1. Polygon Triangulation
전진우
손명배

2. Art Gallery Problem Want to cover all areas
Assume that the art gallery is a simple polygon
Does not intersect itself
How many cameras do we need?
Where do we put it?
Getting an exact solution is NP-hard
We just want to get a lower bound
Simple polygon: a polygon that does not intersect itself.
Num. of camera depends on the complexity of polygon, which means that it depends on n.

3. Triangulation Easy to guard Draw diagonals
An open line segment that connects two vertices of P and lies in the interior of P
A decomposition of a polygon into triangles by a maximal set of non-intersecting diagonals is called a triangulation of the polygon
Why maximal set: colinear vertex 3개가 있을 경우, “triangle”을 만족해도 삼각형의 edge에 vertex가 박혀있는 경우가 있다. 그런 경우 방지용.

4. Triangulation Thm 3.1 Induction on n
Every simple polygon admits a triangulation, and any triangulation of a simple polygon with n vertices consists of exactly n−2 triangles.
Induction on n
n = 3: Trivial
n > 3: Assume that the theorem is true for all m<n ……

5. Triangulation
Pick a vertex v and neighboring vertices w and u. Draw 𝑤𝑢 .
Case 1: vwu does not contain any vertices
Then 𝑤𝑢 is a diagonal; P consists of vwu and the rest is triangulated into n – 3 triangles.
Case 2: vwu contains some vertices
Let v’ be the farthest point from 𝑤𝑢 . Then 𝑣𝑣′ is a diagonal. Why? If 𝑣𝑣′ intersects any edges, one of its endpoint is farther than v’ ↯
Case 2: rest is the same; vv’ separates polygon into two part; whose sum of number of triangles after triangulation is n – 2

6. Triangulation Methods
Recursive triangulation algorithm
Based on the argument stated previously
Draw a line segment and find the diagonal
O(n2) at worst
Split into convex pieces and triangulate
Easy to triangulate - just draw diagonals from a single vertex
Hard to split the polygon into convex pieces
Split into monotone pieces and triangulate
A little bit harder to triangulate, but O(n) anyway
Takes O(nlogn) to split the polygon into monotone pieces
Convex split부터가 triangulation에 기반하므로 cyclic reference

7. Monotone Polygon
A simple polygon is called monotonic with respect to l if for any line l’ perpendicular to l, the intersection of the polygon with l’ is connected.
If we walk from a topmost to a bottommost vertex along one side boundary chain, then we always move downwards or horizontally, never upwards.

8. Monotone Polygon Split the polygon by getting rid of turn vertex
Walk through either left or right chain from topmost to downmost vertex
Our walking direction changes on turn vertex
(either downward to upward or vice versa)

9. Monotone Polygon Regular vertex Start vertex End vertex Split vertex
When v is none of following types
Start vertex
When v is above its neighbors
interior angle < π
End vertex
When v is below its neighbors
Split vertex
interior angle > π
Merge vertex v4 v6 v1 v9 v7 v2 v8
v14
v10
v12
v15
For vertices who resides on same y coord, we make the definition of “above” and “below” to be ordered by x coord as well, like what happens when P is slightly turned clockwise.
Why “split” and “merge”? We’re going to use sweep line algorithm – sweep line is “splitted” and “merged” at those places.
v11
v13

10. Monotone Polygon Lemma 3.4
A polygon is y-monotone if it has no split vertices or merge vertices.
Suppose P is not monotone. Then there’s a horizontal line l that intersects P in more than one connected components.
∴ Either a split vertex or a merge vertex exists
Choose l s.t. leftmost component is a segment
Let pq be that segment.
If we traverse upward from q, and meet l at some point r s.t. p!=r, then there should be a highest vertex located in path from q to r which is a split vertex
If p = r, go downward and lowest vertex is a merge vertex.

11. Monotone Polygon How to get rid of split vertex and merge vertex?
Add diagonal split vertex / merge vertex
Use plane sweep method!

12. Monotone Polygon
Let 𝑣 1 , 𝑣 2 , …, 𝑣 𝑛 be a counterclockwise enumeration of the vertices, and 𝑒 1 , …, 𝑒 𝑛 be a enumeration of the edges along 𝑣 1 ,…, 𝑣 𝑛 starting from 𝑣 1 𝑣 2
Sweep line goes from topmost vertex to bottommost vertex
Event points: every vertices
No more addition/removal on event queue We can just sort vertices on y coord in O(nlogn) and use it
Add diagonals to split / merge vertices along sweep
As stated above, points with same y coord is sorted in such “above”/”below” condition

13. Plane Sweep Handling a split vertex vi
Find edge located on left and right of vi – ej and ek
Choose the lowest vertex between ej and ek
We call it the helper of ej
Draw a diagonal to the helper
Formal defintion: helper(ej) is the lowest vertex above sweep line s.t. the horizontal segment connecting the vertex to ej lies inside P

14. Plane Sweep Handling a merge vertex vi
Draw a diagonal toward some vertex vm which is lower than the sweep line??
vm is a vertex whose left edge (ej) has its helper vi
When we meet vm, draw a diagonal
It is done when

15. Plane Sweep
We need to know the left edge (ej) of vertex and its helper.
Only left edges are needed; store the edges whose interior toward P is right side.
Save the list of left edges intersecting with sweep line
Use dynamic binary search tree T
Save a helper for each edges
Output should be the divided subpolygons with doubly connected edge list.
We also assume that initial input was also doubly connected edge list.

16. Event Points Start vertex When v is above its neighbors
interior angle < π e5

17. Event Points End vertex When v is below its neighbors
interior angle < π v6 e8
이전 예제에서는 helper가 merge vertex인 경우가 없어 예제를 만들었습니다 v9

18. Event Points Split vertex When v is above its neighbors
interior angle > π v9 v8
v14 e9
e14
v11
v13

19. Event Points Merge vertex When v is below its neighbors
interior angle > π v9 v7 v2 e7 e9 v8

20. Event Points Regular vertex When v is none of following types v4 e5 v6
Case 1. Left vertex

21. Event Points Regular vertex When v is none of following types v12 v16
Case 2. Right vertex

22. Monotone Polygon Pretty standard line sweep algorithm
As I told above, you can just sort the list of vertices instead of using PriorityQ

23. Monotone Polygon Lemma 3.5
Algorithm MakeMonotone adds a set of non-intersecting diagonals that partitions P into monotone subpolygons.
It is easy to see that P is partitioned in a way that no split/merge vertices exists
∴By Lemma 3.4, it is monotone
But, are diagonals valid?
Each of them should not intersect with another or edges
We’ll prove that for HandleSplitVertex
Rest are similar

24. Monotone Polygon
Consider a segment 𝑣 𝑚 𝑣 𝑖 added by HandleSplitVertex( 𝑣 𝑖 )
Let Q be the quadrilateral bounded by ej, ek, vi, vm
There are no vertices in Q; otherwise vm couldn’t be the helper of ej
Any “edge” that intersects 𝑣 𝑚 𝑣 𝑖 should go through horizontal segment that connects ej to either vm or vi – contradicting that ej lies immediately to the left for them.
For previously added diagonals, their endpoints should be over vi, and over Q (as no vertices are in Q)
It can’t intersect 𝑣 𝑚 𝑣 𝑖
ej: left edge, ek: right edge

25. Analysis Priority queue construction: O(n) Total n events:
By sort: O(nlogn)
Total n events:
At most one insertion / deletion on T – O(logn)
At most two diagonal insertion on D – O(1)
Space complexity: O(n)
At most n (|V|) items on Q
At most n (|E|) items on T
Thm 3.6 A simple polygon with n vertices can be partitioned into y-monotone polygons in O(nlogn) time with an algorithm that uses O(n) storage

26. Example v5

27. Example v3 e5

28. Example e3 e5 v4

29. Example e5 v6

30. Example v7 e6

31. Example v1 e7

32. Example v9 e7 e1

33. Example e7 e1 v2 e9

34. Example e7 e9 v8

35. Example e9
v14

36. Example e9
v15
e14

37. Example e9
v10

38. Example
e10
v12

39. Example
e10
v11

40. Example
v13

41. Example

42. So Far…
Our ultimate goal is to solve the polygon triangulation, in order to solve the art gallery problem
And for that, we’ve monotonized the polygon to ease the problem
Done in O(nlogn)
Now it’s time to triangulate each monotone polygons!
Done in O(n)

43. Outline of Triangulation
Let polygon P be a y-monotone with n vertices
We always go down when we walk on the left or right boundary chain of P
Stack contains vertices that have been encountered but may still need more diagonals
Add diagonals with vertices in stack whenever this is possible

44. Stack
Stack stores vertices that have been visited but not yet connected with a diagonal
𝑣 1 is the bottom and 𝑣 𝑖 is the top of the stack
Stack must meet the two invariant conditions
The vertices on the stack lie on the same chain on the boundary of P
If i ≥ 3, ∠ 𝑣 𝑗 𝑣 𝑗+1 𝑣 𝑗+2 ≥𝜋 for 1 ≤ j ≤ i - 2

45. Initialization
Sort N vertices of monotone polygon P in order by decreasing y coordinate
If two vertices have the same y-coordinate, then the leftmost one is handled first
𝑢1, 𝑢2, …, 𝑢𝑁 be the sorted sequence of vertices
y(𝑢1)>y(𝑢2)>…>y(𝑢𝑁)
Edge 𝑢𝑖 𝑢𝑗 is an edge of monotone polygon P

46. Cases
Case 1 : u is adjacent to v1 but not vi :

47. Cases
Case 2 : u is adjacent to vi but not v1

48. Cases
Case 3 : u is adjacent to both v1 and vi

49. Pseudocode

50. Example Sorting on decreasing y-coordinate Stack Empty Diagonal u1 u2

51. Example
Initialization
push u1 and u2 onto stack v1 v2

52. Example
u3 : case 2
push u3 v1 v2 u3

53. Example
u4 : case 2
push u4 v1 v2 v3 u4

54. Example u5 : case 1 pop v1, v2, v3, v4 push v4, u5
add diagonal 𝑢 5 𝑣 2 , 𝑢 5 𝑣 3 , 𝑢 5 𝑣 4 v1 v2 v3 v4 u5 v1 v2

55. Example u6 : case 2 u6v2v1 < π add diagonal 𝑢 6 𝑣 1 pop v2 push u6
V1, v2, u6이 꺽엿으니까

56. Example
u7 : case 2
u7v2v1 > π
push u7 v1 v2 u7
V1, v2, u6이 꺽엿으니까

57. Example u8 : case 2 u8v3v2 > π push u8 v1 v2 V1, v2, u6이 꺽엿으니까 v3

58. Example u9 : case 2 u9v4v3 < π add 𝑢9𝑣3 , pop v4 u9v3v2 < π
push u9 v1 v2 v3 u9 v1 v2 v3 v4 u9 v1 v2 v3
V1, v2, u6이 꺽엿으니까

59. Example
u10 : case 3
add 𝑢 10 𝑣 2
exit v1 v2 v3
V1, v2, u6이 꺽엿으니까

60. Analysis
Thm 3.7
A strictly y-monotone polygon with n vertices can be triangulated in linear time
Proof
Step 1 takes linear time
Step 2 takes constant time
For loop is executed n-3 times
One execution may take linear time
At most two vertices are pushed at every execution
total number of pushes : 2n – 4
Number of pops cannot exceed the number of pushes
One execution may take linear time -> pop하고 push하는 한번의 for문이 동작하는 과정을 의미
2m-4 : 처음 2개를 push한 것까지 포함

61. For Planar Subdivision
Thm 3.8
A simple polygon with n vertices can be triangulated in O(nlogn) time with an algorithm that uses O(n) storage
Thm 3.9
A planar subdivision with n vertices in total can be triangulated in O(nlogn) time with an algorithm that uses O(n) storage
증명하기..
각각 증명 과정 너놓고 3페이지로 쪼개기도 가능.
3.7 책의 내용
3.8 85
Polygon cutting theorem
Decompose a polygon into monotone pieces -> o(nlogn) time, o(n) space
split subdivision into monotone pices
(Trapezoid??)

62. 3-coloring
Recall Thm 3.1
Every simple polygon admits a triangulation, and any triangulation of a simple polygon with n vertices consists of exactly n−2 triangles.
Thus n-2 cameras needed- any better solution?
Place camera on a vertex v: every triangle that shares v will be covered!
Lower bound: 𝑛/3
How to get this bound? v

63. 3-coloring Do 3-coloring on triangulation - assign three colors B/G/W
No neighboring two vertices have same color
In triangle, all three vertices have different color!
Choose a color, then put a camera on each vertex with that color (e.g. Black)
The fewest number of vertices is no more than 𝑛/3
Do we have 3-colorings?

64. 3-coloring We can visit each triangle in DFS order on the dual graph
In simple polygon, notice that the dual graph is a tree
Then we can always make a valid assignment of 3-coloring on every triangle

65. Art Gallery Theorem Is 𝒏/𝟑 tight? Yes.
No single camera can see two prongs at a time!
Thm 3.2
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.

66. Art Gallery Theorem 3-coloring takes one DFS which is done in O(n)
We claim that the triangulation of a simple polygon takes O(nlogn) time
If it is true, then…
Thm 3.3
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(nlogn)time.
O(nlogn) is not tight, though the optimal method (O(n)) is complex as hell and practically slower.

67. Reference
M. Garey, D. Johnson, F. Preparata and R, Tarjan, Triangulating a simple polygon, Inf. Process. Lett. 7,
(1978). B. Chazelle. A theorem on polygon cutting with applications. In Proc. 23rd Annu. IEEE Sympos. Found. Comput. Sci., pages 339–349, 1982.
R. Seidel. A simple and fast incremental randomized algorithm for computing trapezoidal decompositions and for triangulating polygons. Comput. Geom. Theory Appl., 1:51–64, 1991.
 Reference 부분에서, 책에서 O(n) optimal 알고리즘의 존재여부를 말했었는데 (Chazelle [92, 94]), 그런게 있지만 대신 대신 드럽게 복잡하고 실행속도도 실용적으로 느려터졌다 정도 잠깐 언급하고 넘어가시면 좋을 것 같습니다. (스크립트 구술도 괜찮습니다)