Feb 2011 1

Polygon Triangulation

Shmuel Wimer

Bar Ilan Univ., School of Engineering

Feb 2011 2

Definitions

0 1 1 1, , , , vertices of an -vertex polygon.n nv v v v n

0 0 1 1 1 1 0

1 1

Edges , , , , are

bounding a polygon by a simple closed curve iff:

, 0, , 1, and , 1.

i i i n n

i i i i j

e v v e v v e v v

e e v i n e e j i

Every simple closed plane

curve divides the plane into two components.

Jordan Curve theorem :

Feb 2011 3

There are no holes in a simple polygon.

Simple polygon

Nonsimple polygons

Feb 2011 4

The Art Gallery Problem

How many stationary guards are needed to guard a room?

Guards are located at fixed points and have 2π range of

visibility. A guard cannot see through walls.

Given a polygon , a point can see a point ( is visible

from ) iff [ , ] .

P x y y

x x y P

A set of guards covers a polygon if every point of the

polygon is visible to some guard.

Problem: Find the max over all n-vertex polygons of the

min number of guards needed to cover the polygons.

Feb 2011 5

This 12-vertex polygon

needs 4 guards to cover it

This 12-vertex polygon

needs 3 guards to cover it.

Feb 2011 6

- smallest number of guards needed to cover .

min | : covers | .S

g P P

g P S S P

- maximum of over all -vertex polygons.

max .n

n

nP

G n g P n

G n g P

G nWhat is ?

A trivial lower bound:

A trivial upper bound:

1G n

.

Put a guard at each vertex.

G n n

Feb 2011 7

What is 3 ?G 3 1.G

What is 4 ?G

4 1. Put a guard on internal diagonal.G

5 1. Put a guard at overlap of triangles.G

Feb 2011 8

Hexagons may require two guards, 6 2.G

3 ?G n n Is the conjecture true

A shape with prongs, 3 , is covered by guards.k n k k

12, 4

4

n k

g P

Feb 2011 9

3G n n Proving

First Proof of sufficiency is due to Chvátal in 1975 (very

complex). We’ll present a simple, four-step proof by Fisk

from 1978.

Necessity follows from the prong’s example.

1. Triangulation of polygon.

2. 3-coloring of triangulation graph.

3. Placing guards at a single color guarantees cover.

4. “Pigeon-hole principle”: placing n objects in k holes,

one at least contains no more than └n/k┘ objects.

Feb 2011 10

Diagonals and Triangulation

A diagonal of a polygon is a line segment connecting two

visible vertices.

[ , ] is a diagonal of iff:

[ , ] , and ( , ) \ .

a b P

a b P a b a b P P

Diagonals [ , ] and [ , ] are called noncrossing if

( , ) ( , ) .

a b c d

a b c d

Triangulation: the partition of a polygon by maximal non

crossing diagonals into triangles (not unique!).

Feb 2011 11

n=14

Interesting, both

triangulations have

same number of

diagonals!

Notice that number

of diagonals equals

n-3.

Feb 2011 12

3-Coloring

G(V,E) is a graph defined on a triangulated polygon P.

E consists of P’s edges and the triangulation diagonals,

while V consists of P’s vertices.

k-coloring is an assignment of k colors to the vertices

of V such that no adjacent vertices have same color.

We’ll show later that a triangulation graph is 3-colorable.

Feb 2011 13

5

4

5

6

3

5

The coloring of a

triangulation

graph is unique!

Assigning color to

first vertex

enforces the rest

color assignments.

Feb 2011 14

Guards Placement

It is sufficient to place guards at one color only (say red).

This follows since every triangle must have 3 distinct

colors at its corners, hence one must be red. Therefore

every triangle is covered.

But the triangles cover the entire polygon P.

Pigeon-hole principle: There must be one color whose

number of assigned vertices doesn’t exceed └n/3┘.

3n Conclusion : guards are sufficient.

Feb 2011 15

Triangulation Theory

Need to prove that every polygon can be triangulated.

Though intuitive, proof is not trivial.

Lemma: Every polygon must have at least one strictly

convex vertex.

Proof: Let the vertices be

counterclockwise ordered.

Traversing the boundary, a

convex vertex corresponds

to a left turn.v L

Feb 2011 16

Pick the lowest vertex (pick the rightmost If there are few).

L is a line passing through v.

The edge following v must lie above L. ■

Lemma: Every polygon of n≥4 vertices has a diagonal.

Proof: There exists a strictly convex vertex v. Let a and b

be vertices adjacent to v.

If [a,b] is a diagonal we are done. So assume it is either

exterior or intersecting with the boundary of P.

Feb 2011 17

v

ab

L

x

must contain at

least one vertex of .

avb

P

Let be the closest

vertex to , measured

orthogonal to the line

passing through .

x

v

ab

The interior of cannot contain any point of .cvd P

dc

Therefore [ , ] , , hence a diagonal. x v P x v ■

Feb 2011 18

Theorem: Every n-vertex polygon P can be triangulated.

Proof: By induction on n.

If n=3 P is a triangle.

Let n≥4. By lemma, P has a

diagonal d which divides P into

two polygons P1 and P2, having

n1<n and n2<n vertices, resp.

P1 and P2 can be triangulated by

induction hypothesis. ■

d

Feb 2011 19

Question: Are all triangulations have the same

number of diagonals and triangles?

Lemma: Every triangulation of an n-vertex polygon P

has n-3 diagonals and n-2 triangles.

Proof: By induction on n.

For n=3 the claims are trivially true.

For n≥4, divide P by a diagonal d into P1 and P2, having

n1<n and n2<n vertices, resp.

n1+n2 = n+2 since the end vertices of d are duplicated.

Feb 2011 20

It follows from the induction hypothesis that the claims

hold for n1-vertex and n2-vertex polygons.

The total number of diagonals (n1-3)+(n2-3)+1=n-3,

where +1 stands for counting d.

The number of resulting triangles is (n1-2)+(n2-2)=n-2.■

Corollary: The sum of internal angles in an n-vertex

polygon is (n-2)π.

Proof: Polygon is covered by n-2 triangles, whose

corners are placed at polygon’s vertices.

Feb 2011 21

Triangulation Dual

An arc exists between two vertices iff the corresponding

triangles are adjacent.

A vertex is placed in every triangle.

Feb 2011 22

Lemma: The dual T of a triangulation of a simple

polygon P is a tree whose vertex degree is 3 at most.

Proof: Degree claim follows immediately from the fact

that a triangle has three sides at most to share.

Assume in contrary that T is not a tree. It then must

have a closed cycle C, completely contained in P.

C must enclose a vertex of P, which is the end point

of a diagonal crossing C.

Feb 2011 23

P must therefore have a hole, thus contradicting its

simplicity. ■

CBorder vertices

Completely in P

Feb 2011 24

Vertices of degree 1 are leaves of T. Vertices of degree

2 lie on T’s paths. Vertices of degree 3 are branches.

When rooted at a vertex of degree 1 or 2, T is a binary.

Definition: Three successive vertices a, b, c of a polygon

form an ear if ac is a diagonal, while ab and bc are not.

Theorem: A polygon with n≥4 has two nonoverlapping

ears at least.

Proof: A leaf of the dual tree T implies an ear. T has

n-2 ≥2 vertices, at least 2 of which are leaves. ■

Feb 2011 25

3-Coloring Theorem

Theorem: The triangulation graph of a simple n-vertex

polygon P is 3-colorable.

Proof: By induction on n.

P has an ear. Ear removal obtains an (n-1)-vertex

polygon, which by induction hypothesis is 3-colorable.

The end vertices of the ear’s base (diagonal) are adjacent

vertices of P and thus have two distinct colors. The tip of

the ear is then colored with the third color. ■

Feb 2011 26

We are done!

Feb 2011 27

Area of Polygons

3Let , , , and ,

, be two vectors. The

area of the parrallelogram

defined by and is .

a b c b a

c a

Α

B

Α B Α B

R

a b

c

A

B

Determinant

notation:.

y z z y xx y z

x y z z x x z y

x y z x y y x z

A B A B

A A A A B A B

B B B A B A B

uu u u

u

u

Feb 2011 28

2In 0, so the area of a triangle is:

1

21

21

.2

z z

x y y x

x x y y y y x x

x y y x y x x y x y y x

A B T

T A B A B

b a c a b a c a

a b a b a c a c b c b c

R

A

Area of polygons can be found by triangulation and

calculation of triangles areas.

We’d like to obtain a closed-form formula, independent

of triangulation.

Feb 2011 29

0 1 2 0 2 3 0 2 1, , , , , ,n nP v v v v v v v v v A A A A

A(P)’s expression depends on triangulation.

We’ll consider first quadrilaterals and show its area is

independent of triangulation.

0v 6v

5v

4v3v

2v

1v

7v

P

Feb 2011 30

d

ba

c

Q

d

ba

c

Q

, , , , , , , ,Q a b c a c d d a b d b c A A A A A

x y y x y x x y x y y x

x y y x y x x y x y y x

Q a b a b a c a c b c b c

a c a c a d a d c d c d

2A

Diagonal related terms are canceled out!

11 10

For any convex polygon: n

i i i iiP x y x y

2A

Feb 2011 31

What happens for non-convex

quadrilateral ?

v2

0 1 2 0 3 2

31 10

0 2

, , , ,

.

The terms involving cancel out.

i i i ii

A Q A v v v A v v v

x y x y

v v

v3

v1

v0

Let be a triangle whose vertices are

counterclockwise oriented, and a point in the plane,

then , , , , , , .

T abc

p

T p a b p b c p c a

Lemma :

A A A A

Q

Feb 2011 32

We’d like to obtain the area of a general simple polygon

in terms of its vertices only.

0 1 1

1 11 1 10 0

11 10

Let the vertices , , , , ,

be labled counterclockwise, defining a polygon ,

and let be an arbitrary point. Then

, ,

n i i i

n ni i i i i ii i

ni i ii

v v v v x y

P

p

P p v v x y x y

x x y y

Theorem :

A A

.i

T

a

b

c

p

Feb 2011 33

p

P is a polygon and p an arbitrary

point (internal or external). p’s

vertices are cyclically ordered

counterclockwise.

We can traverse the edges

along the border and consider

the triangles viewed from p.

The areas of rectangles counterclockwise oriented are

added while the clockwise ones are subtracted. This is

integration along the border (Green’s Theorem).

Feb 2011 34

2 1 0 1

By induction on . The case 3 was

shown before. has an ear. Assume w.l.o.g it is

, , , and obtain by \ .n n n

n n

P

E v v v P P E

Proof :

21 10

2 1 1 0 0 2

By induction hypothesis , , ,

and by lemma

, , , , , , .

nn i ii

n n n n

P p v v

E p v v p v v p v v

A A

A A A A

1

0 2 2 0

Since , and

, , , , , the theorem follows.

n

n n

P P E

p v v p v v

A A A

A A

Feb 2011 35

11 10

1

The proof of is obtained by

expanding the determinant expression of ' area

and canceling out the terms of and common

edge.

ni i i ii

n

x y x y

E s

E P

■

Feb 2011 36

Triangulation Complexity

There are 3 2 diagonal candidates.n n

4This is repeated 3 times, yielding total time.n O n

4

Implementing triangulation as in the existence theorem

requires time.O n

Cheking validity of a diagonal requires intersection test

against all edges and previously defined diagonals, which

takes time.O n

Feb 2011 37

Run time is reduced to O(n2) (Lennes, 1911) as follows.

Pick the leftmost vertex v of P and connect its two

neighbors u and w. Checking whether uw is a diagonal

takes O(n). If it is, the rest is (n-1)-vertex polygon.

If uw is not a diagonal, get x, the farthest vertex from uw

inside Δuvw. This takes O(n) time. vx is a diagonal dividing

P into P1 and P2, having n total number of vertices.

Recursive application of the above procedure consumes

total O(n2) time.

Can we do better?

Feb 2011 38

There is a type of polygons whose triangulation takes

linear time, called monotone.

Definition: P is monotone w.r.t to a line l if P intersects

with any line l’ perpendicular to l in a single segment, a

point or it doesn’t intersect.

ll’

l’

l’

l’

monotone

l’

l’

l’

l’

not monotone

Feb 2011 39

P

We’ll partition P into y-monotone pieces.

start vertex

end vertex

merge vertex

split vertex

regular vertex

On horizontal edge the

left vertex is considered

above the right one.

Feb 2011 40

Lemma: A polygon is y-monotone if it has neither split

vertices nor merge vertices.

Proof: If P is not monotone, there must exist a line l

intersecting P in more than a single segment. Let [p,q] be

its leftmost sub segment.

Follow the boundary of P starting

at q, where P is on the left. At

some point r we must cross l.

If r ≠ p then the highest vertex must be a split one.

P

p q r l

split vertex

Feb 2011 41

If r = p we follow the boundary

from q in opposite direction.

At some point r’ we must

cross l. r’ ≠ p as otherwise

it contradicts that P is not

monotone.

This implies that the lowest

encountered vertex must be

a merge one. ■

P

p r q r l

merge vertex

Feb 2011 42

P

Getting Rid of Split and Merge Vertices

Sort P’s vertices from top to

bottom. It takes O(nlogn)

time.

Apply scan line from top to

bottom to encounter vertices.

Diagonals are introduced at split and merge vertices.

Feb 2011 43

P

Scan line intersection with P’s boundary is stored in a left to right ordered balanced binary tree.

Insertion into and deletion from a tree takes O(logn) time. Neighbor query takes O(1) time.

Total time consumed by scanning is O(nlogn) time.

Feb 2011 44

At a split point we get the nearest left and right edges in O(1) time.

The last upper vertex is also stored (can be the top end of right or left edge).

This data suffices to decide on diagonal in O(1) time.P

It takes O(nlogn) time to split

P into monotone polygons.

Merge vertices are handled

similarly.

Feb 2011 45

The added diagonals are non intersecting

and partition into monotone polygons.P

Lemma :

The resulting partitions contain neither split

nor merge vertices, hence are monotone polygons.

Proof :

To see that the added diagonals are valid, we'll show

that an added segment niether intersects 's edges nor

previously added diagonals.

P

We'll consider a split vertex. Merge vertex is similar.

Feb 2011 46

iv

mv

je ke

Q

When was added it

couldn't intersect 's edge.m iv v

P

If it intersects, there are two cases. If an end vertex

of 's edge is inside it contradics is nearest to .m iP Q v v

Otherwise, the edge must cross , contrudiction to

and being the nearest left and right 's edges.j k i

Q

e e v

Feb 2011 47

Finally, a previously added diagonal must have its

end vertices above . We've showed that cannot

have any vertex of , hence intersection with

is impossible.

i

m i

v Q

P v v

■

Theorem: An n-vertex simple polygon can be partitioned

into y-monotone polygons in O(nlogn) time and O(n)

storage.

Feb 2011 48

P

1

3

2

4

5

6

7

89

1011

Triangulating y-Monotone Polygon (Sketch)

's vertices are sorted in

descending .

P

y

's boundary is traversed with

one leg on and one leg on

, by poping and pushing

vertices from

right

a st .

left

ack

P

Diagonals insertions are decided

according to stack's top status.

Feb 2011 49

Theorem: Strictly y-monotone n-vertex polygon can be

triangulated in linear time.

Theorem: (Gary et. al. 1978) A simple n-vertex polygon

can be triangulated in O(nlogn) time and O(n) storage.

The problem has been studied extensively between

1978 and 1991, when in 1991 Chazelle presented an

O(n) time complexity algorithm.