# Projection on Polyhedral Cone

(Difference between revisions)
 Revision as of 15:13, 18 June 2009 (edit)← Previous diff Revision as of 15:16, 18 June 2009 (edit) (undo) (→Fast projection on monotone nonnegative cone)Next diff → Line 123: Line 123: == Fast projection on monotone nonnegative cone == == Fast projection on monotone nonnegative cone == - Requires [http://www.stanford.edu/~boyd/cvx CVX] + Demo requires [http://www.stanford.edu/~boyd/cvx CVX] to produce estimate of error + by solving the projection problem in polynomial time. + + [http://www.stanford.edu/~boyd/cvx CVX] is not required for piso3() which performs projection by Sandor's method about 100 times faster.

%demo: Sandor's projection on monotone nonnegative cone                                                                                                               %demo: Sandor's projection on monotone nonnegative cone
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## Revision as of 15:16, 18 June 2009

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.

Isotone projection cones are pointed closed convex cones $LaTeX: \mathcal{K}$ for which projection $LaTeX: P_\mathcal{K}$ onto the cone is isotone; that is, monotone with respect to the order $LaTeX: \preceq$ induced by the cone: $LaTeX: \forall\,x\,,\,y\in\mathbb{R}^n$

$LaTeX: x\preceq y\Rightarrow P_\mathcal{K}(x)\preceq P_\mathcal{K}(y),$

or equivalently

$LaTeX: y-x\in\mathcal K\Rightarrow P_\mathcal{K}(y)-P_\mathcal{K}(x)\in\mathcal K.$

In $LaTeX: \mathbb R^n,$ the isotone projection cones are polyhedral cones generated by $LaTeX: n$ linearly independent vectors (i.e., cones that define a lattice structure; called latticial or simplicial cones) such that the generators of the polar cone form pairwise nonacute angles. For example, the nonnegative monotone cone (Example 2.13.9.4.2 in CO&EDG) is an isotone projection cone. The nonnegative monotone cone is defined by

$LaTeX: \{\mathbf x=(x_1,\dots,x_n)^\top\in\mathbb R^n \mid x_1\geq\dots\geq x_n\geq 0\}.$

Projecting onto nonnegative monotone cones (see Section 5.13.2.4 in CO&EDG) is important for the problem of map-making from relative distance information (see Section 5.13 in CO&EDG); e.g., stellar cartography. The isotone projection cones were introduced by George Isac and A. B. Németh in the mid-1980's and then applied by George Isac, A. B. Németh, and S. Z. Németh to complementarity problems. For simplicity we shall call a matrix $LaTeX: \mathbf E ,$ whose columns are generators of an isotone projection cone, an isotone projection cone generator matrix. Recall that an L-matrix is a matrix whose diagonal entries are positive and off-diagonal entries are nonpositive (see more at Z-matrix). A matrix $LaTeX: \mathbf E$ is an isotone projection cone generator 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 in Euclidean space we know a proper face of the cone onto which that point projects. Then we could find the projection, recursively, by projecting onto linear subspaces of decreasing dimension. In case of isotone projection cones, the isotonicity property provides information required about such a proper face. The information is provided by geometrical structure of the polar cone.

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

Matlab code for 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, error('The input cone must be an isotone projection cone'); end
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
error('The input cone must be an isotone projection cone');
end
end
end
I=[1:n];
n1=n-1;
cont=1;
for k=0:n1
[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';

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


$LaTeX: -$Sándor Zoltán Németh

## Fast projection on monotone nonnegative cone

Demo requires CVX to produce estimate of error by solving the projection problem in polynomial time.

CVX is not required for piso3() which performs projection by Sandor's method about 100 times faster.

%demo: Sandor's projection on monotone nonnegative cone
%-Jon Dattorro, June 16, 2009
clear all
clc

n = 500;
a = randn(n,1);

if n < 5000
cvx_quiet('true')
cvx_precision('high')
tic
cvx_begin
variable s(n,1);
variable b(n,1);
minimize(norm(s-a));
for i=1:n
s(i) == sum(b(i:n));
end
b >= 0;
cvx_end
toc
end

tic
t = piso3(a);
fprintf('\n')
toc
if n < 5000
err = sum(abs(s - t))
end


### piso3()

% -Sandor Zoltan Nemeth, with modifications -Jon Dattorro June 14, 2009.
% Project x on monotone nonnegative cone of dimension n.

function p = piso3(x)
n = length(x);
I = [1:n];
cont = 1;
disp('xxxxx')
for k=1:n
fprintf('\b\b\b\b\b\b%6d',k);
l = length(I);
zeroidx = [];  %holds indices <=> columns of E that are zeroed
I1 = I;
if l-1 >= 1
highest = I(l);
if highest < n
for h=n:-1:highest+1
zeroidx = [zeroidx; h];
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
zeroidx = [zeroidx; m];
end
end
lowest = I(1);
if lowest > 1
for w=lowest-1:-1:1
zeroidx = [zeroidx; w];
end
end
end
end
if l==1
zeroidx = (1:n)';
zeroidx(I(1)) = [];
end
V1 = getpinv2(zeroidx,n);
for j=l:-1:1
zz = -x'*V1(I(j),:)';
if zz > 0
I1(j) = [];
end
end
ll = length(I1);
if ~cont
p = x;
return
elseif ~ll
p = sparse(n,1);
return
else
t = V1*x;
for ii=1:n
x(ii) = sum(t(ii:n)) - sum(t(zeroidx(find(zeroidx>=ii))));
end
p = x;
end
if ll==l
cont = 0;
else
cont = 1;
end
I=I1;
end


### getpinv2()

%Assumes Platonic upper triangular ones (generator matrix) with some columns missing.
%Quickly finds pseudoinverse of monotone nonnegative cone generator matrix
%June 14, 2009  -Jon Dattorro

function Y = getpinv2(zeroidx,n);
Y = spdiags([ones(n,1), -ones(n,1)], [0 1], sparse(n,n));
Y(zeroidx,:) = 0;
%find dangling -1
count = 0;
for i=1:n
if ~Y(i,i)
count = count + 1;
if i-1 > 0
Y(i-1,i) = 0;
end
else
if count
if i-count-1 > 0
Y(i-count-1,i-count:i) = -1/(count+1);
end
Y(i,i-count:i) = 1/(count+1);
end
count = 0;
end
end