Math project reconstruction of a two dimensional object from its shadow

Nguyễn Hữu Duy, ID: 2352185 Email: duy.nguyen141205@hemut.edu.vn Role: Information provider and information seeker, study the theory concepts in textbook.. Pham Die Anh, ID: 2352060 Emai

FACULTY OF COMPUTER SCIENCE AND ENGINEERING

BK

TP.HCM

Calculus 2

Math Project

Reconstruction of a two -

dimensional object from its

shadow

Advisor(s): Phùng Trọng Thực Student(s): Trần Minh Duân ID 2352166

Thái Thiên Tường ID 2353303

Nguyễn Hữu Duy ID 2352185 Trần Hùng Sơn ID 2353055

Phạm Dức Anh ID 2352060

HO CHI MINH CITY, APRIL 2024

Nguyễn Hữu Duy, ID: 2352185

Email: duy.nguyen141205@hemut.edu.vn

Role: Information provider and information seeker,

study the theory concepts in textbook

Pham Die Anh, ID: 2352060

Email: anh.phamduc2352060@hcmut.edu.vn

Role: Information provider and information seeker,

study the theory concepts in textbook

Thái Thiên Tường, ID: 2353303

Email: tuong.thaituong3303@hcmut.edu.vn

Role: Facilitator, information provider and information seeker,

study the exercises in textbook

Tran Minh Duan, ID: 2352166

Email: duan.trancatoxg9705Ghcmut.edu.vn

Role: Information provider and information seeker,

study the exercises in textbook

Trần Hùng Sơn, ID: 2353055

Email: son.transvbkk23@hcmut.edu.vn

Role: Wildcard, find materials and provide information in any needs,

typewriter

1 Acknowledgment

Our team extends our deepest gratitude to Mr Phung Trong Thuc from the Ho Chi Minh

City University of Technology - Ho Chi Minh City National University for providing us

with the fundamental knowledge necessary to accomplish our project As we worked to- gether on this significant task, we recognized that our understanding is still developing, and we encountered several challenges in the stages of learning, assessment, and presenta- tion We eagerly await your valuable feedback and guidance to further refine and enhance our skills We sincerely appreciate your support Thank you

Contents

1 Acknowledgment

2 Preface

3 Concepts

3.1.1 A simple problem for image reconstruction .0

3.1.2 The Space Of Lines in the Plane 0.0.00 00s

3.1.3 Reconstructing an object from itsshadows 3.2 Exereise and explaÏlning ch gà gà gà kg kV

2 Preface

We all have to agree that, not much discovery in modern human history have big impact on our life like how the X-Ray does In 1895, when Roentgen first discovered X- Ray, this discovery instantly revolutionized the fields of physics and medicine The X-ray emerged from the laboratory and into widespread use in a startlingly brief leap: within a year of Roentgen’s announcement of his discovery, the application of X-rays to diagnosis and therapy was an established part of the medical profession Sure did it help the doctors diagnose the patients’ problems or help the engineers to figure out what is the problem inside machines or linking parts However, the challenge of rebuilding an image in X- ray tomography is often likened to recreating an object from its “projections”, which are essentially images formed under X-ray illumination In this part, we focus on a related but less complex task: identifying the shape of a convex object based on its shadows Similar

to what is done in medical scenarios, we are dealing with a two-dimensional issue

3 Concepts

3.1 Theoretical basis

3.1.1 A simple problem for image reconstruction

Computed tomography (CT), or computerized x-ray imaging, is a procedure where a narrow x-ray beam is directed at a patient and rapidly rotated around their body The signals produced are processed by a computer to create cross-sectional images, known as

"slices." Let’s consider a two-dimensional problem with a convex region D in a plane

Picture a distant light source illuminating the object, causing parallel rays of light to

form Our goal is to measure the shadow cast by D for various positions of the light

source To achieve this, imagine placing a screen on the opposite side of D, perpendicular

to the direction of the light rays This screen acts as a detector equipped with sensors in practical setups to detect where the shadow begins and ends

Sh

bade h ,

Figure 3.1: The shadow of the convex region

As the light shines, some rays get blocked by the shape, while others pass through Measuring the shadow means finding where the very first and very last rays hit the shape

To fully understand the shape, we’d need to rotate the source and detector through 180°, measuring, at each angle, where the shadow begins and ends The lines where the shadow first touches the shape and where it ends touch the shape’s boundary This is like tracing

the edge of the shape So, figuring out the shape from its shadows is kind of like figuring

out the knowledge of the tangent lines to its boundary To do this well, we'll need a good way to organize all these measurements So, let’s start by describing how all lines work in this setup

3.1.2 The Space Of Lines in the Plane

The line in the plan is a set of point that satisfies an equation:

an+by=c where a? + b2 4 0 We could use (a, b, c), but replacing this equation yields the same points

€

—— b =

The coefficients (g#m: viêm) define a point w on the unit circle, Š! C R? and the

constant Tine can vary Plane lines are parameterized by a unit vector pair

w = (w1;w2) and a real number The line ?¿„ is the set of point in R? satisfying the equation

The vector w is perpendicular to this line (3.2) We can get:

Since cos and sin are 2x — periodic, it’s clear that w(#) and w(@ + 27) are the same

point on the unit circle With this parameterization, the line 4 =/ lwo) is the solution set to the equation

((x,y) , (cos (0) , sin (8))) = t

Using (t,w) as the parameterization is much more efficient than (a, b,c), equation (1) produces the same set of points if (t,w) is replace by (—t, —w)

Laws —_ bp (3)

Tf lis wy = Gg then either t) = t2 and w, = w2 or t) = —ty and w; = —u» This means the pair (¢,w) specifies an oriented line, defining the positive direction The vector

7

os (—we, wr)

is perpendicular to w and therefore parallel to i,,, In fact,both w and —w are parallel to

As : - Wy —2

!,„ The vector @ is selected to ensure the determinant oŸ the 2x2 matrix

Ws) Wy

is +1, determining the line’s positive orientation Thus, the pair (¢,w) specifies the line’s direction This is summarized in a proposition

lho

Unit circle 9+7/2

LY

Figure 3.2: Parameterization of oriented lines in the plane The pairs (t,w) € Rx S; are

in one-to-one correspondence with the set of oriented line in the plane

The vector w is perpendicular to the line, and t is its affine parameter; |t| indicates the distance from the line to the origin

The pair (t,w) defines two half-planes

Hy, = {x € R?| (x, w) > t} and Ay, = {x € R?| (t,w) < t}

the line ;,, is the common boundary of these half-planes Facing along the line ?;„, the half-plane H;,,, lies to the left

3.1.3 Reconstructing an object from its shadows

We can now provide a quantitative description of the shadow Since there are two

lines in each family of parallel lines which are tangent to the boundary of D So we need

a method to select one of them:

e Orientation selection: The positive direction of the boundary D is chosen so that when facing in that direction the region lies to the left By convention, the counter- clockwise direction is considered the positive direction

`

[

SA ⁄

Figure 3.3: The measurement of the shadow

We can see that for any value of 0, there is a fix direction w(@), so there are corre-

sponding 2 values of t, denoted to and t, with to < t, such that the lines J,, (9), and bi, (9) are tangent to the boundary of D

Define the shadow function: Observing figure 3.3, the tangent line /;, @) and

boundaries at the tangent point has direction in agree and opposite compared to the line liy.0(6)- From then, we can define the shadow function of D:

hp(9) = tị and hp(@+7) = —to

It can see that the shadow function is determined for every belonging to the interval

[0; 7] While w(@) = w(@ + 27), the shadow function can be considered as a 27-periodic function defined on the whole real line

Determine the tangent point: The line /;,,(6),.9 is given parametrically by:

{hp (@) (cos (@) ,sin (@)) + s(—sin (A) ,cos (@)) | s € (—ca; 00) }

Or we can represent it by:

hp(@)(w(A)) + s(@) |s € (—co; 00)

To determine the boundary of D we need to determine the point of tangency of In,(4),u(6)

with the boundary of D Therefore, we need to find function s(Ø) so that for each 0:

(z();(0)) = ho(0)¿) + s(0)2(0)

is a point on the boundary of D

Consider Ap is differentiable and using the fact that:

a — © (eos 98 00 0, sin) = (—sin 0, cos 0) = &(0)

55 — -2(—sinØ, cos Ø) — (— cosØ, —sin#) = —w(6)

g0 90

To find s(@) we must notice that at the tangent point the direction of tangent line to

D is &(8), for a curve in the plane given parametrically by (7(@), y(@)) the direction of the tangent line at a point 0) is the same as that of the vector (2'(0), y'(@)):

(2"(8), y'(8)) = hip(A)w(8) + hp(O)@(8) + s/(0)2(0) — s()á(0) =

(hin(9) — s(0))+(0) + (hp(0) + s())6(0)

Since the tangent line at (x/(0), y'(@)) is parallel to @(Ø), the coefficient of œ¿(Ø) in the

derivative expression should be zero:

lip(0) — s(9) =0 —> hp(0) — s(0)

From this we could provide parametric representation for the boundary of a convex region

in terms of its shadow function by subsitude hj,(0) = s(@)

(z(0),(0)) = ho(0)¿(0) + hp(0)20) (4)

This is the boundary D we need to find However, the function above is only valid under the assumption that D is a strictly convex regions and the shadow function /p is differen- tial For example if the region D is a polygon then the shadow function is not everywhere

differentiable and we cannot determine the point of tangency when it is parallel to one of

the polygon’s edges since there are infinitely many of them

Conditions for the shadow function of a convex region:

For objects with smooth, curved boundaries that bulge outwards in all directions (strictly convex), we can understand the complete set of possible shadow functions (4) The tangent vector of the region can be determined by differentiating the equation (4):

(/(05),1/(0a)) — (hp(0) + hp (0))2(0)

As we proof above the tangent line of (z(Øg), (Øa)) and #(8) are parallel, so that (x'(), y(@0}) = kê(0) with k > 0

=> (09) + h(0) >0Vð6€ j0;2zj

This gives a necessary condition for a twice differentiable function h(@) to be the shadow

function for a strictly convex region with a smooth boundary

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3.2 Exercise and explaining

We have exercise 1.2.14:

Suppose that h is a function satisfying (5)

h“(0) + h(0) = OVO € [0; 277] (5)

Show that the area Dy, is given by the:

Area(Dy) = 3 fe" (h (0)? — h'(0)?) do

Explain why this implies that a function satisfying (5) also satisfies the estimate

đệ” M (032 d6 < fo" h (0)? a0 Proof: Using Green’s Theorem, we suppose that P, Q are functions of (x,y) defined on Dy», and has continuous partial derivatives If Cy, is the curve that forms the boundary of

Dy, we will have:

¢.,(Pdx + Qdy) =f fp, (GQ - 92) drdy

Setting Q=x, P=-y, then we have:

$c, (ady —ydx) = ƒ Jo, (1 — (-1)) drdy = J to, 2dady = 2Area (Dy)

With every point on c, is defined parametrically by:

(1(0);4 (0)) = h (0) (8) + Ri (0) (0) =

(h (8) cos? — h’ (O)sin@; h(@)sin@ + Ah’ (@) cos @)

So, we have:

x(0) = h(0) cos? — h’(8) sin? = dx(0) = —(h(0) + h” (0))sin@ dé

(9) = h(Ø)cos0 — h/(0)sin0 = dz(0) = —(h (0) + h”“(0))sin 0 dé

Substitute in, we see that:

+zdụ —= fe" (h (8) cos? 6 — h’ (8) sin @ cos 8) (h(@) + h (8)) dé

2m

ydx = fo" (h(0)sin? 0 + h’ (0) sin cos) (h(O) + h" (0)) dO

11

Therefore, we have:

§o,(edy — ydr) = 5" (h(6)? + A(O)h'()) dO = fo" h(OPdO + fy" h(89)h"(9)40

We call

T= f°" n(o)h"(0)d0

Then we have:

2z

£; (nảy — ydx) = [27 R(O)2d0 +

Integrating by part and using the fact that A is 27 periodic, we have:

” 2z ” ”

T= fo" (8) hh(8)40 = ACA) ROY — fo" RO) 1)46 — 0 — J"h'(0240

v du

T=— f-" h'(0)?d0

Finally, we can come to a conclusion that:

Area(D,) = 4 7" h(0)?d0 + T= 5 fo" (h (0)? — h' (0°) dO ~ 2 JO 0

This is the formula we want to prove

The second part can be obtained by noticing that the area is always a positive number

Thus:

Area(Dn) = 4 f-" (h (0)? — bí (92) d0 >0 = ” h'(0)240 < f°" h(0)240

12

References

[1] Charles L.Epstein Introduction to the Mathematics of Medical Imaging SLAM (Soci-

ety of Industrial and Applied Mathematics) edition, published by Pearson Education, Inc., 2003

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