\documentstyle[proc]{article}

\begin{document}

\title{How to Foobar a Recursive Quux}
\author{Isaac Newton \\ Chichen Itza Divination School \\ Chichen Itza, YU
\and
Albert Einstein \\ W. C. Fields Laboratory \\ Stanford, CA
}
\maketitle

The foobaring of recursive quuxs is a famous problem of computational 
subretrocalimetapreinteranticatainfrapseudoultravacuology. We here present a 
solution based on von Pflunft's algorithm for the pentagonal quadrature
of oscillating attracto-repulsors of semiconverging iterative schemas.

    In Affine geometry, a convex figure is one which contains the segment
    connecting any two of its points.  This definition remains essentially
    valid in the framework, and gives rise to a theory of
    convex sets that is cleaner and richer than the affine version.  

    The only difficulty is the treatment of antipodal pairs of points.
    When checking whether a set is convex, we can either ignore such pairs
    (on the grounds that there is no segment connecting them), or require
    that if there is such a pair, then the set must be the whole plane
    (on the grounds that all half-lines connecting the two points are
    ``segments'' in the limiting sense).  These two alternatives lead to
    two different notions of convexity, each with its own advantages and
    disadvantages.  Each alternative also leads to a different notion of
    duality among convex figures.  Our usage of the word ``segment'' argues
    for the first alternative, but the second seems to give a more useful
    notion of convexity.  Therefore I propose the following nomenclature: 
    
    Threfore, a quasi-convex set is convex iff it is non-empty and contains
    no antipodal pair of points.

    For subsets of the front face of $2$, interpreted as subset of the
    Euclidean plane, the two definitions above are equivalent and agree
    with the usual one.  Examples of convex subsets of $2$ are a single
    point, a proper line segment, a bounded convex polygon on the front
    face.  Other examples of convex sets are an open half-plane, an open
    wedge (the intersection of two half-planes), and half of a straight lin
    without one or both of its endpoints.  
    
    Some examples of quasi-convex sets that are not convex are: a straight
    line, one half of a straight line with both its endpoints, a closed
    half-plane, a pair of antipodal points, and the whole 
    plane.  
    
    Convexity theory provides a dramatic example of the advantages of the
    approach.  The very definition of convex set is meaningless
    in classical projective geometry, since a pair of points usually
    determines { two} segments, which are indinstinguishable by any
    projective criteria.  
    
    One could think of resolving this difficulty by letting some line play
    the role of the line at infinity , which gives affine structure
    to the projective plane.  However, the resulting theory of convex sets
    is no richer that the affine version.  Convexity turns out to be
    preserved only by those projective maps that take $\Omega$ to itself.
    Sets that are convex for one choice of $\Omega$ may be non-convex for
    another.  Many important properties of convex figures (such as closure
    under intersection and the uniqueness of the convex hull) remain true
    only if the line $\Omega$ is fixed.  
    
    In the plane, by contrast, the segment connecting two
    unrelated points is well-defined and unique.  Moreover, it is a purely
    projective concept, which does not depend on any choice of a line at
    infinity.  As a consequence, the family of convex subsets of the
    plane turns out to be closed under { arbitrary}
    projective maps.  That family includes not only all convex subsets of
    the front face that are convex (in the affine sense) for the standard
    choice of $\Omega=[1,0,0]$, but also all such sets for { any} choice
    of $\Omega$, and a few more besides.  

    In spite of this generality, the convex subsets of the 
    plane retain many pleasant properties of their affine counterparts.
    In fact, some important results are much simpler to state and prove in
    the framework, as we show in the following sections.  

      Consider an arbitrary, non-empty family of quasi-convex subsets of 
      , and let $X$ be its intersection.
      Let ${p}{q}$ be a segment whose endpoints are in $X$.
      The points $p$ and $q$ must be in every memebr of the family;
      by definition every member must contain the segment ${p}{q}$, which
      therefore must be contained in $X$.

    The following is an obvious corollary:

      The ``only if'' part is a corollary of theorem {??}, since
      a line of $2$ (considered as the set of points incident to
      it) is quasi-convex.
      
      As for the converse, suppose the intersection of $X$
      a line $m$ is not quasi-convex. Then there is a
      segment ${p}{q} X m$ with $p, q$ in $X m$. 
      Since ${p}{q} m$, we conclude ${p}{q} X$
      contradicting the hypothesis.

    Note that a convex subset of a line has one of the following forms: a
    point; a proper line segment with zero, one, or two of its endpoints;
    or a half-line with at most one of its endpoints.
  
    If $[x]$ and $[y]$ are the homogeneous coordinates of two non-antipodal
    points, then the segment connecting them consists of all points
    $[ x +  y]$, with $, > 0$.  In the model,
    we can see that the segments of $3$ that do not contain the origin
    are mapped by central projection onto the segments of the 
    plane.  This gives us two additional characterizations of convex sets: 
    
      If $Y$ is a convex subset of $3$, the quasi-convexity of its
      projection $X$ follows from the definition and from the observations
      in the preceding paragraph.  

      Conversely, let $X$ be a quasi-convex subset of $2$, represented
      as a subset of the unit sphere.  Let $Y$ be the convex closure of
      $X$ in $3$, i.e.  the set obtained from $X$ by repeatedly adding
      all segments of $3$ whose endpoints are in the set, until it
      becomes convex.  The central projection of $Y$ is initially $X$, and
      is not afected by the closure process (since at every step the
      projection of the added segment is in $X$).  

    The following are trivial corollaries of this result:
    
    Dually, we say that a subset $L$ of  is { quasi-convex} if
    any two non-opposite lines in it determine a fan that is contained in
    $L$.  If $L$ is also non-empty and contains no pair of opposite lines,
    we say it is { convex}.  The image of a (quasi-)convex set of points
    by any duomorphism is a (quasi-)convex set of lines.  
    
    Convex sets of points and lines are related in other less obvious ways.
    The following theorems are worth noting:

      Let $[p0, p1, p2]$ be the homogeneous coordinates of $p$.
      As in theorem {??}, let $Y$ be a convex subset of of $3$ such
      that $X$ is the central projection of $Y-{(0,0,0)}$. 
      The set $L$ can be written as
        $$
          {[p y] y Y { and }
            [y]{p,p,[0,0,0]}},
        $$
      where  is the cross-product operation in $3$. 
      
      The second condition is equivalent to $p y(0,0,0)$.
      If we let $f$ be the 
      the function from $3$ to $3$
      defined by the formula
        $$ f(y) = p y =
             (y1p2 - y2p1, -y0p2 + y2p0, y0p1-y1p0),
        $$
      then we can write
        $$
          L = {[z]  z f(Y)-{(0,0,0)}}
        $$
      But $f$ is a linear (albeit degenerate) map 
      of $3$, and linear maps preserve
      convexity. Therefore $f(Y)$ is a convex subset of $3$; by theorem
      {??}, $L$ is a quasi-convex subset of , and $L$ a
      quasi-convex subset of lines.

    The analogous result for strictly convex sets is
    
      The set $p X$ is non empty, and is quasi-convex by theorem
      {??}.  Suppose it contained a pair of opposite lines $p x1 =
      l = (p x2)$.  This means $p, x1, p$, and $x2$ occur
      in that order on the line $l$.  Then the segment ${x1}{x2}$
      (which is well-defined, since $X$ is convex) must include either
      $p$ or $p$, contradicting the hypothesis.  We conclude that
      $L$ contains no two opposite lines.  

    By duality, it then follows that

    A { perfect half-line} $h$ is a set of points consisting of 
    an open half-line $(h)$ plus exactly one of its endpoints,
    $(h)$. 

    A { perfect hemisphere} is a set 
    $H$ consisting of an open half-plane $(H)$, an open half-line
    $(H)$ on its boundary, and one of the endpoints $(H)$ of that
    half-line (figure {103}).  
    
    Perfect hemispheres are of two types which differ on the relative
    orientation of thair parts.  If we orient the bounding line $l$ of the
    hemisphere so that the $(H)$ lies ahead of $(H)$, we can
    classify the hemisphere as { positive} or { negative} depending
    on whether its interior $(H)$ lies on the left or on the right
    side of $l$.   

    Note that the antipodal image of a perfect hemisphere $H$ is another
    perfect hemisphere of the same sign, and is exactly the same as the
    complement $-H$. 

    Since a perfect hemisphere includes one member of every antipodal pair,
    it cannot be augmented without ceasing to be convex.  The converse is
    also true: every maximal convex set is a perfect hemisphere. This is a
    trivial consequence of the following theorem: 
    
      Let $X$ be a convex set, and $p$ a point on its boundary.
      Since $X$ contains no antipodal pairs, we can always choose $P$ so
      that $p$ is not in $X$. 



