COM S 578X: Optimization for Machine Learning

Transcription

1 COM S 578X: Optimization for Machine Learning Lecture Note 4: Duality Jia (Kevin) Liu Assistant Professor Department of Computer Science Iowa State University, Ames, Iowa, USA Fall 2018 JKL (CS@ISU) COM S 578X: Lecture 4 1/21

2 Outline In this lecture: Lagrange dual problem Weak and strong duality Geometric interpretation Examples in machine learning JKL COM S 578X: Lecture 4 2/21

3 The Lagrangian Standard optimization problem (may or may not be convex): convex Opt : Minimize f(x) it :t convex P : Vi dwdvarormk9471es convex fi g subject to g i (x) apple 0, i =1,,m Ui > o Hi hit 'AE h j (x) =0, j =1,,p Uj EIR 'tj variable Pis x convex 2 R n, opt hike? domain D, optimal value p Lagrangian: L: R n R m R p! R, withdom(l) =D R m R p : mx px L(x, u, v) =f(x)+ u i g i (x)+ v j h j (x) i=1 j=1 weighted sum of objective and constraint functions u i 0 is dual variable (Lagrangian multiplier) associated with g i (x) apple 0 v j 2 R is dual variable (Lagrangian multiplier) associated with h j (x) =0 JKL (CS@ISU) COM S 578X: Lecture 4 3/21

4 ' value ( Lagrangian Dual Function Lagrangian dual function: :R m R p! R: (u, v) = inf L(x, u, v) x2d 0 mx It = u i g i (x)+ (u, v) is concave in u, v, canbe 1 for some u, v bk its ' a pt Lower bound property: If u Proof If x 2Dand u 20<0 =O px t v j h j (x) A x2d i=1 j=1 we EO wise 0, then (u, v) apple p minimum Wirt y 1,1 ) opt of Primal problem by + to ", detyninj, t 4 0, then f( x) L( x, u, v) inf L(x, u, v) = (u, v) x2d Since this holds for any x 2D, minimizing over all x yields p 1 of altmetns DET Lagrangian dual (u, v) JKL (CS@ISU) COM S 578X: Lecture 4 4/21

5 Example: Leastnorm solution of linear equations Minimize x > x subject to Ax = b I EIR " Dual function: Lagrangian is: L(x, v) =x > x + v > (Ax b) To minimize L(x, v) over x, set gradient equal to zero r x L(x, v) =2x + A > v =0 =) x = 1 2 A> v Plug it in L to obtain : (v) =L( which is clearly a concave function of v it 1 2 A> v, v) = 1 4 v> AA > v b > v, NSD Eo Lower bound property: p 1 4 v> AA > v b > v for all v Largest LB JKL (CS@ISU) COM S 578X: Lecture 4 5/21

6 Example: Linear Programming Dual function: Lagrangian is: Minimize c > x subject to Ax = b, x 0 a a IEIR " YEIR "+ L is a ne in x, hence L(x, u, v) =c > x + v > (Ax b) u > x (u, v) =infl(x, u, v) = x = b > v +(c + A > v u) > x 3 w : ni, ( b > v A > v u + c =0 1 otherwise = = at is linear on a ne domain {(u, v) :A > v u + c = 0}, hence concave Lower bound property: p b > v if A > v + c 0 w win + case JKL (CS@ISU) COM S 578X: Lecture 4 6/21

7 Example: Equality Constrained Norm Minimization Minimize kxk subject to Ax = b Dual function: (v) =inf x (kxk v> Ax + b > v)= ( b > v ka > vk apple 1 1 otherwise = bit tenets { MI where kvk, sup kukapple1 u > v (referred to as dual Effeya } = EK see norm of k k), { I # a Proof It follows from the fact that inf x (kxk y > x)=0if kyk apple 1, 1 otherwise If kyk apple 1, thenkxk y > x 0, 8x, withequalityifx = 0 ( Cauchy schwatd, if kyk > 1, choose x = tu, whereu satisfies kuk apple1, u > y = kyk > 1: kxk y > x = t(kuk kyk )! 1 as t!1 Lower bound property: p b > v if ka > vk apple 1 T > 0 = E I > 1 we I EIR " thea JKL (CS@ISU) COM S 578X: Lecture 4 7/21

8 NP Example: Twoway Partitioning Minimize x > Wx = IEFe, c subject to x 2 i =1, i =1,,n Ii E { I, 13 A nonconvex problem; feasible set contains 2 n, ti Hard discrete points Interpretation: partition {1,,n} into two sets; (W) ij is cost of assigning i, j to the same set; (W) ij is cost of assigning to di erent sets Dual function: (v) =inf x (x> Wx + X i Lower bound property: p v i (x 2 i symmetric 1)) = inf x {x> W + Diag(v) x 1 > v} ( 1 > v W+ Diag{v} 0 = 1 otherwise 1 > v if W + Diag{v} 0 Example: v = min(w)1 gives nontrivial bound p n min (W) a " xilhflijnj diagonal matrix I ER " # ( l ;] : t atuitttfivnl =±y YII* ' JKL (CS@ISU) COM S 578X: Lecture 4 8/21

9 Lagrangian Dual and Conjugate Function Dual function: Minimize f(x) subject to Ax apple b, Cx = d erm+ * ytern (u, v) = inf x2dom(f) f(x)+(a > u + C > v) > x b > u d > v = f ( A > u C > v) b > u d > v Legendre Fenehelconjngatekonvex conjugate Definition of conjugate function: f (y) =sup x2dom(f) y > x f(x) Simplifies derivation of dual if conjugate of f is known Example: Entropy maximization f(x) = nx x i log x i, f (y) = i=1 nx exp(y i 1) i=1 JKL (CS@ISU) COM S 578X: Lecture 4 9/21

10 The Lagrangian Dual Problem Maximize (u, v) subject to u 0 if Min EQ OH, v ) Finds largest lower bound on p, obtained from Lagrangian dual function A convex optimization problem; optimal value denoted d u, v are dual feasible if u 0, (u, v) 2 dom( ) Often simplified by making implicit constraint u 0, (u, v) 2 dom( ) explicit Example: Standard form LP and its dual: Minimize c > x subject to Ax = b, x 0 Maximize b > v subject to A > v + c 0 More on this later JKL (CS@ISU) COM S 578X: Lecture 4 10 / 21

11 Weak and Strong Duality Weak duality: d apple p Always holds (for convex and nonconvex problems) Can be used to find nontrivial lower bounds for di cult problems For example, solving SDP semi definite Programming Maximize subject to 1 > v W + Diag{v} 0 yields a lower bound for the twoway partitioning problem Strong duality: d = p Does not hold in general Usually hold for convex problems mild ( under certaihhonditions ) Conditions that guarantee strong duality in convex problems are called constraint qualifications ( CQ ) JKL (CS@ISU) COM S 578X: Lecture 4 11 / 21

12 Slater s Constraint Qualification Strong duality holds for a convex problem Minimize f(x) subject to g i (x) apple 0, Ax = b i =1,,m if it is strictly feasible, ie, 9x 2 intd : g i (x) < 0, i =1,, m, Ax = b Also guarantees that the dual optimum is attained (if p > 1) Can be further relaxed: eg, can replace intd with relintd (interior relative to a ne hull); linear inequalities do not need to hold with strict inequality, There exist many other types of constraint qualifications (see [BSS, Ch 5]) JKL (CS@ISU) COM S 578X: Lecture 4 12 / 21

13 Other WellKnown CQs LICQ (Linear independence CQ): Gradients of active inequality constraints and gradients of equality constraints are linearly independent at x MFCQ (ManasarianFromovitz CQ): Gradients of equality constraints are LI at x, 9d 2 R n such that rg > i (x )d < 0 for active inequality constraints, rh > j (x )d =0for equality constraints CRCQ (Constant rank CQ): For each subset of gradients of active inequality constraints & gradients of equality constraints, the rank at x s vicinity is constant CPLD (Constant positive linear dependence CQ): For each subset of gradients of active inequality constraints & gradients of equality constraints, if it s positivelinear dep at x then it s positivelinear dependent in x s vicinity QNCQ (Quasinormality CQ): If gradients of active inequality constraints and gradients of equality constraints are positivelylinearly dependent at x with duals u i for inequalities and v j for equalities, then there is no sequence x k! x such that v j 6=0) v j h j (x k ) > 0 and u i 6=0) u i g i (x k ) < 0 JKL (CS@ISU) COM S 578X: Lecture 4 13 / 21

14 value 9 " value Geometric Interpretation Mi?fg function " c Bssr op g) peg, sit, p*=pco, For ease of exposition, consider problem with one constraint g(x) apple 0 Interpretation of dual function P sent lines I Perturbed (u) = inf (uq + p), where G = {(g(x),f(x)) x 2D} (q,p)2g t ) p hp =p # obj T q pcobj ppc ) E HI, ptxaicm I * y " c RAS ) q uq + p = (u) is an supporting hyperplane to set G with slope u Hyperplane intersects paxis at p = (u) JKL (CS@ISU) COM S 578X: Lecture 4 14 / 21

15 Geometric Interpretation Epigraph variation: Same interpretation if G is replaced with A = {(q, p) g(x) apple q, f(x) apple p for some x 2D} v : Strong duality Holds if there is a nonvertical supporting hyperplane to A at (0,p ) For convex problem, A is convex, hence has supporting hyperplane at (0,p ) Slater s condition: If there exist ( q, p) 2Awith q <0, then supporting hyperplanes at (0,p ) must be nonvertical JKL (CS@ISU) COM S 578X: Lecture 4 15 / 21

16 vertical Strong Duality Geometric Interpretation p I A Existence i " " of Teo forcing the appointing hyperplane to be non clearly, Slater 's condition does NOT hold JKL (CS@ISU) COM S 578X: Lecture 4 16 / 21 q

17 e Example: Inequality Form of LP Primal problem: Dual function: Dual problem: (u) =inf x Minimize c > x subject to Ax apple b T = LCE, 4) Eta t a I A a I s ZE so to ( (c + A > u) > x b > u = :* Ta : e Maximize b > u Is E subject to A > u + c = 0, u 0 From Slater s condition: p = d if A x < b for some x b > u A > u + c = 0 1 otherwise In fact, p = d except when primal and dual are infeasible I 's ) L JKL (CS@ISU) COM S 578X: Lecture 4 17 / 21

18 , set to O LES of Example: Quadratic Program Primal problem (P 2 S n ++): Minimize x > Px subject to Ax < b 130 Dual function: ( )=inf x Dual problem: take x > Px + u > (Ax b) = 1 4 u> AP 1 A > u b > u = quadratic fu k = e der ZAZ the Maximize 1 4 u> AP 1 A > u b > u subject to u 0 From Slater s condition: p = d if A x < b for some x In fact, p = d always JKL (CS@ISU) COM S 578X: Lecture 4 18 / 21

19 obj! large Example: Support Vector Machine " i ::* i itb s ),, FITYi III Given labels y 2 { 1, 1} n,featurevectorsx 1,,x m LetX,[x 1,,x m ] > hint be I Recall from Lecture 1 that the support vector machine problem: 1 mx Minimize w,b, 2 w> w + C i guess : I=o b = 0, i=1 choose E suft subject to y i (w > x i + b) 1 i, i =1,,m i 0, i =1,,m Ui 30 Vi 30 Introducing dual variables u, v 0 to obtain the Lagrangian: L(w,b,, u, v) = 1 mx nx 2 kwk2 2 + C i + u i (1 i y i (w > x i + b)) i=1 i=1 I orig penalty Minimizing over w,b, yields the Lagrangian dual function terms nx v i i i=1 JKL (CS@ISU) COM S 578X: Lecture 4 19 / 21

20 Example: Support Vector Machine (u, v) = ( 1 2 u> X> Xu + 1 > u if u = C1 v, u > y =0 1 otherwise where X, XDiag{y 1,,y m } The dual problem, after eliminating v, becomes: 1 Minimize 2 u> X> Xu 1 > u subject to u > y =0, 0 apple u i apple C, i =1,, m Slater s condition is satisfied ) strong duality In ML literature, more common to work with the dual: Quadratic having PSD Hessian with simple bounds, plus a single linear & simple box constraints At optimality, we can verify that w = Xu (this is not a coincidence! will be proved by KKT conditions in next class) e WEE g, b o, e sufficiently large JKL (CS@ISU) COM S 578X: Lecture 4 20 / 21

21 Next Class Optimality Conditions JKL COM S 578X: Lecture 4 21 / 21