1. 1 Formation et Analyse d’Images Session 5 Daniela Hall 4 November 2004

2. 2 Course Overview Session 1: –Homogenous coordinates and tensor notation –Image transformations –Camera models Session 2: –Camera models –Reflection models –Color spaces Session 3: –Review color spaces –Pixel based image analysis Session 4: –Gaussian filter operators –Scale Space

3. 3 Course overview Session 5: –Contrast description –Hough transform Session 6: –Kalman filter –Tracking of regions, pixels, and lines Session 7: –Stereo vision –Epipolar geometry Session 8: exam

4. 4 Session overview 1.Contrast description 2.Hough transform

5. 5 Feature extraction in computer vision The goal of feature extraction is –Reduce the amount of information –Simplify the interpretation –Preserve the information necessary for further processing (task dependent) The ideal feature extraction alogrithm is invariant –Invariance: the observation is the same independent of the view point and illumination. Contours are good candidates for invariant feature extraction

6. 6 Contrast Contrast can be caused by –Change in surface orientation (discontinuity in depth) useful for description of polyhedrical objects –Limit of a shadow useful to recognize 3D object form –Texture on a surface independent of surface, can be used to recover the texture –Reflections depends on view point and illumination

7. 7 Contrast detection process Image processing Image analysis image intermediate image representation image description

8. 8 Contrast detection methods Search for extremum in 1 st derivative Search for zero crossing in 2 nd derivative

9. 9 Contrast detection Step edge Smoothed step edge First derivative Second derivative (white pos, yellow negative)

10. 10 Image processing tools Let s(i), f(i) be two real discrete signals s(i) has Ns discrete values on the interval 0<=i<=Smax f(i) has Nf discrete values on the interval 0<=i<=Fmax Pmax=min{Smax,Fmax} s(i)=0 for all i<0, Smax<i f(i)=0 for all i<0, Fmax<i The scalar product of s(i) and f(i) is a scalar p

11. 11 Correlation and convolution Correlation is a scalar product with an offset k over all values of k. Convolution is a correlation where one of the signals is flipped (i-k)->(Fmax-1)-(i-k)

12. 12 Contrast detection Roberst contrast detector Sobel contrast detector Difference filters

13. 13 Roberts contrast detector Consists of two masks m1(i,j)= and m2(i,j)= 10 0 -1 0 1 -1 0 Compute correlation for n=1,2: Contrast magnitude Contrast direction

14. 14 Roberst contrast detector The detector is very sensitive to high frequency noise. This sensitivity can be reduced by a low pass filtering.

15. 15 Sobel detector very popular detector (Duda-Hart 1972) m0(i,j)= m90(i,j)= This filter can be seen as a convolution of two components: m0(i,j)= * m90(i,j)= * 10 -1 20 -2 1 0 -1 12 1 0 0 0 -1 -2 -1 121121 -1 0 1 0 1 1 2 1

16. 16 Sobel detector As for Roberst the amplitude and orientation are computed by: Compute correlation for n=1,2: Contrast magnitude Contrast direction

17. 17 Examples Original image Roberts detector Sobel detector Smoothing by binomial filters7 Difference filter -1 0 1 Difference filter -1 1

18. 18 Contrast detection by derivatives Steps: –Smoothing: suppression of noise –Computation of gradient magnitude and orientation –Local extremum detection by hysteresis thresholding –Edge segment extraction

19. 19 Smoothing by binomial filters Binomial filters are the best approximation of a discrete Gaussian with integer values. Binomial filters are generated by n convolutions of [1 1] Binomial series: The binomial series is composed of the coefficients of the polynom

20. 20 The binomial series The coefficients of the binomial series can be generated by Pascal’s triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 29 56 70 56 29 8 1 Level(n) 0 1 2 3 4 5 6 7 8 Sum 1 2 4 8 16 32 64 128 256 Variance 0 ¼ ½ ¾ 1 5/4 6/4 7/4 2

21. 21 Binomial filters n convolutions of [1 1] Gain: 2 n Variance: var{b n (m)}=n var{b 1 (m)} = n/4 discrete Gaussian with sigma = 1: 1 4 6 4 1 sigma= 2: 1 8 29 56 70 56 29 8 1 sigma=0.5: 1 2 1

22. 22 Discrete Gaussian filter discrete Gaussian with sigma = 1: 1 4 6 4 1 Gain: 2 4 =16 G(x,1.0) |0.004|0.054|0.242|0.399|0.242|0.054|0.004| Example of sampling a 1D Gaussian (sigma=1, R=3) session 4 16 G(x,1.0) |0.064|0.864|3.872|6.384|3.872|0.864|0.064|

23. 23 Contrast detection by derivatives Steps: –Smoothing: suppression of noise –Computation of gradient magnitude and orientation –Local extremum detection by hysteresis thresholding –Edge segment extraction

24. 24 Difference filters For a discrete signal lim does not exist This is equivalent to s(n)*[1/2, 0, ½] This is equivalent to s(n)*[1,-1] To compute derivatives of a discrete signal, do convolution with the Gaussian derivative kernel (more details in session 4)

25. 25 Difference filters Alternative method to convolution with Gaussian derivative kernel are the difference filters. Smoothing followed by a difference filter is equivalent to the Sobel operator. -1 0 1 0 1

26. 26 Examples Original image Roberts detector Sobel detector Smoothing by binomial filters Difference filter -1 0 1 Difference filter -1 1

27. 27 Examples Original image Roberts detector Sobel detector Smoothing by binomial filters7 Difference filter -1 0 1 Difference filter -1 1

28. 28 Contrast detection by derivatives Steps: –Smoothing: suppression of noise –Computation of gradient magnitude E(i,j) and direction theta –Local extremum detection and hysteresis thresholding –Edge segment extraction

29. 29 Local extremum detection Find direction of maximum gradient. Without loss of generality, let the direction be (1, 0) T Check if E(i,j) is a maximum in this direction This means: –Find contrast points C(i,j) that fulfill

30. 30 Hysteresis thresholding Problem: some detectors produce weak edges which should be eliminated. Solution: hysteresis thresholding Properties: eliminates weak edges below a low threshold, but not if they are connected to an edge above a high threshold. weak edge strong edge C(i,j) C(i,j) after chaining

31. 31 Hysteresis thresholding Uses a high and a low threshold –low threshold th0: sensitive to noise, but produces contour candidates –high threshold th1: indicates strong edges 1.Compute map F(i,j) with two thresholds th0, th1

32. 32 Hysteresis filtering 2. Computing connected components of edge elements An edge is selected when it contains at least one strong edge element (with label 2). Edges consisting of only weak edge elements are discarded.

33. 33 Hysteresis filtering The thresholding and chaining can be computed in one step. Start with the high threshold image. For each strong edge continue adding (8 or 4) connected pixels with E(i,j)>=th0.

34. 34 Example original image Contour image Low threshold image High threshold image

35. 35 Contrast detection methods Search for extremum in 1 st derivative: ok Search for zero crossing in 2 nd derivative

36. 36 Contrast detection by 2 nd derivatives The Laplacien is a isotropic filter that is used for the contrast detection. –Lap(i,j)= Lxx(i,j) + Lyy(i,j) –with Lxx(i,j)= I(i,j) * Gxx(sigma) As for Gaussians there exists a integer approximation of the Laplacian. –Lxx(i,j) : [ 1 -1] * [1 -1] = [-1 2 -1] –Lxx(i,j)+Lyy(i,j): * + * = 0 -1 0 -1 4 -1 0 -1 0 0 0 0 0 1 0 0 0 0 -1 2 -1 0 0 0 0 1 0 0 0 0 2

37. 37 Gradient norm and Laplacian

38. 38 Example image Lap Difference image Lx Difference image Ly

39. 39 Example Difference image Lx Laplacian

40. 40 Example image Lap Difference image Lx Difference image Ly

41. 41 Contour detection by 2 nd derivatives In theorie: –Contours can be detected more precisely using zero-crossing of 2nd derivative. –Sub-pixel precision is possible by interpolation. –zero crossing of 2 nd derivatives produce closed contours. In real life: –The zero crossing of the 2 nd derivative detect also many small unstable regions.

42. 42 Detection of zero crossing The extrema of the 1 st derivative correspond to zero crossing of the 2 nd derivative. To generate a contour image C(i,j) proceed as follows: –consider 4-connected neighbors (N,S,W,E)

43. 43 Detection of zero crossing For a stabler detection add a term that is related to the strength of the edge. Disadvantage: this measure may break a closed contour. The contours are no longer closed.

44. 44 Session overview 1.Contrast description 2.Hough transform

45. 45 Hough transform The Hough transform is a standard tool in image analysis that allows recognition of global patterns in an image space by recognition of local patterns in a transformed parameter space. It is particularly useful when the patterns one is looking for are sparsely digitized, have ``holes'' and/or the pictures are noisy. The basic idea of this technique is to find curves that can be parameterized like straight lines, polynomials, circles, etc., in a suitable parameter space. Hough59 P.V.C. Hough, Machine Analysis of Bubble Chamber Pictures, International Conference on High Energy Accelerators and Instrumentation, CERN, 1959.

46. 46 Hough transform Although the transform can be used in higher dimensions the main use is in two dimensions to find, e.g. straight lines, centres of circles with a fixed radius, parabolas y = ax2 + bx + c with constant c, etc. As an example consider the detection of straight lines in an image. We assume them parameterized in the form: y=mx+c Each image point (x,y) gives rise to a certain number of possible values for m and c. This set of values forms a line c=-mx+y in the Hough space (m,c) When the contour points are aligned, the corresponding lines in the Hough space have the same intersection point (m0,c0). A point in hough space indicates a line in the image space. source: http://rkb.home.cern.ch/rkb/AN16pp/node122.html

47. 47 Hough transform Computation: –Initialise histogram h(m,c) (this represents the Hough space) –for each image point (i,j) if C(i,j)=1, then compute all possible couples (m,c) that pass through the point (i,j). increment all corresponding cells by 1. –A local maximum (m0,c0) in h(m,c) indicates that the image points are aligned to the line y=m0 x+c0

48. 48 Hough transform Problems: –Care has to be taken when one quantizes the parameter space. When the bins table are chosen too fine, the local maximum can be dispersed over several bins. When the quantization is not fine enough, on the other hand, nearly parallel lines which are close together will lie in the same bin. –The coefficients m,c are not uniform in orientation Solution: use a different parameterization (p,theta): x sin(theta)-y cos(theta) + p =0

49. 49 Hough transform –Use a different parameterization (p,theta): x sin(theta)-y cos(theta) + p =0 –p is the distance from the origin and theta the angle with the normal –Here each point in the image space corresponds to a sinusoidal curve in the Hough space. The intersection of the sinusoidal curves in the Hough space correspond to a line in the image space.

50. 50 Hough transform Computation: –Initialise histogram h(p,theta) (this represents the Hough space) –for each image point (i,j) if C(i,j)=1, then –for all 0<theta<2pi compute p=-i sin(theta)+j cos(theta) –set h(p,theta) = h(p,theta)+1 –A local maximum (p0,theta0) in h(p,theta) indicates that the image points are aligned to the line p0=-x sin(theta0)+y cos(theta)

51. 51 Generalised Hough transform You can compute the hough transform for any pattern that can be parameterized, such as circles, lines, parabolas. For circles: (x-a) 2 + (y-b) 2 = r 2 We use a hough space h(a,b,r). Each image point corresponds to a cone in space (a,b,r) For a fixed r, each image point corresponds to a circle in hough space. Algorithm: –for each image point and each radius r>0 compute the corresponding circle in hough space. Increment all cells. –After filling the histogram, locate the local maximum (a0,b0,r0). This maximum corresponds to the circle in the image with center (a0, b0) and radius r0.