% program which calculates the minimal rank for every interval
% matrix with 3 columns


function a=mrkpx3(n,N)

  if rk01(n,N)==0 
  a=0;
  return
  elseif rk01(n,N)==1
    a=1;
  return
else


  
  X=reduce(n,N)  ;
[p,r]=size(X)
 q=r/2;

if q==2
  a=2;
return
end

  
m=X(:,1:q);
M=X(:,q+1:r);




v=zeros(1,0);
for i=1:p
		if m(i,1)<=0 & M(i,1)>=0 & -m(i,1)+M(i,1)>0
		v=[v i];
		end
		end
		
		% v is the set of the indices i st of [m,M]_1i has
% min negative max positive or
% min zero max positive or
% min negative max zero



      % now we construct two tridimensional matrices A and B such that
      % for every i the slice [A(:,:,i) B(:,:,i)] is an interval matrix
      % whose entries of the first column
      % are all either contained in the set of the
      % nonnegative real numbers or in the set of nonpositive real numbers
      % and the union of [A(r,s,i),B(r,s,i)] with i varying is the entry
      % r,s of [m,M]

		A=zeros(p,q,1);
A(:,:,1) =m;
  B=zeros(p,q,1);
B(:,:,1) =M;
    for i=1:size(v,2)
	    A=cat(3,A,A);
B=cat(3,B,B);
for l=1:size(A,3)/2 
	B(v(i),1,l)=0;
A(v(i),1,l+size(A,3)/2)=0;
  end
		    end

% Now we change A and B in such way that the entries of the
  % first columns of every slice [A(:,:,i) B:,:,i)] are contained
  %in the set of nonnegative real numbers


	for k=1:size(A,3)
		for i=1:size(A,1)
			if A(i,1,k)<0
			X=A(i,:,k);
Y=B(i,:,k);
			A(i,:,k)=min(-X,-Y);
                        B(i,:,k)=max(-X,-Y);
			end
			end
			
	  
				end
		end


				


	     

	


	      
	for i=1:size(A,3)
		if rklesseq2(A(:,:,i),B(:,:,i))==1
		a=2;
		return
		end
		end
		a=3;

end