# Projection on Polyhedral Cone

This is an open problem in Convex Optimization. At first glance, it seems rather simple; the problem is certainly easily understood:

We simply want a formula for projecting a given point in Euclidean space on a cone described by the intersection of an arbitrary number of halfspaces;
we want the closest point in the polyhedral cone.

By "formula" I mean a closed form; an equation or set of equations (not a program, algorithm, or optimization).
A set of formulae, the choice of which is conditional, is OK as long as size of the set is not factorial (prohibitively large).

This problem has many practical and theoretical applications. Its solution is certainly worth a Ph.D. thesis in any Math or Engineering Department.

## Projection onto isotone projection cones

Together with my coauthor A. B. Németh we recently discovered a very simple algorithm for projecting onto a special class of cones, the isotone projection cones.

The isotone projection cones are cones for which the projection onto the cone is isotone; that is, monotone with respect to the order induced by the cone. In $LaTeX: \mathbb R^n$ the isotone projection cones are polyhedral cones generated by $LaTeX: n$ linearly independent vectors (i.e., cones which define a lattice structure, or shortly latticial cones) such that the generators of the polar of the cone form pairwise nonacute angles. For example the nonnegative monotone cone (Example 2.13.9.4.2 in http://meboo.convexoptimization.com/Meboo.html) is an isotone projection cone. The isotone projection cones were introduced by George Isac and A. B. Németh at the middle of 1980's and used for solving complementarity problems by George Isac, A. B. Németh and S. Z. Németh. For simplicity we shall call a matrix $LaTeX: \mathbf E$ whose columns are the generators of an isotone projection cone isotone projection matrix. Recall that an L-matrix is a matrix whose diagonal entries are positive and off-diagonal entries are nonpositive (see more at http://en.wikipedia.org/wiki/Z-matrix_(mathematics)). A matrix $LaTeX: \mathbf E$ is an isotone projection matrix if and only if it is of the form

$LaTeX: \mathbf E=\mathbf O\mathbf T^{-\frac12},$

where $LaTeX: \mathbf O$ is an orthogonal matrix and $LaTeX: \mathbf T$ is a positive definite L-matrix.

The algorithm is a finite one that stops in at most $LaTeX: n$ steps and finds the exact projection point. Suppose that we want to project onto a latticial cone and for each point we know a proper face of the cone onto which the point projects. Then, we could find the projection recursively, by projecting onto linear subspaces of decreasing dimensions. In case of isotone projection cones the isotonicity property provides us with the required information about such a proper face. The information is provided by the geometrical structure of the polar cone.

If a polyhedral cone can be written as a union of reasonably small number of isotone projection cones, then we could project a point onto the polyhedral cone by projecting onto the isotone projection cones and taking the minimum distance of the point from these cones. Due to the simplicity of the method of projecting onto an isotone projection cone, it is a challenging open question to find polyhedral cones which are a union of reasonably small number of isotone projection cones and for which the latter cones can be easily determined. We conjecture that the latticial cones which are subsets of the nonnegative orthant (and their isometric images) are such cones.

Scilab script of the algorithm:

// You are free to use, redistribute and modifiy this code if you include
// as a comment the author of the original code
// (c) Sandor Zoltan Nemeth, 2009.
function p=piso(x,E)
[n,n]=size(E)
if det(E)==0 then
error ('The input cone must be an isotone projection cone'); return
else
V=inv(E)
U=-V'
F=-V*U
G=F-diag(diag(F))
for i=1:n
for j=1:n
if G(i,j)>0 then
error('The input cone must be an isotone projection cone'); return
end
end
end
I=[1:n]
n1=n-1
cont=1
for k=0:n1
//disp(cont,"cont=")
[q1,l]=size(I)
E1=E
I1=I
if l-1>=1
highest=I(l)
if highest<n
for h=n:-1:highest+1
E1(:,h)=zeros(n,1)
end
end
for j=l-1:-1:1
low=I(j)+1
high=I(j+1)-1
if high>=low
for m=high:-1:low
E1(:,m)=zeros(n,1)

end
end
lowest=I(1)
if lowest>1
for w=lowest-1:-1:1
E1(:,w)=zeros(n,1)
end
end
end
end
if l==1
E1=zeros(n,n)
E1(:,I(1))=E(:,I(1))
end

V1=pinv(E1)
U1=-V1'
//disp(x,"start x=")
//disp(I,"I=")

//disp(U1,"U1=")
//disp(V1,"V1=")
for j=l:-1:1
zz=x'*U1(:,I(j))
if zz>0 I1(j)=[]
end
end
//disp(I1,"I1=")
[q2,ll]=size(I1)
if cont==0 then p=x; return
elseif ll==0 then p=zeros(n,1); return
else
//disp(E1,"E1=")
//disp(V1,"V1=")
A=E1*V1
//disp(A,"A=")
x=A*x
//disp(x,"x=")
p=x;
end
if ll==l cont=0
else cont=1
end
I=I1
end
end
endfunction


$LaTeX: -$Sándor Zoltán Németh