1. 3. Polygon Triangulation
Iwama lab M1
Naoki Hatta

2. How to minimize # of cameras….
Introduction
We consider about the security in the art gallery
Owner wants to minimize the number of cameras
How to minimize # of cameras….

3. preliminary simple polygon We consider about only simple polygons.
regions enclosed by a single closed polygonal chain
that does not intersect itself
a simple polygon
NOT a simple polygon
We consider about only simple polygons.

4. example A convex polygon can always be guarded with one camera.
Finding the minimum number of cameras for a given polygon is NP-hard
For any simple polygon with n vertices, we will give a worst-case bound.
convex polygon
1 camera
3 camera

5. Triangulation A simple polygon may be a complicated shape.
So, we decompose it into triangles.
Drawing diagonals between pairs of vertices without intersecting.
Many different ways of triangulation exist.

6. Theorem 3.1
Every simple polygon admits a triangulation, and any triangulation
of a simple polygon with n vertices consists of exactly n-2 triangles.
n=5 T= 3
n=19 T= 17

7. Proof We prove this by induction. case n=3 Clearly, 1 triangle exists.
n=19 T= 17
We prove this by induction.
case n=3
Clearly, 1 triangle exists.
case n=m (m<n)
We assume that the theorem is true.
case n=n
Any diagonal cuts P into 2 simple subpolygons(P1,P2).
Let m1 and m2 be # of vertices of P1 and P2.
Total vertices increase 2 by cutting,
thus m1 + m2 = n + 2.
However, # of triangles do not change,
so (m1 - 2)+(m2 - 2)= n-2 triangles.
decompose
n=16 T= 14
n=5 T= 3

8. Does a diagonal always exist?v u
Choose one vertex (v) and connect two neighbors(w, u), we get a diagonal wu.
If there are vertices inside △vuw,
wu is not a diagonal,
but vv’ cannot intersect an edge of a polygon, so vv’ is a diagonal . w v u v’
In conclusion, a diagonal always exists.
v’: the vertex farthest from uw

9. By the way, how many cameras are needed?
on each triangle
on well-chosen diagonals
on well-chosen vetices
n=19 T= 17
17 cameras
9 cameras
6 cameras
n-2
roughly n/2
roughly n/3

10. Therorem 3.2
For a simple polygon with n vertices, cameras are occasionally necessary and always sufficient to have every point in the polygon visible from at least one of the cameras.
3-Coloring
The coloring will be such that any two vertices connected by an edge or a diagonal have different colors.
any triangles have each 3 color vertices
Set cameras on one color vertices,
all triangles are covered.
Thus, cameras are sufficient.

11. Does 3-coloring always exist?
Consider about dual graph.
The edge of dual graph corresponds to a diagonal of the original graph.
Dual graph is a tree.
Coloring algorithm
Step 1: Choose one vertex on dual graph and color the three vertices of the triangle.
Step 2: Depth first search (O(n) time)
Step 3: color the remining vertex on the next triangle without same colored vertex.
Step 4: Go to step2

12. Is this limit? needles and n vertices Look at this example.
・・・
An instance which needs
cameras to cover exists.
Thus, the worst case bound is
and it is optimal.
cameras are needed.

13. ? How to triangulation?? We notice that cameras is sufficient,
but how to triangulation actually?
And how long does it take to compute this?
3-coloring
INPUT
OUTPUT ?
doubly connected edge list
triangulated list

14. ? Trivial algorithm In worst case computation time is O(n^2)
n vertices
1 vertex + n-1 vertices
O(n)
In worst case computation time is O(n^2)

15. Outline of efficient algorithms?
Make Monotone Algorithm
Triangulation Algorithm
O(nlogn)
O(n)

16. Monotone
y–monotone : A simple polygon is y-monotone if for any line perpendicular to
y-axis the intersection of the polygon with the line is connected.
y-monotone
NOT y-monotone

17. Make Monotone Algorithm
INPUT: A simple polygon
OUTPUT: Some simple polygons which are y-monotone
We distinguish five types of vertices.
start vertex : ① and ③
end vertex : ① and ④
regular vertex : else
split vertex : ② and ③
merge vertex : ② and ④
①　interior angle is less than π
②　interior angle is greater than π
③　neighbors lie below
④　neighbors lie above
※if two vertices have same y-coordinate, let the left one be the “above” vertex.

18. Lemma 3.4
A polygon is y-monotone if it has no split vertices or merge vertices.
Proof
Suppose a polygon is not y-monotone, scan edges from p(second vertex from most left), then have to meet a split or merge vertex, until reaching the next intersection.
split vertex q p r q p r
merge vertex

19. detail of algorithm Step1 : Sort vertices by y-coordinate and
By Lemma3.4, we should remove a split and merge vertices.
sweep line v5 e4 v3 v6 e3
Find diagonals to decompose a polygon into
y-monotone polygons without such vertices. v4 e2 v7 v9 e1 v1 v2 v8
Step1 : Sort vertices by y-coordinate and
set event queue
Step2 : Check type of each vertices by neighbors
Step3 : Plane sweep algorithm using the queue
Step4 : At a vertex, work the algorithm which
corresponds to type of it
v14
v10
v15
v11
v12
v13
event queue v5 v3 v4 v6 …
start
merge
regular

20. definition of helper
helper(ej) : The lowest vertex above the sweep line such that
the horizontal segment connecting the vertex
to ej lies inside a polygon.
sweep line v1 v2 v3 v4 v5 v6 v7 v8 v9
v10
v11
v12
v13
v14
v15 e1 e2 e3 e4 1 3 1
helper(e3) = v3
helper(e5) = v5
helper(e7) = v2
helper(e9) = v8 2 2
helper(e3) = v3
helper(e5) = v4 3
When sweep line is on a split or merge vertex,
if we connect the vertex to helper(ek) (ek is the left edge of it), the line is a good diagonal.

21. work like this
We use a binary search tree in order to save the way to the edges against vertices v1 v2 v3 v4 v5 v6 v7 v8 v9
v10
v11
v12
v13
v14
v15 e1 e2 e3 e4 1
binary search tree 1
helper(e5) = v5 2
helper(e3) = v3 3 2
helper(e5) = v4 4
helper(e3) = v3 5 7 5 6 3 3
Get a diagonal !!
helper(e6) = v6
helper(e3) = v3 9 7 3 1 4 ……

22. Computation time Computation time is O(nlogn) Preprocessing :
binary search tree
Preprocessing :
Sorting in O(nlogn) time
At one event
search in O(logn) time
insert and delete in O(1) time
n vertices 9 7 3 1
Computation time is O(nlogn)

23. Triangulation Algorithm
INPUT: A y-monotone polygon
OUTPUT: Some triangles
Scan the vertices from above, and cut out the triagnles
using a stack. v1 v2 v3 v4 v5
v12
v11
v10 v9 v8 v7 v6
Step 1 : Search a vertex and check the stacked vertices
If we get three vertices
then If the triangle is outer of the polygon,
then stack the vertices.
else add a diagonal as much as possible
Step 2 : Goto step 1

24. Work like this v1 v2 v3 v4 v5 v12 v11 v10 v9 v8 v7 v6 stack v8 v9 v10

25. Computation time Computation time is O(n) Vertices are already sorted
At one vertex
push happens at most twice
pop happens at most 2n times
Computation time is O(n)

27. algorithm of each type
start vertex(vi) : Insert ei in T and set helper(ei) to vi
end vertex(vi) : Insert ei in T and set helper(ei) to vi v1 v2 v3 v4 v5 v6 v7 v8 v9
v10
v11
v12
v13
v14
v15 e1 e2 e3 e4