1:rc,:);
b=ones(size(A,1),1);
b=b+A*c';

% eliminate duplicate constraints:
[A,b]=rownormalize(A,b);
[discard,I]=unique( round([A,b]*1e6),'rows');

A=A(I,:); % NOTE: rounding is NOT done for actual returned results
b=b(I);
return


function [A,b]=rownormalize(A,b)
%Modifies A,b data pair so that norm of rows of A is either 0 or 1

  if isempty(A), return; end

  normsA=sqrt(sum(A.^2,2));
  idx=normsA>0;
  A(idx,:)=bsxfun(@rdivide,A(idx,:),normsA(idx));
  b(idx)=b(idx)./normsA(idx);


function [r,idx,Xsub]=lindep(X,tol)
%Extract a linearly independent set of columns of a given matrix X
%
% [r,idx,Xsub]=lindep(X)
%
%in:
%
%  X: The given input matrix
%  tol: A rank estimation tolerance. Default=1e-10
%
%out:
%
% r: rank estimate
% idx:  Indices (into X) of linearly independent columns
% Xsub: Extracted linearly independent columns of X

if ~nnz(X) %X has no non-zeros and hence no independent columns

   Xsub=[]; idx=[];
   return
end

if nargin<2, tol=1e-10; end


   [Q, R, E] = qr(X,0);

   diagr = abs(diag(R));

   %Rank estimation
   r = find(diagr >= tol*diagr(1), 1, 'last'); %rank estimation

   if nargout>1
   idx=sort(E(1:r));
      idx=idx(:);
   end


   if nargout>2
   Xsub=X(:,idx);
   end

[ Last edited by daicong on 2012-7-16 at 02:07 ]

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代碼有點(diǎn)長，麻煩各位耐心查看。

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8樓2012-07-17 22:41:54

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daicong

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9樓2012-07-17 22:43:40

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18樓: Originally posted by liang8887 at 2012-07-19 14:57:39

有點(diǎn)長，是的。前面不是重點(diǎn)，重點(diǎn)在后面的計算，我有點(diǎn)看不懂。

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21樓2012-07-20 21:25:57

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3樓: Originally posted by xiabiaojun at 2012-07-16 03:38:23

any idea?

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4樓2012-07-16 23:39:12

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11樓: Originally posted by strive123123 at 2012-07-17 13:49:19

求討論。。。

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12樓2012-07-20 09:07:15

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眼被看花了、、、

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14樓2012-07-20 09:39:09

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14樓: Originally posted by 假大空 at 2012-07-19 13:39:09

眼被看花了、、、

耐心。。。求討論算法。

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16樓2012-07-20 10:45:10

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漠漠輕寒上小寒

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您好，我現(xiàn)在也在看這個，請問您當(dāng)時看懂了嗎

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