1. Object Recognition Using Genetic Algorithms CS773C Advanced Machine Intelligence Applications Spring 2008: Object Recognition

2. 2D Case Recover the geometric transformation that aligns the model(s) with the scene. (affine transformation)

3. Image-Space Approaches Identify a set of features from the unknown scene which approximately match a set of features from a model object. Recover the geometric transformation that the model object has undergone. Examples: –Interpretation tree (Grimson & Lozano-Perez, 1987) –Alignment (Huttenlocher and Ullman, 1990) –Geometric hashing (Lamdan et al., 1990)

4. Transformation-Space Approaches Search the transformation space. Find a transformation which aligns a large number of model features with the scene. Examples: –Hough-transform based methods (Ballard, 1981). –Pose clustering techniques (Cass, 1988)

5. Why Using GAs for Object Recognition? Genetic algorithms were designed to efficiently search large solution spaces. Both the image and transformation spaces are very large! Image space: M 3 S 3 possible alignments Transformation space: much larger! G. Bebis. S. Louis, Y. Varol, and A. Yfantis, "Genetic Object Recognition Using Combinations of Views", IEEE Transactions on Evolutionary Computation, vol 6, no. 2, pp. 132-146, April 2002.

6. Genetic Algorithms (GAs) Review What is a GA? –An optimization technique for searching very large spaces. –Inspired by the biological mechanisms of natural selection and reproduction. What are the main characteristics of a GA? –Global optimization technique. –Uses objective function information, not derivatives. –Searches probabilistically using a population of structures (i.e., candidate solutions using some encoding). selection crossovermutation –Structures are modified at each iteration using selection, crossover, and mutation.

7. Structure of GA 10010110… 10010110… 01100010… 01100010… 10100100... 10100100… 10010010… 01111001… 01111101… 10011101… Evaluation and Selection Crossover Mutation Current GenerationNext Genaration

8. Encoding and Fitness Evaluation Encoding scheme –Transforms solutions in parameter space into finite length strings (chromosomes) over some finite set of symbols. Fitness function –Evaluates the goodness of a solution.

9. Selection Operator exploitProbabilistically filters out solutions that perform poorly, choosing high performance solutions to exploit. –Chromosomes with high fitness are copied over to the next generation. fitness

10. Crossover and Mutation Operators explorationGenerate new solutions for exploration. Crossover –Allows information exchange between points. Mutation –Its role is to restore lost genetic material. Mutated bit

11. Object Recognition Using Gas: Two Approaches Genetic search in the image space (GA-IS) Genetic search in the transformation space (GA-TS) Important issues: –How to encode solutions? –How to modify solutions ? –How to evaluate solutions?

12. GA-IS: Encoding At least three model-scene point matches are need to compute the affine transformation. Chromosome contains the binary encoded identities of the three pairs of points. –Model points: 19 (5 bits) –Scene points: 19 - 45 (6 bits) –Chromosome length: 3 x 5 + 3 x 6 = 33 bits

13. GA-IS: Decoding Some encoded solutions might be invalid –5 bits can encode at most 32 values. –[0, 31] was linearlymapped to [0, 18] –6 bits can encode at most 64 values. –[0, 63] was linearly mapped to [0, 44]

14. GA-IS: Fitness Evaluation Compute affine transformation. Apply the transformation on all the model points. Compute the back-projection error (BE) between the model and scene. (d j min distance between the j-th model point and the scene)

15. GA-TS: Encoding Need to estimate the range of values that the parameters of affine transformation can assume.

16. GA-TS: Encoding (cont’d) Each chromosome contains six fields. Only the range of each coefficient needs to be represented. –e.g., a 11 assumes values in [-0.408, 0.408] –Its range is: 0.408 - (-0.408) = 0.816 –2 decimal digit accuracy: 82 values must be encoded. –7 bits are needed to encode 82 values. –Chromosome length: 6 x 7 = 42 bits

17. GA-TS: Decoding Some encoded solutions might be invalid –7 bits can encode at most 128 values. –[0, 127] should be mapped to [0, 81]

18. GA-TS: Fitness Evaluation Same as before Less expensive to compute in this case …

19. Experiments Three scenes (S 1, S 2, S 3 ) of increasing complexity. S 2, S 3 are shown below (S 1 was the same as model). 10 trials per scene – tried to find model1 each time

20. Experiments (cont’d) We searched for the model below in a number of scenes.

21. GA Parameters Two-point crossover (prcoss: 0.95). Point mutation (pmut: 0.05). Cross generational selection strategy. Fitness scaling (scaling factor: 1.2).

22. Results – Scene 1 Correct solutions were found in all 10 trials.

23. Results – Scene 2 Correct solutions were found in all 10 trials.

24. Results – Scene 3 Correct solution was missed once!

25. Efficiency: GA-IS

26. Efficiency: GA-TS

27. 3D case – GA-TS Apply Genetic Search in the space of the AFoVs parameters. We used 2 views per model

28. Experiments Four scenes (S 1, S 2, S 3,S 4 ) of increasing complexity. S 1 was the same as model 10 trials per scene

29. Experiments (cont’d) We searched for the model below in a number of scenes.

30. GA Parameters Two-point crossover (prcoss: 0.95). Point mutation (pmut: 0.05). Cross generational selection strategy. Fitness scaling (scaling factor: 1.2). Population size: 200 Number of generations: 150

31. Results - Scene 1 Correct solutions were found in all 10 trials.

32. Results – Scene 2 Correct solutions were found in all 10 trials.

33. Results – Scene 3 Correct solutions were found in all 10 trials.

34. Results – Scene 4 Correct solutions were found in all 10 trials.

35. Efficiency: GA-TS ~ 2 x 10 15

36. More Challenging Scene GAs can find near-exact matches. Could be used as input to more sophisticated recognition algorithms.