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Yılan: Aktif çevrit model

Başlık çevirisi mevcut değil.

  1. Tez No: 75418
  2. Yazar: BAHAR ÇOLAKOĞLU
  3. Danışmanlar: DOÇ. DR. MUHİTTİN GÖKMEN
  4. Tez Türü: Yüksek Lisans
  5. Konular: Bilgisayar Mühendisliği Bilimleri-Bilgisayar ve Kontrol, Computer Engineering and Computer Science and Control
  6. Anahtar Kelimeler: Belirtilmemiş.
  7. Yıl: 1998
  8. Dil: Türkçe
  9. Üniversite: İstanbul Teknik Üniversitesi
  10. Enstitü: Fen Bilimleri Enstitüsü
  11. Ana Bilim Dalı: Kontrol ve Bilgisayar Mühendisliği Ana Bilim Dalı
  12. Bilim Dalı: Belirtilmemiş.
  13. Sayfa Sayısı: 57

Özet

Günümüzde endüstriyel parça tanıma, karakter tanıma, hedef tesbit etme gibi birçok konuda ayrıt saptamadan ve çevrit tesbitinden yararlanılmaktadır. Bu çalışmada çevrit tesbit yöntemlerinden aktif çevrit model bir diğer adıyla yılan incelenmiştir. Farklı yöntemlerle gerçeklenen yılan modelleri kıyaslanmış, sonuçlar gösterilmiştir. Gerçekleme sırasında karşılaşılan problemlere çözümler üretilmeye çalışılmıştır.

Özet (Çeviri)

1. Introduction In many image interpretation tasks, the correct interpretation of low-level events can require high-level kowledge. Low-level processings can provide sets of alternative organizations among which higher level processes may choose. Energy functions constructed on images comprise these alternative solutions available to higher level processes. Selection of an answer from this set is accomplised by the addition of the energy terms that push the model toward the desired solution. The result is an active model that falls into the desired solution when placed near it. Active contour model can be used for finding salient image contours, including the tasks of finding edges, lines and subjective contours in images, as well as tracking those contours during motion and matching them in stereopsis. Unlike most other techniques for finding salient contours, this model is active. It is allways minimizing its energy functional and therefore exhibits dynamic behavior. Because of the way the contours slither while minimizing their energy, they are called snakes. Changes in high-level interpretation can exert forces on a snake as it continues its minimization. Snakes do not try to solve the entire problem of finding salient image contours. They rely on other mechanisms to place them somewhere near the desired contour. If an expert user pushes a snake close to an intended contour, its energy minimization will carry it the rest of the way. Snakes are an example of a more general technique of matching a deformable to an image by means of energyminimization. From any starting point, the snake deforms itself into conformity with the nearest salient contour. 2. Snake's Energy Our basic snake model is a controlled continuity spline under the influence of image forces and external constraint forces. The internal spline forces serve to impose a smoothness constraint. The image forces push the snake towards salient image features like lines, edges and subjective contours. The external constraint forces are responsible for putting the snake near the desired local minimum. These forces can, for example, come from a user interface, automatic attentional mechanisms or high level interpretations. Representing the position of a snake parametrically by v(s)=(x(s),y(s)), its energy functional can be written as = fX.W)+^(v(^))+£-(v(# (1) where Eint represent the internal energy of the spline due to bending, Eimage gives rise to the image forces and Econ gives rise to the external constraint forces. 2.1. Internal Energy The internal spline energy can be written Eint = (a(s)/Vs(s)/2)/2The spline energy is composed of a first order term controlled by a(s) and and a second order term controlled by ?(s). The first order term makes the snake act like a membrane and the second order term makes it act like a thin plate. Adjusting the weights a(s) and ?(s) controls the relative importance of the membrane and thin plate terms. 2.2. External Constraint Energy External forces are applied in order to push or pull a snake to a desired location. Springs are used to pull a snake to a location where image gradients exist, volcanos are usefull for pushing a snake out of one local minmum into another. External constraint energy of a spring is: Espring = -k(x1.-x2) (3) where k is the spring constant, x1 and x2are the end points of the spring. One end point of the spring is attached to the snake while the other is a point on the image. The energy of a volvano is: E volcano =1 /2v 4 where r is the distance between the center of the volcano and the points of the snake. 2.3. Image Energy Image forces attract snakes to salient features in images. Three different energy functional which attract a snake to lines, edges, and terminations can bestated. The total image energy can be expressed as a weighted combination of the three energy functionals E image = WlineE!ine + WedgeEedge + WternrElerm VV By adjusting the weights, a wide range of snake behavior can be created. (L)Line Functional ?. The simplest useful image functional is the image intensity itself. E». = Kx,y) (6) depending on the sign of wline, the snake will be attracted either to light lines or dark lines. Subject to its other constraints, the snake will try to align itself with the lightest or darkest nearby contour. b.)Edge Functional Finding edges in an image can also be done with a very simple energy functional. Eedge - ~r Hx,y)\ » w^ make the snake attract to contours with large image gradients. c.)Termination Functional The curvature of level lines in a slightly smooted image is used to find terminations of line segments and corners. Let C(x,y) = Ga(x,y) *I(x,y) be a slight smoothed version of the image. Let 6 = tan'1 (Cy/Cx) be the gradient angle and let// =(cosB,sin9) and ni= (-sinG,cos0) be unit vectors along and perpendicular to the gradient direction. Then the curvature of the level contours in C(x,y) can be written E =- P) â2C/ân2 âC/ân - (8) c c2 - or re a-c c2 (C2 + C2yn W 3. Energy minimization The energy equations of the snake can be solved by two methods. First one is Euler Lagrange equations of the snake energy, the second one is the dynamic programming method. A Variational Method -Euler Eagrange Equations Let E^, = Eimage + E^. When oc(s)=a, and {3(s)=P are constants, minimizing the energy functional of equation 1 gives rise to the following two independent Euler equations âE «xss+^xssss+-^ = 0 (10) âE = o (ii)When oc(s) and P(s) are not constant, it is simpler to go directly to a discrete formulation of the energy functional in equation 2. £L=2X(0+£-(0 (i2) 1=1 Approximating the derivatives with finite differences and converting to vector notation with v, = (xjtyf) = (x(ih),y(ih)), we expand Eml(i) Ejf) = a,\vt - vj/2h2 +^i|v,_I - 2v, + vj /2h4 (13) where we define v(0)=v(n). Let fx(i)=dEex/dxi zndify(i)=âEex/âyi where the derivatives are approximated by a finite difference if they can not be computed analytically. Now the corresponding Euler equations are ai+(\Yvi.i)-ai+,(vi+rvi)+pi.,[vi.r2vi.l+vi]-2Pi[vi.r2vi+vi+1] +P,1rvr2vi+]+vi+2]+(fx(i)lfyO))=0 (14) The above Euler equations can be written in matrix form as Ax+fx(x,y)=0 (15) Ay+fy(x,y)=0 (16) To solve the equationsl5 and 16 the right hand side of the equations are set equal to the product of a step size and the negative time derivatives of the left hand sides.When a(s) and p(s) are not constant, it is simpler to go directly to a discrete formulation of the energy functional in equation 2. £L.=2X(0+£-(0 (i2) i=l Approximating the derivatives with finite differences and converting to vector notation with v,. = (*,,.)>,) = (x(ih),y(ih)), we expand E^t(0 £*('") = «Jv.-vJ2 /2//2+/?i|v,_,-2v,+v,.+1|2 /2h4 (13) where we define v(0)=v(n). Let fx(i)=âEex/â)Ci ®h.ûfy(i)=âEex/efyi where the derivatives are approximated by a finite difference if they can not be computed analytically. Now the corresponding Euler equations are ai+(vrvi.1)-ai+ı(vt+rVi)-Pi-ı[vi.z-2vi.,+vi]-2pi[vi.r2vi+vH,] +J3,1[vr2v,+1+vi+2J+(fx(0Jy(0J=0 (14) The above Euler equations can be written in matrix form as Ax+fx(x,y)=0 (15) Ay+fy(x,y)=0 (16) To solve the equations 15 and 16 the right hand side of the equations are set equal to the product of a step size and the negative time derivatives of the left hand sides.Ax, + fx (x,.,,y,.i) = -7 fa - x,.i) ( 1 7) 4v« + />. (x'-i. yt-ı) = -r(yt- y>û 0 8) where y is a step size, equations 17 and 18 can be solved by matrix inversion; xt = (A + yl/'fx,, -fx (x,, -fx(xt.uyui)) (19) y, = (A + yl/'fy,., -fy (y,, -/A-/, y,-,)) (19)

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