(a) Only u
3
intersects polygon(1, 2).
(b) Both u
3
and l
3
intersect
polygon(1, 2).
(c) Only l
3
intersects polygon(1, 2). (d) polygon(1, 3) is included in
band(4).
Fig. 2. The possible relationship between band( j+ 1) and polygon(i, j) when
band( j + 1) ∩ polygon(i, j) , Ø.
When the first data point (1, x[1]) arrives, in order to satisfy
the error-bounded requirement for (1, x[1]), we have
x[1] − ε ≤ a · 1 + b ≤ x[1] + ε, (4)
where f (t) = a · t + b is the representation function for the
current segment, and the area described by Equation (4) is
band(1), as shown in Figure 1(a). When (2, x[2]) arrives, for
guaranteeing that the vertical distance of (1, x[1]), (2, x[2]) to
the representation function f (t) = a · t + b are within ε, we
obtain
(
x[1] − ε ≤ a · 1 + b ≤ x[1] + ε,
x[2] − ε ≤ a · 2 + b ≤ x[2] + ε.
(5)
The area represented by the above two inequalities (
polygon(1, 2)) is bounded by a parallelogram (as shown in
Figure 1(b)), and the equations of four edges on the parallel-
ogram are as follows.
a + b = x[1] − ε,
a + b = x[1] + ε,
2a + b = x[2] − ε,
2a + b = x[2] + ε,
(6)
When (3, x[3]) arrives, we check whether band(3) intersects
polygon(1, 2). If so, it means that the three data points
(i, x[i]), 1 ≤ i ≤ 3 can be approximated by a representation
function. If not, it means that the three data points (i, x[i]), 1 ≤
i ≤ 3 cannot be approximated by a representation function. For
the former, there are three cases shown as follows.
1) Only the line u
3
(3a + b = x[2] + ε) intersects
polygon(1, 2), as shown in Figure 2(a).
2) Only the line l
3
(3a + b = x[2] − ε) intersects
polygon(1, 2), as shown in Figure 2(c).
3) Both u
3
and l
3
intersect polygon(1, 2), as shown in
Figure 2(b).
When ((4, x[4])) arrives, there exists the fourth case except
the above three similar cases, that is, polygon(1, 3) ⊂ band(4)
Fig. 3. v
1
, v
3
and v
4
are all the vertices of polygon(i, j), and v
3
, v
4
are two
vertices adjacent to v
1
.
as shown in Figure 2(d). When subsequent data points arrives,
we cannot find the fifth different case for band( j + 1) ∩
polygon(i, j). Therefore, there are at most four different cases
when band( j + 1) ∩ polygon(i, j) , Ø. For the previous three
cases, as shown in Figures 2(a), 2(b) and 2(c), we observe
that when checking whether the FSS for (3, x[3]) intersects
polygon(1, 2), we only decide whether the points A and D (as
shown in Figure 2(a)) are on different sides on the line u
3
or
l
3
, where the point A and D are both vertices of polygon(1, 2),
and A is the vertex with the maximum b-coordinates of the
vertices of the parallelogram ABCD, while D is the vertex with
the minimum b-coordinates of the vertices of the parallelogram
ABCD. Based on the above observation, we have a guess that
we only need to check two points A and D to decide whether
the newly arrived data point can be compressed together with
the data points seen so far but not compressed, where A is
the vertex with the maximum b-coordinates of the vertices of
the polygon R (the boundary of the FSS for the data points
seen so far but not compressed), and D is the vertex with
the minimum b-coordinates of the vertices of the polygon R.
Before validating our guess, we first present two special points
shown as follows.
Definition 3 (Upper Vertex). Given the boundary of
polygon(i, j), a vertex v
1
is the upper vertex(UV) of
polygon(i, j), if it satisfies that the b-coordinate of v
1
is greater
than that of any other vertex of polygon(i, j).
Definition 4 (Lower Vertex). Given the boundary of
polygon(i, j), a vertex v
2
is the lower vertex (LV) of
polygon(i, j), if it satisfies that the b-coordinate of v
2
is less
than that of any other vertex of polygon(i, j).
For the upper vertex and the lower vertex with respect to
polygon(i, j), we have the following lemma.
Lemma 2. Given the polygon(i, j) (i < j), the upper vertex
v
1
and the lower vertex v
2
of polygon(i, j). Then, when ( j +
1, x[ j + 1]) arrives, for a line L
0
: b = −a( j + 1) + m (when
m = x[ j + 1] + ε, L
0
= u
j+1
; m = x[ j + 1] − ε, L
0
= l
j+1
) in
ab-coordinate system, if m increases from −∞ to +∞, then
v
1
is the f irst point that L
0
encounters in polygon(i, j), and
v
2
is the last point that L
0
encounters in polygon(i, j), and
therefore we have
L
0
∩ polygon(i, j) , Ø ⇔ L
0
∩ v
1
v
2
, Ø, (7)
where v
1
v
2
is the line segment connecting v
1
and v
2
.
Proof: It is easy to show that polygon(i, j) is a convex
polygon by mathematical induction. Then, if m increases from