1 / 36

3-D Computer Vision CSc 83020

3-D Computer Vision CSc 83020. Binary Images Horn (Robot Vision): pages 46-64. Binary Images. Binary Image b(x,y): Obtained from Gray-Level (or other image) g(x,y) By Thresholding. Characteristic Function: 1 g(x,y) < T b(x,y)= 0 g(x,y) >= T Geometrical Properties

sydnee
Télécharger la présentation

3-D Computer Vision CSc 83020

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 3-D Computer VisionCSc 83020 Binary Images Horn (Robot Vision): pages 46-64. 3-D Computer Vision CSc 83020 – Ioannis Stamos

  2. Binary Images • Binary Image b(x,y): Obtained from Gray-Level (or other image) g(x,y) • By Thresholding. • Characteristic Function: • 1 g(x,y) < T • b(x,y)= • 0 g(x,y) >= T • Geometrical Properties • Discrete Binary Images • Multiple objects • Sequential and Iterative Processing. 3-D Computer Vision CSc 83020 – Ioannis Stamos

  3. Selecting the Threshold (T) g(x,y) h(i) BIMODAL HISTOGRAM T i (intensity) b(x,y) 3-D Computer Vision CSc 83020 – Ioannis Stamos

  4. Histogram Thresholding T 3-D Computer Vision CSc 83020 – Ioannis Stamos

  5. Geometric Properties b(x,y) Assume: 1) b(x,y) is continuous 2) Only one object y x 3-D Computer Vision CSc 83020 – Ioannis Stamos

  6. Geometric Properties b(x,y) Assume: 1) b(x,y) is continuous 2) Only one object y x Area: (Zeroth Moment) 3-D Computer Vision CSc 83020 – Ioannis Stamos

  7. Geometric Properties b(x,y) Assume: 1) b(x,y) is continuous 2) Only one object y x Area: (Zeroth Moment) Position: “Center” (x,y) of Area (First Moment) 3-D Computer Vision CSc 83020 – Ioannis Stamos

  8. Orientation Difficult to define! We use axis of least second moment y axis (x,y) r ρ x θ 3-D Computer Vision CSc 83020 – Ioannis Stamos

  9. Orientation Difficult to define! We use axis of least second moment For mass distribution: Axis of minimum Inertia. y axis (x,y) r ρ x θ Minimize: 3-D Computer Vision CSc 83020 – Ioannis Stamos

  10. Orientation Difficult to define! We use axis of least second moment For mass distribution: Axis of minimum Inertia. y axis (x,y) r ρ x θ Minimize: Equation of Axis: y=mx+b ??? 0<= m <= infinity We use: (ρ, θ) finite. Find (ρ, θ) that minimize E for a given b(x,y)

  11. We can show that: So: 3-D Computer Vision CSc 83020 – Ioannis Stamos

  12. We can show that: So: Using dE/dρ=0 we get: 3-D Computer Vision CSc 83020 – Ioannis Stamos

  13. We can show that: So: Using dE/dρ=0 we get: Note: Axis passes through center (x,y)!! So, change coordinates: x’=x-x, y’=y-y 3-D Computer Vision CSc 83020 – Ioannis Stamos

  14. We can show that: So: Using dE/dρ=0 we get: Note: Axis passes through center (x,y)!! So, change coordinates: x’=x-x, y’=y-y We get: Second Moments w.r.t. (x, y) 3-D Computer Vision CSc 83020 – Ioannis Stamos

  15. We can show that: So: Using dE/dρ=0 we get: Note: Axis passes through center (x,y)!! So, change coordinates: x’=x-x, y’=y-y We get: Second Moments w.r.t. (x, y) Mistake in your handout. 3-D Computer Vision CSc 83020 – Ioannis Stamos

  16. Using dE/dθ=0, we get: tan(2θ)=b/(a-c) b 2θ a-c 3-D Computer Vision CSc 83020 – Ioannis Stamos

  17. Using dE/dθ=0, we get: tan(2θ)=b/(a-c) So: b 2θ a-c 3-D Computer Vision CSc 83020 – Ioannis Stamos

  18. Using dE/dθ=0, we get: tan(2θ)=b/(a-c) So: Solutions with positive sign may be used to find θ that minimizes E (why??). Emin/Emax -> (ROUDNESS of the OBJECT). b 2θ a-c 3-D Computer Vision CSc 83020 – Ioannis Stamos

  19. Discrete Binary Images bij: value at pixel in row i & column j. j (y) m O i (x) n 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3-D Computer Vision CSc 83020 – Ioannis Stamos

  20. Discrete Binary Images bij: value at pixel in row i & column j. Assume pixel area is 1. Area: Position (Center of Area): j (y) m O i (x) n 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3-D Computer Vision CSc 83020 – Ioannis Stamos

  21. Discrete Binary Images bij: value at pixel in row i & column j. Assume pixel area is 1. Area: Position (Center of Area): j (y) m O i (x) n 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Second Moments. 3-D Computer Vision CSc 83020 – Ioannis Stamos

  22. Discrete Binary Images (cont) Note: a’,b’,c’ are second moments w.r.t ORIGIN a,b,c (w.r.t “center”) can be found from (a’,b’,c’,x,y,A). Note: UPDATE (a’,b’,c’,x,y,A) during RASTER SCAN. 3-D Computer Vision CSc 83020 – Ioannis Stamos

  23. Multiple Objects Need to SEGMENT image into separate COMPONENTS (regions). Connected Component: Maximal Set of Connected Points: B A Points A and B are connected: Path exists between A and B along which b(x,y) is constant. 3-D Computer Vision CSc 83020 – Ioannis Stamos

  24. Connected Components 4-way connected 8-way connected Label all pixels that are connected 3-D Computer Vision CSc 83020 – Ioannis Stamos

  25. Region Growing Algorithm(Connected Component Labeling) • Start with “seed” point where bij=1. • Assign LABEL to seed point. • Assign SAME LABEL to its NEIGHBORS (b=1). • Assign SAME LABEL to NEIGHBORS of NEIGHBORS. Terminates when a component is completely labeled. Then pick another UNLABELED seed point. 3-D Computer Vision CSc 83020 – Ioannis Stamos

  26. What do we mean by NEIGHBORS? Connectedness 8-Connectedness (8-c) Neither is perfect! 4-Connectedness (4-c) 3-D Computer Vision CSc 83020 – Ioannis Stamos

  27. What do we mean by NEIGHBORS? Jordan’s Curve Theorem: Closed curve -> 2 connected regions B1 O1 B1 0 1 0 1 0 1 O2 B2 O3 0 1 (4-c) Hole without closed curve! 0 B1 O4 B1 3-D Computer Vision CSc 83020 – Ioannis Stamos

  28. What do we mean by NEIGHBORS? Jordan’s Curve Theorem: Closed curve -> 2 connected regions B1 O1 B1 B O B 0 1 0 1 0 1 O2 B2 O3 O B O (8-c) Connected background with closed ring! 0 (4-c) Hole without closed curve! 1 0 B1 O4 B1 B O B

  29. Solution: Introduce Assymetry Use: or (a) (b) Using (a) Two separate line segments. B O1 B 0 1 0 1 0 1 O2 B O1 0 1 0 B O2 B Hexagonal Tesselation Above assymetry makes SQUARE grid like HEXAGONAL grid. 3-D Computer Vision CSc 83020 – Ioannis Stamos

  30. Sequential Labeling Algorithm Raster Scan D B C Note: B,C,D are already labeled A 3-D Computer Vision CSc 83020 – Ioannis Stamos

  31. Sequential Labeling Algorithm Raster Scan D B C Note: B,C,D are already labeled A Label(A) = background a. X X X 0 Label(A) = label(D) b. D X X 1 3-D Computer Vision CSc 83020 – Ioannis Stamos

  32. Sequential Labeling Algorithm Raster Scan D B C Note: B,C,D are already labeled A Label(A) = label(C) c. 0 0 C 1 Label(A) = label(B) d. 0 B 0 1 If Label(B) = label(C ), then Label(A)=Label(B)= Label(C) d. 0 B C 1 3-D Computer Vision CSc 83020 – Ioannis Stamos

  33. Sequential Labeling Algorithm Raster Scan D B C Note: B,C,D are already labeled A Label(A) = background a. X X X 0 Label(A) = label(D) b. D X X 1 Label(A) = label(C) c. 0 0 C 1 Label(A) = label(B) d. 0 B 0 1 If Label(B) = label(C ), then Label(A)=Label(B)= Label(C) d. 0 B C 1 3-D Computer Vision CSc 83020 – Ioannis Stamos

  34. Sequential Labeling (Cont.) What if B & C are labeled but label(B) label(C) ? 3-D Computer Vision CSc 83020 – Ioannis Stamos

  35. Sequential Labeling (Cont.) What if B & C are labeled but label(B) label(C) ? 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 1 1 1 1 1 1 1 1 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ? 3-D Computer Vision CSc 83020 – Ioannis Stamos

  36. Sequential Labeling (Cont.) Solution: Let: Label(A)=Label(B) =2 & create an EQUIVALENCE TABLE. Resolve Equivalences in SECOND PASS. 2 1 7 3,6,4 … 3-D Computer Vision CSc 83020 – Ioannis Stamos

More Related