Complex Support Vector Machines For Quaternary Classification

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Presentation for the IEEE International Workshop on MACHINE LEARNING FOR SIGNAL PROCESSING (MLSP) 2013 conference

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Complex Support Vector Machines For Quaternary Classification

  1. 1. Introduction Support Vector Machines The Complex Case Complex Support Vector Machines For Quaternary Classification P. Bouboulis, E. Theodoridou, S. Theodoridis Department of Informatics and Telecommunications University of Athens Athens, Greece 23-09-2013 P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 1 / 47
  2. 2. Introduction Support Vector Machines The Complex Case Outline 1 Introduction Reproducing Kernel Hilbert Spaces Complex RKHS 2 Support Vector Machines Linear SVMs Non-linear SVM 3 The Complex Case Complex Hyperplanes Problem formulation Experiments P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 2 / 47
  3. 3. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Outline 1 Introduction Reproducing Kernel Hilbert Spaces Complex RKHS 2 Support Vector Machines Linear SVMs Non-linear SVM 3 The Complex Case Complex Hyperplanes Problem formulation Experiments P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 3 / 47
  4. 4. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Reproducing Kernel Hilbert Spaces. Consider a linear class H of real (complex) valued functions f defined on a set X (in particular H is a Hilbert space), for which there exists a function (kernel) κ : X × X → R(C) with the following two properties: P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 4 / 47
  5. 5. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Reproducing Kernel Hilbert Spaces. Consider a linear class H of real (complex) valued functions f defined on a set X (in particular H is a Hilbert space), for which there exists a function (kernel) κ : X × X → R(C) with the following two properties: 1 For every x ∈ X, κ(x, ·) belongs to H. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 4 / 47
  6. 6. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Reproducing Kernel Hilbert Spaces. Consider a linear class H of real (complex) valued functions f defined on a set X (in particular H is a Hilbert space), for which there exists a function (kernel) κ : X × X → R(C) with the following two properties: 1 For every x ∈ X, κ(x, ·) belongs to H. 2 κ has the so called reproducing property, i.e., f(x) = f, κ(x, ·) H, for all f ∈ H, x ∈ X. (1) P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 4 / 47
  7. 7. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Reproducing Kernel Hilbert Spaces. Consider a linear class H of real (complex) valued functions f defined on a set X (in particular H is a Hilbert space), for which there exists a function (kernel) κ : X × X → R(C) with the following two properties: 1 For every x ∈ X, κ(x, ·) belongs to H. 2 κ has the so called reproducing property, i.e., f(x) = f, κ(x, ·) H, for all f ∈ H, x ∈ X. (1) Then H is called a Reproducing Kernel Hilbert Space (RKHS) associated to the the kernel κ. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 4 / 47
  8. 8. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Kernel Trick The notion of RKHS is a popular tool for treating non-linear learning tasks. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 5 / 47
  9. 9. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Kernel Trick The notion of RKHS is a popular tool for treating non-linear learning tasks. Usually this is attained by the so called “kernel trick”. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 5 / 47
  10. 10. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Kernel Trick The notion of RKHS is a popular tool for treating non-linear learning tasks. Usually this is attained by the so called “kernel trick”. If X ∋ x → Φ(x) := κ(x, ·) ∈ H X ∋ y → Φ(y) := κ(y, ·) ∈ H, P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 5 / 47
  11. 11. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Kernel Trick The notion of RKHS is a popular tool for treating non-linear learning tasks. Usually this is attained by the so called “kernel trick”. If X ∋ x → Φ(x) := κ(x, ·) ∈ H X ∋ y → Φ(y) := κ(y, ·) ∈ H, then the inner product in H is given as a function computed on X: κ(x, y) = κ(x, ·), κ(y, ·) H kernel trick P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 5 / 47
  12. 12. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Developing Learning Algorithms in RKHS The black box approach. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 6 / 47
  13. 13. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Developing Learning Algorithms in RKHS The black box approach. Develop the learning Algorithm in X. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 6 / 47
  14. 14. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Developing Learning Algorithms in RKHS The black box approach. Develop the learning Algorithm in X. Express it, if possible, in inner products. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 6 / 47
  15. 15. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Developing Learning Algorithms in RKHS The black box approach. Develop the learning Algorithm in X. Express it, if possible, in inner products. Choose a kernel function κ. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 6 / 47
  16. 16. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Developing Learning Algorithms in RKHS The black box approach. Develop the learning Algorithm in X. Express it, if possible, in inner products. Choose a kernel function κ. Replace inner products with kernel evaluations according to the kernel trick. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 6 / 47
  17. 17. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Developing Learning Algorithms in RKHS The black box approach. Develop the learning Algorithm in X. Express it, if possible, in inner products. Choose a kernel function κ. Replace inner products with kernel evaluations according to the kernel trick. Work directly in the RKHS, assuming that the data have been mapped and live in the RKHS H, i.e., X ∋ x → Φ(x) := κ(x, ·) ∈ H. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 6 / 47
  18. 18. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Advantages Advantages of kernel-based learning tasks: P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 7 / 47
  19. 19. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Advantages Advantages of kernel-based learning tasks: The original nonlinear task is transformed into a linear one. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 7 / 47
  20. 20. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Advantages Advantages of kernel-based learning tasks: The original nonlinear task is transformed into a linear one. Different types of nonlinearities can be treated in a unified way. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 7 / 47
  21. 21. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Outline 1 Introduction Reproducing Kernel Hilbert Spaces Complex RKHS 2 Support Vector Machines Linear SVMs Non-linear SVM 3 The Complex Case Complex Hyperplanes Problem formulation Experiments P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 8 / 47
  22. 22. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Complex RKHS Although the theory of RKHS holds for complex spaces too, most of the kernel-based learning techniques were designed to process real data only. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 9 / 47
  23. 23. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Complex RKHS Although the theory of RKHS holds for complex spaces too, most of the kernel-based learning techniques were designed to process real data only. Moreover, in the related literature the complex kernel functions have been ignored. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 9 / 47
  24. 24. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Complex RKHS Although the theory of RKHS holds for complex spaces too, most of the kernel-based learning techniques were designed to process real data only. Moreover, in the related literature the complex kernel functions have been ignored. Recently, however, a unified kernel-based framework, which is able to treat complex data, has been presented. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 9 / 47
  25. 25. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Complex RKHS Although the theory of RKHS holds for complex spaces too, most of the kernel-based learning techniques were designed to process real data only. Moreover, in the related literature the complex kernel functions have been ignored. Recently, however, a unified kernel-based framework, which is able to treat complex data, has been presented. This machinery transforms the input data into a complex RKHS, i.e., Φ(z) = κC(·, z). P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 9 / 47
  26. 26. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Complex RKHS Although the theory of RKHS holds for complex spaces too, most of the kernel-based learning techniques were designed to process real data only. Moreover, in the related literature the complex kernel functions have been ignored. Recently, however, a unified kernel-based framework, which is able to treat complex data, has been presented. This machinery transforms the input data into a complex RKHS, i.e., Φ(z) = κC(·, z). and employs the Wirtinger’s Calculus to derive the respective gradients. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 9 / 47
  27. 27. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Complex RKHS Definitions: H denotes a complex RKHS. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 10/ 47
  28. 28. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Complex RKHS Definitions: H denotes a complex RKHS. H denotes a real RKHS. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 10/ 47
  29. 29. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Complex RKHS Definitions: H denotes a complex RKHS. H denotes a real RKHS. The complex RKHS can be expressed as H = H + iH. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 10/ 47
  30. 30. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Complex RKHS Definitions: H denotes a complex RKHS. H denotes a real RKHS. The complex RKHS can be expressed as H = H + iH. H is isomorphic to the doubled real space H2. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 10/ 47
  31. 31. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Complex Kernels The complex Gaussian kernel: κ(z, w) = exp − d i=1(zi −w∗ i )2 σ2 , P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 11/ 47
  32. 32. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Complex Kernels The complex Gaussian kernel: κ(z, w) = exp − d i=1(zi −w∗ i )2 σ2 , The Szego kernel: κ(z, w) = 1 1−wH z , P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 11/ 47
  33. 33. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Complex Kernels The complex Gaussian kernel: κ(z, w) = exp − d i=1(zi −w∗ i )2 σ2 , The Szego kernel: κ(z, w) = 1 1−wH z , Bergman kernel: κ(z, w) = 1 (1−wH z)2 . P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 11/ 47
  34. 34. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Wirtinger Calculus Complex differentiability is a very strict notion. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 12/ 47
  35. 35. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Wirtinger Calculus Complex differentiability is a very strict notion. In learning tasks that involve complex data, we often encounter functions (e.g., the cost functions, which are defined in R) that ARE NOT complex differentiable. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 12/ 47
  36. 36. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Wirtinger Calculus Complex differentiability is a very strict notion. In learning tasks that involve complex data, we often encounter functions (e.g., the cost functions, which are defined in R) that ARE NOT complex differentiable. Example: f(z) = |z|2 = zz∗. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 12/ 47
  37. 37. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Wirtinger Calculus Complex differentiability is a very strict notion. In learning tasks that involve complex data, we often encounter functions (e.g., the cost functions, which are defined in R) that ARE NOT complex differentiable. Example: f(z) = |z|2 = zz∗. In these cases one has to express the cost function in terms of its real part fr and its imaginary part fi , and use real derivation with respect to fr , fi . P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 12/ 47
  38. 38. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Wirtinger’s Calculus This approach leads usually to cumbersome and tedious calculations. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 13/ 47
  39. 39. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Wirtinger’s Calculus This approach leads usually to cumbersome and tedious calculations. Wirtinger’s Calculus provides an alternative equivalent formulation. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 13/ 47
  40. 40. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Wirtinger’s Calculus This approach leads usually to cumbersome and tedious calculations. Wirtinger’s Calculus provides an alternative equivalent formulation. It is based on simple rules and principles. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 13/ 47
  41. 41. Introduction Support Vector Machines The Complex Case Reproducing Kernel Hilbert Spaces Complex RKHS Wirtinger’s Calculus This approach leads usually to cumbersome and tedious calculations. Wirtinger’s Calculus provides an alternative equivalent formulation. It is based on simple rules and principles. These rules bear a great resemblance to the rules of the standard complex derivative. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 13/ 47
  42. 42. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM Outline 1 Introduction Reproducing Kernel Hilbert Spaces Complex RKHS 2 Support Vector Machines Linear SVMs Non-linear SVM 3 The Complex Case Complex Hyperplanes Problem formulation Experiments P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 14/ 47
  43. 43. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM The primal problem Suppose we are given training data, which belong to two separate classes C+, C−,i.e., {(xn, dn); n = 1, . . . , N} ⊂ X × {±1}, where if dn = +1, then the n-th sample belongs to C+, while if dn = −1, then the n-th sample belongs to C−. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 15/ 47
  44. 44. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM The primal problem Suppose we are given training data, which belong to two separate classes C+, C−,i.e., {(xn, dn); n = 1, . . . , N} ⊂ X × {±1}, where if dn = +1, then the n-th sample belongs to C+, while if dn = −1, then the n-th sample belongs to C−. For example: Figure: Training points belonging to two classes. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 15/ 47
  45. 45. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM The primal problem The goal of the SVM task is to estimate the maximum margin hyperplane (wT x + c = 0), that separates the points of the two classes as best as possible P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 16/ 47
  46. 46. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM The primal problem The goal of the SVM task is to estimate the maximum margin hyperplane (wT x + c = 0), that separates the points of the two classes as best as possible minimize w∈X,c∈R 1 2 w 2 H + C N N n=1 ξn subject to dn wT xn + c ≥ 1 − ξn ξn ≥ 0 for n = 1, . . . , N, P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 16/ 47
  47. 47. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM The primal problem The goal of the SVM task is to estimate the maximum margin hyperplane (wT x + c = 0), that separates the points of the two classes as best as possible minimize w∈X,c∈R 1 2 w 2 H + C N N n=1 ξn subject to dn wT xn + c ≥ 1 − ξn ξn ≥ 0 for n = 1, . . . , N, Note that C is chosen a priori. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 16/ 47
  48. 48. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM Physical justification minimize w∈X,c∈R 1 2 w 2 H + C N N n=1 ξn subject to dn wT xn + c ≥ 1 − ξn ξn ≥ 0 for n = 1, . . . , N, P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 17/ 47
  49. 49. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM Physical justification minimize w∈X,c∈R 1 2 w 2 H + C N N n=1 ξn subject to dn wT xn + c ≥ 1 − ξn ξn ≥ 0 for n = 1, . . . , N, Figure: Linear SVM P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 17/ 47
  50. 50. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM The dual problem To solve this task, usually we consider the dual problem derived by the Lagrangian: maximize a∈RN N n=1 an − 1 2 N n,m=1 anamdndmxT mxn subject to N n=1 andn = 0 and an ∈ [0, C/N]. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 18/ 47
  51. 51. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM Outline 1 Introduction Reproducing Kernel Hilbert Spaces Complex RKHS 2 Support Vector Machines Linear SVMs Non-linear SVM 3 The Complex Case Complex Hyperplanes Problem formulation Experiments P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 19/ 47
  52. 52. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM The kernel trick Choose a positive definite kernel κR. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 20/ 47
  53. 53. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM The kernel trick Choose a positive definite kernel κR. In the dual problem, replace the inner products xT n xm with the respective kernel evaluations, i.e., κR(xn, xm). P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 20/ 47
  54. 54. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM The kernel trick Choose a positive definite kernel κR. In the dual problem, replace the inner products xT n xm with the respective kernel evaluations, i.e., κR(xn, xm). The application of the kernel trick leads to the nonlinear SVM: P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 20/ 47
  55. 55. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM The kernel trick Choose a positive definite kernel κR. In the dual problem, replace the inner products xT n xm with the respective kernel evaluations, i.e., κR(xn, xm). The application of the kernel trick leads to the nonlinear SVM: maximize a∈RN N n=1 an − 1 2 N n,m=1 anamdndmκR(xm, xn) subject to N n=1 andn = 0 and an ∈ [0, C/N]. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 20/ 47
  56. 56. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM Mapping to the feature space The application of the kernel trick to the dual problem is equivalent to the following procedure: P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 21/ 47
  57. 57. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM Mapping to the feature space The application of the kernel trick to the dual problem is equivalent to the following procedure: Choose a positive definite kernel κR, that is associated to a specific RKHS H. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 21/ 47
  58. 58. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM Mapping to the feature space The application of the kernel trick to the dual problem is equivalent to the following procedure: Choose a positive definite kernel κR, that is associated to a specific RKHS H. Map the points xn to Φ(xn) ∈ H, n = 1, . . . , N. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 21/ 47
  59. 59. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM Mapping to the feature space The application of the kernel trick to the dual problem is equivalent to the following procedure: Choose a positive definite kernel κR, that is associated to a specific RKHS H. Map the points xn to Φ(xn) ∈ H, n = 1, . . . , N. Solve the linear SVM task on the infinite dimensional RKHS H, for the training data {(Φ(xn), dn); n = 1, . . . , N}. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 21/ 47
  60. 60. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM A toy example −4 −3 −2 −1 0 1 2 3 4 5 6 −6 −4 −2 0 2 4 6 −1 −1 −1 −1 0 0 0 0 0 0 0 1 1 1 Figure: Non linear SVM classification, C = 2, gaussian kernel (σ = 2). P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 22/ 47
  61. 61. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM A toy example −4 −3 −2 −1 0 1 2 3 4 5 6 −6 −4 −2 0 2 4 6 −1 −1 −1 −1 −1 0 0 0 0 0 0 0 1 1 1 1Figure: Non linear SVM classification, C = 5, gaussian kernel (σ = 2). P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 23/ 47
  62. 62. Introduction Support Vector Machines The Complex Case Linear SVMs Non-linear SVM A toy example −4 −3 −2 −1 0 1 2 3 4 5 6 −6 −4 −2 0 2 4 6 −1 −1 −1 −1 −1 −1 −1 −1 0 0 0 0 0 0 0 0 1 1 1 1 1 Figure: Non linear SVM classification, C = 15, gaussian kernel (σ = 2). P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 24/ 47
  63. 63. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Outline 1 Introduction Reproducing Kernel Hilbert Spaces Complex RKHS 2 Support Vector Machines Linear SVMs Non-linear SVM 3 The Complex Case Complex Hyperplanes Problem formulation Experiments P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 25/ 47
  64. 64. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Real Hyperplanes Recall that in any real Hilbert space H, a hyperplane consists of all the elements f ∈ H that satisfy f, w H + b = 0, (2) for some w ∈ H, b ∈ R. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 26/ 47
  65. 65. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Real Hyperplanes Recall that in any real Hilbert space H, a hyperplane consists of all the elements f ∈ H that satisfy f, w H + b = 0, (2) for some w ∈ H, b ∈ R. Moreover, any hyperplane of H divides the space into two parts, H+ = {f ∈ H; f, w H + b > 0} and H− = {f ∈ H; f, w H + b < 0}. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 26/ 47
  66. 66. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Generalization Due to this constraint all the efforts so far to generalize real SVMs to more generic Algebras (quaternions, Clifford algebras, e.t.c.) has been explored so that: P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 27/ 47
  67. 67. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Generalization Due to this constraint all the efforts so far to generalize real SVMs to more generic Algebras (quaternions, Clifford algebras, e.t.c.) has been explored so that: The output variable y is retained to be real. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 27/ 47
  68. 68. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Generalization Due to this constraint all the efforts so far to generalize real SVMs to more generic Algebras (quaternions, Clifford algebras, e.t.c.) has been explored so that: The output variable y is retained to be real. The set of functions considered is in one way or another of a special structure, so that the inner product is real. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 27/ 47
  69. 69. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Generalization Due to this constraint all the efforts so far to generalize real SVMs to more generic Algebras (quaternions, Clifford algebras, e.t.c.) has been explored so that: The output variable y is retained to be real. The set of functions considered is in one way or another of a special structure, so that the inner product is real. The difficulty is that the set of complex numbers is not an ordered one, and thus one may not assume that a complex version of the aforementioned relation divides the space into two parts, as H+ and H− cannot be defined. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 27/ 47
  70. 70. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Generalization In order to be able to generalize the SVM rationale to complex spaces, we need first to develop an appropriate definition for a complex hyperplane. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 28/ 47
  71. 71. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Generalization In order to be able to generalize the SVM rationale to complex spaces, we need first to develop an appropriate definition for a complex hyperplane. We begin by considering the following two relations, Re ( f, w H + c) = 0, Im ( f, w H + c) = 0, for some w ∈ H, c ∈ C, where f ∈ H. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 28/ 47
  72. 72. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Generalization In order to be able to generalize the SVM rationale to complex spaces, we need first to develop an appropriate definition for a complex hyperplane. We begin by considering the following two relations, Re ( f, w H + c) = 0, Im ( f, w H + c) = 0, for some w ∈ H, c ∈ C, where f ∈ H. It is not difficult to see, that this couple of relations represent two orthogonal hyperplanes of the doubled real space, i.e., H2. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 28/ 47
  73. 73. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Generalization To overcome this constraint and be able to define arbitrarily placed hyperplanes, we need to employ the so called widely linear estimation functions, i.e., Re ( f, w H + f∗ , v H + c) = 0, Im ( f, w H + f∗ , v H + c) = 0, for some w, v ∈ H, c ∈ C, where f ∈ H. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 29/ 47
  74. 74. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Generalization To overcome this constraint and be able to define arbitrarily placed hyperplanes, we need to employ the so called widely linear estimation functions, i.e., Re ( f, w H + f∗ , v H + c) = 0, Im ( f, w H + f∗ , v H + c) = 0, for some w, v ∈ H, c ∈ C, where f ∈ H. Depending on the values of w, v, these hyperplanes may be placed arbitrarily on H2. We define this complex couple of hyperplanes as the set of all f ∈ H that satisfy either one of the two relations, for some w, v ∈ H, c ∈ C. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 29/ 47
  75. 75. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Generalization The aforementioned arguments demonstrate the significant difference between complex linear estimation and widely linear estimation functions, which has been pointed out by many other authors, in the context of regression tasks. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 30/ 47
  76. 76. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Generalization The aforementioned arguments demonstrate the significant difference between complex linear estimation and widely linear estimation functions, which has been pointed out by many other authors, in the context of regression tasks. In the current context of classification, we have just seen that confining to complex linear modeling is quite restrictive, as the corresponding couple of complex hyperplanes are always orthogonal. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 30/ 47
  77. 77. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Generalization The aforementioned arguments demonstrate the significant difference between complex linear estimation and widely linear estimation functions, which has been pointed out by many other authors, in the context of regression tasks. In the current context of classification, we have just seen that confining to complex linear modeling is quite restrictive, as the corresponding couple of complex hyperplanes are always orthogonal. On the other hand, the widely linear case is more general and covers all cases. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 30/ 47
  78. 78. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Generalization The complex couple of hyperplanes divides the space H (or H2) into four parts, i.e., H++ = f ∈ H; Re ( f, w H + f∗, v H + c) > 0, Im ( f, w H + f∗, v H + c) > 0 , H+− = f ∈ H; Re ( f, w H + f∗, v H + c) > 0, Im ( f, w H + f∗, v H + c) < 0 , H−+ = f ∈ H; Re ( f, w H + f∗, v H + c) < 0, Im ( f, w H + f∗, v H + c) > 0 , H−− = f ∈ H; Re ( f, w H + f∗, v H + c) < 0, Im ( f, w H + f∗, v H + c) < 0 . P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 31/ 47
  79. 79. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Generalization Figure: A complex couple of hyperplanes separates the space of complex numbers (i.e., H = C) into four parts. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 32/ 47
  80. 80. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Outline 1 Introduction Reproducing Kernel Hilbert Spaces Complex RKHS 2 Support Vector Machines Linear SVMs Non-linear SVM 3 The Complex Case Complex Hyperplanes Problem formulation Experiments P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 33/ 47
  81. 81. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments The problem Suppose we are given training data, which belong to four separate classes C++, C+−, C−+, C−−, i.e., {(zn, dn); n = 1, . . . , N} ⊂ X × {±1 ± i)}. If dn = +1 + i, then the n-th sample belongs to C++, i.e., zn ∈ C++, if dn = 1 − i, then zn ∈ C+−, e.t.c. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 34/ 47
  82. 82. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments The problem Suppose we are given training data, which belong to four separate classes C++, C+−, C−+, C−−, i.e., {(zn, dn); n = 1, . . . , N} ⊂ X × {±1 ± i)}. If dn = +1 + i, then the n-th sample belongs to C++, i.e., zn ∈ C++, if dn = 1 − i, then zn ∈ C+−, e.t.c. As zn is complex, we denote by xn its real part and by yn its imaginary part respectively, i.e., zn = xn + iyn, n = 1, . . . , N. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 34/ 47
  83. 83. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments The problem Suppose we are given training data, which belong to four separate classes C++, C+−, C−+, C−−, i.e., {(zn, dn); n = 1, . . . , N} ⊂ X × {±1 ± i)}. If dn = +1 + i, then the n-th sample belongs to C++, i.e., zn ∈ C++, if dn = 1 − i, then zn ∈ C+−, e.t.c. As zn is complex, we denote by xn its real part and by yn its imaginary part respectively, i.e., zn = xn + iyn, n = 1, . . . , N. Our objective is to develop an SVM rationale for the complex training data. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 34/ 47
  84. 84. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Complex formulation Consider the complex RKHS, H, with respective kernel κC. Following a similar rationale to the real case, we transform the input data from X to H, via the feature map ΦC. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 35/ 47
  85. 85. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Complex formulation Consider the complex RKHS, H, with respective kernel κC. Following a similar rationale to the real case, we transform the input data from X to H, via the feature map ΦC. The goal of the SVM task is to estimate a complex couple of maximum margin hyperplanes, that separates the points of the four classes as best as possible. To this end, we formulate the primal complex SVM as P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 35/ 47
  86. 86. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Complex formulation Consider the complex RKHS, H, with respective kernel κC. Following a similar rationale to the real case, we transform the input data from X to H, via the feature map ΦC. The goal of the SVM task is to estimate a complex couple of maximum margin hyperplanes, that separates the points of the four classes as best as possible. To this end, we formulate the primal complex SVM as min w,v,c 1 2 w 2 H + 1 2 v 2 H + C N N n=1 (ξr n + ξi n) s. to    dr n Re ( ΦC(zn), w H + Φ∗ C(zn), v H + c) ≥ 1 − ξr n di n Im ( ΦC(zn), w H + Φ∗ C(zn), w H + c) ≥ 1 − ξi n ξr n, ξi n ≥ 0 for n = 1, . . . , N. (3) for some C > 0. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 35/ 47
  87. 87. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Complex formulation Consequently, the Lagrangian function becomes L(w, v, a, ˆa, b, ˆb) = 1 2 w 2 H + 1 2 v 2 H + C N N n=1 (ξr n + ξi n) − N n=1 an (dr n Re ( ΦC(zn), w H + Φ∗ C(zn), v H + c) − 1 + ξr n) − N n=1 bn di n Im ( ΦC(zn), w H + Φ∗ C(zn), w H + c) − 1 + ξi n − N n=1 ηnξr n − N n=1 θnξi n, where an, bn, ηn, θn are the positive Lagrange multipliers of the respective inequalities, for n = 1, . . . , N. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 36/ 47
  88. 88. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Complex formulation Employing the notion of Wirtinger’s calculus to derive the respective gradients and exploiting the saddle point conditions of the Lagrangian function, it turns out that the dual problem can be split into two separate maximization tasks: maximize a N n=1 an − 1 2 N n,m=1 anamd r nd r mκ r C(zm, zn) subject to    N n=1 and r n = 0 0 ≤ an ≤ C N for n = 1, . . . , N (4a) and maximize ˆa N n=1 bn − 1 2 N n,m=1 bnbmd i nd i mκ r C(zm, zn) subject to    N n=1 bnd i n = 0 0 ≤ bn ≤ C N for n = 1, . . . , N, (4b) P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 37/ 47
  89. 89. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Complex formulation We observe that these problems are equivalent with two distinct real SVM (dual) tasks employing the induced real kernel κr C: κr C(z, z′ ) = 2 Re(κC(z, z′ )), (5) P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 38/ 47
  90. 90. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Complex formulation We observe that these problems are equivalent with two distinct real SVM (dual) tasks employing the induced real kernel κr C: κr C(z, z′ ) = 2 Re(κC(z, z′ )), (5) One may split the (output) data to their real and imaginary parts, P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 38/ 47
  91. 91. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Complex formulation We observe that these problems are equivalent with two distinct real SVM (dual) tasks employing the induced real kernel κr C: κr C(z, z′ ) = 2 Re(κC(z, z′ )), (5) One may split the (output) data to their real and imaginary parts, solve two real SVM tasks employing any one of the standard algorithms and, finally, P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 38/ 47
  92. 92. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Complex formulation We observe that these problems are equivalent with two distinct real SVM (dual) tasks employing the induced real kernel κr C: κr C(z, z′ ) = 2 Re(κC(z, z′ )), (5) One may split the (output) data to their real and imaginary parts, solve two real SVM tasks employing any one of the standard algorithms and, finally, combine the solutions to take the complex labeling function: g(z) = sign i N n=1 (andr n + ibndi n)κr C(zn, z) + cr + ici , where sign i (z) = sign(Re(z)) + i sign(Im(z)). P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 38/ 47
  93. 93. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Pure Complex SVM Figure: Pure Complex Support Vector Machines. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 39/ 47
  94. 94. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Complexification An alternative path is the so called complexification procedure. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 40/ 47
  95. 95. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Complexification An alternative path is the so called complexification procedure. We employ a real kernel κR and transform the input data from X to the complexified space H, i.e., x → ΦR(x) + iΦR(x). P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 40/ 47
  96. 96. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Complexification An alternative path is the so called complexification procedure. We employ a real kernel κR and transform the input data from X to the complexified space H, i.e., x → ΦR(x) + iΦR(x). We can similarly deduce that the dual of the complexified SVM task is equivalent to two real SVM tasks employing the kernel 2κR. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 40/ 47
  97. 97. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Complexification An alternative path is the so called complexification procedure. We employ a real kernel κR and transform the input data from X to the complexified space H, i.e., x → ΦR(x) + iΦR(x). We can similarly deduce that the dual of the complexified SVM task is equivalent to two real SVM tasks employing the kernel 2κR. We conclude that, in both cases, we end up with two real SVM tasks (although employing different types of kernels). P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 40/ 47
  98. 98. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Binary Classification Although both scenarios are developed naturally for quaternary classification, they can be easily adapted to the binary case also. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 41/ 47
  99. 99. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Binary Classification Although both scenarios are developed naturally for quaternary classification, they can be easily adapted to the binary case also. This can be done by considering that the labels of the data are real numbers (i.e., dn ∈ R) taking the values ±1. In this case we solve one problem instead of two. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 41/ 47
  100. 100. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Outline 1 Introduction Reproducing Kernel Hilbert Spaces Complex RKHS 2 Support Vector Machines Linear SVMs Non-linear SVM 3 The Complex Case Complex Hyperplanes Problem formulation Experiments P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 42/ 47
  101. 101. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments MNIST We use the popular MNIST database of handwritten digits. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 43/ 47
  102. 102. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments MNIST We use the popular MNIST database of handwritten digits. Each digit is encoded as an image file with 28 × 28 pixels. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 43/ 47
  103. 103. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments MNIST We use the popular MNIST database of handwritten digits. Each digit is encoded as an image file with 28 × 28 pixels. MNIST contains 60000 handwritten digits (from 0 to 9) for training and 10000 handwritten digits for testing. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 43/ 47
  104. 104. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments MNIST We use the popular MNIST database of handwritten digits. Each digit is encoded as an image file with 28 × 28 pixels. MNIST contains 60000 handwritten digits (from 0 to 9) for training and 10000 handwritten digits for testing. To exploit the structure of complex numbers, we perform a Fourier transform to each training image and keep only the 100 most significant coefficients. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 43/ 47
  105. 105. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments First Experiment We compare a standard one-versus-all SVM scenario that exploits the original (real) data (images of 28 × 28 = 784 pixels) with P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 44/ 47
  106. 106. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments First Experiment We compare a standard one-versus-all SVM scenario that exploits the original (real) data (images of 28 × 28 = 784 pixels) with a complex one versus all variant exploiting the complexified binary SVM, where we use only the 100 most significant (complex) Fourier coefficients of each picture. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 44/ 47
  107. 107. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments First Experiment We compare a standard one-versus-all SVM scenario that exploits the original (real) data (images of 28 × 28 = 784 pixels) with a complex one versus all variant exploiting the complexified binary SVM, where we use only the 100 most significant (complex) Fourier coefficients of each picture. In both scenarios we use the first 6000 digits of the MNIST training set to train the learning machines and test their performances using the 10000 digits of the testing set. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 44/ 47
  108. 108. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments First Experiment We compare a standard one-versus-all SVM scenario that exploits the original (real) data (images of 28 × 28 = 784 pixels) with a complex one versus all variant exploiting the complexified binary SVM, where we use only the 100 most significant (complex) Fourier coefficients of each picture. In both scenarios we use the first 6000 digits of the MNIST training set to train the learning machines and test their performances using the 10000 digits of the testing set. We used the gaussian kernel with t = 1/64 and t = 1/1402 respectively. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 44/ 47
  109. 109. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments First Experiment We compare a standard one-versus-all SVM scenario that exploits the original (real) data (images of 28 × 28 = 784 pixels) with a complex one versus all variant exploiting the complexified binary SVM, where we use only the 100 most significant (complex) Fourier coefficients of each picture. In both scenarios we use the first 6000 digits of the MNIST training set to train the learning machines and test their performances using the 10000 digits of the testing set. We used the gaussian kernel with t = 1/64 and t = 1/1402 respectively. The SVM parameter C has been set equal to 100. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 44/ 47
  110. 110. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments First Experiment The error rate of the standard real-valued scenario is 3.79%, while the error rate of the complexified (one-versus-all) SVM is 3.46%. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 45/ 47
  111. 111. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments First Experiment The error rate of the standard real-valued scenario is 3.79%, while the error rate of the complexified (one-versus-all) SVM is 3.46%. In both learning tasks we used the SMO algorithm to train the SVM. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 45/ 47
  112. 112. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments First Experiment The error rate of the standard real-valued scenario is 3.79%, while the error rate of the complexified (one-versus-all) SVM is 3.46%. In both learning tasks we used the SMO algorithm to train the SVM. The total amount of time needed to perform the training of each learning machine is almost the same for both cases (the complexified task is slightly faster). P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 45/ 47
  113. 113. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Second Experiment - Quaternary Classification This is a quaternary classification problem. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 46/ 47
  114. 114. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Second Experiment - Quaternary Classification This is a quaternary classification problem. Using the complex approach, such a problem can be solved using only 2 distinct SVM tasks, instead of the 4 SVM tasks needed by the standard 1-versus-all or the 1-versus-1 strategies. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 46/ 47
  115. 115. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Second Experiment - Quaternary Classification This is a quaternary classification problem. Using the complex approach, such a problem can be solved using only 2 distinct SVM tasks, instead of the 4 SVM tasks needed by the standard 1-versus-all or the 1-versus-1 strategies. We compare a complex quaternary SVM task with the 1-versus-all scenario. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 46/ 47
  116. 116. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Second Experiment - Quaternary Classification This is a quaternary classification problem. Using the complex approach, such a problem can be solved using only 2 distinct SVM tasks, instead of the 4 SVM tasks needed by the standard 1-versus-all or the 1-versus-1 strategies. We compare a complex quaternary SVM task with the 1-versus-all scenario. To this end we use the first 6000, 0, 1, 2 and 3 digits of the MNIST training set and compare the performances of the two algorithms using the respective digits of the MNIST training set. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 46/ 47
  117. 117. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Second Experiment - Quaternary Classification The error rate of the 1-versus-all SVM was 0.721%, P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 47/ 47
  118. 118. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Second Experiment - Quaternary Classification The error rate of the 1-versus-all SVM was 0.721%, while the error rate of the complex SVM was 0.866%. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 47/ 47
  119. 119. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Second Experiment - Quaternary Classification The error rate of the 1-versus-all SVM was 0.721%, while the error rate of the complex SVM was 0.866%. In terms of speed the 1-versus-all SVM task required about double the time for training, compared to the complex SVM. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 47/ 47
  120. 120. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Second Experiment - Quaternary Classification The error rate of the 1-versus-all SVM was 0.721%, while the error rate of the complex SVM was 0.866%. In terms of speed the 1-versus-all SVM task required about double the time for training, compared to the complex SVM. This is expected, as the latter solves half as many distinct SVM tasks as the first one. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 47/ 47
  121. 121. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Second Experiment - Quaternary Classification The error rate of the 1-versus-all SVM was 0.721%, while the error rate of the complex SVM was 0.866%. In terms of speed the 1-versus-all SVM task required about double the time for training, compared to the complex SVM. This is expected, as the latter solves half as many distinct SVM tasks as the first one. In both experiments we used the gaussian kernel with t = 1/49 and t = 1/1602 respectively. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 47/ 47
  122. 122. Introduction Support Vector Machines The Complex Case Complex Hyperplanes Problem formulation Experiments Second Experiment - Quaternary Classification The error rate of the 1-versus-all SVM was 0.721%, while the error rate of the complex SVM was 0.866%. In terms of speed the 1-versus-all SVM task required about double the time for training, compared to the complex SVM. This is expected, as the latter solves half as many distinct SVM tasks as the first one. In both experiments we used the gaussian kernel with t = 1/49 and t = 1/1602 respectively. The SVM parameter C has been set equal to 100 in this case also. P. Bouboulis, E. Theodoridou, S. Theodoridis CSVR 47/ 47

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