The Two-Ears Theorem
The Two-Ears Theorem was developed and proven by Gary H. Meisters at the
University of Nebraska in 1975
[4].
The Two-Ears Theorem:
Proof #1: By Gary H. Meisters
The proof by Meisters is by induction on the number of vertices, n,
in the
simple polygon
P. It is quite elegant.
Base Case:
n = 4. The simple polygon P is a quadrilateral, which has
two ears.
Induction:
Let P be a simple polygon
with at least 4 vertices. Vertex pi is a vertex of P
where the interior angle formed by pi-1,
pi, and pi+1 is less than
180o.
Case 1:
Polygon P has an ear at pi. If this
ear is removed, the remaining polygon P' is a simple polygon with more
than three vertices, but with one less vertex than P. By the induction
hypothesis, there are two non-overlapping ears E1 and
E2 for P'. Since they are non-overlapping, at least
one of these two ears, say E1, is not at either of the
vertices pi-1 or pi+1.
Since all ears of
P' are also ears of P, polygon P has two ears:
E1 and the ear at pi.
Case 2:
Polygon P does not have an ear at
pi. So, the triangle formed by
pi-1,
pi, and pi+1 contains at least
one vertex
of P in its interior. Let q be the vertex in the interior of
this triangle such that the line L through q and parallel to
the line
through pi-1 and pi+1 is
as close to pi as possible.
Let a and b be the points of intersection of line L
with the polygon edges (pi,
pi+1) and (pi-1,
pi),
respectively. The
triangle formed by pi, a, and b does not
contain in its interior any vertex of P (or else our choice of vertex
q would be incorrect).
The line segment Q from vertex q to pi
can be
constructed without intersecting any edges of P. Line segment Q
divides P into two simple polygons: P1 (that contains
vertices pi, pi+1,..., q)
and P2 (that contains vertices pi,
pi-1,..., q).
There are now two sub-cases to consider:
Case 2a:
Polygon P1 is a triangle. So, P1 is
an ear for polygon P. By the induction hypothesis, polygon
P2 must have at least two non-overlapping ears,
E1 and E2 (or else it too would be a
triangle and polygon P would have only four vertices).
Since they are non-overlapping, at least one, say E1, is at
neither vertices pi nor q. E1 does
not overlap with the ear formed by polygon P1, so it is the
second ear in polygon P.
Case 2b:
Polygon P1 is not a triangle. So, by the induction
hypothesis, polygon P1 has two ears, E11
and E12 and polygon P2 has two ears,
E21 and E22. Since they are
non-overlapping, at least one of the ears in P1, say
E11, is not at
vertex pi or q. Similarly, at least one of the ears
in P2, say E21, is not at vertex
pi or q. These two ears, E11 and
E21 will be non-overlapping ears for polygon P
Q.E.D.
It is known that a simple polygon can always be
triangulated.
Leaves in the
dual-tree
of the triangulated polygon correspond to ears and
every tree of two or more nodes must have at least two leaves.
Q.E.D.
Some Examples:
This page was last updated on Wednesday, December 10th,
1997.
© 1997 Ian Inc.