Polygonal estimation of planar convex-set perimeter
from its two projections I
Étienne Baudriera,b,c , Mohamed Tajinea,b,c , Alain Daurat
a
Laboratoire des Sciences de l’Image, de l’Informatique et de la Télédétection, Illkirch,
France
b University of Strasbourg, Strasbourg, France
c Centre National de la Recherche Scientifique, UMR 7005, Illkirch, France
Abstract
This paper deals with the problem of extracting qualitative and quantitative information from few tomographic projections of an object without reconstructing
this object. It focuses on the extraction of quantitative information, precisely
the border perimeter estimation for a convex set from horizontal and vertical
projections. In the case of a multiple reconstruction, lower and upper bounds
for the perimeter are established. In the case of a unique reconstruction, we
give conditions and a method for constructing an inscribed polygon in a convex
set only from the convex-set projections. An inequality on border perimeter
is proved when a convex set is included in another one. The convergence of
the polygon perimeter, when the number of vertices increases, is established for
such polygons.
Keywords: perimeter estimation, convex set, tomography, two projections,
polygonal reconstruction
1. Introduction
This paper addresses geometrical property estimation from convex-set projections without reconstructing the planar convex set. It is known that a convex
I This work was supported by the Agence Nationale de la Recherche through contract
ANR-2010-BLAN-0205-01.
Email addresses: [email protected] (Étienne Baudrier), [email protected]
(Mohamed Tajine)
y
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f
O
Figure 1: A and B have the same projection f but different perimeters.
set can be reconstructed from seven projections and from four projections with
conditions on the angles [2]. So the knowledge of four projections is sufficient for
geometrical property estimation. As the projection preserves area, one projection is sufficient for area estimation. On the other hand, the perimeter cannot
be estimated from only one projection as illustrated Fig. 1: The parallelograms
A and B have the same projection against the vertical axis Oy (a rectangle)
and B has a perimeter arbitrarily large, depending on the shearing between the
sets A and B. Thus we focus on the perimeter estimation from two projections
in this paper. Kuba demonstrates in [4] that the general case of two arbitrary
directions (non collinear) for the projections comes down to the case of orthogonal projections with an affine transformation. Then, without loss of generality,
we focus on the case of orthogonal projections.
A property well studied in this case is the uniqueness of the reconstruction. The characterization of unique reconstruction from two projections has
been studied theoretically in [6, 5]. In [5], Kuba and Volčič give a reconstruction
formula in the unique-reconstruction case and make a link between the multiplereconstruction case and switching elements. Huang and Takiguchi present in [3]
a stability result in the unique-reconstruction case for orthogonal projections.
They also give a reconstruction algorithm in this case. This work has been extended by Takiguchi in [8] to non necessary orthogonal projections. Estimation
of properties (hv-convexity, 4 and 8-connectedness) without reconstruction has
been studied in [1] from a learning method point of view. The paper shows
that the tested properties can be estimated rather properly after training of the
2
learning methods. This piece of information is interesting even if a learning
method cannot give the theoretical reasons that lead its classification. Nevertheless, the perimeter estimation from two projections without reconstruction
is not studied in the literature. Our approach to this problem is based on
polygons because they offers interesting properties for the inclusion and for the
reconstruction. Tomographic reconstruction of polygons has been studied with
a statistical point of view in [7]. Nevertheless, Milanfar et al. use more than
two projections so as to be able to exploit projection moments in their polygon
estimations.
In this paper, we study the convex-set perimeter estimation in both of the
cases: unique reconstruction and multiple reconstructions. The layout is as
follows: in Section 2, the necessary definitions and notations are introduced.
In Section 3, we prove that, for the class of convex sets, the perimeter is an
increasing function relatively to the set inclusion. In Section 4, the multiple
reconstruction case is discussed: a polygon reconstruction algorithm is shown
and bounds for the perimeter are computed. In Section 5, the unique case
is detailed: we propose a perimeter estimation based on approximation of the
convex set by a polygon, the conditions on the projections that imply the convexset inclusion are detailed; polygon reconstruction and a convex-set perimeter
estimation based on a polygon is shown (construction and convergence).
2. Definitions and notations
Our work is based on the notions introduced in [6, 5], thus we define basic
notions and recall the main results of these articles in this section.
Remark 2.1 In all this article, the sets, equalities and inequalities are defined
modulo a set of measure zero (in the sense of the usual Lebesgue measure) and
all the considered functions are measurable. Moreover, all the topological notions
used here are those of the topologies induced by the Euclidean distances.
In the following, we consider a function f : Df 7→ R+ where Df is the definition
domain of f and Df is included in R.
Definition 2.1 (Hypograph, epigraph) The hypograph of f is the set
HG(f ) = {(x, y) | x ∈ Df and y 6 f (x)} and the epigraph of f is the set
EG(f ) = {(x, y) | x ∈ Df and y > f (x)}.
3
Definition 2.2 (Support) The support of the function f : Df 7→ R+ is the
set supp(f ) = {x ∈ Df | f (x) > 0}.
Definition 2.3 (Convexity) Let C be a set of R2 . C is called convex if
∀x, y ∈ C, ∀t ∈ [0, 1], (1 − t)x + ty ∈ C.
A function f is convex if EG(f ) is a convex set.
A function f is concave if −f is convex, which is equivalent to HG(f ) is a
convex set.
Definition 2.4 (Convex hull) Let A ⊆ R2 . The convex hull of A, CH(A) is
the minimal convex set (under inclusion) containing A.
In the following, we consider a second function g : Dg → R+ where Dg is the
definition domain of g and Dg is included in R.
Definition 2.5 (Function comparison) We denote f g if Df ⊆ Dg and
∀x ∈ Df , f (x) 6 g(x).
In sec. 5.1, our work employs the theorems of characterization and reconstruction proposed in [5]. Let us introduce notations and recall the main characterization theorems from [5].
Let C ⊆ R2 such that λ2 (C) < ∞, where λ2 is the usual two-dimensional
Lebesgue measure. Let χC be the characteristic function of C. Let λ1 be
the one-dimensional usual Lebesgue measure. From the Fubini’s theorem, the
projections of χC along the horizontal direction:
∫ ∞
C
(y) =
χC (x, y)dx = λ1 ({x | (x, y) ∈ C})
fX
−∞
and along the vertical direction:
∫ ∞
fYC (x) =
χC (x, y)dy = λ1 ({y | (x, y) ∈ C})
−∞
exist almost everywhere on R and are integrable.
Remark 2.2 The projections of a convex set along horizontal and vertical directions are concave functions.
4
We define in the same way the projections of functions as follows: Let g :
Dg 7→ R+ be a positive integrable function and let S = supp(g). As the sets
{z ∈ S | g(z) > x} are measurable sets for almost all x, then
gY (x) = λ1 ({y ∈ S | g(y) > x})
and
gX (y) = λ1 ({x ∈ S | g(x) > y})
exist for almost all x and y.
Remark 2.3 The functions gX , gY are similar, the only difference is that gY is
seen as a function whose domain is in the axis Ox and gX is seen as a function
whose domain is in the axis Oy.
Similarly, one can define the reprojection functions
gXY (x) = (gX )Y (x) = λ1 ({y | gX (y) > x})
and
gY X (y) = (gY )X (y).
An illustration of the reprojection functions is given Figure 2.
Definition 2.6 A pair of functions v : I1 → R+ , h : I2 → R+ , where I1 (respectively I2 ) is the definition domain of v (respectively h), is called
D
• unique if there exists a unique set D such that D ⊆ I1 × I2 , fX
= v and
D
fY = h,
D
• non-unique if there exists different sets D such that D ⊆ I1 × I2 , fX
=v
D
and fY = h,
D
• inconsistent if there exists no set D such that D ⊆ I1 × I2 , fX
= v and
D
fY = h.
Then we have the following theorem that characterizes these different situations (detail and proofs can be found in [5]):
Theorem 1 (Characterization of unique, non unique and inconsistent projections)
Let v and h be two integrable positive functions such that
∫ ∞
∫ ∞
v(y)dy =
h(x)dx,
−∞
−∞
5
y
y
C
fX
x
x
y
fXY
y
fY
fYXY
x
x
Figure 2: Corollary 2 uses the reprojections fXY X and fY X to characterize the different cases
of reconstruction.
1. v and h are unique if and only if
∫ z
∫
∀z > 0,
hY (x)dx =
0
z
vXY (x)dx,
0
2. v and h are non-unique if and only if
∫ z
∫
∀z > 0,
hY (x)dx >
0
z
vXY (x)dx,
0
and there exists a z for which strict inequality holds
3. v and h are inconsistent if and only if
∫ z
∫ z
∃z > 0,
hY (x)dx <
vXY (x)dx.
0
0
Another characterization of the unique case which will be useful, is first given
in [6]. It is a direct consequence of the first item of Theorem 1:
Corollary 2 (Characterization of the unique case) the set C is determined
C
uniquely by its projections if and only if fXY
and fYCXY are equal to each other.
Example 2.1 Let aX , aY ∈ R+ and a, b, c, d ∈ R such that a ≤ b and c ≤ d.
Consider the functions v = aX χ[a,b] and h = aY χ[c,d] then
gXY = (b − a)χ[0,aX ] ,
and
hY XY = aY χ[0,d−c] .
So, Theorem 2 implies that the pair of functions (v, h) is unique if and only if
R
b − a = aY and d − c = aX and in this case, v = fX
and h = fYR where R is the
rectangle [c, d] × [a, b].
6
In the case of unique reconstruction, there is an explicit formula of reconstruction [5]:
Theorem 3 (Convex-set reconstruction) If C is uniquely determined by its
C
projections fX
, fYC , then
C
C
C = {(x, y) | fYC (x) > fXY
(fX
(y))}.
(1)
In all the following, a planar curve is considered as a function γ : [0, 1] 7→ R2 .
Let define the perimeter length.
Definition 2.7 (Planar curve length) Let γ : [0, 1] 7→ R2 be a planar curve.
The length of γ, l(γ), is defined by
n−1
∑
l(γ) = sup({
d(γ(ti ), γ(ti−1 )) | n ∈ N and
i=0
0 = t0 < t1 < t2 < ... < tn−1 < tn = 1}). (2)
where d is the euclidean distance.
The perimeter of a planar set is then defined as follows:
Definition 2.8 (Perimeter of a planar set) Let C be a planar set such that
there exists γC : [0, 1] → R2 with ∂(C) = γC ([0, 1]) (i.e. γC is a parametrization
of the boundary ∂(C) of C). The perimeter of C is noted l(C) and is defined as
l(C) = l(γC ).
3. Convex inclusion and perimetrical inequality
In the following, straight lines are described by using two parameters (r, θ) ∈
P = R+ × [0, 2π).
So, let D be a straight line and consider the straight line D0 through the origin
and perpendicular to D and the point M such that {M } = D ∩ D0 . Then,
the parameter r (respectively θ) is d2 ((0, 0), M ), the distance of the origin to D
(respectively the angle between D0 and the X-axis) (see Fig. 3 for illustration).
As D depends on the pair (r, θ) in a unique way, then D will be denoted in the
following by D(r, θ).
Theorem 4 (Crofton’s formula) Let γ : [0, 1] 7→ R2 be a planar curve.
Then,
∫
1
nγ (r, θ)drdθ.
l(γ) =
2 P
7
Figure 4: Number of intersections of the
line D = D(r, θ) with the border of the sets
C and K: NγC (r, θ) = 2 and NγK (r, θ) = 4
where γC : [0, 1] → R2 (respectively γK :
[0, 1] → R2 ) is such that ∂(C) = γC ([0, 1])
(respectively ∂(K) = γK ([0, 1])).
Figure 3: Representation of a straight line
where P = R+ ×[0, 2π) and for all (r, θ) ∈ P, nγ (r, θ) = card(γ([0, 1])∩D(r, θ)) ∈
N ∪ {∞} which is the number of intersection points of the curve γ with the
straight line D(r, θ).
Corollary 5 (Perimeter inequality) Let K be a compact subset of R2 such
that there exists γK : [0, 1] → R2 with ∂(K) = γK ([0, 1] and C be a convex
subset in R2 such that C ⊆ K. Then,
l(C) ≤ l(K).
Proof. For any straight line D(r, θ), as C is a convex set, nγC (r, θ) = 0, 1, 2 or ∞.
nγC (r, θ) = ∞ means that C ∩ D(r, θ) is a segment of positive length. As C has
a finite border perimeter, this situation occurs at most a countable number of
times. Thus this situation has measure zero in P.
In the other cases, we have nγC (r, θ) ≤ 2 ≤ nγK (r, θ) (See Fig. 4 for illustration). So, Crofton’s formula (Theorem 4) implies the result.
4. Case of multiple reconstructions
There is no reconstruction formula for the multiple reconstruction case. The
formula of the Theorem 3 used for the reconstruction in the unique case is not
available for the multiple reconstruction case. Without formula, we propose in
Subsection 4.1 a lower and upper bound for the perimeter length in the general
case. Then, in Subsection 4.2 the particular case of polygonal convex sets is
studied. In this case, an algorithm that reconstructs all the convex solutions is
8
y
f
X
D
d
a
C
c
1
b
O
f
O
Y
1
x
Figure 5: Counterexample showing two distinct reconstructions C (red solid line) and
D (green dashed
The perimeters of the parallelograms
are
√ line).
√
√
√ not equal since l(C) =
2(a + b) = 2( 2 + 13) ≈ 10.1 and l(D) = 2(c + d) = 2( 5 + 10) ≈ 10.8.
shown. The maximum and minimum perimeter of these solutions give an upper
and a lower bound for the perimeter of the considered polygonal convex set.
4.1. Estimation of convex-set perimeter
The first question is to know if two projections are sufficient to determine
the perimeter length of a convex set. The answer is negative. We indeed exhibit
Fig. 5 one example where both of the reconstructions are convex and where
these reconstructions do not have the same perimeter. The convex sets C and
D have same horizontal and vertical projections. We detail the computation of
their perimeter hereafter
a
b
c
d
√
√
22 + 32 = 13
√
√
=
12 + 12 = 2
√
√
22 + 1 = 5
=
√
√
=
32 + 1 = 10
=
The perimeter length of the set C and D are:
√
√
2(a + b) = 2( 2 + 13) ≈ 10.1,
√
√
l(D) = 2(c + d) = 2( 5 + 10) ≈ 10.8,
l(C) =
then one has l(C) 6= l(D).
9
(3)
(4)
Thus, one can say that
• all the convex reconstructions do not have generally the same perimeter.
• Moreover, some reconstructions can be non convex sets even if one of them
is a convex set.
Thus, we propose for this case upper and lower bounds.
4.1.1. Lower bound
Definition 4.1 (Steiner’s symmetrized set) Let C be a plane measurable
convex set and fX its horizontal projection. The Steiner’s symmetrized set
symX (C) against Oy is defined by:
1
1
symX (C) = {(x, α) | α ∈ supp(fX ) and − fX (α) 6 x 6 fX (α)}
2
2
We can define similarly the Steiner’s symmetrized set symY (C) against Ox.
Theorem 6 (Perimeter lower bound) Let C be a measurable convex set such
that there exists γ : [0, 1] 7→ R2 a piecewise C 1 parametrization of ∂C and
symX (C) be its Steiner’s symmetrized set, thus one has:
l(symX (C)) 6 l(C)
Proof. One can parametrize C with two concave functions x 7→ f1 and x 7→ f2 .
As C belongs to the class piecewise C 1 , f1 and f2 are concave and piecewise C 1 ,
and thus, the perimeter of C can be defined as
∫ 1
∫ 1√
√
0
||C || =
1 + f102 + 1 + f202
−1
0
For the symmetrized set, its two concave functions are equal and equal to f3 =
(f1 + f2 )/2 and belong to piecewise C 1 . So its perimeter is
∫ 1√
l(symX (C)) = 2
1 + f302
∫
i.e. l(symX (C))
4 + (f10 + f20 )2
=
(5)
0
1√
(6)
0
The following result is well-known:
√
√
√
4 + (a + b)2 6 1 + a2 + 1 + b2 ,
which concludes the proof.
10
Remark 4.1 We can prove similarly that:
l(symY (C)) 6 l(C).
So we have:
max(l(symX (C)), l(symY (C))) 6 l(C).
4.1.2. Upper bound
C
Assume that fX
, fYC belong to the class piecewise C 1 on their support Dx , Dy .
As fYC is equal to zero outside its support Dy and with positive value, there exists
x0 ∈ Dy | fY0C (x0 ) = 0. With a parametrization of C with two concave functions
0
f1 et f2 , one has fYC = f1 − f2 and as fYC (x0 ) = 0, one has f10 (x0 ) = f20 (x0 ).
Moreover, as f1 and f2 are concave, their hypographs are under their tangents.
It implies that C is between two straight lines
D1 : y = f1 (x0 ) + f10 (x0 )(x − x0 )
and
D2 : y = f2 (x0 ) + f20 (x0 )(x − x0 ).
We note d = fYC (x0 ), c = f10 (x0 ) = f20 (x0 ), a = λ1 (Dy ), b = λ1 (Dx ) and we
define a parallelogram
P = {(x, y) | y ∈ Dx and D1 (x) 6 y 6 D2 (x)}.
(
)
√
Then l(P ) = 2 d + a2 + (ac)2 , where c is the only unknown value. By
construction of the parallelogram P , there is necessarily a point of C belonging
to both of the vertical sides of P . So the slope c satisfies |c| 6 (b + d)/a. The
function
(
)
√
g : c 7→ 2 d + a2 + (ac)2
is even and growing on [0, b+d
a ], thus the maximum perimeter is
)
(
√
l(P ) = 2 d + a2 + (b + d)2 .
As C ⊆ P , l(C) 6 l(P ), so one has the following proposition:
11
A
a
y
f1
B
b
O
d
D
f2
C
d
0
x0
C , f C and the
Figure 6: Example of a convex set C, its two orthogonal projections fX
Y
circumscribed parallelogram ABCD. The upper bound of the perimeter is given by the
perimeter of the parallelogram ABCD.
Proposition 1 (Perimeter upper bound) For a measurable convex set C
whose frontier is piecewise C 1 , the perimeter has an upper bound
√
l(C) 6 2(d + a2 + (b + d)2 )
(7)
C
where d = max(fYC ), a = λ1 (supp(fYC )), b = λ1 (supp(fX
)).
One can easily see that the inequality comes down to the circumscribed rectangle
perimeter when d = λ1 (Dx ) (and the slope c = 0) thus this upper bound
is interesting only when d < λ1 (Dx ). Now a unique reconstruction implies
that d = λ1 (Dx ) so Proposition 1 is interesting only in the case of a multiple
reconstruction. An illustration is given Figure 4.1.2.
4.2. Polygonal reconstruction
The underlying idea in Algorithm 4.2 is to construct alternatively the upper
side and the lower side of a polygon solution at a given abscissa thanks to the
value of the vertical projection at this abscissa.
More precisely, the principle of Algorithm 4.2 is to reconstruct sequentially
the vertices of the polygons that are potential solution. The data used are the
coordinates of the breakpoints of the projections provided on input. This is
done through a list of candidate polygons which is separated into two lists: a
list containing the vertices of the upper edge and a list containing the lower-edge
vertices. These lists are divided according to the steps:
12
• (list U vert, list V vert) containing the incomplete polygons,
• (list U vert X, list V vert X) containing the complete polygons and whose
projection X is by construction the one of the input data,
• (list U vert XY, list V vert XY) containing the complete solution polygons (whose projections X and Y are those of the input data).
After the list initialization (lines 1-8), there is a while loop (lines 10-29) in which
the polygons are constructed. The current step of the loop for a candidate polygon is to look, from its vertices already constructed and the vertical projection
of the last built summit, vertices satisfying the following constraints:
• the projections of the vertices already built are equal to those of the input
vertical projection,
• the built polygon is convex.
For each vertex satisfying these constraints, a polygon is created in the current
list. By cons, if no vertex is appropriate, the corresponding candidate polygon
is removed from the current list. When all breakpoints of the vertical projection
have been run, it is verified that the built polygons have a horizontal projection
equal to the input data.
Remark 4.2 Algorithm 4.2 is designed to reconstruct only solutions which are
convex polygons and thus all the non convex solutions can not be reconstructed
with this algorithm.
At each step of Algorithm 4.2, new partial polygons are created and existent
ones are deleted. The maximal number of polygons created at each step is
bounded by m × (n − i). So, the complexity is a O(mn n!). In practice, there is
a complexity collapse after one or two steps, in general, due to the fact that few
partial polygons can be increased under the constraints of the vertex position
and the projection equality to projx . Thus the experiments on studied examples
show a complexity close to O(m2 n2 ).
The estimation of a convex set perimeter from a multiple pair of functions can
be done by constructing all the possible convex reconstructions with Algorithm
13
Algorithm 1 Multiple case: polygonal reconstruction from vertex projections
1: (a,b)= leftmost breakpoint of projy
2: for sk ordinate(projx ) do
. Initialization of the U and V vertex lists
3:
if sk+1 − sk = b then
4:
list U vert={(sk ,a)}
5:
list V vert={(sk+1 ,a)}
6:
end if
7: end for
8: list U vert X={}, list V vert X={} . Initialization of the U and V vertex
lists having projX as vertical projection
9: list U vert XY={}, list V vert XY={}
. Initialization of the U and
V vertex lists having projX as vertical projection and projY as horizontal
projection
10: count=0
11: while list U vert not empty do
12:
if max(absc{U-vertices})=max(absc{L-vertices})=max(absc{projX })
then
13:
list U vert X=list U vert X∪list U vert(count)
14:
list V vert X=list V vert X∪list V vert(count)
15:
count=count+1
16:
if max(abscissa{U-vertices})>max(abscissa{L-vertices}) then
17:
find all the vertices v satisfying the convexity and such that all
the projections of the polygon U ∪ V are equal to projX
18:
if no v satisfying these conditions then
. suppress the
corresponding lists
19:
list U vert=list U vert\ {list U vert(count)}
20:
list V vert=list V vert\ {list V vert(count)}
21:
end if
22:
for each v do
. add the new vertices
23:
list V vert(count)=list V vert(count)∪{v}
24:
count=count+1
25:
end for
26:
else
27:
reverse list U vert and list V vert and run the former case
28:
end if
29:
end if
30: end while
31: for U in list U vert X do
32:
compute the horizontal projection PY of the polygon from U and the
corresponding V in list V vert X
33:
if PY = projY then
34:
list U vert XY=list U vert XY∪U
35:
list V vert XY=list V vert XY∪V
36:
end if
37: end for
14
x
b
D
f XD
O
f XC
C
f XC
f XD
a
aY
DC
aX
fXYC
fXYD(aX)=aY
f YC
f YD
O
aX
Figure 8: Example of a rectangle inscribed
in a convex set. In the case of unique reconstruction, its projection and reprojecD f C and f D
C
tion satisfy fX
X
XY fXY .
Figure 7: Counterexample showing that inequalities on projections do not imply the
convex-set inclusion
4.2 and by computing the maximum and the minimum perimeter lengths. A
perspective is to study whether all the perimeter have equal length or not.
5. Case of unique reconstruction
5.1. Projection inclusion and convex-set inclusion
The question addressed here is to know whether there is a link between the
variation of horizontal and vertical projections and the inclusion of convex sets
themselves on the other. One implication is immediate:
Proposition 2 (inclusion, forward direction) Let C and D be two measurable convex sets of R2 . Then,
D
C
D ⊆ C =⇒ (fX
fX
∧ fYD fYC ).
The converse of Proposition 2 is not true in general. If we also assume that
C
D
one has concave projections satisfying fX
fX
and fYD fYC , one cannot
deduce that C and D are convex and D ⊂ C. Fig. 7 gives a counterexample.
D
C
Although projections are concave and satisfy fX
fX
and fYD fYC , D is not
included in C.
In the following, we prove that the converse is true under certain conditions
given in Theorem 8.
15
P
C
Figure 9: A polygon P inscribed in a convex set C.
To prove this theorem, we will first consider the case where the convex set
D
C
D is a rectangle such that fX
fX
or fYD fYC . Next, we will prove the result
in the case where C and D are both rectangles. Then, we use a decomposition
of D in union of rectangles in order to prove the last result.
Remark 5.1 The idea of using rectangles as a first step is similar to the use
of measurable inscribed sets made in [4] and in [3]. The properties obtained in
[3] are about a L1 L∞ stability and do not directly lead to the convergence of the
perimeter. Indeed, set-convergence doe not imply a priori the convergence of the
set characteristics. Thus it cannot be used straightly in our study.As rectangles
are particular polygons, the use of rectangles is nevertheless more consistent with
the following propositions using polygons than the use of measurable inscribed
sets.
Let us first define an inscribed polygon.
Definition 5.1 (Inscribed polygon) A polygon P is called inscribed in a convex set F if its vertices belong to ∂(F ). A piecewise affine function f is called
inscribed in a concave function g if HG(f ), which is a polygon, is inscribed in
the convex set HG(g).
Figure 9 gives an illustration of an inscribed polygon. In the following proposition, we consider a particular polygon, the rectangle.
Proposition 3 (Case of the rectangle with an included projection) Let
C
fX
, fYC be unique projections of a convex set C and g = aX χ[a,b] be an inscribed
C
function in the function fX
. Then there exists a rectangle D whose sides are
D
parallel to the projection directions such that fX
= g and fYD is inscribed in fYC .
Moreover the rectangle D is inscribed in C.
Fig. 8 illustrates Proposition 3.
C
Proof (Proposition 3). Let note CX = supp(fX
) and CY = supp(fYC ). The
conditions of Proposition 3 are equivalent to
16
C
• fX
, fYC are unique,
C
• g fX
,
C
C
• ∃a, b ∈ CX , a < b and g(a) = fX
(a) = g(b) = fX
(b) = aX > 0,
As projections are unique, according to Corollary 2, we must have:
C
fXY
= fYCXY .
By definition,
C
C
fXY
(aX ) = λ1 ({y | fX
(y) > aX }),
then
C
fXY
(aX ) = b − a
C
because fX
is a concave function. Put
aY = b − a
then,
fYCXY (aX ) = aY .
This means that
λ1 ({y | fYCX (y) > aX }) = aY
and as fYCX is decreasing, it means
fYCX (aY ) = aX ,
i.e.
λ1 ({x ∈ CX | fYC (x) > aY }) = aX .
(8)
As fYC is concave, ∃c, d ∈ CY such that c 6 d and fYC (c) = fYC (d) = aY . By (8),
d − c = aX . Therefore the rectangular function h : x 7→ aY χ[c,d] (x) is inscribed
in fYC . Thus, the rectangle D = [a, b] × [c, d] is inscribed in C and its projections
D
are fX
= g and fYD = h. Moreover fYD is inscribed in fYC .
We can see that (under the hypothesis of Proposition 3), fYD is inscribed
into fYC by construction. We introduce a definition reflecting this property:
17
C
D
Definition 5.2 (P-inscribed) Let fX
, fYC and fX
, fYD be projections respectively of a measurable convex set C and a rectangle D whose sides are parallel
D
to the directions of projection. We say that D is p-inscribed in C if fX
(resp.
D
C
C
fY ) is inscribed in fX (resp. fY ).
The letter ”p” in p-inscribed comes from ”projection”. A consequence of Proposition 3 concerns the projection support length:
C
, fYC be unique projections of a convex set C, then
Corollary 7 Let fX
C
λ1 (supp(fX )) = sup(fYC ) and reciprocally.
n
C
Proof. Define a sequence of rectangular functions (gX
)n∈N∗ inscribed in fX
with a height
1
n
then
n
C
lim λ1 (supp(gX
)) = λ1 (supp(fX
)).
n→∞
(9)
n
From Proposition 3, each gX
for n ∈ N∗ corresponds to a rectangle Dn such that
fYDn is inscribed in fYC . Now λ1 (supp(fYDn )) =
then
max(fYDn )
1
n
tends to zero when n → ∞,
C
tends to max(fYC ). With (9), we deduce that λ1 (supp(fX
)) =
C
sup(fYC ). The result holds if fX
and fYC are switched.
We can now extend Proposition 3 to the case where D is a subset of a
rectangle inscribed in C. For this, we introduce the following definition
Definition 5.3 (P-sub-inscribed) D is called p-sub-inscribed in C if there is
a rectangle D0 such that
• D ⊆ D0 and
• D0 is p-inscribed in C.
This definition implies (Proposition 2) that the projection of D is less than those
of D0 .
D
C
Proposition 4 (Inclusion and p-sub-inscribed set) Let fX
, fYC , fX
, fYD be
unique projections such that D is p-sub-inscribed in C. Then
D ⊆ C.
18
Proof (Proposition 4). From Def. 5.3, there is D0 such that
0
0
0
0
D
D
D
C
fX
fX
and fX
fX
,
fYD fYD and fYD fYC .
Thus, by Proposition 3, we know that D0 ⊂ C. On the other hand, as projections
of D0 are rectangular functions satisfying
0
0
D
D
fX
fX
, fYD fYD ,
one has D ⊆ D0 so D ⊆ C
It is now possible to write a general theorem for the reciprocal of Proposition
2.
C
D
Theorem 8 (Reciprocal inclusion) Let fX
, fYC , fX
, fYD be unique projections
such that the set D is the convex hull of a union of sets (Di )i∈I p-sub-inscribed
in C, then
D ⊆ C.
In the counterexample shown Fig. 7, the convex set D in not the convex
hull of union of p-sub-inscribed sets: none of the points of C\D can be included
in a p-sub-inscribed set.
Proof (thm 8). As Di is p-sub-inscribed in C for all i ∈ I, we have by Proposition 4 Di ⊆ C for all i ∈ I so,
∪
Di ⊆ C,
i∈I
thus,
(
D = CH
∪
)
Di
⊂ CH(C) = C.
i∈I
Corollary 9 (Inscribed polygon) Let D1 , . . . , Dn be rectangles p-sub-inscribed
in a measurable convex set C and
Di
fX = CH(∪i∈I HG(fX
)), and fY = CH(∪i∈I HG(fYDi )).
Then these projections are unique and reconstruct a polygon inscribed in C.
19
Figure 10: from the projections of the convex set C, Theorem 9 allows one to say that the
convex hull of two p sub-inscribed sets D1 , D2 is included in C.
In other words, the projections of the generated polygon can be computed
straightly from the projections of the rectangles D1 , . . . , Dn .
Figure 10 illustrates Corollary 9 by showing the construction of a polygon
included in a convex set C only from the projections of C. The build polygon
is the convex hull of two p sub-inscribed sets D1 and D2 .
Proof. From Theorem 8, the rectangles D1 , . . . , Dn are unique and included
in the convex set C and their convex hull D = CH(∪i∈I Di ) is included in C.
Let us show that the projections of D are unique. As D1 , . . . , Dn are unique,
the set ∪i∈I Di is unique. Then, by Corollary 2, one has
∪i∈I Di
i∈I Di
fXY
= fY∪XY
.
The projections of this set ∪i∈I Di are piecewise constant. The projections of
D are piecewise linear and share the same breakpoints as the projections of
∪i∈I Di . Then, one has
D
fXY
= fYDXY ,
D
and by Corollary 2, the projection pair fX
, fYD of D is unique.
As a projection of a convex set is a concave function, its hypograph is a
convex set. In order to have the equivalent of the convex hull for the projections,
we focus on the hypograph of the projections. As the projection keeps the
20
straight lines, the projection of the convex hull of sets is the convex hull of the
hypographs of the sets. Then one has
CH(∪i∈I Di )
fX
CH(∪i∈I Di )
Di
= CH(∪i∈I HG(fX
)) and fY
= CH(∪i∈I HG(fYDi )),
i.e.
Di
D
)) and fYD = CH(∪i∈I HG(fYDi )).
= CH(∪i∈I HG(fX
fX
The results of this section show that the convex polygons can be used for
approximating the perimeter. The next subsection exhibits efficient algorithms
to construct convex polygons.
5.2. Polygonal reconstruction
Remark 5.2 It is possible to identify a projection of a polygonal convex set to
a convex polygon.
Assume that the convex set P is a polygon with n vertices. Its vertical and
horizontal projections fY and fX are then concave linear piecewise functions.
Moreover, they are equal to zero out of a finite domain. Then, their hypographs
EG(fY ), EG(fX ) are convex polygons to which the projections can be identify.
Thus, the projection fX (respectively fY ) can be represented by its set of
breakpoints sfX (respectively sfY ):
sfX
= {(y1 , q1 ), . . . , (ym , qm )},
(10)
sfY
= {(x1 , p1 ), . . . , (xk , pk )}.
(11)
A breakpoint of a projection is the projection of at least one vertex of the
polygon and at most two vertices of the polygon if they are aligned along the
projection direction. The alignment of two vertices in a direction is not linked
to the alignment of vertices in the other direction. Then the numbers of breakpoints m, k of the projections sfX , sfY satisfy n2 ≤ m, k ≤ n.
Remark 5.3 sfX , sfY can be seen as sets of vertices if fX , fY are seen as polygons
Let’s see if the vertex coordinates allows us to express Theorem 3.
21
For fXY : the function fXY is the projection of fX on the x-axis. So the
set of its breakpoints is (qi )16i6m−1 reordered: α1 = qσ(1) , . . . , αk = qσ(k) with
k 6 m − 1 and the corresponding values r1 , . . . , rk can be calculated from fX .
We assume them to be known.
fXY : {(α1 , r1 ), . . . , (αk , rk )}.
For the composed function fXY (fX (y)): fX takes values in the set of (qi )16i6m
which are the abscissas of the breakpoints of fXY . Thus there is a permutation
σ2 ∈ Sm such that
fXY ◦ fX : {(y1 , qσ2 (1) ), . . . , (ym , qσ2 (m) )}.
(12)
This allows us to reformulate Theorem 3:
Theorem 10 (polygon vertex reconstruction) Let sfX : {(y1 , q1 ), . . . , (ym , qm )},
sfY : {(x1 , p1 ), . . . , (xn , pn )} be the given vertices of the unique projections of a
convex polygon. Then there exists a permutation σ2 ∈ Sm such that
sfXY ◦ sfX : {(y1 , qσ2 (1) ), . . . , (ym , qσ2 (m) )}
and the polygon is then given by:
CH{(xi1 , yj1 ), . . . , (xil , yjl )},
where
∀k ∈ {1, . . . , l}, sfY (xjk ) > sfXY ◦ sfX (yjk ).
Thus we can define a reconstruction algorithm (Algorithm 5.2) based on the
given vertices.
Algorithm 2 Unique case: polygonal reconstruction from vertex projections
1: from vertices sfX of the projection fX , compute vertices of sfXY ◦ sfX
2: for i = 1 to m do
3:
for j = 1 to n do
4:
t ← sfY (xi ) − sfXY ◦ sfX (yj )
5:
if t > 0 then
6:
L ← L ∪ (xi , yj )
7:
end if
8:
end for
9: end for
10: P ← CH(L)
. Computation of the convex hull
The computational complexity of Algorithm 5.2 is detailed here:
22
• the computation of sfXY ◦ sfX can be done in O(m)
• the points in the polygon can be computed in O(mn)
• the convex hull can be computed in O(mn log mn)
This gives an overall complexity of O(mn log mn).
It is possible to improve the complexity by looking for vertices in the neighbourhood of those already found, through the projections values (with the notations of (10) and (11)) (pi )16i6m , (qi )16i6m . Using the projections values
allows us to find the points on the polygon border straightly. This suppresses
the step of convex hull research, so the computational complexity can be reduce
to a O(mn).
5.3. Estimation of convex-set perimeter
C
Let C be a convex set such that fX
, fYC are unique projections and P be a
convex polygon which is inscribed in C. Let V(P ) = {p0 , . . . , pn−1 } be the set
of vertices of P with pn = p0 . We first give an upper bound of the difference
between the perimeters of C and P then we present a limit for the perimeter of P
when the number of vertices of P tends to infinity. Let us introduce the following
notations: let γK : [0, 1] → R2 a parametrization of ∂K where K = C or P . As
P is inscribed in C we can define (t0 , . . . , tn ) such that 0 = t0 < t1 < . . . < tn = 1
and γK (t0 ) = p0 , . . . , γK (tn ) = pn for K = C, P . We define
• Ki = γK ([ti , ti+1 ]) for i 6 n − 1
• IX = {i | Pi parallel to Ox}.
• IY = {i | Pi parallel to Oy}.
• I = {1, . . . , n}\(IX ∪ IY )
Let i ∈ (I ∪ IX ), we define
K
K
• FX,i
= {(y, fX
(y)) | y ∈ [ypi , ypi+1 ]} where K = C, P
P
• the affine transformation AX,i determined by A(FX,i
) = Pi .
23
Figure 11: The notations for Theorem 11 are presented on a part of the convex set C and the
convex P
• the convex set Di = CH(Pi ∪ Ci ))
C
• the convex set MX,i = CH(Pi ∪ A(FX,i
))
K
In the same way, we also define IY and for i ∈ (I ∩ IY ), FY,i
, Ki , AY,i , MY,i , Di .
Fig. 11 illustrates the notations introduced above.
Theorem 11 We have the following inequality:
∑
∑
(
) ∑
C
C
C
C
l(C) 6
min l(A(FX,i
)), l(A(FY,i
)) +
l(FX,i
)+
l(FY,i
).
i∈I
i∈IX
i∈IY
Remark 5.4 This upper bound does not depend on reconstruction of the convex
set C but only on its projections.
Lemma 12 ∀i ∈ I ∪ Iaxis , Di ⊆ Maxis,i with axis = X, Y
Proof (Lemma12). To simplify the notation, we take axis = Y . Let i ∈ I ∪IY
and x0 ∈ supp(fYP ). As P ⊆ C and C is convex, (C\P ) ∩ {(x0 , y) ∈ R2 | y ∈ R}
is composed of at most two segments (which can be empty) that we note du0
and dl0 : (C\P ) ∩ {(x0 , y) ∈ R2 | y ∈ R} = du0 ∪ dl0 ; u (resp. l) denotes the
upper (resp. lower) segment position. Then the projection can be decomposed
as follows:
fYC (x0 ) = fYP (x0 ) + l(du0 ) + l(dl0 ).
24
For sake of simplicity, we suppose that the vertical position of the polygon edge
Pi is on the upper side of the polygon. By construction,
P
(Di ∩ {(x0 , y) ∈ R2 | y ∈ R}) = {(x0 , y) | y ∈ (du0 + A(fX
(x0 )))}
and
P
(MY,i ∩ {(x0 , y) ∈ R2 | y ∈ R}) = {(x0 , y) | y ∈ (du0 ∪ dl0 + A(fX
(x0 )))},
then
∀x0 ∈ supp(fYP ), Di ∩ {(x0 , y) ∈ R2 | y ∈ R} ⊆ MY,i ∩ {(x0 , y) ∈ R2 | y ∈ R},
so
Di ⊆ MY,i .
Let us prove the following theorem:
Proof (Theorem 11). Let i ∈ I ∪ IY . Lemma 12 implies Di ⊆ MY,i . Then
by Cor. 5,
l(Di ) 6 l(MY,i ),
i.e.
l(Ci ) + l(Pi ) 6 l(A(FY,i )) + l(Pi ),
then
l(Ci ) 6 l(A(FY,i )).
Thus if i ∈ I, this inequality holds with axis = X and so
l(γC ([ti , ti+1 ]) 6 min (l(A(FY,i )), l(A(FX,i )))
For i ∈
/ I, the edge γP ([ti , ti+1 ]) is parallel to an axis, e.g. Ox. Then the
transformation A comes down to a translation that conserves the length. Thus
l(A(FY,i )) = l(FY,i )
To demonstrate the polygon perimeter convergence toward the convex-set
perimeter, we have to establish the following lemma.
25
C
Lemma 13 Let fX
, fYC be two unique projections of a convex set C and let
P
P
fX , fY be the unique projections of polygon P included in C. Then it is possible
C
P
to construct only from fX
, fYC , fX
, fYP a polygon Q such that P ⊆ Q ⊆ C.
Proof (Lemma 13). Let px , py the coordinates of a breakpoint p of the pieceP
P
C
wise linear function fX
. As P is included in C, fX
fX
. Let us define fX a
C
C
rectangular function inscribed in fX
such that fX (px ) = fX
(px ). From PropoR
sition 3, there exists an inscribed rectangle Rp in C such that fX p = fX . As
Rp is inscribed in C, it contains all the points of C whose abscissa is px and in
particular, Rp contains the point p.
Let us define Q = CH(∪p Rp ) where p are the breakpoints of P . Then Q
is a convex polygon inscribed in C containing all the breakpoints of P. As Q is
convex, it also contains P .
C
Theorem 14 (Perimeter convergence) Let fX
, fYC be a unique pair of projections of a convex set C then it exists (Pn )n∈N a sequence of inscribed polygons
in C such that
lim l(Pn ) = l(C)
n→∞
Proof (theo. 14). There exists γC : [0, 1] 7→ R2 with ∂(C) = γC ([0, 1]).Then,
l(γC ) = l(∂(C))
and by Def. 2.7,
l(γC ) = sup({
n−1
∑
d(γC (ti ), γC (ti−1 )) | n ∈ N and
i=0
0 = t0 < t1 < t2 < . . . < tn−1 < tn = 1}). (13)
For each (t0 , . . . , tn ) such that 0 = t0 < t1 < . . . < tn = 1, the points
{γC (t0 ), . . . , γC (tn )} define the vertices of an inscribed polygon Pt0 ,...,tn in C
which satisfies
l(Pt0 ,...,tn ) =
n−1
∑
d(γC (ti ), γC (ti−1 ))
i=0
thus
l(C) = sup(l(Pt0 ,...,tn ) | n ∈ N and 0 = t0 < t1 < ... < tn = 1}).
26
Now, by Lem. 13,
∀(t0 , . . . , tn ) such that 0 = t0 < . . . < tn = 1,
∃Qt0 ,...,tn | Pt0 ,...,tn ⊆ Qt0 ,...,tn ⊆ C
(14)
where Qt0 ,...,tn is constructible from the projections of C and Pt0 ,...,tn
Then, by Cor. 5,
l(Pt0 ,...,tn ) 6 l(Qt0 ,...,tn )
As Qt0 ,...,tn is a polygon inscribed in C, one has also
l(Qt0 ,...,tn ) 6 l(C)
Thus
l(C) =
sup
(0=t0 <...<tn =1),n∈N∗
l(Pt0 ,...,tn ) 6
sup
(0=t0 <...<tn =1),n∈N∗
l(Qt0 ,...,tn ) 6 l(C)
(15)
This theorem leads to the following corollary.
C
Corollary 15 (polygonal convergence) Let fX
, fYC be two unique projections of a convex set C then it exists (Pn )n∈N a sequence of inscribed polygons
in C such that
lim dH (Pn , C) = 0
n→∞
where dH is the Hausdorff distance
Proof. Let P an inscribed polygon in C. We compute first a distance for an
edge of P . Let a = [pi , pi + 1] with i ∈ {0, . . . , n − 1}, an edge of P . As P is
inscribed in C, pi and pi+1 are included in C. Then there exists ca an arc of
C between the points pi and pi+1 satisfying ca ∩ P = {pi , pi+1 }. Let us find
h the upper bound of the distance between the edge a and the arc ca . This
upper bound is obtained when ca is composed of two straight lines and a quick
computation gives the upper bound h
√
l(ca )2 − l(a)2
h=
.
2
27
We can now compute the Hausdorff distance for the whole polygon P . Let us
introduce for this Edg(P ) = {[pi , pi+1 ], i ∈ {0, . . . , n − 1} the set of the edges of
the polygon P . By definition,
dH (P, C) = max(max(d(x, C)), max(d(x, P )))
x∈P
x∈C
As P is inscribed in C, it comes down to
dH (P, C)
=
max
dH (a, ca )
√
l(ca )2 − l(a)2
<
max
2
a∈Edg(P )
√
√
(l(ca ) + l(a))
<
max
(l(ca ) − l(a))
2
a∈Edg(P )
(16)
a∈Edg(P )
(17)
(18)
As P is inscribed in C, one has
l(C) − l(P ) =
∑
l(ca ) − l(a)
a∈Edg(P )
From Theorem 14, it exists (Pn )n∈N a sequence of inscribed polygons in C such
that
l(C) − l(Pn ) =⇒ 0
n→∞
then, for such a sequence,
l(can ) − l(an ) =⇒ 0
n→∞
(19)
where an is an edge of Pn . From (18) and (19), one deduces
dH (Pn0 , C) =⇒ 0
n→∞
Thus, in the case of unique reconstruction, an estimation of the convex-set
perimeter is made thanks to a polygonal approximation.
6. Conclusion
In this paper, we present results regarding perimeter inequality and convexset inclusion. In the case of unique reconstruction, we find conditions on the
28
projections to construct an inscribed polygon. This has led to an algorithm that
constructs polygon projection such that the polygon is included in the convex
set. The polygon reconstruction has also been detailed and gives interesting
computational complexity for few vertices. An upper bound is exhibited for the
difference between the perimeter of the convex set and the polygon. Moreover,
the convergence toward the convex set perimeter has been proved. The case
of multiple reconstruction is more difficult to tackle because no reconstruction
formula is available. In this case, we propose lower and upper bounds for the
perimeter.
We show in this paper that two projections are not sufficient in general to
determine the perimeter of a convex set. An open question is: are three projections sufficient to determine the perimeter of a convex set?
Acknowledgements: Alain Daurat, co-author of this article, died on June
the 25th, 2010. This article is dedicated to his memory.
[1] M. Gara, T. S. Tasi, and P. Balázs. Learning connectedness and convexity of
binary images from their projections. Pure Mathematics and Applications,
20:1-2:27–48, 2009.
[2] R.J. Gardner. Geometric tomography, volume 58 of Encyclopedia of Math
and its App. Cambridge university press, 1995.
[3] L. Huang and T. Takiguchi. The reconstruction of uniquely determined
plane sets from two projections. Journal of Inverse and Ill-Posed Problems,
6(3):217242, 1998.
[4] A Kuba. Reconstruction of measurable plane sets from their two projections
taken in arbitrary directions. Inverse Problems, 7(1):101, 1991.
[5] A. Kuba and A. Volčič. Characterisation of measurable plane sets which
are reconstructable from their two projections. Inverse Problems, 4(2):513,
1988.
29
[6] G. G. Lorentz. A problem of plane measure. Am J Math, 71(2):pp. 417–426,
1949.
[7] P. Milanfar, W. C. Karl, and A. S. Willsky. Reconstructing binary polygonal
objects from projections: A statistical view. CVGIP: Graph. Mod. and Im.
Proc., 56(5):371–391, 1994.
[8] T. Takiguchi. Reconstruction of the measurable sets in the two dimensional plane by two projections. Journal of Physics: Conference Series,
73(1):012022, 2007.
30