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- 1. Introduction to SVMs
- 2. SVMs• Geometric – Maximizing Margin• Kernel Methods – Making nonlinear decision boundaries linear – Efficiently!• Capacity – Structural Risk Minimization
- 3. SVM History• SVM is a classifier derived from statistical learning theory by Vapnik and Chervonenkis• SVM was first introduced by Boser, Guyon and Vapnik in COLT-92• SVM became famous when, using pixel maps as input, it gave accuracy comparable to NNs with hand-designed features in a handwriting recognition task• SVM is closely related to: – Kernel machines (a generalization of SVMs), large margin classifiers, reproducing kernel Hilbert space, Gaussian process, Boosting
- 4. αLinear Classifiers x f yest f(x,w,b) = sign(w. x - b) denotes +1 denotes -1 How would you classify this data?
- 5. αLinear Classifiers x f yest f(x,w,b) = sign(w. x - b) denotes +1 denotes -1 How would you classify this data?
- 6. αLinear Classifiers x f yest f(x,w,b) = sign(w. x - b) denotes +1 denotes -1 How would you classify this data?
- 7. αLinear Classifiers x f yest f(x,w,b) = sign(w. x - b) denotes +1 denotes -1 How would you classify this data?
- 8. αLinear Classifiers x f yest f(x,w,b) = sign(w. x - b) denotes +1 denotes -1 Any of these would be fine.. ..but which is best?
- 9. αClassifier Margin x f yest f(x,w,b) = sign(w. x - b) denotes +1 denotes -1 Define the margin of a linear classifier as the width that the boundary could be increased by before hitting a datapoint.
- 10. αMaximum Margin x f yest f(x,w,b) = sign(w. x - b) denotes +1 denotes -1 The maximum margin linear classifier is the linear classifier with the maximum margin. This is the simplest kind of SVM (Called an LSVM) Linear SVM
- 11. αMaximum Margin x f yest f(x,w,b) = sign(w. x - b) denotes +1 denotes -1 The maximum margin linear classifier is the linear classifierSupport Vectors with the, um,are thosedatapoints that maximum margin.the margin This is thepushes upagainst simplest kind of SVM (Called an LSVM) Linear SVM
- 12. Why Maximum Margin? 1. Intuitively this feels safest. f(x,w,b) = sign(w. x - b) 2. If we’ve made a small error in the denotes +1 location of the boundary (it’s been denotes -1 The maximum jolted in its perpendicular direction) this gives us least chance linear margin of causing a misclassification. classifier is the linear classifier 3. There’s some theory (using VCSupport Vectors dimension) that is related to um, not with the, (butare those the same as) the proposition that thisdatapoints that maximum margin. is a good thing.the margin This is the 4. Empirically it works very very well.pushes upagainst simplest kind of SVM (Called an LSVM)
- 13. A “Good” Separator O O X X X OX O O X X OX O X O
- 14. Noise in the Observations O O X X X OX O O X X OX O X O
- 15. Ruling Out Some Separators O O X X X OX O O X X O X O X O
- 16. Lots of Noise O O X X X OX O O X X OX O X O
- 17. Maximizing the Margin O O X X X OX O O X X OX O X O
- 18. Specifying a line and margin ” = +1 Plus-Plane s C las e Classifier Boundary dict zon “ Pre Minus-Plane = -1” s C l as dict zone “ Pre• How do we represent this mathematically?• …in m input dimensions?
- 19. Specifying a line and margin ” = +1 Plus-Plane s C las e Classifier Boundary ict zon “Pred Minus-Plane = -1” s =1 C l as wx +b =0 dict zone + b -1 w x b= “ Pre + wx• Plus-plane = { x : w . x + b = +1 }• Minus-plane = { x : w . x + b = -1 } Classify +1 if w . x + b >= 1 as.. -1 if w . x + b <= -1 Universe if -1 < w . x + b < 1 explodes
- 20. Computing the margin width ” = +1 M = Margin Width s C las e ict zon “Pred = -1” How do we compute s =1 C l as M in terms of w wx +b =0 dict zone + b -1 w x b= “ Pre and b? + wx• Plus-plane = { x : w . x + b = +1 }• Minus-plane = { x : w . x + b = -1 }Claim: The vector w is perpendicular to the Plus Plane. Why?
- 21. Computing the margin width ” = +1 M = Margin Width s C las e ict zon “Pred = -1” How do we compute s =1 C l as M in terms of w wx +b =0 dict zone + b -1 w x b= “ Pre and b? + wx • Plus-plane = { x : w . x + b = +1 } • Minus-plane = { x : w . x + b = -1 } Claim: The vector w is perpendicular to the Plus Plane. Why? Let u and v be two vectors on the Plus Plane. What is w . ( u – v ) ?And so of course the vector w is also perpendicular to the Minus Plane
- 22. Computing the margin width ” = +1 M = Margin Width s x+ C las e ict zon “Pred =- - 1” How do we compute x =1 l as s M in terms of w +b i ct C zone wx =0 + b -1 “Pred and b? w x b= + wx• Plus-plane = { x : w . x + b = +1 }• Minus-plane = { x : w . x + b = -1 }• The vector w is perpendicular to the Plus Plane in Any location R : not A : not m m• Let x- be any point on the minus plane necessarily a datapoint• Let x+ be the closest plus-plane-point to x-.
- 23. Computing the margin width ” = +1 M = Margin Width s x+ C las e ict zon “Pred =- - 1” How do we compute x =1 l as s M in terms of w +b i ct C zone wx =0 + b -1 “Pred and b? w x b= + wx• Plus-plane = { x : w . x + b = +1 }• Minus-plane = { x : w . x + b = -1 }• The vector w is perpendicular to the Plus Plane• Let x- be any point on the minus plane• Let x+ be the closest plus-plane-point to x-.• Claim: x+ = x- + λ w for some value of λ. Why?
- 24. Computing the margin width ” = +1 M = Margin Width s x+ C las e ict zon The line from x- to x+ is “Pred -- 1” How do we compute perpendicular to the = ss x planes.terms of w M in b= 1 Cla e ct zon + i wx =0 + b -1 “Pred So to getbfrom x- to x+ and ? w x b= + wx travel some distance in• Plus-plane = { x : w . x + b =direction w. +1 }• Minus-plane = { x : w . x + b = -1 }• The vector w is perpendicular to the Plus Plane• Let x- be any point on the minus plane• Let x+ be the closest plus-plane-point to x-.• Claim: x+ = x- + λ w for some value of λ. Why?
- 25. Computing the margin width ” = +1 M = Margin Width s x+ C las e ict zon “Pred 1” x =- - =1 l as s +b i ct C zone wx =0 + b -1 “Pred w x b= + wxWhat we know:• w . x+ + b = +1• w . x- + b = -1• x+ = x- + λ w• |x+ - x- | = MIt’s now easy to get M in terms of w and b
- 26. Computing the margin width ” = +1 M = Margin Width s x+ C las e ict zon “Pred 1” x =- - =1 l as s +b i ct C zone wx =0 + b -1 “Pred w . (x - + λ w) + b = 1 w x b= + wx =>What we know: w . x + b + λ w .w = 1 -• w . x+ + b = +1 =>• w . x- + b = -1 -1 + λ w .w = 1• x+ = x- + λ w => 2 λ=• |x+ - x- | = M w.wIt’s now easy to get M in terms of w and b
- 27. Computing the margin width ” 2 = +1 M = Margin Width = s x+ C las e w.w ict zon “Pred 1” x =- - =1 l as s +b i ct C zone wx =0 + b -1 “Pred M= |x+ - x- | =| λ w |= w x b= + wxWhat we know: = λ | w | = λ w.w• w . x+ + b = +1 2 w.w 2• w . x- + b = -1 = = w.w w.w• x+ = x- + λ w• |x+ - x- | = M 2• λ= w.w
- 28. Learning the Maximum Margin Classifier ” 2 = +1 M = Margin Width = s x+ C las e w.w ict zon “Pred 1” x =- - =1 l as s +b i ct C zone wx =0 + b -1 “Pred w x b= + wxGiven a guess of w and b we can• Compute whether all data points in the correct half-planes• Compute the width of the marginSo now we just need to write a program to search the space of w’s and b’s to find the widest margin that matches all the datapoints. How?Gradient descent? Simulated Annealing? Matrix Inversion? EM? Newton’s Method?
- 29. Don’t worry… it’s good for you…• Linear Programming find w argmax c⋅w subject to w⋅ai ≤ bi, for i = 1, …, m wj ≥ 0 for j = 1, …, nThere are fast algorithms for solving linear programs including thesimplex algorithm and Karmarkar’s algorithm
- 30. Learning via Quadratic Programming• QP is a well-studied class of optimization algorithms to maximize a quadratic function of some real-valued variables subject to linear constraints.
- 31. Quadratic Programming uT Ru Find arg max c + dT u + Quadratic criterion u 2Subject to a11u1 + a12u 2 + ... + a1mum ≤ b1 a21u1 + a22u 2 + ... + a2 mu m ≤ b2 n additional linear inequality : constraints an1u1 + an 2u 2 + ... + anmu m ≤ bnAnd subject to e additional linear a( n +1)1u1 + a( n +1) 2u2 + ... + a( n +1) mu m = b( n +1) constraints equality a( n + 2)1u1 + a( n + 2 ) 2u2 + ... + a( n + 2 ) mu m = b( n + 2 ) : a( n + e )1u1 + a( n + e ) 2u2 + ... + a( n + e ) mu m = b( n + e )
- 32. Quadratic Programming uT Ru Find arg max c + dT u + Quadratic criterion u 2Subject to a11u1 + a12u 2 + ... + a1mum ≤ b1g in for find re ex u lg ... h e s qu m ≤ ic a21u1 + a22ist2a+ orit+ md2 muadratb2 a n additional linear Th e constra in tl y s uch : ore efficien inequality much m n gradient constraints optima a lia...y th u ≤ b bl an1u1 + ann2d re+ ascennm m a u2 + a t. n ouAnd subject to very f iddly…y e e additional linear are+ ... + a t to writ = b a( n +1)1u1 (+ ut( nh1)y u2 n’t wan( n +1) mu m B a t + e 2 do ( n +1) constraints equality ly probab yoursel f) a u +a ( n + 2 )1 1 uon+ ... + a e ( n+ 2) 2 2 u =b ( n+ 2) m m ( n+2) : a( n + e )1u1 + a( n + e ) 2u2 + ... + a( n + e ) mu m = b( n + e )
- 33. Learning the Maximum Margin Classifier ” M= = +1 2 Given guess of w , b we s C las e w.w can dict zon “ Pre • Compute whether all = -1” s =1 C l as data points are in thewx +b 0 dict zone + b= wx b=-1 “ Pre correct half-planes + wx • Compute the margin widthWhat should our How many constraints Assume R datapoints, quadratic optimization each (xhave? R yk = will we k,yk) where criterion be? What 1 +/- should they be?Minimize w.w w . xk + b >= 1 if yk = 1 w . xk + b <= -1 if yk = -1
- 34. This is going to be a problem! Uh-oh! What should we do?denotes +1 Idea 1:denotes -1 Find minimum w.w, while minimizing number of training set errors. Problem: Two things to minimize makes for an ill-defined optimization
- 35. This is going to be a problem! Uh-oh! What should we do?denotes +1 Idea 1.1:denotes -1 Minimize w.w + C (#train errors) Tradeoff parameter There’s a serious practical problem that’s about to make us reject this approach. Can you guess what it is?
- 36. This is going to be a problem! Uh-oh! What should we do?denotes +1 Idea 1.1:denotes -1 Minimize w.w + C (#train errors) Tradeoff parameter Can’t be expressed as a Quadratic Programming problem. Solving it may be too slow. There’s a serious practical (Also, doesn’t distinguish between problem that’s abouteyr make disastrous errors and near misses) n So… a to o th us reject this approach. Can ideas? you guess what it is?
- 37. This is going to be a problem! Uh-oh! What should we do?denotes +1 Idea 2.0:denotes -1 Minimize w.w + C (distance of error points to their correct place)
- 38. Learning Maximum Margin with Noise M= 2 Given guess of w , b we w.w can • Compute sum of b =1 distances of points tow x+ 0 + b= wx b=-1 their correct zones + wx • Compute the margin widthWhat should our Assume R datapoints, quadratic optimization each (xk,yk) where yk = criterion be? +/- 1 How many constraints will we have? What should they be?
- 39. Large-margin Decision Boundary• The decision boundary should be as far away from the data of both classes as possible – We should maximize the margin, m – Distance between the origin and the line wtx=k is k/||w|| Class 2 m Class 1 05/18/12 39
- 40. Finding the Decision Boundary• Let {x1, ..., xn} be our data set and let yi ∈ {1,-1} be the class label of xi• The decision boundary should classify all points correctly ⇒• The decision boundary can be found by solving the following constrained optimization problem• This is a constrained optimization problem. Solving it requires some new tools – Feel free to ignore the following several slides; what is important is 05/18/12 constrained optimization problem above the 40
- 41. Back to the Original Problem • The Lagrangian is – Note that ||w||2 = wTw • Setting the gradient of w.r.t. w and b to zero, we have05/18/12 41
- 42. • The Karush-Kuhn-Tucker conditions, 05/18/12 42
- 43. The Dual Problem• If we substitute to , we have• Note that• This is a function of αi only 05/18/12 43
- 44. The Dual Problem• The new objective function is in terms of αi only• It is known as the dual problem: if we know w, we know all αi; if we know all αi, we know w• The original problem is known as the primal problem• The objective function of the dual problem needs to be maximized!• The dual problem is therefore: Properties of αi when we introduce The result when we differentiate the the Lagrange multipliers original Lagrangian w.r.t. b 44
- 45. The Dual Problem• This is a quadratic programming (QP) problem – A global maximum of αi can always be found• w can be recovered by 05/18/12 45
- 46. Characteristics of the Solution• Many of the αi are zero – w is a linear combination of a small number of data points – This “sparse” representation can be viewed as data compression as in the construction of knn classifier• xi with non-zero αi are called support vectors (SV) – The decision boundary is determined only by the SV – Let tj (j=1, ..., s) be the indices of the s support vectors. We can write• For testing with a new data z – Compute and classify z as class 1 if the sum is positive, and class 2 otherwise – Note: w need not be formed explicitly05/18/12 46
- 47. A Geometrical Interpretation Class 2 α10=0 α8=0.6 α7=0 α2=0 α5=0 α1=0.8 α4=0 α6=1.4 α9=0 α3=0 Class 105/18/12 47
- 48. Non-linearly Separable Problems• We allow “error” ξi in classification; it is based on the output of the discriminant function wTx+b• ξi approximates the number of misclassified samples Class 2 Class 1 05/18/12 48
- 49. Learning Maximum Margin with Noise M= ε11 2 Given guess of w , b we ε2 w.w can • Compute sum of +b =1 distances of points towx 0 ε7 + b= wx b=-1 their correct zones + wx • Compute the margin widthWhat should our How many R datapoints, Assume constraints will quadratic optimization we have?k,yk) where yk = each (x R criterion be? R What should they be? +/- 1 1Minimize w.w + C ∑ εk w . x + b >= 1-ε if y = 1 2 k =1 k k k w . xk + b <= -1+εk if yk = -1
- 50. Learning Maximum Margin with Noise m = # input M= dimensions 2 Given guess of w , b we ε11 ε2 w.w can • Compute sum m+1 Our original (noiseless data) QP had of 1 variables: w1, w2, … wm, and of points to distances b. b=wx + 0 ε7 Our new (noisy data) QP correct zones their has m+1+R + b= wx b=-1 + wx variables: w1, • 2, … wm, b, εthe1margin w Compute k , ε ,… εR widthWhat should our How many R datapoints, Assume constraints records R= # will quadratic optimization we have?k,yk) where yk = each (x R criterion be? R What should they be? +/- 1 1Minimize w.w + C ∑ εk w . x + b >= 1-ε if y = 1 2 k =1 k k k w . xk + b <= -1+εk if yk = -1
- 51. Learning Maximum Margin with Noise M= ε11 2 Given guess of w , b we ε2 w.w can • Compute sum of +b =1 distances of points towx 0 ε7 + b= wx b=-1 their correct zones + wx • Compute the margin widthWhat should our How many R datapoints, Assume constraints will quadratic optimization we have?k,yk) where yk = each (x R criterion be? R What should they be? +/- 1 1Minimize w.w + C ∑ εk w . x + b >= 1-ε if y = 1 2 k =1 k k k There’s a bug in this QP. Can you kspot it? -1+εk if yk = -1 w . x + b <=
- 52. Learning Maximum Margin with Noise M= ε11 2 Given guess of w , b we ε2 w.w can • Compute sum of +b =1 distances of points towx 0 ε7 + b= wx b=-1 their correct zones + wx • Compute the margin width How many constraints willWhat should our Assume R 2R we have? datapoints, quadratic optimization What should they be? yk = each (xk,yk) where criterion be? 1 R w . xk +1b >= 1-εk if yk = 1 +/-Minimize w.w + C ∑ εk 2 k =1 w . xk + b <= -1+εk if yk = -1 εk >= 0 for all k
- 53. Learning Maximum Margin with Noise M= ε11 2 Given guess of w , b we ε2 w.w can • Compute sum of +b =1 distances of points towx 0 ε7 + b= wx b=-1 their correct zones + wx • Compute the margin width How many constraints willWhat should our Assume R 2R we have? datapoints, quadratic optimization What should they be? yk = each (xk,yk) where criterion be? 1 R w . xk +1b >= 1-εk if yk = 1 +/-Minimize w.w + C ∑ εk 2 k =1 w . xk + b <= -1+εk if yk = -1 εk >= 0 for all k
- 54. An Equivalent Dual QPMinimize R w . xk + b >= 1-εk if yk = 1 1 w.w + C ∑ εk w . xk + b <= -1+εk if yk = -1 2 k =1 εk >= 0 ， for all k R 1 R RMaximize ∑ αk − ∑∑ αk αl Qkl where Qkl = yk yl ( x k .x l ) k =1 2 k =1 l =1 RSubject to these constraints: 0 ≤ αk ≤ C ∀k ∑α k=1 k yk = 0
- 55. An Equivalent Dual QP R 1 R RMaximize ∑ αk − ∑∑ αk αl Qkl where Qkl = yk yl ( x k .x l ) k =1 2 k =1 l =1 R Subject to these constraints: 0 ≤ αk ≤ C ∀k ∑α k=1 k yk = 0Then define: Rw = ∑ k yk x k α Then classify with: k=1 f(x,w,b) = sign(w. x - b)b = y K (1 −ε K ) −x K .w K where K = arg max αk k
- 56. Example XOR problem revisited:Let the nonlinear mapping be : φ(x) = (1,x12, 21/2 x1x2, x22, 21/2 x1 , 21/2 x2)TAnd: φ(xi)=(1,xi12, 21/2 xi1xi2, xi22, 21/2 xi1 , 21/2 xi2)TTherefore the feature space is in 6D with input data in 2D x1 = (-1,-1), d1= - 1 x2 = (-1, 1), d2 = 1 x3 = ( 1,-1), d3 = 1 x4 = (-1,-1), d4= -1
- 57. Q(a)= Σ ai – ½ Σ Σ ai aj di dj φ(xi) Tφ(xj) =a1 +a2 +a3 +a4 – ½(9 a1 a1 - 2a1 a2 -2 a1 a3 +2a1 a4 +9a2 a2 + 2a2 a3 -2a2 a4 +9a3 a3 -2a3 a4 +9 a4 a4 )To minimize Q, we only need to calculate ∂Q ( a ) ∂a i = 0, i = 1,...,4(due to optimality conditions) which gives 1 = 9 a1 - a2 - a3 + a4 1 = -a1 + 9 a2 + a3 - a4 1 = -a1 + a2 + 9 a3 - a4 1 = a1 - a2 - a3 + 9 a4
- 58. The solution of which gives the optimal values:a0,1 =a0,2 =a0,3 =a0,4 =1/8w0 = Σ a0,i di φ(xi) = 1/8[φ(x1)- φ(x2)- φ(x3)+ φ(x4)] 1 1 1 1 0 1 1 1 1 0 1 2 − 2 − 2 2 − 1 = − + − − = 2 8 1 1 1 1 0 − 2 − 2 2 2 0 − 2 2 − 2 2 0 Where the first element of w0 gives the bias b
- 59. From earlier we have that the optimal hyperplane is defined by: w0T φ(x) = 0That is: 1 2 x1 2x x 0 0 0 w0T φ(x) = 0 0 − 1 = − x1 x2 = 0 1 2 2 2 x2 2x 1 2x 2 which is the optimal decision boundary for the XOR problem. Furthermore we note that the solution is unique since the optimal decision boundary is unique
- 60. Output for polynomialRBF
- 61. Harder 1-dimensional dataset Remember how permitting non- linear basis functions made linear regression so much nicer? Let’s permit them here too x=0 z k = ( xk , xk ) 2
- 62. For a non-linearly separable problem we have to first map data onto feature space so that they are linear separable xi φ(xi)Given the training data sample {(xi,yi), i=1, …,N}, find theoptimum values of the weight vector w and bias b w = Σ a0,i yi φ(xi)where a0,i are the optimal Lagrange multipliers determined bymaximizing the following objective function N 1 N N Q ( a ) = ∑ai − ∑∑ai a j d i d j φ ( x i )φ( x j ) T i=1 2 i =1 j =1subject to the constraints Σ ai yi =0 ; ai >0
- 63. SVM building procedure:• Pick a nonlinear mapping φ• Solve for the optimal weight vectorHowever: how do we pick the function φ?• In practical applications, if it is not totally impossible to find φ, it is very hard• In the previous example, the function φ is quite complex: How would we find it?Answer: the Kernel Trick
- 64. Notice that in the dual problem the image of input vectorsonly involved as an inner product meaning that theoptimization can be performed in the (lower dimensional)input space and that the inner product can be replaced byan inner-product kernel N N Q ( a ) = ∑ i − 1 ∑ai a j d i d jϕj ( x )T ϕj ( x i ) a i=1 2 i , j =1 N N = ∑ i − 1 ∑ai a j d i d j K ( xi , x j ) a i=1 2 i , j =1How do we relate the output of the SVM to the kernel K?Look at the equation of the boundary in the feature space and use theoptimality conditions derived from the Lagrangian formulations
- 65. Hyperplane is defined by m1 ∑w ϕ ( x ) + b = 0 j =1 j jor m1 ∑w ϕ ( x ) = 0; j =0 j j where ϕ0 ( x ) = 1writing : ϕ( x ) = [ϕ0 ( x ), ϕ1 ( x ),..., ϕm1 ( x )] w ϕ( x ) = 0 Twe get : Nfrom optimality conditions : w = ∑ai d i ϕ( x i ) i =1
- 66. N ∑a d ϕ ( x i )ϕ( x ) = 0 TThus : i i i=1 Nand so boundary is : ∑ i d i K ( x, xi ) = 0 a i=1 Nand Output = w ϕ( x ) = ∑ i d i K ( x, xi ) T a i=1 m1where : K ( x , xi ) = ∑ ϕj ( x )ϕj ( x i ) j =0
- 67. In the XOR problem, we chose to use the kernel function: K(x, xi) = (x T xi+1)2 = 1+ x12 xi12 + 2 x1x2 xi1xi2 + x22 xi22 + 2x1xi1 ,+ 2x2xi2Which implied the form of our nonlinear functions: φ(x) = (1,x12, 21/2 x1x2, x22, 21/2 x1 , 21/2 x2)TAnd: φ(xi)=(1,xi12, 21/2 xi1xi2, xi22, 21/2 xi1 , 21/2 xi2)THowever, we did not need to calculate φ at all and could simplyhave used the kernel to calculate: Q(a) = Σ ai – ½ Σ Σ ai aj di dj Κ(xi, xj)Maximized and solved for ai and derived the hyperplane via: N ∑a d K ( x, x ) = 0 i =1 i i i
- 68. We therefore only need a suitable choice of kernel function cf:Mercer’s Theorem: Let K(x,y) be a continuous symmetric kernel that defined in theclosed interval [a,b]. The kernel K can be expanded in the form Κ (x,y) = φ(x) T φ(y)provided it is positive definite. Some of the usual choices for K are:Polynomial SVM (x T xi+1)p p specified by userRBF SVM exp(-1/(2σ2) || x – xi||2) σ specified by userMLP SVM tanh(s0 x T xi + s1)
- 69. An Equivalent Dual QP R 1 R RMaximize ∑ αk − ∑∑ αk αl Qkl where Qkl = yk yl ( x k .x l ) k =1 2 k =1 l =1 R Subject to these constraints: 0 ≤ αk ≤ C ∀k ∑α k=1 k yk = 0Then define: Datapoints with αk > 0 R will be the supportw = ∑ k yk x k α vectors classify with: Then k=1 f(x,w,b) = sign(w. x - b) ..so this sum only needsb = y K (1 −ε K ) −x K .w K to be over the support vectors. where K = arg max αk k
- 70. 1 Constant Term Quadratic 2 x1 2 x2 Linear Terms Basis : 2 xm Functions x12 2 Pure Number of terms (assuming m input x2 Quadratic dimensions) = (m+2)-choose-2 : Terms xm2 = (m+2)(m+1)/2 Φ(x) = 2 x1 x2 = (as near as makes no difference) m2/2 2 x1 x3 : 2 x1 xm You may be wondering what those Quadratic 2 x2 x3 Cross-Terms 2 ’s are doing. : •You should be happy that they do no 2 x1 xm harm : 2x x •You’ll find out why they’re there m −1 m soon.
- 71. QP with basis functions R 1 R RMaximize ∑ αk − 2 ∑∑ αk αl Qkl where Qkl = yk yl (Φ(x k ).Φ(xl )) k =1 k =1 l =1 R Subject to these constraints: 0 ≤ αk ≤ C ∀k ∑α k=1 k yk = 0 Then define: w= ∑ k y k Φ( x k ) α Then classify with: k s.t. α k >0 f(x,w,b) = sign(w. φ (x) - b) b = y K (1 −ε K ) −x K .w K where K = arg max αk k
- 72. QP with basis functions R 1 R RMaximize ∑ αk − 2 ∑∑ αk αl Qkl where Qkl = yk yl (Φ(x k ).Φ(xl )) k =1 k =1 l =1 We must do R2/2 dot products to get this matrix ready. R Subject to these 0 ≤ αk ≤ C ∀k product ∑ k m k = 0 Each dot α y requires /2 2 constraints: k= 1 additions and multiplications The whole thing costs R2 m2 /4. Then define: Yeeks! w= ∑ k y k Φ( x k ) α Then classify with: or does it? … k s.t. α k >0 f(x,w,b) = sign(w. φ (x) - b) b = y K (1 −ε K ) −x K .w K where K = arg max αk k
- 73. 1 1 1 2a1 2b1 + 2 a2 2b2 m : : ∑ 2a b i i 2 am 2bm i =1 a12 b1 2 + 2 2 a2 b2 m : : ∑ ai2bi2 2 2 i =1 am bm Φ(a) • Φ(b) = 2a1a2 • 2b1b2 Quadratic 2a1a3 2b1b3 + : : Dot Products 2a1am 2b1bm m m 2a2 a3 2b2b3 ∑ ∑ 2a a b b i j i j : : i =1 j =i +1 2a1am 2b1bm : : 2a a 2b b m −1 m m −1 m
- 74. Just out of casual, innocent, interest, let’s look at another function of a andQuadratic b: (a.b + 1) 2Dot Products = (a.b) 2 + 2a.b + 1 2 m m = ∑ ai bi + 2∑ ai bi + 1 i =1 i =1Φ(a) • Φ(b) = m m m m m m m = ∑∑ ai bi a j b j + 2∑ ai bi + 11 + 2∑ ai bi + ∑ a b + ∑ 2 2 i i ∑ 2a a b bi j i j i =1 j =1 i =1 i =1 i =1 i =1 j =i +1 m m m m = ∑ (ai bi ) + 2∑ 2 ∑a b a b i i j j + 2∑ ai bi + 1 i =1 i =1 j =i +1 i =1
- 75. Just out of casual, innocent, interest, let’s look at another function of a andQuadratic b: (a.b + 1) 2Dot Products = (a.b) 2 + 2a.b + 1 2 m m = ∑ ai bi + 2∑ ai bi + 1 i =1 i =1Φ(a) • Φ(b) = m m m m m m m = ∑∑ ai bi a j b j + 2∑ ai bi + 11 + 2∑ ai bi + ∑ a b + ∑ 2 2 i i ∑ 2a a b bi j i j i =1 j =1 i =1 i =1 i =1 i =1 j =i +1 m m m m = ∑ (ai bi ) + 2∑ 2 ∑a b a b i i j j + 2∑ ai bi + 1 i =1 i =1 j =i +1 i =1 They’re the same! And this is only O(m) to compute!
- 76. QP with Quadratic basis functions R 1 R RMaximize ∑ αk − 2 ∑∑ αk αl Qkl where Qkl = yk yl (Φ(x k ).Φ(xl )) k =1 k =1 l =1 We must do R2/2 dot products to get this matrix ready. R Subject to these 0 ≤ αk ≤ C ∀k product ∑ k y k = 0 Each dot α now only requires constraints: k=1 m additions and multiplications Then define: w= ∑ k y k Φ( x k ) α Then classify with: k s.t. α k >0 f(x,w,b) = sign(w. φ (x) - b) b = y K (1 −ε K ) −x K .w K where K = arg max αk k
- 77. Higher Order PolynomialsPolynomial φ (x) Cost to build Cost if 100 φ (a).φ (b) Cost to Cost if Qkl matrix inputs build Qkl 100 traditionally matrix inputs efficientlyQuadratic All m2/2 m2 R2 /4 2,500 R2 (a.b+1)2 m R2 / 2 50 R2 terms up to degree 2Cubic All m3/6 m3 R2 /12 83,000 R2 (a.b+1)3 m R2 / 2 50 R2 terms up to degree 3Quartic All m4/24 m4 R2 /48 1,960,000 (a.b+1)4 m R2 / 2 50 R2 terms up to R2 degree 4
- 78. QP with Quintic basis functions R R RMaximize ∑ α + ∑∑ α α Q k =1 k k =1 l =1 k l kl where Qkl = yk yl (Φ(x k ).Φ(x l )) R Subject to these constraints: 0 ≤ αk ≤ C ∀k ∑α k=1 k yk = 0 Then define: w= ∑α k s.t. α k >0 k y k Φ( x k ) b = y K (1 −ε K ) −x K .w K Then classify with: where K = arg max αk k f(x,w,b) = sign(w. φ (x) - b)
- 79. QP with Quintic basis functionsWe must do RR/2 dot products to get this 2 R R ∑ α + ∑∑ α α Qmatrix ready.Maximize k k l klIn 100-d, each 1 product now needs 103 k = dot k =1 l =1 where Qkl = yk yl (Φ(x k ).Φ(x l ))operations instead of 75 million RBut there are still worrying things lurking away. Subject to theseWhat are they? constraints: 0 ≤ α ≤ C ∀k k ∑α ky =0 k •The fear of overfittingk = this enormous with 1 number of terms Then define: •The evaluation phase (doing a set of predictions on a test set) will be very ∑ expensive (why?) w= α k y k Φ( x k ) k s.t. α k >0 b = y K (1 −ε K ) −x K .w K Then classify with: where K = arg max αk k f(x,w,b) = sign(w. φ (x) - b)
- 80. QP with Quintic basis functionsWe must do RR/2 dot products to get this 2 R R ∑ α + ∑∑ α α Qmatrix ready.Maximize k k l klIn 100-d, each 1 product now needs 103 k = dot k =1 l =1 where Qkl = yk yl (Φ(x k ).Φ(x l )) The use of Maximum Marginoperations instead of 75 million magically makes this not a problem RBut there are still worrying things lurking away. Subject to theseWhat are they? constraints: 0 ≤ α ≤ C ∀k k ∑α ky =0 k •The fear of overfittingk = this enormous with 1 number of terms Then define: •The evaluation phase (doing a set of predictions on a test set) will be very ∑ expensive (why?) w= α k y k Φ( x k ) k s.t. α k >0 Because each w. φ (x) (see below) needs 75 million operations. What b = y K (1 −ε K ) −x K .w K can be done? Then classify with: where K = arg max αk k f(x,w,b) = sign(w. φ (x) - b)
- 81. QP with Quintic basis functionsWe must do RR/2 dot products to get this 2 R R ∑ α + ∑∑ α α Qmatrix ready.Maximize k k l klIn 100-d, each 1 product now needs 103 k = dot k =1 l =1 where Qkl = yk yl (Φ(x k ).Φ(x l )) The use of Maximum Marginoperations instead of 75 million magically makes this not a problem RBut there are still worrying things lurking away. Subject to theseWhat are they? constraints: 0 ≤ α ≤ C ∀k k ∑α k y =0 k •The fear of overfittingk = this enormous with 1 number of terms Then define: •The evaluation phase (doing a set of predictions on a test set) will be very ∑ expensive (why?) w= α k y k Φ( x k ) k s.t. α k >0 Because each w. φ (x) (see below) needs 75 million operations. What b Φ x ( ∑ ) − x ⋅Φ w ⋅=(y) =1 −ε k y k Φ(xk ) .w( x) K α K k s.t. α k >0 K K can be done? Then classify with: where ∑ k yk ( max α =K = argx k ⋅ x +1) 5 α k s.t. α k >0 k k Only Sm operations (S=#support vectors) f(x,w,b) = sign(w. φ (x) - b)
- 82. QP with Quintic basis functionsWe must do RR/2 dot products to get this 2 R R ∑ α + ∑∑ α α Qmatrix ready.Maximize k k l klIn 100-d, each 1 product now needs 103 k = dot k =1 l =1 where Qkl = yk yl (Φ(x k ).Φ(x l )) The use of Maximum Marginoperations instead of 75 million magically makes this not a problem RBut there are still worrying things lurking away. Subject to theseWhat are they? constraints: 0 ≤ α ≤ C ∀k k ∑α k y =0 k •The fear of overfittingk = this enormous with 1 number of terms Then define: •The evaluation phase (doing a set of predictions on a test set) will be very ∑ expensive (why?) w= α k y k Φ( x k ) k s.t. α k >0 Because each w. φ (x) (see below) needs 75 million operations. What b Φ x ( ∑ ) − x ⋅Φ w ⋅=(y) =1 −ε k y k Φ(xk ) .w( x) K α K k s.t. α k >0 K K can be done? Then classify with: where ∑ k yk ( max α =K = argx k ⋅ x +1) 5 α k s.t. α k >0 k k Only Sm operations (S=#support vectors) f(x,w,b) = sign(w. φ (x) - b)
- 83. QP with Quintic basis functions R 1 R RMaximize ∑ αk − 2 ∑∑ αk αl Qkl where Qthink:don’tyoverfit(as kmuch asxl )) k =1 k =1 l =1 you’d kl = Why SVMs y k l (Φ x ).Φ( No matter what the basis function, there are really R up to R only Subject to these constraints: 0 ≤ αk ≤ C ∀k ∑α Ry =0 parameters: α1, α2 .. αk, and usually k most are set to zero by the Maximum k=1 Margin. Then define: Asking for small w.w is like “weight decay” in Neural Nets and like Ridge Regression parameters in Linear w= ∑α k s.t. α k >0 k y Φ( x ) k k regression and like the use of Priors in Bayesian Regression---all designed to smooth the function and reducew ⋅Φ( x) = ∑ b = y (1 −ε ) −x .wαk y k Φ( x k ) ⋅Φ( x) overfitting. K k s.t. α k >0 K K K Then classify with: ∑ =K = argx k ⋅ x +1) 5 where k s.t. α >0k yk ( max αk α f(x,w,b) = sign(w. φ (x) - b) k k Only Sm operations (S=#support vectors)
- 84. SVM Kernel Functions• K(a,b)=(a . b +1)d is an example of an SVM Kernel Function• Beyond polynomials there are other very high dimensional basis functions that can be made practical by finding the right Kernel Function – Radial-Basis-style Kernel Function: (a − b) 2 K (a, b) = exp − σ, κ and δ are magic 2σ 2 parameters that must – Neural-net-style Kernel Function: be chosen by a model K (a, b) = tanh(κ a.b − δ ) selection method such as CV or VCSRM
- 85. SVM Implementations• Sequential Minimal Optimization, SMO, efficient implementation of SVMs, Platt – in Weka• SVMlight – http://svmlight.joachims.org/
- 86. References• Tutorial on VC-dimension and Support Vector Machines: C. Burges. A tutorial on support vector machines for pattern recognition. Data Mining and Knowledge Discovery, 2(2):955-974, 1998. http://citeseer.nj.nec.com/burges98tutorial.html• The VC/SRM/SVM Bible: Statistical Learning Theory by Vladimir Vapnik, Wiley- Interscience; 1998

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