Transcript POLYGON TRIANGULATION - uni

POLYGON
TRIANGULATION
THE ART GALERY PROBLEM
SİNEM GETİR
2008-2009
FEBRUARY 4, 2009
COMPUTER GRAPHIC
SEMINAR
CONTENTS
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Guarding and triangulations
Can we do better?
3-Coloring
The art galery theorem
Partitioning a polygon into monotone pieces
Triangulating a Monotone polygon
The Art Gallery Problem
camera
How many cameras are needed to guard a gallery and
How should they be placed?
Transform to a Geometric Problem
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Floor plan may be sufficient and can be
approximated as a simple polygon.
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A simple polygon is a region enclosed by single closed
polygonal chain that doesn’t self-intersect
A camera’s position corresponds to a point in the
polygon
A camera sees those points in the polygon to
which it can be connected with an open segment
that lies in the interior of the polygon
• assuming we have omni-cam that sees all directions
Guards have a viewport of 360°
 A polygon is completely guarded, if every point within the
polygon is guarded by at least one of the watchmen
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Simple Polygon Model
Model the art gallery as a region bounded by some simple
polygon (no self-crossing).
Regions with holes are not allowed.
convex polygon
one camera
an arbitrary n-gon (n vertices)
Bad news: finding the minimum number of cameras for a
given polygon is NP-hard (exponential time).
Triangulation
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Let P a simple polygon with n vertices
We first decompose P into pieces that are
easy to guard, namely triangles.
DEFİNİTİON:A decomposition of a polygon
into triangles by a maximal set of nonintersecting diagonals is called a
triangulation of the polygon.
Triangulations are usually NOT unique
Many applications: visibility, robotics,
mesh generation, point location etc.
Triangulation
To make things easier, we decompose a polygon into pieces that are
easy to guard.
Draw diagonals between pair of vertices.
an open line segment that connects two
vertices and lie in the interior of the polygon.
Guard the polygon
by placing a camera in
every triangle …
THE QUESTION…
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Does every simple polygon admit a triangulation?
If yes, what is the number of triangles?
Does any triangulation lead to the same number of triangles?
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Theorem 1 Every simple polygon has a triangulation.
Any triangulation of a simple polygon with n
vertices consists of exactly n – 2 triangles.
Proof By induction.
Trivial for n = 3.
Assume true for all m < n.
Existence
w
v
P
Let v be the leftmost vertex and u and w its two neighbors.
 uw in the interior of P  it is a diagonal.
 Otherwise, the triangle determined
by u, v, w contains at least one
vertex. Let v be the one closest
to v. Then vv is a diagonal.
w
v
v
u
u
The diagonal splits the polygon into two (which by induction can be triangulated).
Proof
# triangles = n – 2
Any diagonal splits P into two simple polygons with k and m
vertices, respectively.
By induction these two subpolygons can be triangulated.
They are decomposed into k – 2 and m – 2 triangles, resp.
Vertices defining the diagonal occur in each subpolygon
once.
Other vertices of P each occurs in exactly on one
subpolygon.
Thus k + m = n + 2.
By induction, the triangulation of P has (k – 2) + (m – 2) = n – 2
triangles.
Theorem1 shows that any simple polygon with n vertices can be
guarded with n-2 cameras.
But placing a camera inside every triangle seems OVERKILL.
A camera on a diagonal guards two triangles.
 # cameras can be reduced to roughly n/2.
A vertex is adjacent to many triangles.
So placing cameras at vertices can do even better …
3-Coloring
Idea: Select a set of vertices, such that any triangle has at least one
selected vertex.
Assign each vertex a color:
pink, green, or yellow.
Any two vertices connected by
an edge or a diagonal must be
assigned different colors.
Thus the vertices of every triangle
will be in three different colors.
If 3-coloring exists, place cameras at all vertices of the same color.
Choose the smallest color class to place the cameras.
 n/3 cameras.
A 3-Coloring Algorithm
A 3-coloring can be found through a graph traversal (such as DFS).
During DFS, maintain the invariant:
All polygon vertices of encountered
triangles have been colored such that
no adjacent two have the same color.
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Start DFS at any node of G.
Color the three vertices of the
corresponding triangle.
Suppose node v is visited from u.  Their triangles T(v) and T(u) are adjacent.
Only one vertex of T(v) is not colored.  Its color is uniquely determined.
Since G is a tree, the other nodes adjacent
 to v have not been visited yet.
Otherwise there exists a cycle (which contradicts that G is a tree.)
Apply the color to v.
Art gallery theorem
Theorem: For a simple polygon with n vertices,
n/3 cameras are occasionally necessary and
always sufficient to have every point in the polygon
visible from at least one of the cameras.
Can we do better?
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THE WORST CASE
SCENARIO…
A Worst Case
A triangulated polygon can always be 3-colored.
 Any simple polygon can be guarded with n/3 cameras.
n/3 prongs
a
There exists no position at which a
camera can oversee two prongs.
n/3 cameras are needed.
The 3-coloring approach is optimal in the worst case.
Solution to the Art Gallery
Problem
1. Triangulate a simple polygon with a fast algorithm.
DCEL representation for the simple polygon so we can
visit a neighbor from a triangle in constant time.
2. Generate a 3-coloring by DFS (as presented earlier).
3. Take the smallest color class to place the cameras.
THEOREM
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Let P be a simple polygon with n
vertices.A set of [n/3] camera
positions in P such that any point
inside P is visible from at least
one of the cameras can be
computed in O(n logn) time.
PARTITIONING A POLYGON INTO
MONOTONE PİECES
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We have just seen that a
triangulation of P always exists.the
proof of that theorem is constructive
and leads to a recursive triangulation
algorithm:
Find a diagonal and triuanglate the
two resulting subpolygons
recursively.
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…  The triangulation algorithm will
take quadratic time in the worst
case.
Can we do better?
Convex polygon
Pick one vertex of the
Polygon and draw diagonals
From this vertex to all other
vertices except
its neighbors.
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This takes only linear time.
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For a non-convex polygon
unfortunately ,it is as difficult to
partition a polygon into convex
pieces as it is to trianglate
it.Therefore we shall decompose
P into so-called monotone
pieces.
Monotone polygon
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A polygon P in the plane is called
monotone with respect to a straight
line L, if every line orthogonal to L
intersects P at most twice.
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Line sweep(top down)
Vertices where the
direction changes
downward<> are called
turn vertices.
To have y-monotone
pieces ,we need to get rid
of turn vertices:
When we encounter a turn
vertex,it might be necessary
to introduce a diagonal and
split the polygon into pieces
We will not add diagonals at
all turn vertices
Types of Turn Vertices
If we want to define the different types of turn vertices carefully,we should pay special
attention to vertices with equal y-coordinate.We do this by defining the notions of
“below” and “above” as follows ; a point p is below another point q if
p <q
y
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y
or
p q
y
y
and p  q , and p is above q if
x
x
p q
y
y
or p  q and p  q
y
y
x
x
Start Vertex - its two neighbors lie below it and
the interior angle < 180°
End Vertex - its two neighbors lie above it and
the interior angle < 180°
Split Vertex - its two neighbors lie below it and
the interior angle > 180°
Merge Vertex - its two neighbors lie above it and
the interior angle > 180°
VERTEX ONTOLOGY
p q
y
y
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The split and merge vertices are
sources of local non-monotonicity.The
following stronger stament is true.
LEMMA:A polygon is y-monotone if it
has no split vertices or merge vertices.
Adding diagonals
* The partition p into ymonotone pieces , get rid of
split and merge vertices
1. Add a diagonal going upward
each split vertex.
2. Add a diagonal going
downward from each merge
vertex.
* Where do the edges go?
Helpers
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Let helper helper(ej)
be the lowest vertex
above the sweep-line
Such that the
horizontal segment
connecting the vertex
to ej lies inside P
Removing Split Vertices
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For a split vertex vi ,let
be the edge
immediately to the left
of it.
Add a diagonal from vi
to helper ej
Removing merge vertex
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For a merge vertex vi let ej
be edge immediately to the
left of it
vi becomes helper(ej ) ones
we reach it
Whenever the helper(ej ) is
replaced by some vertex vm
,add a diagonal from vm to vi
If vi is never replaced as
helper(ej ), we can connect
it to the lower endpoint of ej
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Removing Split/Merge Vertices
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v1 … vn: a counter-clock enumeration of vertices of P
e1 … en: a set of edges of P, where ei = segment (vi , vi+1)
Events are stored in event queue, ordered by y-coord.
If a split vertex, connect it to the lowest vertex (helper of its
left edge) between the edges to its left and right
If a merge vertex, connect it to the highest vertex between the
edges to its left and right
Store the edges (and their helpers) of P in the leaves of
dynamic binary search tree T, left-to-right order reflects in
order of leaves. Helpers may be replaced. Store only edges
that have P to their right (or the left edges).
THE ALGORITHM…
MakeMonotone(P)
Input: A simple polygon P stored in a doubly-connected edge
list D
Output: A partitioning of P into monotone subpolygons stored
in D
1. Construct a priority queue Q on the vertices of P, using
their y-coordinates as priority. If two points have the same
y-coordinates, the one with smaller x has higher priority
2. Initialize an empty binary search tree T
3. while Q is not empty
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5.
do Remove vi with the highest priority from Q
Call the appropriate procedure to handle the vertex,
depending on its type
HandleStartVertex(Vi)
1. Insert ei in T and set helper(ei) to vi
HandleEndVertex(Vi)
if helper(ei-1) is a merge vertex
2. then Insert diagonal connecting vi to helper(ei-1) in D
3. Delete ei-1 from T
1.
HandleSplitVertex(Vi)
1. Search in T to find the edge ej directly left of
vi
2. Insert diagonal connecting vi to helper(ej ) in D
3. helper(ej )  vi
4. Insert ei in T and set helper(ei ) to vi
HandleMergeVertex(Vi)
1.
2.
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if helper(ei-1) is a merge vertex
then Insert diagonal connecting vi to helper(ei-1) in D
Delete ei-1 from T
Search in T to find the edge ej ej directly left of vi
if helper(ej) is a merge vertex
then Insert diagonal connecting vi to helper(ej) in D
helper(ej )  vi
HandleRegularVertex(Vi)
1. if the interior of P lies to the right of vi
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then if helper(ei-1) is a merge vertex
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then Insert diag. connect vi to helper(ei-1) in D
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Delete ei-1 from T
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Insert ei in T and set helper(ei) to vi
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else Search in T to find the edge ej directly left of vi
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if helper(ej) is a merge vertex
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then Insert diag. connect vi to helper(ej) in D
7.
helper(ej )  vi
Partitioning Algorithm Analysis
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Construct priority queue: O(n)
Initialize T: O(1)
Handle an event: O(log n)
• one operation on Q: O(logn)
• at most 1 query, 1 insertion & 1 deletion on T:
O(logn)
• insert at most 2 diagonals into D: O(1)
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Total run time: O(n log n)
Storage: O(n)
An attempt was made to triangulate the following polygon, this
time with respect to the horizontal, can you guess why the
triangulation wasn't successful?
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Monotonic triangulation that failed because the polygon is not monotonic
After a slight modification to the polygon (it doesn't look like a donkey
anymore), the polygon now looks like this and can be triangulated using
the concept.
A successful monotonic triangulation of the modified polygon
LEMMA
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Algorithm MAKE MONOTONE
adds set of non-intersecting
diagonals that partitions P
into monotone subpolygons.
Proof:
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We shall prove for the segment added
in HANDLE-SPLİT-VERTEX ; the proof
for the segments added in HANDLEEND-VERTEX,HANDLE-REGULARVERTEX,and HANDLE-MERGE-VERTEX
is similar.
We assume that no two vertices have
the same y-coordinate.
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Consider a segment vmvi that is added by
HANDLE-SPLİT-VERTEX when vi is reached.
Let ej be the edge to the left of vi and let
Be the edge to right of vi .Thus helper(ej)= vm
When we reach vi .
We first argue that vmvi does not intersect an edge
of P.
To see this, consider the quadrilateral
Q bounded by the horizantel lines through vm and
vi, and by ej and ek.There are no vertices of P
inside Q, otherwise vm would not be the helper of
ej .
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It is supposed that there would be an
edge of P intersecting vmvi.Since the edge
cannot have an endpoint inside Q and
polygon edges do not intersect each other,
it would have to intersect the horizontal
segment connecting vm to ej or the
horizontel segment connecting vi to ej
.Both are impossible, since for both vm and
vi, the edge ej lies immediately to the
left.Hence, no edge of P can intersect vmvi.
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Since there are no
vertices of P inside
Q, and any
previously added
diagonal must
have both of its
endpoints above vi
it can can not
intersect vmvi.
Triangulating a Monotone Polygon
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The main idea behind the algorithm is quite
simple. First the vertices are sorted with respect
to the line of monotonicity, which in the first case
is the y-axis (which involves sorting the vertices
by their y-coordinate), and in the second case it
is the x-axis. Sorting becomes more complex
when the line of monotonicity is neither the x nor
the y-axis, yet it is not impossible. However,
finding the line of monotonicity can be quite
difficult to implement. This issue requires more
effort and it lies beyond the scope of this report.
After sorting the vertices from top to bottom, the
triangles are cut off from the top.
Triangulation Algorithm Analysis
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A strictly y-montone polygon with n
vertices can be triangulated in linear time
A simple polygon with n vertices can be
triangulated in O(n log n) time with an
algorithm that uses O(n) storage
A planar subdivision with n vertices can be
triangulated in O(n log n) time with an
algorithm that uses O(n) storage
Triangulation History
1. A really naive algorithm is O(n*4): check all
n*2 choices for a diagonal, each in O(n)
time. Repeat this n ¡ 1 times.
2. A better naive algorithm is O(n*2); find an
ear in O(n) time; then recurse.
3. First non-trivial algorithm: O(n log n)
[GJPT-78]
4. A long series of papers and algorithms in
80s until Chazelle produced an optimal
O(n) algorithm in 1991.
5. Linear time algorithm insanely
complicated; there are randomized,
expected linear time that are more
accessible.
6. We content ourselves with O(n log n)
MORE…
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Computational Geometry-algorithms
and applications
http://www.me.cmu.edu/faculty1/shi
mada/cg97/bader/index.html
http://cgm.cs.mcgill.ca/~godfried/te
aching/cgprojects/97/Thierry/thierry507webprj
/artgallery.html
DANKE SCHÖN…
TEŞEKKÜRLER…