COMP 551 Applied Machine Learning Lecture 9: Support Vector Machines (cont d)
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1 COMP 551 Applied Machine Learning Lecture 9: Supprt Vectr Machines (cnt d) Instructr: Herke van Hf Slides mstly by: Class web page: Unless therwise nted, all material psted fr this curse are cpyright f the instructrs, and cannt be reused r repsted withut the instructrs written permissin.
2 Quizzes Quizzes fr: - Instance based learning - Supprt vectr machines Nw available fr self-test COMP-598: Applied Machine Learning 2
3 Cnsider bth slutins: Recap primal and dual p* = min w max α:αi 0 L(w,α) Primal d* = max α:αi 0 min w L(w,α) Dual The ptimal w*, α* are the same, as the fllwing hld: f and g i are cnvex and the g i can all be satisfied simultaneusly (fr linearly seperable data) Thus we can chse whichever is easier t slve Dual cubic in n (#examples) Primal cubic in m (#features) 3
4 Recap primal and dual d* = max α:αi 0 min w L(w,α) Dual Slutin strategy: First step: slve the inner prblem (min w L(w,α)) : Taking derivatives f L(w, α) wrt w, setting t 0, and slving fr w : L(w, α) = ½ w 2 i α i (1 y i (w T x i ) ) δl/δw = w - i α i y i x i = 0 w* = i α i y i x i Just like fr the perceptrn with zer initial weights, the ptimal slutin w* is a linear cmbinatin f the x i. We can plug this in, and nw find the α that maximize the uter prblem 4
5 Recap primal and dual Result f first step: The Lagrangian and slutin fr inner prblem are: d* = max α:αi 0 min w L(w,α) L(w, α) = ½ w 2 i α i (1 y i (w T x i ) ) w* = i α i y i x i Slutin strategy: Secnd step: Plug this back int L t get the dual: max α:αi 0 L(w*,α) max α:αi 0 ½ w* 2 i α i (1 y i (w* T x i ) ), use a bit f algebra t get max α:αi 0 i α i ½ i,j y i y j α i α j (x it x) with cnstraints α i 0 and i α i y i = 0. Quadratic prgramming prblem. 5
6 Cmparisn with lgistic regressin In lgistic regressin, the lss was: lg(σ(bw T x i )) fr samples frm the psitive class lg(1- σ(bw T x i )) fr samples frm the negative class lg(1- σ(bw T x i )) = lg(σ(-(bw T x i ))) Fr labels y i = 1/-1 can write: i=1:n lg(σ(y i (bw T x i ))) 6
7 Cmparisn with lgistic regressin In lgistic regressin (with labels 1/-1), the lss was: i=1:n lg(σ(y i (b w T x i ))) λ w 2 (we can include regularizatin) Fr supprt vectr machines, we have the lss i=1:n E (y i (bw T x i )-1) λ w 2 Where E (z) is 0 if z 0, r therwise. λ des nt matter in this case! Only wrks fr linearly separable data (fr nw!) 7
8 SVM frmulatin SVM prblem: Min ½ w 2 w.r.t. w s.t. y i w T x i 1 M x i 0 γ i x i w This can be slved with quadratic prgramming. 8
9 Nn-linearly separable data A linear bundary might be t simple t capture the data. Optin 1: Relax the cnstraints and allw sme pints t be misclassified by the margin. Optin 2: Allw a nnlinear decisin bundary in the input space by finding a linear decisin bundary in an expanded space (similar t adding plynmial terms in linear regressin.) Here x i is replaced by ɸ(x i ), where ɸ is called a feature mapping. 9
10 Sften the primal bjective We wanted t slve: min w ½ w 2 s.t. y i w T x i 1 This can be re-written: min w i E (y i (bw T x i )-1) ½ w 2 where i E (w T x i, y i ) = ( fr a pints inside margin, 0 therwise) Sften misclassificatin cst: min w i L 0-1 (w T x i, y i ) ½ w 2 where i L 0-1 (w T x i, y i ) = (1 fr a misclassificatin, 0 crrect classificatin) But this is a nn-cnvex bjective! 10
11 Sften the primal bjective Nn-cnvex bjective? Functin is cnvex if: average f f(x 1 ) and f(x 2 ) > f(x 1 /2x 2 /2) fr all x 1, x 2. In a graph, this means: when yu cnnect tw lines n the graph, the cnnecting segment is never belw the riginal line Zer-One lss z i =y i (b w T x i ) 11
12 Sften the primal bjective Nn-cnvex bjective? Functin is cnvex if: average f f(x 1 ) and f(x 2 ) > f(x 1 /2x 2 /2) fr all x 1, x 2. In a graph, this means: when yu cnnect tw lines n the graph, the cnnecting segment is never belw the riginal line Zer-One lss z i =y i (b w T x i ) 12
13 Apprximatin f the L 0-1 functin L 0-1 z i =y i (b w T x i ) Cpyright: C. Bishp, Pattern recgnitin and machine learning 13
14 Apprximatin f the L 0-1 functin L 0-1 z i =y i (b w T x i ) Hinge Lss Cpyright: C. Bishp, Pattern recgnitin and machine learning 14
15 Apprximatin f the L 0-1 functin L 0-1 Lgistic regressin z i =y i (b w T x i ) Hinge Lss Cpyright: C. Bishp, Pattern recgnitin and machine learning 15
16 Apprximatin f the L 0-1 functin L 0-1 Quadratic lss Lgistic regressin z i =y i (b w T x i ) Hinge Lss Cpyright: C. Bishp, Pattern recgnitin and machine learning 16
17 SVM with hinge lss Hinge lss: L hin (w T x i, y i ) = max {1-y i w T x i, 0} Sften misclassificatin cst: min w C i L hin (w T x i, y i ) ½ w 2 where C cntrls trade-ff between slack penalty and margin. The hinge lss upper-bunds the 0-1 lss. L hin (w T x i, y i ) L 0-1 (w T x i, y i ) 17
18 Primal Sft SVM prblem Define slack variables ξ i = L hin (w T x i, y i ) = max {1-y i (w T x i b) 0} Slve: ŵ sft = argmin w,ξ C i :1:n ξ i ½ w 2 Add Lagrange mult: s. t. y i (w T x i b) 1 ξ i, i = 1,, n <= α i ξ i 0, i = 1,, n <= β i where w ε R m, ξ ε R n Intrduce Lagrange multipliers: α = ( α 1, α 2,, α n ) T, 0 α i β = ( β 1, β 2,, β n ) T, 0 β i 18
19 Sft SVM prblem: Adding Lagrange multipliers Primal bjective: (w, ξ, α, β) = arg min w,ξ max α,β L(w, ξ, α, β) where L(w, ξ, α, β) = ½ w 2 C i :1:n ξ i - i :1:n α i (y i (w T x i b) -1ξ i ) - i :1:n β i ξ i 19
20 Sft SVM prblem: Adding Lagrange multipliers Primal bjective: (w, ξ, α, β) = arg min w,ξ max α,β L(w, ξ, α, β) where L(w, ξ, α, β) = ½ w 2 C i :1:n ξ i - i :1:n α i (y i w T x i -1ξ i ) - i :1:n β i ξ i Dual (invert min and max): (w, ξ, α, β) = arg max α,β min w,ξ L(w, ξ, α, β) 20
21 Sft SVM prblem: Adding Lagrange multipliers Primal bjective: (w, ξ, α, β) = arg min w,ξ max α,β L(w, ξ, α, β) where L(w, ξ, α, β) = ½ w 2 C i :1:n ξ i - i :1:n α i (y i w T x i -1ξ i ) - i :1:n β i ξ i Dual (invert min and max): (w, ξ, α, β) = arg max α,β min w,ξ L(w, ξ, α, β) Slve: δl/δw = w - i α i y i x i = 0 => w* = i α i y i x i δl/δξ = C1 n α β = 0 => β = C1 n α Lagrange multipliers are psitive, s we have: 0 β i, 0 α i C 21
22 Sft SVM prblem: Adding Lagrange multipliers Primal bjective: (w, ξ, α, β) = arg min w,ξ max α,β L(w, ξ, α, β) where L(w, ξ, α, β) = ½ w 2 C i :1:n ξ i - i :1:n α i (y i w T x i -1ξ i ) - i :1:n β i ξ i Dual (invert min and max): (w, ξ, α, β) = arg max α,β min w,ξ L(w, ξ, α, β) Slve: δl/δw = w - i α i y i x i = 0 => w* = i α i y i x i δl/δξ = C1 n α β = 0 => β = C1 n α Lagrange multipliers are psitive, s we have: 0 β i, 0 α i C Plug int dual : max α i α i ½ i,j y i y j α i α j (x i x) with cnstraints 0 α i C and i α i y i = 0. This is a quadratic prgramming prblem (similar t Hard SVM). 22
23 Sft SVM slutin Sft-SVM has ne mre cnstraint 0 α i C (vs 0 α i in Hard SVM). When C=>, then Sft-SVM=>Hard-SVM. w T x=0 23
24 Sft SVM slutin Sft-SVM has ne mre cnstraint 0 α i C (vs 0 α i in Hard SVM). When C=>, then Sft-SVM=>Hard-SVM. Pints away frm margin have α i = 0, ξ i =0. The fllwing have α i > 0 Pints n the margin, ξ i =0. Pints within the margin, 0 < ξ i < 1 Pints n the decisin line, ξ i = 1. Misclassified pints, ξ i > 1. α j =0 w T x=0 24
25 Sft SVM slutin Sft-SVM has ne mre cnstraint 0 α i C (vs 0 α i in Hard SVM). When C=>, then Sft-SVM=>Hard-SVM. Pints away frm margin have α i = 0, ξ i =0. The fllwing have α i > 0 Pints n the margin, ξ i =0. Pints within the margin, 0 < ξ i < 1 Pints n the decisin line, ξ i = 1. Misclassified pints, ξ i > 1. α j >0, ξ j =0 α j =0 w T x=0 25
26 Sft SVM slutin Sft-SVM has ne mre cnstraint 0 α i C (vs 0 α i in Hard SVM). When C=>, then Sft-SVM=>Hard-SVM. Pints away frm margin have α i = 0, ξ i =0. The fllwing have α i > 0 Pints n the margin, ξ i =0. Pints within the margin, 0 < ξ i < 1 Pints n the decisin line, ξ i = 1. Misclassified pints, ξ i > 1. α j >0, ξ j =0 α j =0 0< ξ j <1 w T x=0 26
27 Sft SVM slutin Sft-SVM has ne mre cnstraint 0 α i C (vs 0 α i in Hard SVM). When C=>, then Sft-SVM=>Hard-SVM. Pints away frm margin have α i = 0, ξ i =0. The fllwing have α i > 0 Pints n the margin, ξ i =0. Pints within the margin, 0 < ξ i < 1 Pints n the decisin line, ξ i = 1. Misclassified pints, ξ i > 1. α j >0, ξ j =0 α j =0 ξ j =1 0< ξ j <1 w T x=0 27
28 Sft SVM slutin Sft-SVM has ne mre cnstraint 0 α i C (vs 0 α i in Hard SVM). When C=>, then Sft-SVM=>Hard-SVM. Pints away frm margin have α i = 0, ξ i =0. The fllwing have α i > 0 Pints n the margin, ξ i =0. Pints within the margin, 0 < ξ i < 1 Pints n the decisin line, ξ i = 1. Misclassified pints, ξ i > 1. α j >0, ξ j =0 α j =0 ξ j =1 0< ξ j <1 ξ j >1 ξ j >1 w T x=0 28
29 Sft SVM slutin Sft-SVM has ne mre cnstraint 0 α i C (vs 0 α i in Hard SVM). When C=>, then Sft-SVM=>Hard-SVM. Pints away frm margin have α i = 0, ξ i =0. The fllwing have α i > 0 Pints n the margin, ξ i =0. Pints within the margin, 0 < ξ i < 1 Pints n the decisin line, ξ i = 1. Misclassified pints, ξ i > 1. T predict n test data: h w (x) = sign( i=1:n α i y i (x i x) ) Only need t stre the supprt vectrs (i.e. pints with α i > 0 ) t predict. α j >0, ξ j =0 α j =0 ξ j =1 0< ξ j <1 ξ j >1 w T x=0 ξ j >1 29
30 Sft SVM, Bias & Variance ŵ sft = argmin w,ξ C i :1:n ξ i ½ w 2 C=>, then Sft-SVM=>Hard-SVM What des lwering C d t bias and variance? Increase bias and decrease variance? Decrease bias and increase variance? 30
31 Nn-linearly separable data A linear bundary might be t simple t capture the data. Optin 1: Relax the cnstraints and allw sme pints t be misclassified by the margin. Optin 2: Allw a nnlinear decisin bundary in the input space by finding a linear decisin bundary in an expanded space (similar t adding plynmial terms in linear regressin.) Here x i is replaced by ɸ(x i ), where ɸ is called a feature mapping. 31
32 Margin ptimizatin in feature space Replacing x i by ɸ(x i ), the ptimizatin prblem fr w becmes: Primal frm: Min ½ w 2 w.r.t. w s.t. y i w T ɸ(x i ) 1 32
33 Margin ptimizatin in feature space Replacing x i by ɸ(x i ), the ptimizatin prblem fr w becmes: Primal frm: Min ½ w 2 w.r.t. w s.t. y i (w T ɸ(x i )b) 1 Dual frm: Max i α i ½ i,j y i y j α i α j (ɸ(x i ) ɸ(x)) w.r.t. s.t. α i α i 0 i α i y i = 0 33
34 Feature space slutin The ptimal weights, in the expanded feature space, are w = i=1:n α i y i ɸ(x i ) Classificatin f an input x is given by: h w (x) = sign( i=1:n α i y i (ɸ(x i ) ɸ(x))b ) 34
35 Feature space slutin The ptimal weights, in the expended feature space, are w = i=1:n α i y i ɸ(x i ) Classificatin f an input x is given by: h w (x) = sign( i=1:n α i y i (ɸ(x i ) ɸ(x))b ) Nte that t slve the SVM ptimizatin prblem in dual frm and t make a predictin, we nly ever need t cmpute dtprducts f feature vectrs. 35
36 Feature space slutin Ptential prblem: Let s say we might need many feature 3 rd rder? If we had m riginal features, nw we have 1 m m 2 m 3 If we had 10 riginal features, we nw have mre than 1000! In primal frm this will be very cstly t slve Even in dual frm we will need t cmpute the inner prducts! 36
37 Kernel functins Whenever a learning algrithm (such as SVMs) can be written in terms f dt-prducts, it can be generalized t kernels. A kernel is any functin K: R m x R m R, which crrespnds t a dt prduct fr sme feature mapping ɸ: K(x 1, x 2 ) = ɸ(x 1 ) ɸ(x 2 ) fr sme ɸ 37
38 Kernel functins Whenever a learning algrithm (such as SVMs) can be written in terms f dt-prducts, it can be generalized t kernels. A kernel is any functin K: R m x R m R, which crrespnds t a dt prduct fr sme feature mapping ɸ: K(x 1, x 2 ) = ɸ(x 1 ) ɸ(x 2 ) fr sme ɸ Cnversely, by chsing feature mapping ɸ, we implicitly chse a kernel functin. Recall that ɸ(x 1 ) ɸ(x 2 ) = ɸ(x 1 ) ɸ(x 2 ) cs (ɸ(x 1 ),ɸ(x 2 ) ), where dentes the angle between the vectrs, s a kernel functin can be thught f as a ntin f similarity. 38
39 Example: Quadratic kernel Let K(x, z) = (x z) 2. Is this a kernel? K(x, z) = ( i=1:m x i z i ) ( j=1:m x j z j ) = i,jε{1..m} x i z i x j z j = i,j ε {1..m} ( x i x j ) ( z i z j ) 39
40 Example: Quadratic kernel Let K(x, z) = (x z) 2. Is this a kernel? K(x, z) = ( i=1:m x i z i ) ( j=1:m x j z j ) = i,jε{1..m} x i z i x j z j = i,j ε {1..m} ( x i x j ) ( z i z j ) We see it is a kernel, with feature mapping: ɸ(x) = < x 12, x 1 x 2,, x 1 x m, x 2 x 1, x 22,, x m 2 > Feature vectr includes all squares f elements and all crss terms. 40
41 Example: Quadratic kernel Let K(x, z) = (x z) 2. Is this a kernel? K(x, z) = ( i=1:m x i z i ) ( j=1:m x j z j ) = i,jε{1..m} x i z i x j z j = i,j ε {1..m} ( x i x j ) ( z i z j ) We see it is a kernel, with feature mapping: ɸ(x) = < x 12, x 1 x 2,, x 1 x m, x 2 x 1, x 22,, x m 2 > Feature vectr includes all squares f elements and all crss terms. Imprtant: Cmputing ɸ takes O(m 2 ) but cmputing K nly takes O(m). 41
42 Plynmial kernels Mre generally, K(x, z) = (x z) d is a kernel, fr any psitive integer d: K(x, z) = ( i=1:m x i z i ) d If we expanded the sum abve in the naïve way, we get m d terms. Terms are mnmials (prducts f x i ) with ttal pwer equal t d. 42
43 Plynmial kernels Mre generally, K(x, z) = (x z) d is a kernel, fr any psitive integer d: K(x, z) = ( i=1:m x i z i ) d If we expanded the sum abve in the naïve way, we get m d terms. Terms are mnmials (prducts f x i ) with ttal pwer equal t d. If we use the primal frm f the SVM, each term gets a weight. Curse f dimensinality: it is very expensive bth t ptimize and t predict with an SVM in primal frm. Hwever, evaluating the dt-prduce f any tw feature vectrs can be dne using K in O(m). 43
44 The kernel trick If we wrk with the dual, we d nt have t ever cmpute the feature mapping ɸ. We just cmpute the similarity kernel K. 44
45 The kernel trick If we wrk with the dual, we d nt have t ever cmpute the feature mapping ɸ. We just cmpute the similarity kernel K. We just replace any inner prduct ɸ(x 1 ) ɸ(x 2 ) with K(x 1, x 2 ) This is justified as any valid kernel defines an inner prduct with sme feature expansin even if we d nt want t actually frm this feature expansin (smetimes it is even impssible) 45
46 The kernel trick We can slve the dual fr the α i with the kernel trick (just replace ɸ(x 1 ) ɸ(x 2 ) with K(x 1, x 2 ) ) Max i=1:n α i ½ i,j=1:n y i y j α i α j K(x i,x j ) w.r.t. α i s.t. α i 0 and i:1..n α i y i = 0 The class f a new input x is cmputed as: h w (x) = sign( i=1:n α i y i K(x i,x ) ) where x i are the supprt vectrs (defining the margin). Remember, K(, ) can be evaluated in O(m) time = big savings! 46
47 Sme ther kernel functins K(x, z) = (1 x z) d - feature expansin has all mnmial terms f ttal pwer. Radial basis / Gaussian kernel: K(x, z) = exp ( - x-z 2 / 2σ 2 ) This kernel has an infinite-dimensinal feature expansin, but dtprducts can still be cmputed in O(m) (where m=#riginal features) It wuld be impssible t use this representatin in a primal frmulatin! Sigmidal kernel: K(x, z) = tanh(c 1 x z c 2 ) 47
48 Example: Gaussian kernel Example: Gaussian kernel Nte the nn-linear decisin bundary 48
49 Nn-parametric Remember that we call mdels that cannt be expressed with a fixed (finite) number f parameters nn-parametric With Gaussian kernels ur feature expansin is infinitedimensinal Thus, SVM with a Gaussian kernel is anther example f a nnparametric mdel Same advantages and disadvantages we discussed in Lecture 7 (Instance-based learning) 49
50 Kernels beynd SVMs A lt f research related t defining kernel functins suitable t particular tasks / kinds f inputs (e.g. wrds, graphs, images). Many kernels are available: Infrmatin diffusin kernels (Lafferty and Lebann, 2002) Diffusin kernels n graphs (Kndr and Jebara, 2003) String kernels fr text classificatin (Ldhi et al, 2002) String kernels fr prtein classificatin (Leslie et al, 2002) and thers! 50
51 Example: String kernels Very imprtant fr DNA matching, text classificatin, Often use a sliding windw f length k ver the tw strings that we want t cmpare. Within the fixed-size windw we can d many things: Cunt exact matches. Weigh mismatches based n hw bad they are. Cunt certain markers, e.g. AGT. The kernel is the sum f these similarities ver the tw sequences. 51
52 Kernelizing ther ML algrithms Many ther machine learning algrithms have a dual frmulatin, in which dt-prducts f features can be replaced by kernels. Examples: Perceptrn Lgistic regressin Linear regressin (We ll d this later in the curse!) 52
53 Multiple classes One-vs-All: Learn K separate binary classifiers. Can lead t incnsistent results. Training sets are imbalanced, e.g. assuming n examples per class, each binary classifier is trained with psitive class having 1*n f the data, and negative class having (K-1)*n f the data. 53
54 Multiple classes One-vs-All: Learn K separate binary classifiers. Can lead t incnsistent results. Training sets are imbalanced, e.g. assuming n examples per class, each binary classifier is trained with psitive class having 1*n f the data, and negative class having (K-1)*n f the data. Multi-class SVM: Define the margin t be the gap between the crrect class and the nearest ther class. 54
55 SVMs fr regressin Minimize a regularized errr functin: ŵ = argmin w C i :1:n ( y i - w T x i ) 2 ½ w 2 Intrduce slack variables t ptimize tube arund the regressin functin. 55
56 SVMs fr regressin Minimize a regularized errr functin: ŵ = argmin w C i :1:n ( y i - w T x i ) 2 ½ w 2 Intrduce slack variables t ptimize tube arund the regressin functin. Typically, relax t ε-sensitive errr n the linear target t ensure sparse slutin (i.e. few supprt vectrs): ŵ = argmin w C i :1:n E ε ( y i - w T x i ) 2 ½ w 2 where E ε = 0 if ( y i - w T x i )<ε, (y i - w T x i ) ε therwise 56
57 What yu shuld knw Frm last class and frm tday: Perceptrn algrithm. Margin definitin fr linear SVMs. Use f Lagrange multipliers t transfrm ptimizatin prblems. Primal and dual ptimizatin prblems fr SVMs. Feature space versin f SVMs. The kernel trick and examples f cmmn kernels. 57
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