Submodular Functions, Optimization, and Applications to Machine Learning

Transcription

1 Submodular Functions, Optimization, and Applications to Machine Learning Spring Quarter, Lecture 17 Prof. Jeff Bilmes University of Washington, Seattle Department of Electrical Engineering May 23st, 2018 Clockwise from top left:v Lásló Lovász Jack Edmonds Satoru Fujishige George Nemhauser Laurence Wolsey András Frank Lloyd Shapley H. Narayanan Robert Bixby William Cunningham William Tutte Richard Rado Alexander Schrijver Garrett Birkhoff Hassler Whitney Richard Dedekind Prof. Jeff Bilmes + f (A) + f (B) f (A [ B) = f (Ar ) + 2f (C ) + f (Br ) = f (Ar ) +f (C ) + f (Br ) f (A \ B) = f (A \ B) EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F1/54 (pg.1/192)

2 Logistics Review Announcements, Assignments, and Reminders Next homework will be posted tonight. Rest of the quarter. One more longish homework. Take home final exam (like a long homework). As always, if you have any questions about anything, please ask then via our discussion board ( Can meet at odd hours via zoom (send message on canvas to schedule time to chat). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F2/54 (pg.2/192)

3 Logistics Class Road Map - EE563 Review L1(3/26): Motivation, Applications, & Basic Definitions, L2(3/28): Machine Learning Apps (diversity, complexity, parameter, learning target, surrogate). L3(4/2): Info theory exs, more apps, definitions, graph/combinatorial examples L4(4/4): Graph and Combinatorial Examples, Matrix Rank, Examples and Properties, visualizations L5(4/9): More Examples/Properties/ Other Submodular Defs., Independence, L6(4/11): Matroids, Matroid Examples, Matroid Rank, Partition/Laminar Matroids L7(4/16): Laminar Matroids, System of Distinct Reps, Transversals, Transversal Matroid, Matroid Representation, Dual Matroids L8(4/18): Dual Matroids, Other Matroid Properties, Combinatorial Geometries, Matroids and Greedy. L9(4/23): Polyhedra, Matroid Polytopes, Matroids! Polymatroids L10(4/29): Matroids! Polymatroids, Polymatroids, Polymatroids and Greedy, L11(4/30): Polymatroids, Polymatroids and Greedy L12(5/2): Polymatroids and Greedy, Extreme Points, Cardinality Constrained Maximization L13(5/7): Constrained Submodular Maximization L14(5/9): Submodular Max w. Other Constraints, Cont. Extensions, Lovasz Extension L15(5/14): Cont. Extensions, Lovasz Extension, Choquet Integration, Properties L16(5/16): More Lovasz extension, Choquet, defs/props, examples, multiliear extension L17(5/21): Finish L.E., Multilinear Extension, Submodular Max/polyhedral approaches, Most Violated inequality, Still More on Matroids, Closure/Sat L18(5/23): L (5/28): Memorial Day (holiday) L19(5/30): L21(6/4): Final Presentations maximization. Last day of instruction, June 1st. Finals Week: June 2-8, Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F3/54 (pg.3/192)

4 Logistics One slide review of concave relaxation Review convex closure ˇf(x) =min p24 n (x) E S p [f(s)], wherewhere4 n (x) = n p 2 R 2n : P S V p S =1,p S 08S V, & P o S V p S1 S = x Edmonds extension f(w) = max(wx : x 2 B f ) Lovász extension f LE (w) = P m i=1 if(e i ),with i such that w = P m i=1 i1 Ei f(w) = max 2 [m] w c, [m] set of m! permutations of [m], 2 [m] apermutation,c vector with c i = f(e i ) f(e i 1 ), E i = {e 1,e 2,...,e i }. Choquet integral C f (w) = P m i=1 (w e i w ei+1 )f(e i ) f(w) = R +1 1 ˆf( )d, ˆf( ) = All the same when f is submodular. ( f({w }) if 0 f({w }) f(e) if <0 Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F4/54 (pg.4/192)

5 Logistics Lovász extension properties Using the above, have the following (some of which we ve seen): Theorem Let f,g :2 E! R be normalized (f(;) =g(;) =0). Then 1 Superposition of LE operator: Given f and g with Lovász extensions f and g then f + g is the Lovász extension of f + g and f is the Lovász extension of f for 2 R. 2 If w 2 R E + then f(w) = R +1 0 f({w })d. 3 For w 2 R E,and 2 R, we have f(w + 1 E )= f(w)+ f(e). 4 Positive homogeneity: I.e., f( w) = f(w) for 0. 5 For all A E, f(1 A )=f(a). 6 f symmetric as in f(a) =f(e \ A), 8A, then f(w) = f( w) ( f is even). 7 Given partition E 1 [ E 2 [ [E k of E and w = P k i=1 i1 Ek with 1 2 k, and with E 1:i = E 1 [ E 2 [ [E i, then f(w) = P k i=1 if(e i E 1:i 1 )= P k 1 i=1 f(e1:i )( i i+1 )+f(e) k. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F5/54 (pg.5/192). 0 Review

6 Example: m =3, E = {1, 2, 3} In order to visualize in 3D, we make a few simplifications. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F6/54 (pg.6/192)

7 Example: m =3, E = {1, 2, 3} In order to visualize in 3D, we make a few simplifications. # # I -ate Consider any submodular f 0 and x 2 B f 0.Thenf(A) =f 0 (A) is submodular x(a) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F6/54 (pg.7/192)

8 Example: m =3, E = {1, 2, 3} + ( EKFCE ) In order to visualize in 3D, we make a few simplifications. Consider any submodular f 0 and x 2 B f 0.Thenf(A) =f 0 (A) is submodular, andmoreoverf(e) =f 0 (E) x(e) =0. / x(a) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F6/54 (pg.8/192)

9 Example: m =3, E = {1, 2, 3} In order to visualize in 3D, we make a few simplifications. Consider any submodular f 0 and x 2 B f 0.Thenf(A) =f 0 (A) is submodular, andmoreoverf(e) =f 0 (E) x(e) =0. Hence, from f(w + 1 E )= f(w)+ f(e), wehavethat f(w + 1 E )= f(w) when f(e) =0. x(a) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F6/54 (pg.9/192)

10 Example: m =3, E = {1, 2, 3} In order to visualize in 3D, we make a few simplifications. Consider any submodular f 0 and x 2 B f 0.Thenf(A) =f 0 (A) is submodular, andmoreoverf(e) =f 0 (E) x(e) =0. Hence, from f(w + 1 E )= f(w)+ f(e), wehavethat f(w + 1 E )= f(w) when f(e) =0. x(a) Thus, n we can look down on the contour plot of the Lovász extension, w : f(w) o =1,fromavantagepointrightontheline {x : x = 1 E, >0} since moving in direction 1 E changes nothing. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F6/54 (pg.10/192)

11 Example: m =3, E = {1, 2, 3} In order to visualize in 3D, we make a few simplifications. Consider any submodular f 0 and x 2 B f 0.Thenf(A) =f 0 (A) is submodular, andmoreoverf(e) =f 0 (E) x(e) =0. Hence, from f(w + 1 E )= f(w)+ f(e), wehavethat f(w + 1 E )= f(w) when f(e) =0. x(a) Thus, n we can look down on the contour plot of the Lovász extension, w : f(w) o =1,fromavantagepointrightontheline {x : x = 1 E, >0} since moving in direction 1 E changes nothing. I.e., consider 2D plane perpendicular to the line {x : 9, x = 1 E } at any point along that line, then Lovász extension is surface plot with coordinates on that plane (or alternatively we can view contours). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F6/54 (pg.11/192)

12 Example: m =3, E = {1, 2, 3} Example 1 (from Bach-2011): f(a) =1 A 2{1,2} =min{ A, 1} +min{ E \ A, 1} 1 is submodular, and f(w) = max k2{1,2,3} w k min k2{1,2,3} w k. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F7/54 (pg.12/192)

13 Example: m =3, E = {1, 2, 3} a Example 1 (from Bach-2011): f(a) =1 A 2{1,2} =min{ A, 1} +min{ E \ A, 1} 1 is submodular, and f(w) = max k2{1,2,3} w k min k2{1,2,3} w k. :* } w 1> w 3>w2 w > w >w 3 (1,0,1)/F({1,3}) 1 2 w =w 1 2 (0,0,1)/F({3}) w > w >w w > w >w (0,1,1)/F({2,3}) (1,0,0)/F({1}) w =w 2 3 w > w >w w 2> w 1>w3 (0,1,0)/F({2}) w =w 1 3 Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F7/54 (pg.13/192)

14 Example: m =3, E = {1, 2, 3} Example 2 (from Bach-2011): f(a) = 1 12A 1 22A A 1 32A (1,0,1)/2 (0,0,1) (0,1,1) (1,0,0) (0,1,0)/2 (1,1,0) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F8/54 (pg.14/192)

15 Example: m =3, E = {1, 2, 3} Example 2 (from Bach-2011): f(a) = 1 12A 1 22A A 1 32A This gives a total variation function for the Lovász extension, with f(w) = w 1 w 2 + w 2 w 3. (1,0,0) (1,0,1)/2 (0,0,1) (0,1,1) (0,1,0)/2 (1,1,0) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F8/54 (pg.15/192)

16 Example: m =3, E = {1, 2, 3} Example 2 (from Bach-2011): f(a) = 1 12A 1 22A A 1 32A This gives a total variation function for the Lovász extension, with f(w) = w 1 w 2 + w 2 w 3. When used as a prior, prefers piecewise-constant signals (e.g., P i w i w i+1 ). (1,0,0) (1,0,1)/2 (0,0,1) (0,1,1) (0,1,0)/2 (1,1,0) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F8/54 (pg.16/192)

17 Total Variation Example From Nonlinear total variation based noise removal algorithms Rudin, Osher, and Fatemi, Top left original, bottom right total variation. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F9/54 (pg.17/192)

18 Example: Lovász extension of concave over modular Let m : E! R + be a modular function and define f(a) =g(m(a)) where g is concave. Then f is submodular. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F10/54 (pg.18/192)

19 Example: Lovász extension of concave over modular Let m : E! R + be a modular function and define f(a) =g(m(a)) where g is concave. Then f is submodular. Let M j = P j i=1 m(e i) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F10/54 (pg.19/192)

20 Example: Lovász extension of concave over modular Let m : E! R + be a modular function and define f(a) =g(m(a)) where g is concave. Then f is submodular. Let M j = P j i=1 m(e i) f(w) is given as f(w) = mx w(e i ) g(m i ) g(m i 1 ) (17.1) i=1 Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F10/54 (pg.20/192)

21 Example: Lovász extension of concave over modular Let m : E! R + be a modular function and define f(a) =g(m(a)) where g is concave. Then f is submodular. Let M j = P j i=1 m(e i) f(w) is given as f(w) = mx w(e i ) g(m i ) g(m i 1 ) (17.1) i=1 And if m(a) = A, we get f(w) = mx w(e i ) g(i) g(i 1) (17.2) i=1 Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F10/54 (pg.21/192)

22 Example: Lovász extension and cut functions Cut Function: Given a non-negative weighted graph G =(V,E,m) where m : E! R + is a modular function over the edges, we know from Lecture 2 that f :2 V! R + with f(x) =m( (X)) where (X) ={(u, v) (u, v) 2 E,u 2 X, v 2 V \ X} is non-monotone submodular. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F11/54 (pg.22/192)

23 Example: Lovász extension and cut functions Cut Function: Given a non-negative weighted graph G =(V,E,m) where m : E! R + is a modular function over the edges, we know from Lecture 2 that f :2 V! R + with f(x) =m( (X)) where - (X) ={(u, v) (u, v) 2 E,u 2 X, v 2 V \ X} is non-monotone submodular. Simple way to write it, with m ij = m((i, j)): X f(x) = m ij (17.3) i2x,j2v \X Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F11/54 (pg.23/192)

24 Example: Lovász extension and cut functions Cut Function: Given a non-negative weighted graph G =(V,E,m) where m : E! R + is a modular function over the edges, we know from Lecture 2 that f :2 V! R + with f(x) =m( (X)) where (X) ={(u, v) (u, v) 2 E,u 2 X, v 2 V \ X} is non-monotone submodular. Simple way to write it, with m ij = m((i, j)): X f(x) = m ij (17.3) i2x,j2v \X Exercise: show that Lovász extension of graph cut may be written as: f(w) = X m ij max {(w i w j ), 0} (17.4) i,j2v where elements are ordered as usual, w 1 w 2 w n. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F11/54 (pg.24/192)

25 Example: Lovász extension and cut functions Cut Function: Given a non-negative weighted graph G =(V,E,m) where m : E! R + is a modular function over the edges, we know from Lecture 2 that f :2 V! R + with f(x) =m( (X)) where (X) ={(u, v) (u, v) 2 E,u 2 X, v 2 V \ X} is non-monotone submodular. Simple way to write it, with m ij = m((i, j)): X f(x) = m ij (17.3) i2x,j2v \X Exercise: show that Lovász extension of graph cut may be written as: f(w) = X m ij max {(w i w j ), 0} (17.4) i,j2v where elements are ordered as usual, w 1 w 2 w n. This is also a form of total variation Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F11/54 (pg.25/192)

26 A few more Lovász extension examples Some additional submodular functions and their Lovász extensions, where w(e 1 ) w(e 2 ) w(e m ) 0. LetW k, P k i=1 w(e i). f(a) f(w) A kwk 1 min( A, 1) kwk 1 min( A, 1) max( A m +1, 0) kwk 1 min i w i min( A,k) W k min( A,k) max( A (n k)+1, 1) 2W k W m min( A, E \ A ) 2W bm/2c W m (thanks to K. Narayanan). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F12/54 (pg.26/192)

27 Supervised And Unsupervised Machine Learning Given training data D = {(x i,y i )} m i=1 with (x i,y i ) 2 R n R, perform the following risk minimization problem: 1 mx mask min `(y i,w x i "u@ )+ (w), (17.5) w2r n m.± Q # i=1 where `( ) is a loss function (e.g., squared error) and (w) is a norm. When data has multiple responses (x i,y i ) 2 R n R k,learningbecomes: min w 1,...,w k 2R n kx j=1 1 m mx `(yi k, (w k ) x i )+ (w k ), (17.6) i=1 When data has multiple responses only that are observed, (y i ) 2 R k we get dictionary learning (Krause & Guestrin, Das & Kempe): min min kx x 1,...,x m w 1,...,w k 2R n j=1 1 m mx `(yi k, (w k ) x i )+ (w k ), (17.7) i=1 Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F13/54 (pg.27/192)

28 Norms, sparse norms, and computer vision Common norms include p-norm (w) =kwk p =( P p i=1 wp i )1/p Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F14/54 (pg.28/192)

29 Norms, sparse norms, and computer vision Common norms include p-norm (w) =kwk p =( P p i=1 wp i )1/p 1-norm promotes sparsity (prefer solutions with zero entries). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F14/54 (pg.29/192)

30 Norms, sparse norms, and computer vision Common norms include p-norm (w) =kwk p =( P p i=1 wp i )1/p 1-norm promotes sparsity (prefer solutions with zero entries). Image denoising, total variation is useful, norm takes form: NX (w) = w i w i 1 (17.8) i=2 Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F14/54 (pg.30/192)

31 Norms, sparse norms, and computer vision Common norms include p-norm (w) =kwk p =( P p i=1 wp i )1/p 1-norm promotes sparsity (prefer solutions with zero entries). Image denoising, total variation is useful, norm takes form: NX (w) = w i w i 1 (17.8) i=2 Points of difference should be sparse (frequently zero). (Rodriguez, 2009) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F14/54 (pg.31/192)

32 Submodular parameterization of a sparse convex norm Prefer convex norms since they can be solved. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F15/54 (pg.32/192)

33 Submodular parameterization of a sparse convex norm Prefer convex norms since they can be solved. For w 2 R V, supp(w) 2{0, 1} V has supp(w)(v) =1iff w(v) > 0 Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F15/54 (pg.33/192)

34 Submodular parameterization of a sparse convex norm Prefer convex norms since they can be solved. For w 2 R V, supp(w) 2{0, 1} V has supp(w)(v) =1iff w(v) > 0 Desirable sparse norm: count the non-zeros, kwk 0 = 1 supp(w). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F15/54 (pg.34/192)

35 Submodular parameterization of a sparse convex norm Prefer convex norms since they can be solved. For w 2 R V, supp(w) 2{0, 1} V has supp(w)(v) =1iff w(v) > 0 Desirable sparse norm: count the non-zeros, kwk 0 = 1 supp(w). Using (w) =kwk 0 is NP-hard, instead we often optimize tightest convex relaxation, kwk 1 which is the convex envelope. t.tt Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F15/54 (pg.35/192)

36 Submodular parameterization of a sparse convex norm Prefer convex norms since they can be solved. For w 2 R V, supp(w) 2{0, 1} V has supp(w)(v) =1iff w(v) > 0 Desirable sparse norm: count the non-zeros, kwk 0 = 1 supp(w). Using (w) =kwk 0 is NP-hard, instead we often optimize tightest convex relaxation, kwk 1 which is the convex envelope. With kwk 0 or its relaxation, each non-zero element has equal degree of penalty. Penalties do not interact. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F15/54 (pg.36/192)

37 Submodular parameterization of a sparse convex norm Prefer convex norms since they can be solved. For w 2 R V, supp(w) 2{0, 1} V has supp(w)(v) =1iff w(v) > 0 Desirable sparse norm: count the non-zeros, kwk 0 = 1 supp(w). Using (w) =kwk 0 is NP-hard, instead we often optimize tightest convex relaxation, kwk 1 which is the convex envelope. With kwk 0 or its relaxation, each non-zero element has equal degree of penalty. Penalties do not interact. Given submodular function f :2 V! R +, f(supp(w)) measures the complexity of the non-zero pattern of w; canhavemorenon-zero values if they cooperate (via f) withothernon-zerovalues. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F15/54 (pg.37/192)

38 Submodular parameterization of a sparse convex norm Prefer convex norms since they can be solved. For w 2 R V, supp(w) 2{0, 1} V has supp(w)(v) =1iff w(v) > 0 Desirable sparse norm: count the non-zeros, kwk 0 = 1 supp(w). Using (w) =kwk 0 is NP-hard, instead we often optimize tightest convex relaxation, kwk 1 which is the convex envelope. With kwk 0 or its relaxation, each non-zero element has equal degree of penalty. Penalties do not interact. Given submodular function f :2 V! R +, f(supp(w)) measures the complexity of the non-zero pattern of w; canhavemorenon-zero values if they cooperate (via f) withothernon-zerovalues. f(supp(w)) is hard to optimize, but it s convex envelope f( w ) (i.e., largest convex under-estimator of f(supp(w))) is obtained via the Lovász-extension f of f (Vondrák 2007, Bach 2010). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F15/54 (pg.38/192)

39 Submodular parameterization of a sparse convex norm Prefer convex norms since they can be solved. For w 2 R V, supp(w) 2{0, 1} V has supp(w)(v) =1iff w(v) > 0 Desirable sparse norm: count the non-zeros, kwk 0 = 1 supp(w). Using (w) =kwk 0 is NP-hard, instead we often optimize tightest convex relaxation, kwk 1 which is the convex envelope. With kwk 0 or its relaxation, each non-zero element has equal degree of penalty. Penalties do not interact. Given submodular function f :2 V! R +, f(supp(w)) measures the complexity of the non-zero pattern of w; canhavemorenon-zero values if they cooperate (via f) withothernon-zerovalues. f(supp(w)) is hard to optimize, but it s convex envelope f( w ) (i.e., largest convex under-estimator of f(supp(w))) isobtainedviathe Lovász-extension f 8 of f (Vondrák 2007, Bach 2010). Submodular functions thus parameterize structured convex sparse norms via the Lovász-extension! Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F15/54 (pg.39/192)

40 Submodular parameterization of a sparse convex norm Prefer convex norms since they can be solved. For w 2 R V, supp(w) 2{0, 1} V has supp(w)(v) =1iff w(v) > 0 Desirable sparse norm: count the non-zeros, kwk 0 = 1 supp(w). Using (w) =kwk 0 is NP-hard, instead we often optimize tightest convex relaxation, kwk 1 which is the convex envelope. With kwk 0 or its relaxation, each non-zero element has equal degree of penalty. Penalties do not interact. Given submodular function f :2 V! R +, f(supp(w)) measures the complexity of the non-zero pattern of w; canhavemorenon-zero values if they cooperate (via f) withothernon-zerovalues. f(supp(w)) is hard to optimize, but it s convex envelope f( w ) (i.e., largest convex under-estimator of f(supp(w))) is obtained via the Lovász-extension f of f (Vondrák 2007, Bach 2010). Submodular functions thus parameterize structured convex sparse norms via the Lovász-extension! Ex: total variation is Lovász-ext. of graph cut, but 9 many more! Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F15/54 (pg.40/192)

41 Lovász extension and norms Using Lovász extension to define various norms of the form kwk f = f( w ). Thisrendersthefunctionsymmetricaboutallorthants (meaning, kwk f = kb wk f for any b 2 { 1, 1} m and is element-wise multiplication). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F16/54 (pg.41/192)

42 Lovász extension and norms Using Lovász extension to define various norms of the form kwk f = f( w ). Thisrendersthefunctionsymmetricaboutallorthants (meaning, kwk f = kb wk f for any b 2 { 1, 1} m and is element-wise multiplication). Simple example. The Lovász extension of the modular function f(a) = A is the `1 norm, and the Lovász extension of the modular function f(a) =m(a) is the weighted `1 norm. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F16/54 (pg.42/192)

43 Lovász extension and norms Using Lovász extension to define various norms of the form kwk f = f( w ). Thisrendersthefunctionsymmetricaboutallorthants (meaning, kwk f = kb wk f for any b 2 { 1, 1} m and is element-wise multiplication). Simple example. The Lovász extension of the modular function f(a) = A is the `1 norm, and the Lovász extension of the modular function f(a) =m(a) is the weighted `1 norm. With more general submodular functions, one can generate a large and interesting variety of norms, all of which have polyhedral contours (unlike, say, something like the `2 norm). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F16/54 (pg.43/192)

44 Lovász extension and norms Using Lovász extension to define various norms of the form kwk f = f( w ). Thisrendersthefunctionsymmetricaboutallorthants (meaning, kwk f = kb wk f for any b 2 { 1, 1} m and is element-wise multiplication). Simple example. The Lovász extension of the modular function f(a) = A is the `1 norm, and the Lovász extension of the modular function f(a) =m(a) is the weighted `1 norm. With more general submodular functions, one can generate a large and interesting variety of norms, all of which have polyhedral contours (unlike, say, something like the `2 norm). Hence, not all norms come from the Lovász extension of some submodular function. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F16/54 (pg.44/192)

45 Lovász extension and norms Using Lovász extension to define various norms of the form kwk f = f( w ). Thisrendersthefunctionsymmetricaboutallorthants (meaning, kwk f = kb wk f for any b 2 { 1, 1} m and is element-wise multiplication). Simple example. The Lovász extension of the modular function f(a) = A is the `1 norm, and the Lovász extension of the modular function f(a) =m(a) is the weighted `1 norm. With more general submodular functions, one can generate a large and interesting variety of norms, all of which have polyhedral contours (unlike, say, something like the `2 norm). Hence, not all norms come from the Lovász extension of some submodular function. Similarly, not all convex functions are the Lovász extension of some submodular function. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F16/54 (pg.45/192)

46 Lovász extension and norms Using Lovász extension to define various norms of the form kwk f = f( w ). Thisrendersthefunctionsymmetricaboutallorthants (meaning, kwk f = kb wk f for any b 2 { 1, 1} m and is element-wise multiplication). Simple example. The Lovász extension of the modular function f(a) = A is the `1 norm, and the Lovász extension of the modular function f(a) =m(a) is the weighted `1 norm. With more general submodular functions, one can generate a large and interesting variety of norms, all of which have polyhedral contours (unlike, say, something like the `2 norm). Hence, not all norms come from the Lovász extension of some submodular function. Similarly, not all convex functions are the Lovász extension of some submodular function. Bach-2011 has a complete discussion of this. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F16/54 (pg.46/192)

47 Concave closure The concave closure is defined as: f(x) = max p24 n (x) S V X p S f(s) (17.9) where 4 n (x) = n p 2 R 2n : P S V p S =1,p S 08S V, & P o S V p S1 S = x Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F17/54 (pg.47/192)

48 Concave closure The concave closure is defined as: f(x) = where 4 n (x) = n p 2 R 2n : P S V p S =1,p S max p24 n (x) S V X p S f(s) (17.9) 08S V, & P o S V p S1 S = x This is tight at the hypercube vertices, concave, and the concave envolope for the dual reasons as the convex closure. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F17/54 (pg.48/192)

49 Concave closure The concave closure is defined as: f(x) = where 4 n (x) = n p 2 R 2n : P S V p S =1,p S max p24 n (x) S V X p S f(s) (17.9) 08S V, & P o S V p S1 S = x This is tight at the hypercube vertices, concave, and the concave envolope for the dual reasons as the convex closure. Unlike the convex extension, the concave closure is defined by the Lovász extension iff f is a supermodular function. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F17/54 (pg.49/192)

50 Concave closure The concave closure is defined as: f(x) = where 4 n (x) = n p 2 R 2n : P S V p S =1,p S max p24 n (x) S V X p S f(s) (17.9) 08S V, & P o S V p S1 S = x This is tight at the hypercube vertices, concave, and the concave envolope for the dual reasons as the convex closure. Unlike the convex extension, the concave closure is defined by the Lovász extension iff f is a supermodular function. When f is submodular, even evaluating f is NP-hard (rough intuition: submodular maxmization is NP-hard (reduction to set cover), if we could evaluate f in poly time, we can maximize concave function to solve submodular maximization in poly time). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F17/54 (pg.50/192)

51 Multilinear extension Rather than the concave closure, multi-linear extension is used as a surrogate. For x 2 [0, 1] V =[0, 1] [n] f(x) = X f(s) Y Y x i (1 x i )=E S x [f(s)] (17.10) S V i2s i2v \S Mtt - concave what to do? functions 1) multiline - extension 2) restricted 3) that class of csubmoduln fun - tin ham lasin Conan closure deep. M Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F18/54 (pg.51/192) # Concan submodulu frnctrbs relaxants functions

52 Multilinear extension Rather than the concave closure, multi-linear extension is used as a surrogate. For x 2 [0, 1] V =[0, 1] [n] f(x) = X f(s) Y Y x i (1 x i )=E S x [f(s)] (17.10) S V i2s i2v \S Can be viewed as expected value of f(s) where S is a random set distributed via x, sopr(v 2 S) =x v and is independent of Pr(u 2 S) =x u, v 6= u. KEEQDV Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F18/54 (pg.52/192)

53 Multilinear extension Rather than the concave closure, multi-linear extension is used as a surrogate. For x 2 [0, 1] V =[0, 1] [n] f(x) = X f(s) Y Y x i (1 x i )=E S x [f(s)] (17.10) S V i2s i2v \S Can be viewed as expected value of f(s) where S is a random set distributed via x, sopr(v 2 S) =x v and is independent of Pr(u 2 S) =x u, v 6= u. This is tight at the hypercube vertices (immediate, since f(1 A ) yields only one term in the sum non-zero, namely the one where S = A). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F18/54 (pg.53/192)

54 Multilinear extension Rather than the concave closure, multi-linear extension is used as a surrogate. For x 2 [0, 1] V =[0, 1] [n] f(x) = X f(s) Y Y x i (1 x i )=E S x [f(s)] (17.10) S V i2s i2v \S Can be viewed as expected value of f(s) where S is a random set distributed via x, sopr(v 2 S) =x v and is independent of Pr(u 2 S) =x u, v 6= u. This is tight at the hypercube vertices (immediate, since f(1 A ) yields only one term in the sum non-zero, namely the one where S = A). Why called multilinear (multi-linear) extension? It is linear in each of its arguments (i.e., f(x 1,x 2,..., x k + x 0 k,...,x n)= f(x 1,x 2,...,x k,...,x n )+ f(x 1,x 2,...,x 0 k,...,x n) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F18/54 (pg.54/192)

55 Multilinear extension Rather than the concave closure, multi-linear extension is used as a surrogate. For x 2 [0, 1] V =[0, 1] [n] f(x) = X f(s) Y Y x i (1 x i )=E S x [f(s)] (17.10) S V i2s i2v \S Can be viewed as expected value of f(s) where S is a random set distributed via x, sopr(v 2 S) =x v and is independent of Pr(u 2 S) =x u, v 6= u. This is tight at the hypercube vertices (immediate, since f(1 A ) yields only one term in the sum non-zero, namely the one where S = A). Why called multilinear (multi-linear) extension? It is linear in each of its arguments (i.e., f(x 1,x 2,..., x k + x 0 k,...,x n)= f(x 1,x 2,...,x k,...,x n )+ f(x 1,x 2,...,x 0 k,...,x n) This is unfortunately not concave. However there are some useful properties. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F18/54 (pg.55/192)

56 Multilinear extension Lemma Let f(x) be the multilinear extension of a set function f :2 V! R. Then: Proof. If f is monotone non-decreasing, v 0 for all v 2 V within [0, 1] V (i.e., f is also monotone non-decreasing). If f is submodular, then f has an antitone supergradient, 2 j apple 0 for all i, j 2 V within [0, 1] V. First part (monotonicity). Choose x 2 [0, 1] V and let S x be random where x is treated as a distribution (so elements v is chosen with probability x v independently of any other element).... Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F19/54 (pg.56/192)

57 Multilinear extension...proof continued. Since f is multilinear, derivative is a simple difference when only one argument varies, f r f(x 1,x 2,...,x v1, 1,x v+1,...,x n ) (17.11) v f(x 1,x 2,...,x v1, 0,x v+1,...,x n ) (17.12) = E S x [f(s + v)] E S x [f(s v)] (17.13) 0 (17.14) where the final part follows due to monotonicity of each argument, i.e., f(s + i) f(s i) for any S and i 2 V. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F20/54 (pg.57/192)

58 Multilinear extension...proof continued. Second part of proof (antitone supergradient) also relies on simple consequence of multilinearity, namely multilinearity of the derivative as well. In this 2 f f (x 1,...,x i 1, 1,x i+1,...,x n ) j j (x 1,...,x i 1, 0,x i+1,...,x n ) (17.16) = E S x [f(s + i + j) f(s + i j)] (17.17) E S x [f(s i + j) f(s i j)] (17.18) apple 0 (17.19) since by submodularity, we have f(s + i j)+f(s i + j) f(s + i + j)+f(s i j) (17.20) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F21/54 (pg.58/192)

59 Multilinear extension: some properties Corollary sxr.info#etfmwnu let f be a function and f its multilinear extension on [0, 1] V. if f is monotone non-decreasing then f is non-decreasing along any strictly non-negative direction (i.e., f(x) apple f(y) whenever x apple y, or f(x) apple f(x + 1 v ) for any v 2 V and any 0. If f is submodular, then f is concave along any non-negative direction (i.e., the function g( ) = f(x + z) is 1-D concave in for any z 2 R + ). If f is submodular than f is convex along any diagonal direction (i.e., the function g( ) = f(x + (1 v 1 u )) is 1-D convex in for any u 6= v. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F22/54 (pg.59/192) *

60 Submodular Max and polyhedral approaches We ve spent much time discussing SFM and the polymatroidal polytope, and in general polyhedral approaches for SFM. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F23/54 (pg.60/192)

61 Submodular Max and polyhedral approaches We ve spent much time discussing SFM and the polymatroidal polytope, and in general polyhedral approaches for SFM. Most of the approaches for submodular max have not used such an approach, probably due to the difficulty in computing the concave extension of a submodular function (the convex extension is easy, namely the Lovász extension). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F23/54 (pg.61/192)

62 Submodular Max and polyhedral approaches We ve spent much time discussing SFM and the polymatroidal polytope, and in general polyhedral approaches for SFM. Most of the approaches for submodular max have not used such an approach, probably due to the difficulty in computing the concave extension of a submodular function (the convex extension is easy, namely the Lovász extension). A paper by Chekuri, Vondrak, and Zenklusen (2011) make some progress on this front using multilinear extensions. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F23/54 (pg.62/192)

63 Multilinear extension (review) Definition For a set function f :2 V! R, defineitsmultilinear extension F :[0, 1] V! R by F (x) = X S V f(s) Y i2s x i Y j2v \S (1 x j ) (17.21) Note that F (x) =Ef(ˆx) where ˆx is a random binary vector over {0, 1} V with elements independent w. probability x i for ˆx i. While this is defined for any set function, we have: Lemma Let F :[0, 1] V! R be multilinear extension of set function f :2 V! R, then If f is monotone i 0 for all i 2 V, x 2 [0, 1] V. If f is submodular, i@x j apple 0 for all i, j inv, x 2 [0, 1] V. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F24/54 (pg.63/192)

64 Submodular Max and polyhedral approaches Basic idea: Given a set of constraints I, weformapolytopep I such that {1 I : I 2I} P I We find max x2pi F (x) where F (x) is the multi-linear extension of f, to find a fractional solution x We then round x to a point on the hypercube, thus giving us a solution to the discrete problem. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F25/54 (pg.64/192)

65 Submodular Max and polyhedral approaches In the recent paper by Chekuri, Vondrak, and Zenklusen, they show: Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F26/54 (pg.65/192)

66 Submodular Max and polyhedral approaches classic In the recent am paper by Chekuri, Vondrak, and Zenklusen, they show: 1) constant factor approximation algorithm for max {F (x) :x 2 P } for any down-monotone solvable polytope P and F multilinear extension of any non-negative submodular function. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F26/54 (pg.66/192)

67 Submodular Max and polyhedral approaches In the recent paper by Chekuri, Vondrak, and Zenklusen, they show: 1) constant factor approximation algorithm for max {F (x) :x 2 P } for any down-monotone solvable polytope P and F multilinear extension of any non-negative submodular function. 2) A randomized rounding (pipage rounding) scheme to obtain an integer solution Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F26/54 (pg.67/192)

68 Submodular Max and polyhedral approaches In the recent paper by Chekuri, Vondrak, and Zenklusen, they show: 1) constant factor approximation algorithm for max {F (x) :x 2 P } for any down-monotone solvable polytope P and F multilinear extension of any non-negative submodular function. 2) A randomized rounding (pipage rounding) scheme to obtain an integer solution 3) An optimal (1 1/e) instance of their rounding scheme that can be used for a variety of interesting independence systems, including O(1) knapsacks, k matroids and O(1) knapsacks, a k-matchoid and ` sparse packing integer programs, and unsplittable flow in paths and trees. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F26/54 (pg.68/192)

69 Submodular Max and polyhedral approaches In the recent paper by Chekuri, Vondrak, and Zenklusen, they show: 1) constant factor approximation algorithm for max {F (x) :x 2 P } for any down-monotone solvable polytope P and F multilinear extension of any non-negative submodular function. 2) A randomized rounding (pipage rounding) scheme to obtain an integer solution 3) An optimal (1 1/e) instance of their rounding scheme that can be used for a variety of interesting independence systems, including O(1) knapsacks, k matroids and O(1) knapsacks, a k-matchoid and ` sparse packing integer programs, and unsplittable flow in paths and trees. Also, Vondrak showed that this scheme achieves the 1 c (1 e c ) curvature based bound for any matroid, which matches the bound we had earlier for uniform matroids with standard greedy. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F26/54 (pg.69/192)

70 Submodular Max and polyhedral approaches In the recent paper by Chekuri, Vondrak, and Zenklusen, they show: 1) constant factor approximation algorithm for max {F (x) :x 2 P } for any down-monotone solvable polytope P and F multilinear extension of any non-negative submodular function. 2) A randomized rounding (pipage rounding) scheme to obtain an integer solution 3) An optimal (1 1/e) instance of their rounding scheme that can be used for a variety of interesting independence systems, including O(1) knapsacks, k matroids and O(1) knapsacks, a k-matchoid and ` sparse packing integer programs, and unsplittable flow in paths and trees. Also, Vondrak showed that this scheme achieves the 1 c (1 e c ) curvature based bound for any matroid, which matches the bound we had earlier for uniform matroids with standard greedy. In general, one needs to do Monte-Carlo methods to estimate the multilinear extension (so further approximations would apply). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F26/54 (pg.70/192)

71 Review from lecture 10 The next slide comes from lecture 10. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F27/54 (pg.71/192)

72 Apolymatroidfunction spolyhedronisapolymatroid. Theorem Let f be a polymatroid function defined on subsets of E. Foranyx 2 R E +, and any P + f -basis yx 2 R E + of x, thecomponentsumofy x is y x (E) =rank(x), max y(e) :y apple x, y 2 P + f =min(x(a)+f(e \ A) :A E) (17.10) As a consequence, P + f is a polymatroid, since r.h.s. is constant w.r.t. y x. Taking E \ B =supp(x) (so elements B are all zeros in x), and for b/2 B we make x(b) is big enough, the r.h.s. min has solution A = B. Werecover submodular function from the polymatroid polyhedron via the following: rank 1 1 E\B = f(e \ B) = max n y(e \ B) :y 2 P + f o (17.11) In fact, we will ultimately see a number of important consequences of this theorem (other than just that P + f is a polymatroid) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F28/54 (pg.72/192)

73 Review from lecture 11 The next slide comes from lecture 11. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F29/54 (pg.73/192)

74 Matroid instance of Theorem?? Considering Theorem??, the matroid case is now a special case,where we have that: Corollary We have that: max {y(e) :y 2 P ind. set (M),y apple x} =min{r M (A)+x(E \ A) :A E} (17.21) where r M is the matroid rank function of some matroid. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F30/54 (pg.74/192)

75 000 Most violated inequality problem in matroid polytope case Consider P r + = x 2 R E : x 0,x(A) apple r M (A), 8A E (17.22) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F31/54 (pg.75/192)

76 Most violated inequality problem in matroid polytope case Consider P r + = x 2 R E : x 0,x(A) apple r M (A), 8A E (17.22) Suppose we have any x 2 R E + such that x 62 P r +. # Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F31/54 (pg.76/192)

77 Most violated inequality problem in matroid polytope case Consider P + r = x 2 R E : x 0,x(A) apple r M (A), 8A E (17.22) Suppose we have any x 2 R E + such that x 62 P + r. Hence, there must be a set of W 2 V,eachmemberofwhich corresponds to a violated inequality, i.e., equations of the form x(a) >r M (A) for A 2W. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F31/54 (pg.77/192)

78 Most violated inequality problem in matroid polytope case Consider P + r = x 2 R E : x 0,x(A) apple r M (A), 8A E (17.22) Suppose we have any x 2 R E + such that x 62 P r +. Hence, there must be a set of W 2 V,eachmemberofwhich corresponds to a violated inequality, i.e., equations of the form x(a) >r M (A) for A 2W. The most violated inequality when x is considered w.r.t. P r + corresponds - to the set A that maximizes x(a) r M (A), i.e., the most violated inequality is valuated as: max {x(a) r M (A) :A 2W}= max {x(a) r M (A) :A E} (17.23) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F31/54 (pg.78/192)

79 Most violated inequality problem in matroid polytope case Consider P + r = x 2 R E : x 0,x(A) apple r M (A), 8A E (17.22) Suppose we have any x 2 R E + such that x 62 P r +. Hence, there must be a set of W 2 V,eachmemberofwhich corresponds to a violated inequality, i.e., equations of the form x(a) >r M (A) for A 2W. The most violated inequality when x is considered w.r.t. P r + corresponds to the set A that maximizes x(a) r M (A), i.e., the most violated inequality is valuated as: max {x(a) r M (A) :A 2W}= max {x(a) r M (A) :A E} (17.23) Since x is modular and x(e \ A) =x(e) min as in;: x(a), wecanexpressthisviaa min {r M (A)+x(E \ A) :A E} (17.24) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F31/54 (pg.79/192)

80 Most violated inequality/polymatroid membership/sfm Consider P + f = x 2 RE : x 0,x(A) apple f(a), 8A E (17.25) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F32/54 (pg.80/192)

81 Most violated inequality/polymatroid membership/sfm Consider P + f = x 2 RE : x 0,x(A) apple f(a), 8A E (17.25) Suppose we have any x 2 R E + such that x 62 P + f. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F32/54 (pg.81/192)

82 Most violated inequality/polymatroid membership/sfm Consider P + f = x 2 RE : x 0,x(A) apple f(a), 8A E (17.25) 2 Suppose we have any x 2 R E + such that x 62 P + f. Hence, there must be a set of W 2 V,eachmemberofwhich corresponds to a violated inequality, i.e., equations of the form x(a) >r M (A) for A 2W. x 2 2 P P P x W = {{1}{1, 2}} W = {{2}, {1, 2}} W = {{1, 2}} Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F32/54 (pg.82/192)

83 Most violated inequality/polymatroid membership/sfm The most violated inequality when x is considered w.r.t. P + f corresponds to the set A that maximizes x(a) f(a), i.e., the most violated inequality is valuated as: max {x(a) f(a) :A 2W}= max {x(a) f(a) :A E} (17.26) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F33/54 (pg.83/192)

84 Most violated inequality/polymatroid membership/sfm The most violated inequality when x is considered w.r.t. P + f corresponds to the set A that maximizes x(a) f(a), i.e., the most violated inequality is valuated as: max {x(a) f(a) :A 2W}= max {x(a) f(a) :A E} (17.26) Since x is modular and x(e \ A) =x(e) via a min as in;: x(a), wecanexpressthis min {f(a)+x(e \ A) :A E} (17.27) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F33/54 (pg.84/192)

85 Most violated inequality/polymatroid membership/sfm The most violated inequality when x is considered w.r.t. P + f corresponds to the set A that maximizes x(a) f(a), i.e., the most violated inequality is valuated as: max {x(a) f(a) :A 2W}= max {x(a) f(a) :A E} (17.26) Since x is modular and x(e \ A) =x(e) via a min as in;: x(a), wecanexpressthis min {f(a)+x(e \ A) :A E} (17.27) More importantly, min {f(a)+x(e \ A) :A E} is a form of submodular function minimization, namely min {f(a) x(a) :A E} for a submodular f and x 2 R E +, consisting of a difference of polymatroid and modular function (so f x is no longer necessarily monotone, nor positive). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F33/54 (pg.85/192)

86 Most violated inequality/polymatroid membership/sfm The most violated inequality when x is considered w.r.t. P + f corresponds to the set A that maximizes x(a) f(a), i.e., the most violated inequality is valuated as: max {x(a) f(a) :A 2W}= max {x(a) f(a) :A E} (17.26) Since x is modular and x(e \ A) =x(e) via a min as in;: x(a), wecanexpressthis min {f(a)+x(e \ A) :A E} (17.27) More importantly, min {f(a)+x(e \ A) :A E} is a form of submodular function minimization, namely min {f(a) x(a) :A E} for a submodular f and x 2 R E +, consisting of a difference of polymatroid and modular function (so f x is no longer necessarily monotone, nor positive). We will ultimatley answer how general this form of SFM is. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F33/54 (pg.86/192)

87 Review from Lecture 6 The following three slides are review from lecture 6. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F34/54 (pg.87/192)

88 Matroids, other definitions using matroid rank r :2 V! Z + Definition (closed/flat/subspace) AsubsetA E is closed (equivalently, a flat or a subspace) ofmatroidm if for all x 2 E \ A, r(a [{x}) =r(a)+1. Definition: A hyperplane is a flat of rank r(m) 1. Definition (closure) Given A E, theclosure (or span) ofa, isdefinedby span(a) ={b 2 E : r(a [{b}) =r(a)}. Therefore, a closed set A has span(a) =A. Definition (circuit) AsubsetA E is circuit or a cycle if it is an inclusionwise-minimal dependent set (i.e., if r(a) < A and for any a 2 A, r(a \{a}) = A 1). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F35/54 (pg.88/192)

89 . Matroids by circuits Asetisindependentifandonlyifitcontainsnocircuit.Therefore,itisnot surprising that circuits can also characterize a matroid. Theorem (Matroid by circuits) Let E be a set and C be a collection of subsets of E that satisfy the following three properties: 1 (C1): ; /2 C 2 (C2): if C 1,C 2 2Cand C 1 C 2,thenC 1 = C 2. 3 (C3): if C 1,C 2 2Cwith C 1 6= C 2,ande 2 C 1 \ C 2,thenthereexistsa C 3 2Csuch that C 3 (C 1 [ C 2 ) \{e}. c, µm µf www. -9 Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F36/54 (pg.89/192) cz

90 Matroids by circuits Several circuit definitions for matroids. Theorem (Matroid by circuits) Let E be a set and C be a collection of nonempty subsets of E, suchthat no two sets in C are contained in each other. Then the following are equivalent. 1 C is the collection of circuits of a matroid; 2 if C, C 0 2C,andx 2 C \ C 0,then(C [ C 0 ) \{x} contains a set in C; 3 if C, C 0 2C,andx 2 C \ C 0,andy 2 C \ C 0,then(C [ C 0 ) \{x} contains a set in C containing y; Again, think about this for a moment in terms of linear spaces and matrices, and spanning trees. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F37/54 (pg.90/192)

91 Fundamental circuits in matroids Lemma Let I 2I(M), ande 2 E, theni [{e} contains at most one circuit in M. Proof. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F38/54 (pg.91/192)

92 Fundamental circuits in matroids Lemma Let I 2I(M), ande 2 E, theni [{e} contains at most one circuit in M. Proof. Suppose, to the contrary, that there are two distinct circuits C 1,C 2 such that C 1 [ C 2 I [{e}. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F38/54 (pg.92/192)

93 Fundamental circuits in matroids Lemma Let I 2I(M), ande 2 E, theni [{e} contains at most one circuit in M. Proof. Suppose, to the contrary, that there are two distinct circuits C 1,C 2 such that C 1 [ C 2 I [{e}. Then e 2 C 1 \ C 2,andby(C2),thereisacircuitC 3 of M s.t. C 3 (C 1 [ C 2 ) \{e} I Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F38/54 (pg.93/192)

94 Fundamental circuits in matroids Lemma Let I 2I(M), ande 2 E, theni [{e} contains at most one circuit in M. Proof. Suppose, to the contrary, that there are two distinct circuits C 1,C 2 such that C 1 [ C 2 I [{e}. Then e 2 C 1 \ C 2,andby(C2),thereisacircuitC 3 of M s.t. C 3 (C 1 [ C 2 ) \{e} I This contradicts the independence of I. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F38/54 (pg.94/192)

95 Fundamental circuits in matroids Lemma Let I 2I(M), ande 2 E, theni [{e} contains at most one circuit in M. Proof. Suppose, to the contrary, that there are two distinct circuits C 1,C 2 such that C 1 [ C 2 I [{e}. Then e 2 C 1 \ C 2,andby(C2),thereisacircuitC 3 of M s.t. C 3 (C 1 [ C 2 ) \{e} I This contradicts the independence of I. In general, let C(I,e) be the unique circuit associated with I [{e} (commonly called the fundamental circuit in M w.r.t. I and e). -. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F38/54 (pg.95/192)

96 Matroids: The Fundamental Circuit Define C(I,e) be the unique circuit associated with I [{e} (the fundamental circuit in M w.r.t. I and e, ifitexists). s Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F39/54 (pg.96/192)

97 Matroids: The Fundamental Circuit Define C(I,e) be the unique circuit associated with I [{e} (the fundamental circuit in M w.r.t. I and e, ifitexists). If e 2 span(i) \ I, thenc(i,e) is well defined (I + e creates one circuit). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 17 - May 23st, 2018 F39/54 (pg.97/192)