1. 1 Formation et Analyse d’Images Session 3 Daniela Hall 14 October 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.Review reflectance model 2.Pixel based image analysis 1.Color histograms 2.Example face detection 3.Segmentation 4.Connectivity analysis 5.Morphological operators 6.Moments 3.Corrections

5. 5 Di-chromatic reflectance model the reflected light R is the sum of the light reflected at the surface R s and the light reflected from the material body R L R s has the same spectrum as the light source The spectrum of R l is « filtered » by the material (photons are absorbed, this changes the emitted light) Luminance depends on surface orientation Spectrum of chrominance is composed of light source spectrum and absorption of surface material.

6. 6 Pixel based image analysis Amplitude of R L depends on angle i Amplitude is captured by the luminance axis. Body component of the object is captured by the chrominance axis. camera light N e g i

7. 7 Color space (r,g) Intensity normalised color space (r,g) Properties: –less sensitive to intensity changes –less sensitive to variances of i –preserves chrominance (important for object identification)

8. 8 Object detection by color Let p((r,g)|obj) be the probability density of (r,g) given the object and p(r,g) the global probability of the occurrence of (r,g) Then we can compute for any color (r,g) the probability p(obj|(r,g)) This gives rise to a « probability map » of the image.

9. 9 Color histograms A color histogram is a (multi-dimensional) table We define a linear function that computes for any color the index of the corresponding histogram cell. Example: we have a greyscale image with 256 grey values. We want to fill a histogram with N cells. The index of the histogram cell c(val) of the pixel with value val is:

10. 10 Color histograms A histogram is filled by considering pixels within a ROI (region of interest). –For each pixel val we compute the c(val) and increment it. Histograms approximate probability densities.

11. 11 Object detection by color histograms The prior p(obj) can be estimated as the ratio of the obj size N obj to the size of the image N tot Then: Ratio of histograms

12. 12 Object detection by color histograms Constructing the histograms h obj and h tot is called learning. Important points: –h obj must contain only points of the object –h tot must be sufficiently representative to estimate the color distribution in the world. –You need sufficient number of training examples. All cells of h tot should be > 0 (otherwise division by 0) You have sufficient data when N >= k*number cells, k~5 to 10 Example: for 2D histogram 32x32 cells, you need 10240 pixels. for 5D histogram of 10x10 cells, you need 1 million pixels.

13. 13 Example: face detection 1.Learning 1.select images 2.segment pixels that have skin color (by hand) 3.construct h obj and h tot

14. 14 Example: face detection Detection Compute probability map where each pixel has probability p(obj|(r,g))~h obj (r,g)/h tot (r,g). First and second moments of the high probability pixels give the position and extent of the face.

15. 15 Example: face tracking 1.Learning as before 2.Do detection once for initialisation 3.Continous tracking 1.Compute position estimate for the next frame (using Kalman filter, session 6) 2.Compute probability image 3.Multiply by a Gaussian mask that is centered on the most likely position. This removes outliers and makes system more stable. 4.First and second moments give position and size of face.

16. 16 Example: face tracking input image probability image weighted by Gaussian Tracking result

17. 17 Segmentation 1.Segmentation by thresholding 2.Connected components 3.Improvement by morphological operators

18. 18 Segmentation Segmentation by thresholding 1.Make histogram of the probability image 2.Find threshold values by searching for valleys 3.Apply thresholds probability imageThresholded image Threshold at 0.375

19. 19 Notational convention Image operators require to evaluate the neighboring pixels. The neighbor with coordinates (i-1,j-1) of pixel (i,j) is called the neighbor NW. It has the value I(i-1,j-1)= I(NW). Operators are based on two types of support: a 3*3 block (8-connected) or a 3*3 cross (4-connected). I(NW) I(N) I(NE) I(W) I(C) I(E) I(SW) I(S) I(SE) I(N) I(W) I(C) I(E) I(S) I(i-1,j-1) I(i,j-1) I(i+1, j-1) I(i-1,j) I(i,j) I(i+1, j) I(i-1,j+1) I(i, j+1) I(i+1, j+1) 8 connected 4 connected

20. 20 Connected components Algorithm to segment multiple objects within an image. How it works: 1.Use a binary image 2.Scan along a row until a point p with I(p)=1 is found. 3.Examine the 4 neighbors (N,W,NW,NE) 1.if I(N)=I(NW)=I(NE)=I(E)=0, assign a new label to p 2.if only one neighbor has a label, assign this label to p. 3.if more neighbors have a label, assign one to p and make a note of label equivalence. 4.Scan all rows. 5.Determine which labels are equivalent 6.Replace equivalent labels within the image. source: http://homepages.inf.ed.ac.uk/rbf/HIPR2/label.htm p N W NWNE

21. 21 Example connectivity analysis 1. Scan a row until p with I(p)=1 is found. 2. Examine neighbors N,NW,NE,W aa aaa a aaa a a b b a aaa a a b b b b=a a aaa a a b b b c c c Replace equivalences b b b b b b b c c c

22. 22 More examples binary image Connected components with labels coded as colors

23. 23 Example: count number of objects Original image Thresholded image Connected component algorithm gives 163 labels labels coded as grey values labels coded as colors labels coded as 8 different colors

24. 24 Morphological operators This example shows the basic operators of the mathematic morphology on the binary images. The structuring element is a 3*3 block (8- connected) or a 3*3 cross (4-connected). Max (Dilatation for binary image) Min (Erosion for binary image) Close: Min(Max(Image)) Open: Max(Min(Image)) Convention: for the binary image, black=0, white=1.

25. 25 Dilatation and erosion Dilatation operator I’(C) = max(I(N),I(W),I(C),I(E),I(S)) Erosion operator I’(C) = min(I(N),I(W),I(C),I(E),I(S)) Original imageThresholded image Dilated image Eroded image

26. 26 Close and open operators Close: Min(Max(Image)) Open: Max(Min(Image))

27. 27 Moments In order to describe (and recognize) an object we need a method that is invariant to certain image transformations. Method: moment computation 1 st moment: center of gravity 2 nd moment: spatial extent 1 st and 2 nd moment are invariant to image orientation 2 nd moment is invariant to image translation.

28. 28 Moments Input: a binary image. Let S be the sum of all white pixels. 1 st moment (μi,μj) (center of gravity) 2 nd moments (covariance)

29. 29 Moments The covariances define an ellipse. The direction and length of the major axis of the ellipse are computed by principle component analysis. Find a rotation Φ, such that Φ eigen vectors, Λ eigen values

30. 30 Example Height and width of an object depend on the orientation of the object whereas the eigen values of the covariance C P are invariant h w w h w h λ1λ1 λ2λ2 bounding box

31. 31 Example:pattern recognition You dispose of an example image that contains a particular segmented object. Task: decide for a number of new images of different orientations, if the learned object is within the image. Training example Query images

32. 32 Example:pattern recognition Representation by width and height of the bounding box of the object is not a solution. Representation by moments allows to find the correct images. Training example Query images

33. 33 Corrections Session 1: Translation in homogenous coordinates

34. 34 1D linear interpolation P s0 P s1 P s3 P s2 PsPs position P intensity I(P) P s0 PsPs P s1 Gradient Pixel color

35. 35 2D linear interpolation P s0 P s1 P s3 P s2 PsPs x intensity I(P) P s0 PsPs P s1 Gradient Pixel color I(P s ) y P sx P sy

36. 36 Bi-linear interpolation P s0 P s1 P s3 P s2 PsPs x intensity I(P) P s0 PsPs P s1 I(P s ) y P s3 The bilinear approach computes the weighted average of the four neighboring pixels. P sx P sy Bilinear formula: