Polygon Triangulation
Shmuel Wimer
Bar Ilan Univ., School of Engineering
Feb 2011
1
Definitions
v0 , v1,
,vn1, vn1 vertices of an n-vertex polygon.
Edges e0  v0 v1 ,
, ei  vi vi 1 ,
, en 1  vn 1v0 are
bounding a polygon by a simple closed curve iff:
ei
ei 1  vi 1 , i  0,
, n  1, and ei
e j  , j  i  1.
Jordan Curve theorem : Every simple closed plane
curve divides the plane into two components.
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Simple polygon
Nonsimple
polygons
There are no holes in a simple polygon.
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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 P, a point x can see a point y (y is visible
from x) iff [ 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.
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This 12-vertex polygon
needs 3 guards to cover it.
This 12-vertex polygon
needs 4 guards to cover it
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g  P  - smallest number of guards needed to cover P.
g  P   min | S : S covers P | .
S
G  n  - maximum of g  Pn  over all n-vertex polygons.
G  n   max g  Pn  .
Pn
What is G  n ?
A trivial lower bound: G  n   1
G  n   n.
A trivial upper bound: Put a guard at each vertex.
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What is G 3 ?
G 3  1.
What is G  4 ?
G  4  1. Put a guard on internal diagonal.
G 5  1. Put a guard at overlap of triangles.
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Hexagons may require two guards, G 6  2.
A shape with k prongs, n  3k , is covered by k guards.
n  12, k  4
g  P  4
Is the conjecture G  n  n 3 true?
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Proving G  n  n 3
Necessity follows from the prong’s example.
First Proof of sufficiency is due to Chvátal in 1975 (very
complex). We’ll present a simple, four-step proof by Fisk
from 1978.
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.
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Diagonals and Triangulation
A diagonal of a polygon is a line segment connecting two
visible vertices.
[a, b] is a diagonal of P iff:
[a, b] P  a, b and (a, b)  P \ P.
Diagonals [a, b] and [c, d ] are called noncrossing if
(a, b) (c, d )  .
Triangulation: the partition of a polygon by maximal non
crossing diagonals into triangles (not unique!).
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n=14
Interesting, both
triangulations have
same number of
diagonals!
Notice that number
of diagonals equals
n-3.
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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.
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5
4
5
6
3
5
Feb 2011
The coloring of a
triangulation
graph is unique!
Assigning color to
first vertex
enforces the rest
color assignments.
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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┘.
Conclusion : n 3 guards are sufficient.
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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.
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v
L
15
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.
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avb must contain at
least one vertex of P.
a
b
x
c
d
v
L
Let x be the closest
vertex to v, measured
orthogonal to the line
passing through ab.
The interior of cvd cannot contain any point of P.
Therefore [ x, v] P  x, v, hence a diagonal. ■
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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
d
two polygons P1 and P2, having
n1<n and n2<n vertices, resp.
P1 and P2 can be triangulated by
induction hypothesis. ■
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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.
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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.
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Triangulation Dual
A vertex is placed in every triangle.
An arc exists between two vertices iff the corresponding
triangles are adjacent.
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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.
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C
Border vertices
Completely in P
P must therefore have a hole, thus contradicting its
simplicity. ■
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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. ■
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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. ■
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We are done!
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Area of Polygons
Let a, b, c  R 3 , and Α  b  a,
c
B  c  a, be two vectors. The
B
area of the parrallelogram
a
A
b
ux
Determinant
Ax
notation:
Bx
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defined by Α and B is Α  B .
 Ay Bz  Az By  u x 
uy
uz
Ay
Az   Az Bx  Ax Bz  u y 
By
Bz
 Ax By  Ay Bx  u z .
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In R 2 Az  Bz  0, so the area of a triangle T is:
1
A T    Ax By  Ay Bx 
2
1
  bx  ax   c y  a y    by  a y   cx  ax  
2
1
  ax by  a y bx  a y cx  ax c y  bx c y  by cx  .
2
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.
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v3
v4
v2
v5
P
v1
v0
v7
v6
A  P  A  v0 , v1, v2   A v0 , v2 , v3  
 A v0 , vn2 , vn1 
A(P)’s expression depends on triangulation.
We’ll consider first quadrilaterals and show its area is
independent of triangulation.
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c
d
c
d
Q
Q
b
a
b
a
A Q  A  a, b, c   A  a, c, d   A  d , a, b  A  d , b, c 

2A  Q   ax by  a y bx  a y cx  ax c y  bx c y  by cx

 a x c y  a y cx  a y d x  a x d y  cx d y  c y d x


Diagonal related terms are canceled out!
For any convex polygon: 2A  P   
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n1
i 0
 xi yi1  xi 1 yi 
30
v2
What happens for non-convex
quadrilateral ?
v3
Q
v1
A  Q   A  v0 , v1, v2   A  v0 , v3 , v2 

v0
3
i 0
 xi yi1  xi1 yi .
The terms involving v0v2 cancel out.
Lemma : Let T  abc be a triangle whose vertices are
counterclockwise oriented, and p a point in the plane,
then A T   A  p, a, b   A  p, b, c   A  p, c, a  .
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c
T
b
p
a
We’d like to obtain the area of a general simple polygon
in terms of its vertices only.
Theorem : Let the vertices v0 , v1,
, vn1, vi   xi , yi 
be labled counterclockwise, defining a polygon P,
and let p be an arbitrary point. Then
n 1
n 1
A  P    i 0 A  p, vi , vi 1    i 0  xi yi 1  xi 1 yi 

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n 1
i 0
 xi  xi 1   yi 1  yi  .
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P is a polygon and p an arbitrary
point (internal or external). p’s
vertices are cyclically ordered
counterclockwise.
p
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).
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Proof : By induction on n. The case n  3 was
shown before. P has an ear. Assume w.l.o.g it is
E   vn2 , vn1, v0  , and obtain Pn1 by P \ E.
By induction hypothesis A  Pn1   
n2
A
i 0
 p, vi , vi1 ,
and by lemma
A  E   A  p, vn2 , vn1   A  p, vn1, v0   A  p, v0 , vn2  .
Since A  P   A  Pn1   A  E  , and
A  p, v0 , vn2    A  p, vn2 , v0  , the theorem follows.
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The proof of 
n 1
i 0
 xi yi 1  xi 1 yi  is obtained by
expanding the determinant expression of E ' s area
and canceling out the terms of E and Pn1 common
edge. ■
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Triangulation Complexity
Implementing triangulation as in the existence theorem
 
requires O n4 time.
There are n  n  3 2 diagonal candidates.
Cheking validity of a diagonal requires intersection test
against all edges and previously defined diagonals, which
takes O  n  time.
 
This is repeated n  3 times, yielding total O n4 time.
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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?
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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.
l’
l’
l
l’
l’
not
monotone
monotone
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l’
l’
l’
l’
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We’ll partition P into y-monotone pieces.
start vertex
end vertex
merge vertex
P
split vertex
regular vertex
On horizontal edge the
left vertex is considered
above the right one.
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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.
split vertex
p
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Follow the boundary of P starting
at q, where P is on the left. At
some point r we must cross l.
P
q
r
l
If r ≠ p then the highest
vertex must be a split one.
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If r = p we follow the boundary
from q in opposite direction.
P
pr
merge vertex
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q r
l
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. ■
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Getting Rid of Split and Merge Vertices
Sort P’s vertices from top to
bottom. It takes O(nlogn)
time.
P
Apply scan line from top to
bottom to encounter vertices.
Diagonals are introduced at
split and merge vertices.
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Scan line intersection with P’s
boundary is stored in a left to
right ordered balanced binary
tree.
P
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.
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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).
P
This data suffices to decide
on diagonal in O(1) time.
Merge vertices are handled
similarly.
It takes O(nlogn) time to split
P into monotone polygons.
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Lemma : The added diagonals are non intersecting
and partition P into monotone polygons.
Proof : The resulting partitions contain neither split
nor merge vertices, hence are monotone polygons.
To see that the added diagonals are valid, we'll show
that an added segment niether intersects P's edges nor
previously added diagonals.
We'll consider a split vertex. Merge vertex is similar.
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vm
When vm vi was added it
couldn't intersect P's edge.
Q
ej
vi
ek
If it intersects, there are two cases. If an end vertex
of P's edge is inside Q it contradics vm is nearest to vi .
Otherwise, the edge must cross Q, contrudiction to
e j and ek being the nearest left and right vi 's edges.
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Finally, a previously added diagonal must have its
end vertices above vi . We've showed that Q cannot
have any vertex of P, hence intersection with vm vi
is impossible. ■
Theorem: An n-vertex simple polygon can be partitioned
into y-monotone polygons in O(nlogn) time and O(n)
storage.
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Triangulating y-Monotone Polygon (Sketch)
1
2
3
P's vertices are sorted in
descending y.
4
5
6
7
9
11
10
P
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P's boundary is traversed with
one leg on left and one leg on
right, by poping and pushing
vertices from a stack.
Diagonals insertions are decided
according to stack's top status.
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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.
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