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w. v. w. v. 2. ( n ) time in the worst case. . u. u. Brute-Force Triangulation . 1. Find a diagonal. 2. Triangulate the two resulting subpolygons recursively. How to find a diagonal?. leftmost vertex. case 2. closest to v. case 1.

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brute force triangulation

w

v

w

v

2

(n )time in the worst case.

u

u

Brute-Force Triangulation

1. Find a diagonal.

2. Triangulate the two resulting subpolygons recursively.

How to find a diagonal?

leftmost

vertex

case 2

closest

to v

case 1

O(n) time to find a diagonal at every step.

triangulating a convex polygon
Triangulating a Convex Polygon

(n)time!

Idea:

monotone

Decompose a simple polygon into convex pieces.

Triangulate the pieces.

as difficult

as triangulation

y monotone pieces
y-monotone Pieces

y-axis

y-monotone if any line perpendicular

to the y-axis has a connected

intersection with the polygon.

highest

vertex

Stragegy:

walk always

downward or

horizontal

Partition the polygon into monotone

pieces and then triangulate.

lowest

vertex

turn vertex
Turn Vertex

Turn vertex is where the walk

from highest vertex to the lowest

vertex switches direction.

At vertex v

 Both adjacent edges are below.

v

The polygon interior lies above.

Choose a diagonal that goes up.

five types of vertices

q

p

q

p

Merge vetex: lies below its two neighbors and

has interior angle > .

Split vetex: lies above its two neighbors and

has interior angle > .

End vetex: lies below its two neighbors and

has interior angle < .

Start vetex: lies above its two neighbors and

has interior angle < .

Regular vetex: the remaining vertices (no turn).

Five Types of Vertices

Point p = (x, y) is “below” a different point q = (u, v) if

y < v or y = v and x > u.

Otherwise p is “above” q.

4 types of turn vertices

What happens if we rotate the polygon by ?

start vertices  end vertices split vertices  merge vertices

start vertices  end vertices

local non monotonicity

Proof

Suppose the polygon is not y-monotone. We prove that it contains

a split or merge vertex.

exterior

l

q

p

interior

interior

p = r

q

l

r p

r= p

exterior

Local Non-Monotonicity

Lemma A polygon is y-monotone if it has no split or

merge vertices.

split

vertex

There exists a horizontal line l intersecting the polygon in >1 components,

of which the leftmost is a segment between some vertices

p (left) and q (right).

Start at q, traverse the boundary counterclockwise until

it crosses the line l again at point r.

r

Traverse in the

oppose direction

from q and crosses

line l again at point r’.

The lowest point during this

traversal must be a merge vertex.

r

The highest vertex during

the traversal from q to r

must be a split vertex.

merge vertex

partitioning into monotone pieces

The next event found in time O(log n).

Partitioning into Monotone Pieces

The lemma implies that the polygon will have y-monotone

pieces once its split and merge vertices are removed.

Add a downward diagonal at every merge vertex.

Add an upward diagonal at every split vertex.

Use a downward plane sweep.

No new event point will be created except the vertices.

The event queue is implemented as priority queue (e.g., heap).

removal of a split vertex

v : split vertex

i

e : edge immediately to its left.

j

e : edge immediately to its right.

k

h(e )

i

h(e ): lowest vertex above v

and between e and e .

i

i

j

k

or the upper vertex of

e if such vertex does not

exist.

j

e

e

e

e

v

j

i-1

i

i

k

Connect v to h(e ).

i

i

Removal of a Split Vertex

edge

helper

removal of a merge vertex

Merge vertices can be handled the same way in an upward sweep

as split vertices in a downward sweep.

v : highest vertex below the

sweep line and between

e and e .

v : merge vertex

i

m

But why not all in the same sweep?

e : edge immediately to its left.

j

e : edge immediately to its right.

k

j

k

Connect v to v .

m

In case the helper of e is not replaced any more, connect to its lower endpoint.

v will replace v as the helper of e .

Check if the old helper is a merge vertex and add the diagonal if so.

(The diagonal is always added if the new helper is a split vertex).

m

i

j

v

v

e

e

i

i

k

m

j

j

Removal of a Merge Vertex
sweep line status

Only edges to the left of the polygon interior are stored.

Edges are stored in the left-to-right order.

With every edge e its helper h(e) is also stored.

Sweep-Line Status

Implemented as a binary search tree.

This is because we are only interested in edges to

the leftof split and merge vertices.

dcel representation

Construct a doubly-connected edge list to represent the polygon.

Add in diagonals computed for split and merge vertices.

Edges in the status BST and corresponding ones in DCEL

cross-point each other.

DCEL Representation

n vertices + n edges + 2 faces (initially)

Adding an edge can be done in O(1) time.

the algorithm
The Algorithm

MakeMonotone(P)

Input: A simple polygon P stored in DCEL D.

Output: A partitioning of P into monotone subpolygons stored in D.

1. Q priority queue storing vertices of P

2. T   // initialize the status as a binary search tree.

3. i  0

4. while Q  

5. do v  the highest priority vertex from Q// removal from Q

6. case v of

7. start vertex: HandleStartVertex(v )

8. end vertex: HandleEndVertex(v )

9. split vertex: HandleSplitVertx(v )

10. merge vertex: HandleMergeVertex(v )

11. regular vertex: HandleRegularVertex(v )

12. i  i + 1

i

i

i

i

i

i

i

handling start end vertices

HandleStartVertex(v )

1. T T  { e }

2. h(e )  v // set the helper

i

i

i

i

  • HandleEndVertex(v )
  • ifh(e ) is a merge vertex
  • then insert the diagonal
  • connecting v to h(e )
  • in D
  • 3. T  T – { e }

i

i –1

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i –1

i -1

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Handling Start & End Vertices

insert into T

handling split vertex

HandleSplitVertex(v )

  • Search in T to find e directly left of v
  • insert the diagonal connect v to h(e )
  • into D
  • h(e )  v
  • 4. T  T + { e }
  • 5. h(e )  v

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Handling Split Vertex
handling merge vertex 1

j

i

  • HandleMergeVertex(v )
  • ifh(e ) is a merge vertex
  • then insert the diagonal
  • connecting v to h(e )
  • in D
  • 3. T  T – { e }
  • Search in T to find the edge e
  • directly left of v .
  • 5. ifh(e ) is a merge vertex
  • then insert the diagonal
  • connecting v to h(e )
  • in D
  • 6. h(e ) v

j

i

i –1

i

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i –1

j

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i -1

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Handling Merge Vertex (1)
handling merge vertex 2

h(e )

j

j

i

  • HandleMergeVertex(v )
  • ifh(e ) is a merge vertex
  • then insert the diagonal
  • connecting v to h(e )
  • in D
  • 3. T  T – { e }
  • Search in T to find the edge e
  • directly left of v .
  • 5. ifh(e ) is a merge vertex
  • then insert the diagonal
  • connecting v to h(e )
  • in D
  • 6. h(e ) v

j

i

i –1

i

j

i

i –1

j

i

i -1

v

e

e

i – 1

i

j

Handling Merge Vertex (2)

h(e )

i –1

handling regular vertices 1

HandleRegularVertex(v )

  • if the polygon interior lies to the right
  • of v
  • 2. then ifh(e ) is a merge vertex
  • 3. then insert the diagonal
  • connecting v to h(e )
  • in D
  • 4. T  T – { e }
  • 5. T  T + { e }
  • 6. h(e )  v
  • 7. else search in T to find the edge e
  • directly left of v .
  • 8. ifh(e ) is a merge vertex
  • then insert the diagonal
  • connecting v to h(e )
  • in D
  • 9. h(e ) v

i

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i –1

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i -1

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Handling Regular Vertices (1)
handling regular vertices 2

HandleRegularVertex(v )

  • if the polygon interior lies to the right
  • of v
  • 2. then ifh(e ) is a merge vertex
  • 3. then insert the diagonal
  • connecting v to h(e )
  • in D
  • 4. T  T – { e }
  • 5. T  T + { e }
  • 6. h(e )  v
  • 7. else search in T to find the edge e
  • directly left of v .
  • 8. ifh(e ) is a merge vertex
  • then insert the diagonal
  • connecting v to h(e )
  • in D
  • 9. h(e ) v

i

i

i –1

i

i –1

h(e )

j

i -1

i

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i

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e

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i

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Handling Regular Vertices (2)
correctness
Correctness

Theorem The algorithm adds a set of non-intersecting

diagonals that partitions the polygon into

monotone pieces.

Proof The pieces that result from the partitioning contain no split

or merge vertices. Hence they are monotone by an earlier

lemma.

We need only prove that the added segments are diagonals

that intersect neither the polygon edges nor each other.

Establish the above claim for the handling of each of the

five type of vertices during the sweep. (Read the textbook

on how to do this for the case of a split vertex.)

running time on partitioning
Running Time on Partitioning

MakeMonotone(P)

Input: A simple polygon P stored in DCEL D.

Output: A partitioning of P into monotone subpolygons stored in D.

1. Q priority queue storing vertices of P

2. T  

3. i  0

4. whileQ  

5. dov  the highest priority vertex from Q

6. case vof

7. start vertex: HandleStartVertex(v )

8. end vertex: HandleEndVertex(v )

9. split vertex: HandleSplitVertx(v )

10. merge vertex: HandleMergeVertex(v )

11. regular vertex: HanleRegularVertex(v )

12. i  i + 1

// O(n)

// O(log n)

i

i

// O(log n) each case

i

// queries and updates

// in O(log n) time and

// insertion of a diagonal

// in O(1) time.

i

i

i

i

Total time: O(n log n)

Total storage: O(n)

triangulating a y monotone polygon

One boundary of the funnel is a polygon edge.

The other boundary is a chain of reflex vertices

(with interior angles > ) plus one convex vertex

(the highest) at bottom of the stack.

Triangulating a y-Monotone Polygon

Assumption The polygon is strictly y-monotone (no horizontal edges).

It is for clarity of presentation and will be easily removed later.

Order of processing: in decreasing y-coordinate.

A stack S: vertices that have been encountered

and may still need diagonals.

convex

vertex

lowest vertex on top.

funnel

Idea: add as many diagonals from the current

vertex handled to those on the stack as

possible.

current

vertex

Invariants of iteration:

reflex

vertices

case 1 next vertex on opposite chain

Push the previous top of the stack and

the current vertex back onto the stack.

Pop these vertices from the stack.

Add diagonals from the current vertex

to them (except the bottom one) as they

are popped.

u

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j –3

j –3

j –4

j –2

j –4

j –2

j –1

j –1

j

j –1

j

Case 1: Next Vertex on Opposite Chain

This vertex must be the lower endpoint of the

single edge e bounding the chain.

e

current

vertex

case 2 next vertex on the same chain

Pop one vertex from the stack.

It shares an edge with the current vertex.

Pushed the last popped vertex back onto

the stack followed by the current vertex.

Pop other vertices from the stack as

long as they are visible from the current

vertex.

Draw a diagonal between each of them

and the current vertex.

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j –3

j

j –1

j –4

j –2

j –3

j –2

j –5

j –4

j –1

j

j –5

j –5

j –4

Case 2: Next Vertex on the Same Chain

The vertices that can connect to the current

vertex are all on the stack.

current

vertex

the triangulation algorithm
The Triangulation Algorithm
  • TriangulateMonotonePolygon(P)
  • Input: A strictly y-monotone polygon P stored in DCEL D.
  • Output: A triangulation of P stored in D.
  • Merge the vertices on the left and right chains into one sequence
  • sorted in decreasing y-coordinate. (In case there is a tie, the one
  • with smaller x-coordinate comes first.)
  • 2. Push(u , S)
  • Push(u , S)
  • forj 3 ton – 1
  • do ifu and Top(S) are on different chains
  • then while Next(Top(S))  NULL
  • v Top(S)
  • Pop(S)
  • insert a diagonal from u to v
  • Pop(S)
  • Push(u , S)
  • Push(u , S)
  • else Pop(S)
  • while Top(S) is visible from u inside the polygon
  • v  Top(S)
  • insert a diagonal between u and v
  • Push(v, S)
  • Push(u , S)

1

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j –1

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an example

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An Example

Start:

j =4:

j =3:

j =5:

j =6:

example cont d

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Example (cont’d)

j = 7:

j =9:

j =8:

(result)

j =10:

j =11:

example cont d1

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Example (Cont’d)

j = 15:

j = 16:

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(result)

j = 18:

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removal of strict y monotonicity

#pushes  2n - 4

#pops  #pushes

Treat them from left to right.

The effect of this is equivalent to that of rotating the plane

slightly clockwise and then every vertex will have different

y coordinate.

Removal of Strict y-monotonicity

The running time of TriangulateMonotonePolygon is(n).

What to do if some vertices have the same y-coordinates?

time complexity of triangulation
Time Complexity of Triangulation

1. Partition a simple polygon into monotone pieces.

O(n log n)

2. Triangulate each monotone piece.

(n) for all monotone pieces together

Theorem A simple polygon can be triangulated in

O(n log n) time and O(n) storage.

triangulation of a planar subdivision

The plane sweep for decomposition of a polygon into monotone

pieces takes as input only edges that lie to the left of the interior.

This easily generalizes to a planar subdivision in a bounding box.

Triangulation of a Planar Subdivision

The algorithm for splitting a polygon into monotone pieces

does not use the fact that the polygon was simple.

Thus a planar subdivision with n

vertices can also be triangulated

in O(n log n) time using O(n)

storage.