Geometric Computing and Visualization 8

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Geometric Computing and Visualization 8
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Point Location Problem

• Given a polygon P and a query z, is it z?P?
• Given a planar subdivision P and a query z, is it
  z?P?
• The difficulty of the task will essentially
  depend upon the nature of the space and its
  partition.

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Query Mode

• Two modes of queries
• One shot mode query Whether a point q is
  internal to a simple n-gon P can be determined in
  optimal O(n) time, without preprocessing.
• Repetitive mode queries If we preprocess the
  data structure, the query may be done in o(n)
  time.

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Location of a Point in a Polygon 3

• Given a polygon P and a query z, is q internal to
  P?
• P is convex polygon
• P is star-shaped polygon P is star-shaped if
  there exists a pt q?P st ? points r, r is visible
  from q. (Question How do we determine if a
  polygon is star-shaped?)
• P is general polygon
• Kernel of a simple polygon

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Location of a Point in a Polygon 2

• The inclusion question for an n-vertex
  star-shaped polygon can be answered in
• Q(n) O(log n)
• S(n) O(n)

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Location of a Point in a Polygon 1

• Convex ? Star-shaped ? General
• What can we say for general case?

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Planar-straight-line graph 3

• Planar-straight-line graph(PSLG) G(V,E) is a
  PSLG if each edge is a straight-line segment no
  edges intersect except possibly at the end pts.
• A planar graph can always be embedded in the
  plane so that its edges are mapped to
  straight-line segments.

EO(V) FO(V)
Eulers formula V-EF2
of edges E 3(V-1)-B, of bounded faces
F2(V-1)-B
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Planar-straight-line graph 2

• Representation
• Adjacent list (all vertices link to their
  adjacent vertices, and all regions link to their
  adjacent edges in counterclockwise order)
• Doubly-connected edge list (all edges record 1.
  Two end point, 2. Two adjacent regions, 3. Two
  previous edges in each region in counterclockwise
  order.)

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Planar-straight-line graph 1

• Example

v7
v8
e4
v9
v6
f1
v1
e2
f4
f3
e1
e3
v10
v2
v5
V11
f2
e5
v3
v4
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Location of a point in a planar subdivision

• Slab method
• S(n) O(n2)
• Q(n) O(log n)
• Chain method
• S(n) O(n)
• Q(n) O(log2n) ? O(log n)
• Triangulation refinement method
• S(n) O(n)
• Q(n) O(log n)

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Slab method

• Slab method achieves efficient searching by
  decomposing the original subdivision into
  trapezoids.
• S(n)O(n2), Q(n)O(log n)
• Goal S(n)O(n), Q(n)O(log n)

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Chain method 18

• Chain method decomposes the subdivision into
  monotone polygons.

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Chain method 17

• Definition chain
• A chain C (u1, , up) is a PSLG with vertex
  set u1, , up and edge set (ui, ui1)
  i1,p-1

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Chain method 16

• Given PSLG G. Suppose that in G we find a chain C
  of one of the following types either (1) C is
  cycle (2) both u1 and up of C belong to the
  boundary of the unbounded region.
• In both cases C partitions the subdivision
  into two portions.
• How to decide on which side of C a query point z
  lies called discrimination of z against C?
• To discriminate against an arbitrary chain is
  basically the same problem as to test for
  inclusion in a general polygon. Thus, one should
  look for a restricted class of chains.

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Chain method 15

• Definition monotone
• A chain C (u1, , up) is said to be monotone
  w.r.t. a straight line l if a line orthogonal to
  l intersects C in exactly one point.
• Given a monotone chain C, the orthogonal
  projections l(u1), , l(up) of the vertices of
  C on l are ordered.

l(u5)
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Chain method 14

• Given monotone chain C a query point z
• the projection l(z) of z on l can be located by
  binary search in a unique interval (l(ui),
  l(ui1)).
• An O(1) test determines on which side of the line
  uiui1 the query point z lies.
• ? The discrimination of z against a p-vertex
  monotone chain is effected in O(log p) time

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Chain method 13

• Definition monotone complete set of chains of G
• Given is a PSLG G. A set of chains C C1, ,
  Cr monotone w.r.t the same line l with the
  properties
• Property1 The union of the members of C contains
  G
• Property2 For any two chains Ci and Cj of C,
  the vertices of Ci which are not vertices of Cj
  lie on the same side of Cj.
• is referred to as a monotone complete set of
  chains of G
• Note that Property2 means that the chains of a
  complete set are ordered.

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Chain method 12

• For a monotone complete set C of chains of PSLG
  G we can easily locate a point between two
  consecutive chains and the region in O(log p?log
  r) by binary search, where r is the of chains
  in C, and p is the of vertices in the longest
  chain.

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Chain method 11

• But a crucial question is Given PSLG G, does it
  admit a monotone complete set of chains? How do
  we find it ? In general, the answer is NO.
• A Regular PSLG admits of a monotone complete set
  of chains.
• An arbitrary PSLG can be transformed into a
  regular PSLG.

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Chain method 10

• Definition regular PSLG
• Given PSLG G with vertex set v1, , vn, where
  the vertices are indexed so that i lt j iff either
  y(vi) lt y(vj) or, if y(vi)y(vj), then x(vi) gt
  x(vj).
• A vertex vj is regular if there are integers
  iltjltk such that (vi, vj) and (vj, vk) are edges
  of G.
• Graph G is said to be regular if each vj is
  regular for 1ltjltn (i.e. with the exception of the
  two extreme vertices v1 and vn)

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Chain method 9

• A regular PSLG admits of a monotone complete set
  of chains
• Claim1 Show that a regular graph admits of a
  complete set of chains monotone w.r.t the y-axis.
  
• Claim2 Show edge weight can be chosen so that
• Each edge has a positive weight
• For each vj (j?1,n), WIN(vj)WOUT(vj)

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Chain method 8

• Claim1 Show that a regular graph admits of a
  complete set of chains monotone w.r.t the y-axis.
  
• Proof
• To prove that there is a set CC1, , Cr of
  chains monotone w.r.t y-axis, we only need to
  show that, for any vj(1ltj?n), we can construct a
  y-monotone chain from v1 to vj.
• trivial for j2. Assume the statement is true for
  kltj.
• Since vj is regular, there is some iltj such that
  (vi , vj) is an edge of G. But, by inductive
  hypothesis, there is a chain C from v1 to vi
  monotone w.r.t the y-axis. The concatenation of C
  and (vi , vj) is a y-monotone chain.

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Chain method 7

• Claim2 Show that edge weight can be chosen so
  that
• (1) Each edge has positive weight. Condition (1)
  ensures that each edge belongs to at least one
  chain (Property1)
• (2) For each vj (j?1,n), WIN(vj)WOUT(vj).
  Condition (2) guarantees that WIN(vj) chains pass
  through vj and they can be chosen so that they do
  not cross (Property2)

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Chain method 6

• Proof
• Let W(e), the weight of e, be the of
  chains to which e belongs.
• Let WIN(v) ?e?IN(V) W(e)
• Let WOUT(v) ?e?OUT(V) W(e)
• The realization of WINWOUT is a rather
  classical flow problem, and can be achieved by
  two passes over G.
• Assign W(e)1 for all e.
• In the first pass from v1 to vn we achieve
  WIN ? WOUT for non-extreme vj.
• In the second pass from vn to v1 we achieve
  WIN ? WOUT for non-extreme vj.

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Chain method 5

• Procedure Weight-Balancing In Regular PSLG
• for each edge e do W(e)1 //Initialization
• for i 2 until n-1 do //first pass
• WOUT(vi)sum of weights of outgoing edges
  of vi
• WIN(vi)sum of weights of incoming edges
  of vi
• d1 leftmost incoming edge of vi
• if (WOUT(vi) gt WIN(vi) )
• then W(d1) WOUT(vi) - WIN(vi) W(d1)
• for i n-1 until 2 do //second pass
• WOUT(vi)sum of weights of outgoing edges
  of vi
• WIN(vi)sum of weights of incoming edges
  of vi
• d2 leftmost outgoing edge of vi
• if (WIN(vi) gt WOUT(vi) )
• then W(d2) WIN(vi) - WOUT(vi) W(d2)
  

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Chain method 4
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Chain method 3

• Theorem A n-vertex PSLG can be regularized in
  time O(n log n) and space O(n)
• Proof Regularization of an n-vertex PSLG is
  accomplished in time O(n log n) by plane sweep
  method

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Chain method 2
(b)
(a)
(c)
Complete set of monotone chains (a) is searched
according to the hierarchy expressed by a binary
tree (b). Bypass pointers are shown as broken
lines (c).
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Chain method 1

• Chain method
• (S(n), Q(n))(O(n2), O(log2n)).
• Chain method Bypass pointer technique
• (S(n), Q(n))(O(n), O(log2n)).

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Triangulation Refinement Method 9

• Input triangulated PSLG (Any PSLG can be
  converted into a triangulation in O(nlogn) time)
• Step 0 Inscribe the triangulation by a big
  triangle and obtain a new triangulation G.

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Triangulation Refinement Method 8

• Construct a sequence of triangulations S1, S2, ,
  Sh(n), where S1 G and Si is obtained from Si-1
  as follows
• Step 1 Remove a set of independent (i.e.
  nonadjacent) non boundary vertices of Si-1 and
  their incident edges.
• Step 2 Re-triangulate the polygons arising from
  the removal of vertices and edges.
• Repeat steps 1 2 until Sh(n) is a triangle, the
  inscribing triangle.

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Triangulation Refinement Method 7
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Triangulation Refinement Method 6
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Triangulation Refinement Method 5

• Relation between triangles of Si-1 and
• those of Si. Rk?Si,Rj?Si-1.
• A directed edge from Rk to Rj if
• Rj is eliminated from Sj-1 in Step1.
• Rk is created in Si in Step2.
• Rj ? Rk ? ?.

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Triangulation Refinement Method 4

• Once T is available, we perform point location in
  Sh(n),,S1 in that order
• Procedure Point-Location
• if (z?Triangle(root))
• then z belongs to unbounded region
• else v root
• while (?(v)??) //?(v) a list of
  descendants of a node v
• for each u??(v) do
• if (z?Triangle(u)) then vu
• print v

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Triangulation Refinement Method 3

• The choice of the set of triangulation vertices
  to be removed in constructing Si from Si-1 is
  crucial to the performance of the technique.
• Suppose we are able to choose this set so that
  the following properties hold Let Ni denote the
  of vertices of Si
• Property1 Ni ?iNi-1 with ?i ? ? lt 1 for i 1,
  2, , h(n).
• Property2 Each triangle Rj?Si intersects at most
  H triangles in Si-1 ,and vice versa.

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Triangulation Refinement Method 2

• Property1 implies h(n) ? ?log1/?n? O(logn)
• ? Q(n) O(log n)
• Properties 12 would jointly imply O(n) storage
• Property1 implies O(n) nodes
• Property2 implies O(n) pointers
• ? S(n) O(n)

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Triangulation Refinement Method 1

• How to remove vertices so that both properties
  hold?
• Remove a set of nonadjacent vertices of degree
  less than k.
• Proof
• (1) Since removal of a vertex of degree k gives
  rise to polygon with k edges, each of the
  replaced triangle in Si intersects at most k-2
  new triangles in Si-1.
• (2) Euler formula vf-e2
• vn, f2n-5
• ? e3n-7
• ? The total vertex degrees lt 6n
• This implies that there are at least n/2
  vertices of degree less than 12.
• Let k 12
• Let m denote the of vertices removed.
• ? ? 1 1/24 lt 0.959 lt 1