Analysis of Binary Images

Transcription

1 Analysis of Binary Images Introduction to Computer Vision CSE 52 Lecture 7 CSE52, Spr 07

2 The appearance of colors Color appearance is strongly affected by (at least): Spectrum of lighting striking the retina other nearby colors (space) adaptation to previous views (time) state of mind CSE52, Spr 07

3 Color Reflectance Measured color spectrum is a function of the spectrum of the illumination and reflectance From Foundations of Vision, Brian Wandell, 995, via B. Freeman slides CSE52, Spr 07

4 CSE52, Spr 07 slide Intro from Computer T. Darrel Vision

5 Color matching functions Choose primaries, say P (λ), P 2 (λ), P 3 (λ) For monochromatic (single wavelength) energy function, what amounts of primaries will match it? i.e., For each wavelength λ, determine how much of A, of B, and of C is needed to match light of that wavelength alone. RGB a( ) λ b( λ ) c( λ ) These are color matching functions CSE52, Spr 07 primaries are monochromatic at 645.2nm, 526.3nm, 444.4nm

6 CIE xyy (Chromaticity Space) CSE52, Spr 07

7 Blob Tracking for Robot Control CSE52, Spr 07

8 Basic Steps. Labeling pixels as foreground/background (0,). 2. Morphological operators (sometimes) 3. Find pixels corresponding to a region 4. Compute properties of each region CSE52, Spr 07

9 Histogram-based Segmentation Ex: bright object on dark background: Histogram Number of pixels Select threshold Create binary image: I(x,y) < T O(x,y) = 0 I(x,y) T O(x,y) = Gray value T CSE52, Spr 07 [ From Octavia Camps]

10 How do we select a Threshold? Manually determine threshold experimentally. Good when lighting is stable and high contrast. Automatic thresholding P-tile method Mode method Peakiness detection Iterative algorithm CSE52, Spr 07

11 P-Tile Method If the size of the object is approx. known, pick T s.t. the area under the histogram corresponds to the size of the object: T CSE52, Spr 07 [ From Octavia Camps]

12 Mode Method Model intensity in each region R i as constant + N(0,σ i ): CSE52, Spr 07 [ From Octavia Camps]

13 Example: Image with 3 regions Ideal histogram: μ μ 2 μ 3 Add noise: The valleys are good places for thresholding to separate regions. μ μ 2 μ 3 CSE52, Spr 07 [ From Octavia Camps]

14 Finding the peaks and valleys It is a not trivial problem: CSE52, Spr 07 [ From Octavia Camps]

15 Peakiness Detection Algorithm Find the two HIGHEST LOCAL MAXIMA at a MINIMUM DISTANCE APART: g i and g j Find lowest point between them: g k Measure peakiness : min(h(g i ),H(g j ))/H(g k ) Find (g i,g j,g k ) with highest peakiness g i g k g j CSE52, Spr 07 [ From Octavia Camps]

16 Regions CSE52, Spr 07

17 What is a region? Maximal connected set of points in the image with same brightness value (e.g., ) Two points are connected if there exists a continuous path joining them. Regions can be simply connected (For every pair of points in the region, all smooth paths can be smoothly and continuously deformed into each other). Otherwise, region is multiply connected (holes) CSE52, Spr 07

18 CSE52, Spr 07 Connected Regions What are the connected regions in this binary image? Which regions are contained within which region?

19 Connected Regions What the connected regions in this binary image? Which regions are contained within which region? CSE52, Spr 07

20 Four & Eight Connectedness Four Connected Eight Connected CSE52, Spr 07

21 Jordan Curve Theorem Every closed curve in R 2 divides the plane into two region, the outside and inside of the curve. Almost obvious CSE52, Spr 07

22 Problem of 4/8 Connectedness 8 Connected: s form a closed curve, but background only forms one region. 4 Connected Background has two regions, but ones form four open curves (no closed curve) CSE52, Spr 07

23 To achieve consistency w.r.t. Jordan. Treat background as 4-connected and foreground as 8- connected. 2. Use 6-connectedness Curve Theorem CSE52, Spr 07

24 Recursive Labeling Connected Component Exploration 2 CSE52, Spr 07

25 Recursive Labeling Connected Component Exploration Procedure Label (Pixel) BEGIN Mark(Pixel) <- Marker; FOR neighbor in Neighbors(Pixel) DO IF Image (neighbor) = AND Mark(neighbor)=nil THEN Label(neighbor) END BEGIN Main Marker <- 0; FOR Pixel in Image DO IF Image(Pixel) = AND Mark(Pixel)=nil THEN END BEGIN Marker <- Marker + ; Label(Pixel); END; Globals: Marker: integer Mark: Matrix same size as Image, initialized to NIL CSE52, Spr 07

26 Some notes How deep does stack go? Iterative algorithms (See reading from Horn) Parallel algorithms CSE52, Spr 07

27 Properties extracted from binary image A tree showing containment of regions Properties of a region. Genus number of holes 2. Centroid 3. Area 4. Perimeter 5. Moments (e.g., measure of elongation) 6. Number of extrema (indentations, bulges) 7. Skeleton CSE52, Spr 07

28 B(x,y) Moments 0 The region S is defined as: B Given a pair of non-negative negative integers (j,k) the discrete (j,k) th moment of S is defined as: CSE52, Spr 07 M n m = j, k B( x, y) x= y= x j y k Fast way to implement computation over n by m image or window One object

29 Area: Moment M 00 0 Example: Area of S!! CSE52, Spr 07

30 Moments: Centroid 0 Example: Center of gravity (Centroid) of S!! CSE52, Spr 07

31 Shape recognition by Moments 0? = 0 Recognition could be done by comparing moments However, moments M jk are not invariant under: Translation Scaling Rotation Skewing CSE52, Spr 07

32 Central Moments 0 Given a pair of non-negative negative integers (j,k) the central (j,k) th moment of S is given by: CSE52, Spr 07

33 Central Moments 0 Translation by T = (a,b) : 0 Translation INVARIANT! CSE52, Spr 07

34 Normalized Moments 0 Given a pair of non-negative negative integers (j,k) the normalized (j,k) th moment of S is given by: CSE52, Spr 07

35 Normalized Moments 0 Scaling by (a,c) and translating by T = (b,d) : 0 Scaling and translation INVARIANT! CSE52, Spr 07

36 Region orientation from Second Moment Matrix Second Centralized Moment Matrix μ μ 20 μ μ 02 CSE52, Spr 07 Eigenvectors of Moment Matrix give orientation