Convex hulls Gift wrapping, d > 2

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Convex hullsGift wrapping, d gt 2
Problem definition CONVEX HULL, D gt 2 INSTANCE.
Set S p1, p2, pN of points in d-space
(Ed). QUESTION. Construct the convex hull H(S)
of S. The coordinates of the points pi ? S will
be referred to as pi (x1, x2, , xd). For d
2, the constructed hull was given
(represented) as a sequence of hull vertices. How
is the hull represented for d gt 2? To answer
that, and describe the algorithm, more
definitions are needed.
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Convex hullsGift wrapping, d gt 2
Polyhedron In E3 a polyhedron is defined by a
finite set of planar polygons such that every
edge of a polygon is shared by exactly one other
polygon and no subset of the polygons has the
property. Polyhedra is plural for
polyhedron. The polygons that share an edge are
adjacent. The vertices and edges of the
polygons are the vertices and edges of the
polyhedron. The polygons are the facets of the
polyhedron. A polyhedron is simple if there is
no pair of nonadjacent facets sharing a point. A
simple polyhedron partitions 3-space into two
domains, the interior (bounded) and the exterior
(unbounded). The term polyhedron often means
boundary ? interior. A simple polyhedron is
convex if its interior is a convex set. A
polyhedron is the 3-dimensional equivalent of a
polygon.
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Convex hullsGift wrapping, d gt 2
Polytope A half-space is the portion of Ed lying
on one side of a hyperplane. A polyhedral set in
Ed is the intersection of a finite set of closed
half-spaces. Note that a polyhedral set is
convex, since a half-space is convex, and the
intersection of convex sets is convex. Plane
polygons (d 2) and space polyhedra (d 3)
are 2- and 3-dimensional instances of bounded
polyhedral sets. A bounded d-dimensional
polyhedral set is a polytope. Note that
polytopes are convex by definition. The terms
convex d-polytope, d-polytope, and
polytope are all equivalent. Theorem. The
convex hull of a finite set of points in Ed is
a convex d-polytope. Conversely, a polytope is
the convex hull of a finite set of points. For d
3, the convex hull is a convex polyhedron. For
arbitrary d, the convex hull is a d-polytope.
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Convex hullsGift wrapping, d gt 2
Affine set Given k distinct points p1, p2, , pk
in Ed, the set of points p ?1p1 ?2p2 . .
. ?kpk (?j ? ?, ?1 ?2 . . . ?k
1) is the affine set generated by p1, p2, ,
pk, and p is an affine combination of p1, p2, ,
pk. We have seen this before. If k 2, this is
the parametric equation of a line, i.e., a line
is an affine set. For k 3, the affine set is a
plane. In general, an affine set for given k is a
flat object of k - 1 dimensions. Given a
subset L of Ed, the affine hull aff(L) is the
smallest affine set containing L. If L is 2
points or a segment, aff(L) is a line. If L is 3
points or a planar polygon, aff(L) is a plane. A
set of k points is affinely independent if no
subset of them can generate the same affine
set. The text sometimes refers to affine sets as
hyperplanes.
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Convex hullsGift wrapping, d gt 2
Faces of a polytope A d-polytope is described by
its boundary, which consists of faces. For a
d-polytope, there are faces in all dimensions 1
d. Some have special names. For a
d-polytope Dimension Face Name of
face d d-face d-polytope d - 1 (d-1)-face facet d
- 2 (d-2)-face subfacet 1 1-face edge 0 0-face ver
tex For a 3-polytope (polyhedron) Dimension Fac
e Name of face d 3 3-face 3-polytope,
polyhedron d - 1 2 2-face facet, planar
polygon d - 2 1 1-face subfacet,
edge 0 0-face vertex
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Convex hullsGift wrapping, d gt 2
Simplex A d-polytope P is a d-simplex (or just
simplex) iff it is the convex hull of (d 1)
affinely independent points. Any subset of the d
vertices of the convex hull is itself a simplex
and is a face (in some dimension) of
P. d d-simplex 0 vertex 1 edge 2 triangle 3 tetra
hedron
2-simplex convex hull of 2 1 points
not a 2-simplex convex hull of gt 2 1 points
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Convex hullsGift wrapping, d gt 2
Simplicial A d-polytope is simplicial if each of
its facets is a (d-1)-simplex. For example, for
d 3 The convex hull of a set of points in
3-space (the convex hull is a 3-polytope) is
simplicial iff every facet is a 2-simplex (a
triangular convex hull of exactly 3 points). For
example, the first case below. If any facet of
the hull has gt 3 co-planar points, the hull is
not simplicial. For example, the second and third
cases below.
not a 2-simplex convex hull of gt 2 1 points
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Convex hullsGift wrapping, d gt 2
Beneath A point p is beneath a facet F of a
polytope P if the point p lies in the open
half-space determined by hyperplane aff(F) and
containing P. In other words, aff(F) is a
supporting hyperplane of P, and p and P are in
the same half-space bounded by aff(F). Point p is
beyond facet F if p lies in the open
half-space determined by aff(F) and not
containing P. The figure shows these
relationships for d 2.
aff(F)
p1 is beyond F
F
P
p2 is beneath F
Note error in text, 2nd paragraph of 3.4, the P
should be p.
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Convex hullsGift wrapping, d gt 2
Gift wrapping Proposed by Chand and Kapur
(1970). Analyzed by Bhattacharya
(1982). Specialized for d 2 by Jarvis (1973),
Jarvis march. Key idea Given one facet (a
(d-1)-face) of the convex hull, find a
neighboring facet of the hull by wrapping a
(d-1)-dimensional affine set around the point
set. Continue from each facet to its neighbors
until all facets are found. For example, in d
3, imagine wrapping a sheet of 2-dimensional wrapp
ing paper around a 3-dimensional gift box. In d
2 (Jarvis march), a 1-dimensional line is
wrapped around a 2-dimensional point set. My
presentation will often appeal to d 3 for
expository purposes, but the method is applicable
for any d ? 2.
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Convex hullsGift wrapping, d gt 2
Simplicial assumption As presented (here and in
the text), the algorithm assumes that the
resulting polytope (the convex hull) is
simplicial. Recall that in a simplicial
d-polytope, each facet is a (d-1)-simplex, and is
determined by exactly d vertices. There will be
no points in S coplanar with the d vertices
that determine each facet of the convex
hull. Theorem. In a simplicial d-polytope, a
subfacet is shared by exactly two facets, and two
facets F1 and F2 share a subfacet e iff e is
determined by a common subset, with d - 1
vertices, of the sets determining F1 and F2. F1
and F2 are said to be adjacent on e. Restating
the theorem for d 3 In a simplicial
3-polytope, an edge is shared by exactly two
triangular facets, and two facets F1 and F2 share
an edge e iff e is determined by a common subset,
with 2 vertices, of the 3 vertices determining F1
and F2. F1 and F2 are said to be adjacent on
e. The theorem is the basis of the
algorithm. Given an already constructed facet F1
of the convex hull, a subfacet e of F1 is used to
construct the adjacent facet F2 that shares e
with F1.
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Convex hullsGift wrapping, d gt 2
Finding an adjacent facet, in general Let S
p1, p2, pN be a finite set of points in
d-space (Ed). Assume a facet F1 of H(S) is known,
with all its subfacets. The mechanism to advance
from F to an adjacent facet F?, which shares
subfacet e with F, is to select from among all
the points of S not vertices of F the point p?
such that all other points of S are beneath the
hyperplane aff(e ? p?). In other words, from
among all the hyperplanes determined by e and a
point p? ? S but not in F, the one which forms
the largest angle with aff(F). For example, for
d 2
F?
e
F
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Convex hullsGift wrapping, d gt 2
Finding an adjacent facet, for d 3 Facet F is
known. Consider the set of planes determined by
edge e and the points of S and select the one
which forms the largest angle lt ? (convex angle)
with aff(F). Points p1, p2, p3 determine F,
which determines aff(F). Compare the planes
determined by e and p4, p5, p6, and p7.
p5
p6
p4
p7
e
p1
F
p2
p3
aff(F)
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Convex hullsGift wrapping, d gt 2
Finding the largest angle The angle comparison is
carried out by comparing cotangents for each
point pk ? S not part of F. The details are in
the text, pp. 133-134. The time required for one
advance is O(d3 Nd). O(d3) to compute a vector
needed for the angle comparisons, done once per
advance (gift wrapping step). O(Nd) computing and
comparing cotangents for O(N) points.
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Convex hullsGift wrapping, d gt 2
Overview of the algorithm The algorithm starts
from an initial facet. For each subfacet of it,
construct the adjacent facets. Move to one of the
new facets and continue until all facets have
been constructed. A pool of subfacets which are
candidates for being used is kept. A subfacet e,
shared by facets F and F?, is a candidate to be
used iff F or F? but not both have been
constructed.
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Convex hullsGift wrapping, d gt 2
Algorithm Queue Q stores facets. File ? stores
the pool of subfacets. procedure
GiftWrapping(S) begin 1 Q ? ? ? 2 F
find an initial convex hull facet 3 insert into
? all subfacets of F 4 insert(F,Q) / insert F
into Q / 5 while (Q ? ?) do 6 F first(Q)
/ remove first element from Q, put into F
/ 7 T subfacets of F 8 for each e ? T ? ?
/ e is a gift wrapping candidate / 9 F?
facet sharing e with F / Gift wrapping advance
/ 10 insert into ? all subfacets of F? not
yet present delete from ? all subfacets
already present in F? 11 insert(F?,Q)
endfor 12 output F / output a facet of
H(S) / endwhile end Still to be
seen 2. find an initial convex hull
facet 7. generate the subfacets of facet
F 8. check if subfacet e is a candidate
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Convex hullsGift wrapping, d gt 2
Supporting lines and planes Recall that a
supporting line of a convex polygon intersects
the polygon at a vertex such that the entire
polygon is to one side of the line. A supporting
plane (or hyperplane) has a similar
relationship with a polytope.
E2
E3
intersection is edge
intersection is vertex
intersection is facet
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Convex hullsGift wrapping, d gt 2
Step 2. Find an initial convex hull facet.,
1 The idea is to obtain a hyperplane containing a
facet of the convex hull polytope H(S) by
successive approximations. This is done by
constructing a sequence of d (d is dimension
Ed) supporting hyperplanes ?1, ?2, , ?d, such
that ?i shares one more vertex with the convex
hull than ?i-1 for 1 ? i ? d (we define ?0 as
sharing 0 vertices with H(S)). In other words, ?i
for 1 ? i ? d shares i vertices with H(S). A
supporting hyperplane intersects the
polytope such that the entire polytope is to one
side of the hyperplane. Note that if the
intersection is facet F, then the supporting
hyperplane is aff(F). The supporting hyperplanes
are (d-1)-dimensional.
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Convex hullsGift wrapping, d gt 2
Step 2. Find an initial convex hull facet.,
2 For d 3 (E3) the successive hyperplanes
intersect the convex hull as follows supporting
vertices intersection hyperplane of H(S) on
?i object ?0 0 ? ?1 1 0-face,
vertex ?2 2 1-face, edge, subfacet ?3 3 2-face,
polygon, facet In essence, the technique is an
adaptation of the gift wrapping mechanism, where
at the jth of the d iterations the hyperplane
?j contains a (j-1)-face of the convex hull H(S).
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Convex hullsGift wrapping, d gt 2
Step 2. Find an initial convex hull facet.,
3 Thus we begin by determining a point of least
x1-coordinate (call it p1?) p1? is certainly a
vertex (a 0-face) of the convex hull. Hyperplane
?1 is chosen orthogonal to vector (1, 0, ,
0) and passing by (containing) p1?. In other
words, ?1 passes through p1? and is parallel to
the x2x3xd hyperplane. For example, for d 3
p2?
p3?
x2 y
x3 z
?1
p1?
x1 x
This initializes the process of finding the
initial supporting hyperplane. Once the initial
hyperplane ?1 has been found, each successive
hyperplane ?j 2 ? j ? d is found from the
hyperplane before it ?j-1.
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Convex hullsGift wrapping, d gt 2
Step 2. Find an initial convex hull facet.,
4 At the jth iteration, 2 ? j ? d, the hyperplane
?j-1 has normal vector nj-1 and contains vertices
p1?, p2?, , pj-1?. We need to find pj? to
define ?j. Through vector calculations pj? can
be found such that ?j (defined by p1?, p2?, ,
pj?) has the largest angle with ?j-1 (defined by
p1?, p2?, , pj-1?). Each iteration requires
O(Nd) O(d2) time. There are d iterations, so
finding the initial supporting hyperplane ?d
required by step 2 of the overall
algorithm requires O(Nd2) O(d3) ? O(Nd2) time.
p3?
x2 y
x3 z
p2?
?2
?3
p1?
?1
x1 x
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Convex hullsGift wrapping, d gt 2
Step 7. Generate the subfacets of facet
F. Because we have assumed that the convex hull
polytope H(S) is simplicial, each facet of H(S)
is determined by exactly d vertices, and each
subset of those vertices of size d-1 determines a
subfacet. The subfacets of a facet can be
generated in a straightforward fashion by
considering each of the d vertices in turn and
reading off the remaining d-1 vertices. This
requires O(d2) time. Each facet will be
described by a d-component vector of the indices
of its vertices, while a subfacet will be
described by an analagous (d-1)-component
vector. That implies that the vertices are kept
in an array.
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Convex hullsGift wrapping, d gt 2
Step 8. Check if a subfacet e is a candidate.,
1 A subfacet is a candidate iff it is contained
in just one facet generated by the algorithm. Why
just one? Recall that in a simplicial polytope, a
subfacet is shared by exactly two facets. If a
subfacet is generated twice, then both of its
adjacent facets have been found, and the subfacet
is of no further use.
F2
e
F1
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Convex hullsGift wrapping, d gt 2
Step 8. Check if a subfacet e is a candidate.,
2 Recall that ? is a file of subfacets. Given a
newly generated subfacet e, searching ? for e
will determine if e is a candidate. If e ? ?,
delete e. If e ? ?, add it. (This is step 10 in
the algorithm). A subfacet is represented by a
(d-1)-component vector of vertex indices. Store
? as a height-balanced binary tree, ordered
lexicographically on the vertex
indices. Accessing this structure to test
subfacet e for membership (step 8) or to insert
or delete subfacet e (step 10) requires O((d-1)
log M) time, where (d-1) is the number of indices
to compare and M is the maximum number of
subfacets in ?.
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Convex hullsGift wrapping, d gt 2
Analysis, 1 We analyze the algorithm in
steps. Let ?d-1 be the actual number of facets of
the polytope H(S). Let ?d-2 be the actual number
of subfacets of the polytope H(S). Initialization
(steps 1-4) requries O(Nd2) time. Steps 6, 11,
12 process each facet once each, adding it to the
queue, removing it, and outputing it they
require O(d) ? ?d-1.
number of facets
components of the vector representing the
facet The overall complexity of step 7,
generation of subfacets, is O(d-1) ? 2?d-2, as
each subfacet has d-1 components and is generated
twice. The test in step 8 as well as the file
update in step 10 require O((d-1) log ?d-2) per
subfacet,
number of subfacets
components of the vector representing the
subfacet so the overall time required is O((d-1)
log ?d-2) ? ?d-2. The overall time required for
step 9 (gift wrapping) is O(d3 Nd) ? ?d-1.
number of gift wrapping
advances time required for each
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Convex hullsGift wrapping, d gt 2
Analysis, 2 It can be shown that both ?d-1, ?d-2
? O(N?d/2?). Using this the previous analysis
simplifies to Convex hull construction time in d
dimensions on N points T(d,N) using the Gift
wrapping algorithm, requires O(N?d/2?1)
O(N?d/2? log N). Note that even though d is a
constant, it remains in the final order
expression where it is an exponent of the input
size N.