Submodular Functions, Optimization, and Applications to Machine Learning

Transcription

1 Submodular Functions, Optimization, and Applications to Machine Learning Spring Quarter, Lecture 5 Prof. Jeff Bilmes University of Washington, Seattle Department of Electrical Engineering April 9th, 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 5 - April 9th, 2018 F1/66 (pg.1/187)

2 Logistics Review Cumulative Outstanding Reading Read chapter 1 from Fujishige s book. Read chapter 2 from Fujishige s book. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F2/66 (pg.2/187)

3 Logistics Review Announcements, Assignments, and Reminders Homework 1 out, due Monday, 4/9/ :59pm electronically via our assignment dropbox ( If you have any questions about anything, please ask then via our discussion board ( Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F3/66 (pg.3/187)

4 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, Matroids L6(4/11): L7(4/16): L8(4/18): L9(4/23): L10(4/25): L11(4/30): L12(5/2): L13(5/7): L14(5/9): L15(5/14): L16(5/16): L17(5/21): 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 5 - April 9th, 2018 F4/66 (pg.4/187)

5 Logistics Summary submodular properties Review c(a), numberofconnectedcomponentsinducedbya E(G) is supermodular. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F5/66 (pg.5/187)

6 Logistics Summary submodular properties Review c(a), numberofconnectedcomponentsinducedbya E(G) is supermodular. f(x) =m 1 X X M1 X submodular iff off-diagonal elements of M non-positive. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F5/66 (pg.6/187)

7 Logistics Summary submodular properties Review c(a), numberofconnectedcomponentsinducedbya E(G) is supermodular. f(x) =m 1 X X M1 X submodular iff off-diagonal elements of M non-positive. Weighted set cover f(a) =w( S a2a U a), othercoverfunctions,cut functions. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F5/66 (pg.7/187)

8 Logistics Summary submodular properties Review c(a), numberofconnectedcomponentsinducedbya E(G) is supermodular. f(x) =m 1 X X M1 X submodular iff off-diagonal elements of M non-positive. Weighted set cover f(a) =w( S a2a U a), othercoverfunctions,cut functions. Matrix rank r(a), the dimensionality of the vector space spanned by the set of vectors {x a } a2a. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F5/66 (pg.8/187)

9 Logistics Summary submodular properties Review c(a), numberofconnectedcomponentsinducedbya E(G) is supermodular. f(x) =m 1 X X M1 X submodular iff off-diagonal elements of M non-positive. Weighted set cover f(a) =w( S a2a U a), othercoverfunctions,cut functions. Matrix rank r(a), the dimensionality of the vector space spanned by the set of vectors {x a } a2a. Adding modular functions to submodular functions preserves submodularity. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F5/66 (pg.9/187)

10 Logistics Summary submodular properties Review c(a), numberofconnectedcomponentsinducedbya E(G) is supermodular. f(x) =m 1 X X M1 X submodular iff off-diagonal elements of M non-positive. Weighted set cover f(a) =w( S a2a U a), othercoverfunctions,cut functions. Matrix rank r(a), thedimensionalityofthevectorspacespannedby the set of vectors {x a } a2a. Adding modular functions to submodular functions preserves submodularity. Conic P mixtures: if i 0 and f i :2 V! R is submodular, then so is i if i. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F5/66 (pg.10/187)

11 Logistics Summary submodular properties Review c(a), numberofconnectedcomponentsinducedbya E(G) is supermodular. f(x) =m 1 X X M1 X submodular iff off-diagonal elements of M non-positive. Weighted set cover f(a) =w( S a2a U a), othercoverfunctions,cut functions. Matrix rank r(a), thedimensionalityofthevectorspacespannedby the set of vectors {x a } a2a. Adding modular functions to submodular functions preserves submodularity. Conic P mixtures: if i 0 and f i :2 V! R is submodular, then so is i if i. Restrictions: f 0 (A) =f(a \ S) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F5/66 (pg.11/187)

12 Logistics Summary submodular properties Review c(a), numberofconnectedcomponentsinducedbya E(G) is supermodular. f(x) =m 1 X X M1 X submodular iff off-diagonal elements of M non-positive. Weighted set cover f(a) =w( S a2a U a), othercoverfunctions,cut functions. Matrix rank r(a), thedimensionalityofthevectorspacespannedby the set of vectors {x a } a2a. Adding modular functions to submodular functions preserves submodularity. Conic P mixtures: if i 0 and f i :2 V! R is submodular, then so is i if i. Restrictions: f 0 (A) =f(a \ S) max: f(a) = max j2a c j and facility location. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F5/66 (pg.12/187)

13 Logistics Summary submodular properties Review c(a), numberofconnectedcomponentsinducedbya E(G) is supermodular. f(x) =m 1 X X M1 X submodular iff off-diagonal elements of M non-positive. Weighted set cover f(a) =w( S a2a U a), othercoverfunctions,cut functions. Matrix rank r(a), thedimensionalityofthevectorspacespannedby the set of vectors {x a } a2a. Adding modular functions to submodular functions preserves submodularity. Conic P mixtures: if i 0 and f i :2 V! R is submodular, then so is i if i. Restrictions: f 0 (A) =f(a \ S) max: f(a) = max j2a c j and facility location. Log determinant f(a) = log det( A ) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F5/66 (pg.13/187)

14 - Concave over non-negative modular Let m 2 R E + be a non-negative modular function, and over R. Definef :2 E! R as aconcavefunction f(a) = (m(a)) (5.1) then f is submodular. # E " :v lrt othkcttt ) m ( = L MIAI ) min ) ' III. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F6/66 (pg.14/187)

15 Concave over non-negative modular Let m 2 R E + be a non-negative modular function, and over R. Definef :2 E! R as aconcavefunction then f is submodular. f(a) = (m(a)) (5.1) a #... atc b btc Proof. Given A B E \ v, wehave0applea = m(a) apple b = m(b), and 0 apple c = m(v). Forg concave, we have (a + c) (a) (b + c) (b), and thus (m(a)+m(v)) (m(a)) (m(b)+m(v)) (m(b)) (5.2) Aformofconverseistrueaswell. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F6/66 (pg.15/187)

16 m ( A) =/ A ) lvl =3 OH tea, N µ 1011AM

17 - normchtel - monotone Concave composed with non-negative modular Theorem Given a ground set V.Thefollowingtwoareequivalent: 1 For all modular functions m :2 V! R +,thenf :2 V! R defined as f(a) = (m(a)) is submodular 2 : R +! R is concave. If is non-decreasing concave w. (0) = 0, thenf is polymatroidal. fl 0/7=0 non - dlcmcsihy - Submodulm. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F7/66 (pg.16/187)

18 Concave composed with non-negative modular Theorem Given a ground set V.Thefollowingtwoareequivalent: 1 For all modular functions m :2 V! R +,thenf :2 V! R defined as f(a) = (m(a)) is submodular 2 : R +! R is concave. If is non-decreasing concave w. (0) = 0, thenf is polymatroidal. Sums of concave over modular functions are submodular KX f(a) = i(m i (A)) (5.3) i=1 HAl=Ifvy m( Ethan Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F7/66 (pg.17/187)

19 Concave composed with non-negative modular Theorem Given a ground set V.Thefollowingtwoareequivalent: 1 For all modular functions m :2 V! R +,thenf :2 V! R defined as f(a) = (m(a)) is submodular 2 : R +! R is concave. If is non-decreasing concave w. (0) = 0, thenf is polymatroidal. Sums of concave over modular functions are submodular KX f(a) = i(m i (A)) (5.3) i=1 Very large class of functions, including graph cut, bipartite neighborhoods, set cover (Stobbe & Krause 2011), and feature-based submodular functions (Wei, Iyer, & Bilmes 2014). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F7/66 (pg.18/187)

20 Concave composed with non-negative modular Theorem Given a ground set V.Thefollowingtwoareequivalent: 1 For all modular functions m :2 V! R +,thenf :2 V! R defined as f(a) = (m(a)) is submodular 2 : R +! R is concave. If is non-decreasing concave w. (0) = 0, thenf is polymatroidal. Sums of concave over modular functions are submodular KX f(a) = i(m i (A)) (5.3) i=1 Very large class of functions, including graph cut, bipartite neighborhoods, set cover (Stobbe & Krause 2011), and feature-based submodular functions (Wei, Iyer, & Bilmes 2014). However, Vondrak showed that a graphic matroid rank function over K 4 (we ll define this after we define matroids) are not members. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F7/66 (pg.19/187)

21 Monotonicity Definition Afunctionf :2 V! R is monotone nondecreasing (resp. monotone increasing) ifforalla B, wehavef(a) apple f(b) (resp. f(a) <f(b)). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F8/66 (pg.20/187)

22 Monotonicity Definition Afunctionf :2 V! R is monotone nondecreasing (resp. monotone. O increasing) ifforalla B, wehavef(a) apple f(b) (resp. f(a) <f(b)). Definition Afunctionf :2 V! R is monotone nonincreasing (resp. monotone decreasing) ifforalla B, wehavef(a) f(b) (resp. f(a) >f(b)). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F8/66 (pg.21/187)

23 Composition of non-decreasing submodular and non-decreasing concave Theorem Given two functions, one defined on sets and another continuous valued one: f :2 V! R (5.4) : R! R (5.5) the composition formed as h = f :2 V! R (defined as h(s) = (f(s))) isnondecreasingsubmodular,if is non-decreasing concave and f is nondecreasing submodular. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F9/66 (pg.22/187)

24 Monotone difference of two functions Let f and g both be submodular functions on subsets of V and let (f g)( ) be either monotone non-decreasing or monotone non-increasing Then h :2 V! R defined by is submodular. Proof. h(a) =min(f(a),g(a)) (5.6) If h(a) agrees with f on both X and Y (or g on both X and Y ), and since h(x)+h(y )=f(x)+f(y ) f(x [ Y )+f(x \ Y ) (5.7) or h(x)+h(y )=g(x)+g(y ) g(x [ Y )+g(x \ Y ), (5.8) the result (Equation 5.6 being submodular) follows since f(x)+f(y ) g(x)+g(y ) min(f(x [ Y ),g(x [ Y )) + min(f(x \ Y ),g(x \ Y )) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F10/66 (pg.23/187) (5.9)...

25 Monotone difference of two functions...cont. Otherwise, w.l.o.g., h(x) =f(x) and h(y ) =g(y ), giving h(x)+h(y )=f(x)+g(y) f(x [ Y )+f(x \ Y )+g(y) f(y ) (5.10) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F11/66 (pg.24/187)

26 - Monotone difference of two functions...cont. Otherwise, w.l.o.g., h(x) =f(x) and h(y ) =g(y ), giving h(x)+h(y )=f(x)+g(y) f(x [ Y )+f(x \ Y )+g(y) f(y ) (5.10) f ( XVY ) - g ( xvy ) Z f 14 ) Assume the case where f g is monotone non-decreasing Hence, f(x [ Y )+g(y) f(y ) g(x [ Y ) giving - g ( Y ) h(x)+h(y ) g(x [ Y )+f(x \ Y ) h(x [ Y )+h(x \ Y ) (5.11) What is an easy way to : prove the case where f non-increasing? ring g is monotone Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F11/66 (pg.25/187)

27 non Saturation via the min( ) function non - chlrebsirz Let f :2 V! R be a monotone increasing mm or decreasing - submodular function and let be a constant. Then the function h :2 V! R defined by non - ihcrlaskr h(a) =min(, f(a)) (5.12) is submodular. mink,x ). if f is non. - Mercury h( 4=2 = flu ) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F12/66 (pg.26/187)

28 Saturation via the min( ) function Let f :2 V! R be a monotone increasing or decreasing submodular function and let be a constant. Then the function h :2 V! R defined by is submodular. Proof. h(a) =min(, f(a)) (5.12) For constant k, wehavethat(f k) is non-decreasing (or non-increasing) so this follows from the previous result. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F12/66 (pg.27/187)

29 Saturation via the min( ) function Let f :2 V! R be a monotone increasing or decreasing submodular function and let be a constant. Then the function h :2 V! R defined by is submodular. Proof. h(a) =min(, f(a)) (5.12) For constant k, wehavethat(f k) is non-decreasing (or non-increasing) so this follows from the previous result. Note also, g(a) =min(k, a) for constant k is a non-decreasing concave function, so when f is monotone nondecreasing submodular, we can use the earlier result about composing a monotone concave function with a monotone submodular function to get a version of this. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F12/66 (pg.28/187)

30 More on Min - the saturate trick In general, the minimum of two submodular functions is not submodular (unlike concave functions, closed under min). I *.me#antis@ Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F13/66 (pg.29/187)

31 More on Min - the saturate trick In general, the minimum of two submodular functions is not submodular (unlike concave functions, closed under min). However, when wishing to maximize two monotone non-decreasing submodular functions f,g, wecandefinefunctionh :2 V! R as h (A) = 1 2 min(, f(a)) + min(, g(a)) (5.13) then h is submodular, and h (A) if and only if both f(a) and g(a), forconstant 2 R. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F13/66 (pg.30/187)

32 More on Min - the saturate trick In general, the minimum of two submodular functions is not submodular (unlike concave functions, closed under min). However, when wishing to maximize two monotone non-decreasing submodular functions f,g, wecandefinefunctionh :2 V! R as h (A) = 1 2 min(, f(a)) + min(, g(a)) surrogut for mm ( FIA ), g ( A ) ) (5.13) then h is submodular, and h (A) if and only if both f(a) and g(a), forconstant 2 R. This can be useful in many applications. An instance of a submodular surrogate (where we take a non-submodular problem and find a submodular one that can tell us something about it). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F13/66 (pg.31/187)

33 Arbitrary functions: difference between submodular funcs. Theorem Given an arbitrary set function h, itcanbeexpressedasadifference between two submodular functions (i.e., 8h 2 2 V! R, 9f,g s.t. 8A, h(a) =f(a) g(a) where both f and g are submodular). Proof. Let h be given and arbitrary, and define: If = min X,Y :X6 Y,Y 6 X h(x)+h(y ) h(x [ Y ) h(x \ Y ) 0 then h is submodular, so by assumption <0. (5.14) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F14/66 (pg.32/187)

34 Arbitrary functions: difference between submodular funcs. Theorem Given an arbitrary set function h, itcanbeexpressedasadifference between two submodular functions (i.e., 8h 2 2 V! R, 9f,g s.t. 8A, h(a) =f(a) g(a) where both f and g are submodular). Proof. Let h be given and arbitrary, and define: = min X,Y :X6 Y,Y 6 X h(x)+h(y ) h(x [ Y ) h(x \ Y ) (5.14) If 0 then h is submodular, so by assumption <0. Nowletf be an arbitrary strict submodular function and define = min f(x)+f(y ) f(x [ Y ) f(x \ Y ). (5.15) X,Y :X6 Y,Y 6 X Strict means that >0. FLA ) = MAT... Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F14/66 (pg.33/187)

35 Arbitrary functions as difference between submodular funcs....cont. Define h 0 :2 V! R as h 0 (A) =h(a)+ f(a) (5.16) Then h 0 is submodular (why?), and h = h 0 (A) between two submodular functions as desired. f(a), adifference Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F15/66 (pg.34/187)

36 Gain We often wish to express the gain of an item j 2 V in context A, namely f(a [{j}) f(a). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F16/66 (pg.35/187)

37 Gain We often wish to express the gain of an item j 2 V in context A, namely f(a [{j}) f(a). This is called the gain and is used so often, there are equally as many ways to notate this. I.e., you might see: f(a [{j}) f(a) = j (A) (5.17) = A (j) (5.18) = r j f(a) (5.19) = f({j} A) (5.20) = f(j A) (5.21) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F16/66 (pg.36/187)

38 Gain We often wish to express the gain of an item j 2 V in context A, namely f(a [{j}) f(a). This is called the gain and is used so often, there are equally as many ways to notate this. I.e., you might see: We ll use f(j A). f(a [{j}) f(a) = j (A) (5.17) fljlaltgljltf = A (j) (5.18) = r j f(a) (5.19) = f({j} A) (5.20) = f(j A) (5.21) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F16/66 (pg.37/187)

39 - FIA Gain Vii We often wish to express the gain of an item j 2 V in context A, namely f(a [{j}) f(a). This is called the gain and is used so often, there are equally as many ways to notate this. I.e., you might see: A f(a [{j}) f(a) = j (A) (5.17) = A (j) (5.18) ) flj zfljln ) = r j f(a) (5.19) = f({j} A) (5.20) fl ;) z f ( ) furthµ At ;) f- Ii + ) HA ) Zflatj ) = f(j A) (5.21) We ll use f(j A). Submodularity s diminishing returns definition can be stated as saying that f(j A) is a monotone non-increasing function of A, since f(j A) f(j B) whenever A B (conditioning reduces valuation). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F16/66 (pg.38/187)

40 - Non norm. IAU Cose. fifth'la ) flj )+fla ) Z f ( At ;) tf( ) Doe ) this dyim svbnodulmity? fancy

41 Gain Notation It will also be useful to extend this to sets. Let A, B be any two sets. Then So when j is any singleton f(a B), f(a [ B) f(b) (5.22). f(j B) =f({j} B) =f({j}[b) f(b) (5.23) Bt ; = Busj ) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F17/66 (pg.39/187)

42 Gain Notation It will also be useful to extend this to sets. Let A, B be any two sets. Then So when j is any singleton f(a B), f(a [ B) f(b) (5.22) f(j B) =f({j} B) =f({j}[b) f(b) (5.23) Inspired from information theory notation and the notation used for conditional entropy H(X A X B )=H(X A,X B ) H(X B ). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F17/66 (pg.40/187)

43 . Totally normalized functions Any normalized submodular function g (even non-monotone) can be represented as a sum of a polymatroid (normalized monotone non-decreasing submodular) function ḡ and a modular function m g. min f ) ( VIAI ) E 2 Ha. A E v =f( v EIA mouth 2 function. h (A) A) + A h h(v1a7=f( HA ) +2 Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F18/66 (pg.41/187)

44 Totally normalized functions Any normalized submodular function g (even non-monotone) can be represented as a sum of a polymatroid (normalized monotone non-decreasing submodular) function ḡ and a modular function m g. Given arbitrary normalized submodular g :2 V! R, constructa function ḡ :2 V! R as follows: X ḡ(a) =g(a) g(a V \{a}) =g(a) m g (A) (5.24) a2a where m g (A), P a2a g(a V \{a}) is a modular function. g ( and Eglalt ) ttc via Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F18/66 (pg.42/187)

45 Totally normalized functions Any normalized submodular function g (even non-monotone) can be represented as a sum of a polymatroid (normalized monotone non-decreasing submodular) function ḡ and a modular function m g. Given arbitrary normalized submodular g :2 V! R, constructa function ḡ :2 V! R as follows: X ḡ(a) =g(a) g(a V \{a}) =g(a) m g (A) (5.24) a2a where m g (A), P a2a g(a V \{a}) is a modular function. ḡ is normalized since ḡ(;) =0. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F18/66 (pg.43/187)

46 Totally normalized functions Any normalized submodular function g (even non-monotone) can be represented as a sum of a polymatroid (normalized monotone non-decreasing submodular) function ḡ and a modular function m g. Given arbitrary normalized submodular g :2 V! R, constructa function ḡ :2 V! R as follows: X ḡ(a) =g(a) g(a V \{a}) =g(a) m g (A) (5.24) a2a where m g (A), P a2a g(a V \{a}) is a modular function. ḡ is normalized since ḡ(;) =0. ḡ is monotone non-decreasing since for v/2 A V : ḡ(v A) =g(v A) g(v V \{v}) 0 (5.25) Ty ( v1a7 70 man. nonmoving. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F18/66 (pg.44/187)

47 Totally normalized functions Any normalized submodular function g (even non-monotone) can be represented as a sum of a polymatroid (normalized monotone non-decreasing submodular) function ḡ and a modular function m g. Given arbitrary normalized submodular g :2 V! R, constructa function ḡ :2 V! R as follows: X ḡ(a) =g(a) g(a V \{a}) =g(a) m g (A) (5.24) a2a where m g (A), P a2a g(a V \{a}) is a modular function. ḡ is normalized since ḡ(;) =0. ḡ is monotone non-decreasing since for v/2 A V : ḡ(v A) =g(v A) g(v V \{v}) 0 (5.25) ḡ is called the totally normalized version of g. ) - I ( v1v\{ v } ) = 0 # Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F18/66 (pg.45/187)

48 Totally normalized functions Any normalized submodular function g (even non-monotone) can be represented as a sum of a polymatroid (normalized monotone non-decreasing submodular) function ḡ and a modular function m g. Given arbitrary normalized submodular g :2 V! R, constructa function ḡ :2 V! R as follows: X ḡ(a) =g(a) g(a V \{a}) =g(a) m g (A) (5.24) a2a where m g (A), P a2a g(a V \{a}) is a modular function. ḡ is normalized since ḡ(;) =0. ḡ is monotone non-decreasing since for v/2 A V : ḡ(v A) =g(v A) g(v V \{v}) 0 (5.25) ḡ is called the totally normalized version of g. Then g(a) =ḡ(a)+m g (A). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F18/66 (pg.46/187)

49 Arbitrary function as difference between two polymatroids Any normalized function h (i.e., h(;) =0) can be represented as a difference not only between submodular, but between polymatroid (normalized monotone non-decreasing submodular) functions. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F19/66 (pg.47/187)

50 Arbitrary function as difference between two polymatroids Any normalized function h (i.e., h(;) =0) can be represented as a difference not only between submodular, but between polymatroid (normalized monotone non-decreasing submodular) functions. Given submodular f and g, let f and ḡ be them totally normalized. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F19/66 (pg.48/187)

51 Arbitrary function as difference between two polymatroids Any normalized function h (i.e., h(;) =0) canberepresentedasa difference not only between submodular, but between polymatroid (normalized monotone non-decreasing submodular) functions. Given submodular f and g, let f and ḡ be them totally normalized. Given arbitrary h = f g where f and g are normalized submodular, h = f g = f + m f (ḡ + m g ) (5.26) = f ḡ +(m f m g ) (5.27) = f ḡ + m f h (5.28) = f + m + f g (ḡ +( m f g ) + ) (5.29) where m + i is the positive part of modular function m. Thatis, m + (A) = P a2a m(a)1(m(a) > 0). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F19/66 (pg.49/187)

52 Arbitrary function as difference between two polymatroids Any normalized function h (i.e., h(;) =0) canberepresentedasa difference not only between submodular, but between polymatroid (normalized monotone non-decreasing submodular) functions. Given submodular f and g, let f and ḡ be them totally normalized. Given arbitrary h = f g where f and g are normalized submodular, h = f g = f + m f (ḡ + m g ) (5.26) = f ḡ +(m f m g ) (5.27) = f ḡ + m f h (5.28) = f + m + f g (ḡ +( m f g ) + ) (5.29) where m + is the positive part of modular function m. Thatis, m + (A) = P a2a m(a)1(m(a) > 0). Both f + m + f g and ḡ +( m f g) + are polymatroid functions! Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F19/66 (pg.50/187)

53 Arbitrary function as difference between two polymatroids Any normalized function h (i.e., h(;) =0) canberepresentedasa difference not only between submodular, but between polymatroid (normalized monotone non-decreasing submodular) functions. Given submodular f and g, let f and ḡ be them totally normalized. Given arbitrary h = f g where f and g are normalized submodular, h = f g = f + m f (ḡ + m g ) (5.26) = f ḡ +(m f m g ) (5.27) = f ḡ + m f h (5.28) = f + m + f g (ḡ +( m f g ) + ) (5.29) where m + is the positive part of modular function m. Thatis, m + (A) = P a2a m(a)1(m(a) > 0). Both f + m + f g and ḡ +( m f g) + are polymatroid functions! Thus, any function can be expressed as a difference between two, not only submodular (DS), but polymatroid functions. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F19/66 (pg.51/187)

54 Two Equivalent Submodular Definitions Definition (submodular concave) Afunctionf :2 V! R is submodular if for any A, B V,wehavethat: f(a)+f(b) f(a [ B)+f(A \ B) (5.8) An alternate and (as we will soon see) equivalent definition is: Definition (diminishing returns) Afunctionf :2 V! R is submodular if for any A B V,and v 2 V \ B, wehavethat: f(a [{v}) f(a) f(b [{v}) f(b) (5.9) The incremental value, gain, or cost of v decreases (diminishes) as the context in which v is considered grows from A to B. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F20/66 (pg.52/187)

55 Submodular Definition: Group Diminishing Returns An alternate and equivalent definition is: Definition (group diminishing returns) Afunctionf :2 V! R is submodular if for any A B V,and C V \ B, wehavethat: f(a [ C) f(a) f(b [ C) f(b) (5.30) This means that the incremental value or gain of set C decreases as the context in which C is considered grows from A to B (diminishing returns) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F21/66 (pg.53/187)

56 Submodular Definition Basic Equivalencies We want to show that Submodular Concave (Definition 5.4.1), Diminishing Returns (Definition 5.4.2), and Group Diminishing Returns (Definition 5.4.1) are identical. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F22/66 (pg.54/187)

57 Submodular Definition Basic Equivalencies We want to show that Submodular Concave (Definition 5.4.1), Diminishing Returns (Definition 5.4.2), and Group Diminishing Returns (Definition 5.4.1) are identical. We will show that: Submodular Concave ) Diminishing Returns Diminishing Returns ) Group Diminishing Returns Group Diminishing Returns ) Submodular Concave Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F22/66 (pg.55/187)

58 Submodular Concave ) Diminishing Returns f(s)+f(t ) f(s [ T )+f(s \ T ) ) f(v A) f(v B),A B V \ v. Assume Submodular concave, so 8S, T we have f(s)+f(t ) f(s [ T )+f(s \ T ). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F23/66 (pg.56/187)

59 Submodular Concave ) Diminishing Returns f(s)+f(t ) f(s [ T )+f(s \ T ) ) f(v A) f(v B),A B V \ v. Assume Submodular concave, so 8S, T we have f(s)+f(t ) f(s [ T )+f(s \ T ). Given A, B and v 2 V such that: A B V \{v}, wehavefrom submodular concave that: f(a + v)+f(b) f(b + v)+f(a) (5.31) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F23/66 (pg.57/187)

60 Submodular Concave ) Diminishing Returns f(s)+f(t ) f(s [ T )+f(s \ T ) ) f(v A) f(v B),A B V \ v. Assume Submodular concave, so 8S, T we have f(s)+f(t ) f(s [ T )+f(s \ T ). Given A, B and v 2 V such that: A B V \{v}, wehavefrom submodular concave that: Rearranging, we have f(a + v)+f(b) f(b + v)+f(a) (5.31) f(a + v) f(a) f(b + v) f(b) (5.32) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F23/66 (pg.58/187)

61 Diminishing Returns ) Group Diminishing Returns f(v S) f(v T ),S T V \ v ) f(c A) f(c B),A B V \ C. Let C = {c 1,c 2,...,c k }.Thendiminishing returns implies f(a [ C) = f(a [ C) = f(a) kx 1 f(a [{c 1,...,c i}) i=1 kx f(a [{c 1...c i}) f(a [{c 1...c i 1}) = i=1 kx f(c i B [{c 1...c i 1}) = i=1 = f(b [ C) i=1 kx 1 f(b [{c 1,...,c i}) i=1 = f(b [ C) f(b) - f(a [{c 1,...,c i}) (5.33) f(a) (5.34) kx f(c i A [{c 1...c i 1}) (5.35) i=1 kx f(b [{c 1...c i}) f(b [{c 1...c i 1}) (5.36) f(b [{c 1,...,c i}) f(b) (5.37) (5.38) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F24/66 (pg.59/187)

62 Group Diminishing Returns ) Submodular Concave f(u S) f(u T ),S T V \ U ) f(a)+f(b) f(a [ B)+f(A \ B). Assume group diminishing returns. Assume A 6= B otherwise trivial. Define A 0 = A \ B, C = A \ B, andb 0 = B. ThensinceA 0 B 0, giving or f(a 0 + C) f(a 0 ) f(b 0 + C) f(b 0 ) (5.39) f(a 0 + C)+f(B 0 ) f(b 0 + C)+f(A 0 ) (5.40) f(a \ B + A \ B)+f(B) f(b + A \ B)+f(A \ B) (5.41) which is the same as the submodular concave condition f(a)+f(b) f(a [ B)+f(A \ B) (5.42) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F25/66 (pg.60/187)

63 Submodular Definition: Four Points Definition ( singleton, or four points ) Afunctionf :2 V! R is submodular if f for any A V,andany a, b 2 V \ A, wehavethat: f(a [{a})+f(a [{b}) f(a [{a, b})+f(a) (5.43) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F26/66 (pg.61/187)

64 - f( flats Submodular Definition: Four Points Definition ( singleton, or four points ) Afunctionf :2 V! R is submodular if f for any A V,andany a, b 2 V \ A, wehavethat: f(a [{a})+f(a [{b}) f(a [{a, b})+f(a) (5.43) This follows immediately from diminishing returns. fl Ata ) - A) f 2 ( Atsta ) f ( AIA ) 2 t ( altts ) ) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F26/66 (pg.62/187)

65 ... Submodular Definition: Four Points Definition ( singleton, or four points ) Afunctionf :2 V! R is submodular if f for any A V,andany a, b 2 V \ A, wehavethat: f(a [{a})+f(a [{b}) f(a [{a, b})+f(a) (5.43) This follows immediately from diminishing returns. Toachievediminishing returns, assumea B with B \ A = {b 1,b 2,...,b k }.Then f ( AIA ) 7- f ( al Ath ) Zfka ) Ats, th f(a + a) f(a) f(a + b 1 + a) f(a + b 1 ) (5.44) f(a + b 1 + b 2 + a) f(a + b 1 + b 2 ) (5.45)... ) z z fla / b ) (5.46) f(a + b b k + a) f(a + b b k ) (5.47) = f(b + a) f(b) (5.48) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F26/66 (pg.63/187)

66 Submodular on Hypercube Vertices Test submodularity via values on verticies of hypercube. Example: with V = n =2,thisis easy: With V = n =3,abitharder ana.tt?:tttntyng, D How many inequalities? Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F27/66 (pg.64/187)

67 Submodular Concave Diminishing Returns, in one slide. Theorem Given function f :2 V! R, then f(a)+f(b) f(a [ B)+f(A \ B) for all A, B V (SC) if and only if f(v X) f(v Y ) for all X Y V and v/2 Y (DR) Proof. (SC))(DR): Set A X [{v}, B Y.ThenA [ B = Y [{v} and A \ B = X and f(a) f(a \ B) f(a [ B) f(b) implies (DR). (DR))(SC): Order A \ B = {v 1,v 2,...,v r } arbitrarily. For i 2 1:r, f(v i (A \ B) [{v 1,v 2,...,v i 1 }) f(v i B [{v 1,v 2,...,v i 1 }). Applying telescoping summation to both sides, we get: rx rx f(v i (A \ B) [{v 1,v 2,...,v i 1 }) f(v i B [{v 1,v 2,...,v i 1 }) i=1 i=1 ) f(a) f(a \ B) f(a [ B) f(b) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F28/66 (pg.65/187)

68 Many (Equivalent) Definitions of Submodularity f(a) + f(b) f(a [ B)+f(A \ B), 8A, B V (5.54) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F30/66 (pg.66/187)

69 Many (Equivalent) Definitions of Submodularity f(a) + f(b) f(a [ B)+f(A \ B), 8A, B V (5.54) f(j S) f(j T ), 8S T V, with j 2 V \ T (5.55) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F30/66 (pg.67/187)

70 Many (Equivalent) Definitions of Submodularity f(a) + f(b) f(a [ B)+f(A \ B), 8A, B V (5.54) f(j S) f(j T ), 8S T V, with j 2 V \ T (5.55) f(c S) f(c T ), 8S T V, with C V \ T (5.56) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F30/66 (pg.68/187)

71 Many (Equivalent) Definitions of Submodularity f(a) + f(b) f(a [ B)+f(A \ B), 8A, B V (5.54) f(j S) f(j T ), 8S T V, with j 2 V \ T (5.55) f(c S) f(c T ), 8S T V, with C V \ T (5.56) f(j S) f(j S [{k}), 8S V with j 2 V \ (S [{k}) (5.57) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F30/66 (pg.69/187)

72 Many (Equivalent) Definitions of Submodularity f(a) + f(b) f(a [ B)+f(A \ B), 8A, B V (5.54) f(j S) f(j T ), 8S T V, with j 2 V \ T (5.55) f(c S) f(c T ), 8S T V, with C V \ T (5.56) f(j S) f(j S [{k}), 8S V with j 2 V \ (S [{k}) (5.57) f(a [ B A \ B) apple f(a A \ B)+f(B A \ B), 8A, B V (5.58) Conditional subadihut > Strong subadditivit, Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F30/66 (pg.70/187)

73 . Many (Equivalent) Definitions of Submodularity f(a) + f(b) f(a [ B)+f(A \ B), 8A, B V (5.54) f(j S) f(j T ), 8S T V, with j 2 V \ T (5.55) f(c S) f(c T ), 8S T V, with C V \ T (5.56) f(j S) f(j S [{k}), 8S V with j 2 V \ (S [{k}) (5.57) f(a [ B A \ B) apple f(a A \ B)+f(B A \ B), 8A, B V (5.58) f(t ) apple f(s)+ X X f(j S) f(j S [ T {j}), 8S, T V j2t \S j2s\t LB + ± ) fct = sut HSH ) = f ( ) f ( s ) + f ( Tls ft ) + 1 f Cs ) + us (5.59) ) f ( t ) EFLJ ) + UD - LB Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F30/66 (pg.71/187)

74 Many (Equivalent) Definitions of Submodularity f(a) + f(b) f(a [ B)+f(A \ B), 8A, B V (5.54) f(j S) f(j T ), 8S T V, with j 2 V \ T (5.55) f(c S) f(c T ), 8S T V, with C V \ T (5.56) f(j S) f(j S [{k}), 8S V with j 2 V \ (S [{k}) (5.57) f(a [ B A \ B) apple f(a A \ B)+f(B A \ B), 8A, B V (5.58) f(t ) apple f(s)+ X X f(j S) f(j S [ T {j}), 8S, T V j2t \S j2s\t (5.59) f(t ) apple f(s)+ X f(j S), 8S T V (5.60) j2t \S Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F30/66 (pg.72/187)

75 Many (Equivalent) Definitions of Submodularity f(a) + f(b) f(a [ B)+f(A \ B), 8A, B V (5.54) f(j S) f(j T ), 8S T V, with j 2 V \ T (5.55) f(c S) f(c T ), 8S T V, with C V \ T (5.56) f(j S) f(j S [{k}), 8S V with j 2 V \ (S [{k}) (5.57) f(a [ B A \ B) apple f(a A \ B)+f(B A \ B), 8A, B V (5.58) f(t ) apple f(s)+ X X f(j S) f(j S [ T {j}), 8S, T V j2t \S j2s\t (5.59) f(t ) apple f(s)+ X f(j S), 8S T V (5.60) f(t ) apple f(s) j2t \S X f(j S \{j})+ X f(j S \ T ) 8S, T V j2s\t j2t \S (5.61) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F30/66 (pg.73/187)

76 Many (Equivalent) Definitions of Submodularity f(a) + f(b) f(a [ B)+f(A \ B), 8A, B V (5.54) f(j S) f(j T ), 8S T V, with j 2 V \ T (5.55) f(c S) f(c T ), 8S T V, with C V \ T (5.56) f(j S) f(j S [{k}), 8S V with j 2 V \ (S [{k}) (5.57) f(a [ B A \ B) apple f(a A \ B)+f(B A \ B), 8A, B V (5.58) f(t ) apple f(s)+ X X f(j S) f(j S [ T {j}), 8S, T V j2t \S j2s\t (5.59) f(t ) apple f(s)+ X f(j S), 8S T V (5.60) f(t ) apple f(s) f(t ) apple f(s) j2t \S X f(j S \{j})+ X f(j S \ T ) 8S, T V j2s\t j2t \S (5.61) X f(j S \{j}), 8T S V (5.62) j2s\t Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F30/66 (pg.74/187)

77 Equivalent Definitions of Submodularity We ve already seen that Eq Eq Eq Eq Eq Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F31/66 (pg.75/187)

78 Equivalent Definitions of Submodularity We ve already seen that Eq Eq Eq Eq Eq We next show that Eq ) Eq ) Eq ) Eq Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F31/66 (pg.76/187)

79 Approach To show these next results, we essentially first use: f(s [ T )=f(s)+f(t S) apple f(s)+upper-bound (5.63) and f(t )+lower-bound apple f(t )+f(s T)=f(S [ T ) (5.64) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F32/66 (pg.77/187)

80 Approach To show these next results, we essentially first use: f(s [ T )=f(s)+f(t S) apple f(s)+upper-bound (5.63) and f(t )+lower-bound apple f(t )+f(s T)=f(S [ T ) (5.64) leading to f(t )+lower-bound apple f(s)+upper-bound (5.65) or f(t ) apple f(s) +upper-bound lower-bound (5.66) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F32/66 (pg.78/187)

81 Eq ) Eq Let T \ S = {j 1,...,j r } and S \ T = {k 1,...,k q }. First, we upper bound the gain of T in the context of S: or f(s [ T ) f(s) = = rx f(s [{j 1,...,j t }) f(s [{j 1,...,j t 1 }) t=1 rx f(j t S [{j 1,...,j t t=1 = X j2t \S 1 }) apple (5.67) rx f(j t S) (5.68) t=1 f(j S) (5.69) f(t S) apple X j2t \S f(j S) (5.70) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F33/66 (pg.79/187)

82 Eq ) Eq Let T \ S = {j 1,...,j r } and S \ T = {k 1,...,k q }. Next, lower bound S in the context of T : qx f(s [ T ) f(t )= [f(t [{k 1,...,k t }) f(t [{k 1,...,k t 1 })] t=1 (5.71) qx qx = f(k t T [{k 1,...,k t }\{k t }) f(k t T [ S \{k t }) t=1 t=1 = X j2s\t (5.72) f(j S [ T \{j}) (5.73) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F34/66 (pg.80/187)

83 Eq ) Eq Let T \ S = {j 1,...,j r } and S \ T = {k 1,...,k q }. So we have the upper bound f(t S) =f(s [ T ) f(s) apple X j2t \S f(j S) (5.74) and the lower bound f(s T )=f(s [ T ) f(t ) X j2s\t f(j S [ T \{j}) (5.75) This gives upper and lower bounds of the form f(t )+lower bound apple f(s [ T ) apple f(s)+upper bound, (5.76) and combining directly the left and right hand side gives the desired inequality. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F35/66 (pg.81/187)

84 Eq ) Eq This follows immediately since if S T,thenS \ T = ;, andthelastterm of Eq vanishes. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F36/66 (pg.82/187)

85 Many (Equivalent) Definitions of Submodularity f(a) + f(b) f(a [ B)+f(A \ B), 8A, B V (5.54) f(j S) f(j T ), 8S T V, with j 2 V \ T (5.55) f(c S) f(c T ), 8S T V, with C V \ T (5.56) f(j S) f(j S [{k}), 8S V with j 2 V \ (S [{k}) (5.57) f(a [ B A \ B) apple f(a A \ B)+f(B A \ B), 8A, B V (5.58) f(t ) apple f(s)+ X X f(j S) f(j S [ T {j}), 8S, T V j2t \S j2s\t (5.59) f(t ) apple f(s)+ X f(j S), 8S T V (5.60) f(t ) apple f(s) f(t ) apple f(s) j2t \S X f(j S \{j})+ X f(j S \ T ) 8S, T V j2s\t j2t \S (5.61) X f(j S \{j}), 8T S V (5.62) j2s\t Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F37/66 (pg.83/187)

86 Eq ) Eq Here, we set T = S [{j, k}, j/2 S [{k} into Eq to obtain f(s [{j, k}) apple f(s)+f(j S)+f(k S) (5.77) = f(s)+f(s + {j}) f(s)+f(s + {k}) f(s) (5.78) = f(s + {j})+f(s + {k}) f(s) (5.79) = f(j S)+f(S + {k}) (5.80) giving f(j S [{k}) =f(s [{j, k}) f(s [{k}) (5.81) apple f(j S) (5.82) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F38/66 (pg.84/187)

87 Submodular Concave Why do we call the f(a)+f(b) f(a [ B)+f(A \ B) definition of submodularity, submodular concave? Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F39/66 (pg.85/187)

88 Submodular Concave Why do we call the f(a)+f(b) f(a [ B)+f(A \ B) definition of submodularity, submodular concave? Acontinuoustwicedifferentiablefunctionf : R n! R is concave if f r 2 f 0 (the Hessian matrix is nonpositive definite). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F39/66 (pg.86/187)

89 Submodular Concave Why do we call the f(a)+f(b) f(a [ B)+f(A \ B) definition of submodularity, submodular concave? Acontinuoustwicedifferentiablefunctionf : R n! R is concave if f r 2 f 0 (the Hessian matrix is nonpositive definite). Define a discrete derivative or difference operator defined on discrete functions f :2 V! R as follows: (r B f)(a), f(a [ B) f(a \ B) =f B (A \ B) (5.83) read as: the derivative of f at A in the direction B. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F39/66 (pg.87/187)

90 Submodular Concave Why do we call the f(a)+f(b) f(a [ B)+f(A \ B) definition of submodularity, submodular concave? Acontinuoustwicedifferentiablefunctionf : R n! R is concave if f r 2 f 0 (the Hessian matrix is nonpositive definite). Define a discrete derivative or difference operator defined on discrete functions f :2 V! R as follows: (r B f)(a), f(a [ B) f(a \ B) =f B (A \ B) (5.83) read as: the derivative of f at A in the direction B. Hence, if A \ B = ;, then(r B f)(a) =f(b A). Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F39/66 (pg.88/187)

91 Submodular Concave Why do we call the f(a)+f(b) f(a [ B)+f(A \ B) definition of submodularity, submodular concave? Acontinuoustwicedifferentiablefunctionf : R n! R is concave if f r 2 f 0 (the Hessian matrix is nonpositive definite). Define a discrete derivative or difference operator defined on discrete functions f :2 V! R as follows: : 0 (r B f)(a), f(a [ B) f(a \ B) =f B (A \ B) (5.83) read as: the derivative of f at A in the direction B. Hence, if A \ B = ;, then(r B f)(a) =f(b A). Consider a form of second derivative or 2nd difference: (r B f)(a) z } { (r C r B f)(a) =r C [ f(a [ B) f(a \ B) ] (5.84) =(r B f)(a [ C) (r B f)(a \ C) (5.85) = f(a [ B [ C) f((a [ C) \ B) f((a \ C) [ B)+f((A \ C) \ B) (5.86) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F39/66 (pg.89/187)

92 Submodular Concave If the second difference operator everywhere nonpositive: f(a [ B [ C) f((a [ C) \ B) f((a \ C) [ B)+f(A \ C \ B) apple 0 (5.87) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F40/66 (pg.90/187)

93 Submodular Concave If the second difference operator everywhere nonpositive: f(a [ B [ C) f((a [ C) \ B) then we have the equation: f((a \ C) [ B)+f(A \ C \ B) apple 0 (5.87) f((a [ C) \ B)+f((A \ C) [ B) f(a [ B [ C)+f(A \ C \ B) (5.88) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F40/66 (pg.91/187)

94 Submodular Concave If the second difference operator everywhere nonpositive: f(a [ B [ C) f((a [ C) \ B) then we have the equation: f((a \ C) [ B)+f(A \ C \ B) apple 0 (5.87) f((a [ C) \ B)+f((A \ C) [ B) f(a [ B [ C)+f(A \ C \ B) (5.88) Define A 0 =(A [ C) \ B and B 0 =(A \ C) [ B. Thentheabove implies: f(a 0 )+f(b 0 ) f(a 0 [ B 0 )+f(a 0 \ B 0 ) (5.89) and note that A 0 and B 0 so defined can be arbitrary. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F40/66 (pg.92/187)

95 Submodular Concave If the second difference operator everywhere nonpositive: f(a [ B [ C) f((a [ C) \ B) then we have the equation: f((a \ C) [ B)+f(A \ C \ B) apple 0 (5.87) f((a [ C) \ B)+f((A \ C) [ B) f(a [ B [ C)+f(A \ C \ B) (5.88) Define A 0 =(A [ C) \ B and B 0 =(A \ C) [ B. Thentheabove implies: f(a 0 )+f(b 0 ) f(a 0 [ B 0 )+f(a 0 \ B 0 ) (5.89) and note that A 0 and B 0 so defined can be arbitrary. One sense in which submodular functions are like concave functions. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F40/66 (pg.93/187)

96 Submodular Concave C C A B A B (a) A 0 =(A [ C) \ B (b) B 0 =(A \ C) [ B Figure: A figure showing A 0 [ B 0 = A [ B [ C and A 0 \ B 0 = A \ C \ B. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F41/66 (pg.94/187)

97 Submodular Concave C C A B A B (a) A 0 =(A [ C) \ B (b) B 0 =(A \ C) [ B Figure: A figure showing A 0 [ B 0 = A [ B [ C and A 0 \ B 0 = A \ C \ B. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F41/66 (pg.95/187)

98 Submodularity and Concave This submodular/concave relationship is more simply done with singletons. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F42/66 (pg.96/187)

99 Submodularity and Concave This submodular/concave relationship is more simply done with singletons. Recall four points definition: A function is submodular if for all X V and j, k 2 V \ X f(x + j)+f(x + k) f(x + j + k)+f(x) (5.90) f ( Al + FIB ) z fltub ) + fans ) # Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F42/66 (pg.97/187)

100 Submodularity and Concave This submodular/concave relationship is more simply done with singletons. Recall four points definition: A function is submodular if for all X V and j, k 2 V \ X f(x + j)+f(x + k) f(x + j + k)+f(x) (5.90) This gives us a simpler notion corresponding to concavity. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F42/66 (pg.98/187)

101 Submodularity and Concave This submodular/concave relationship is more simply done with singletons. Recall four points definition: A function is submodular if for all X V and j, k 2 V \ X f(x + j)+f(x + k) f(x + j + k)+f(x) (5.90) This gives us a simpler notion corresponding to concavity. Define gain as r j (X) =f(x + j) f(x), aformofdiscretegradient. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F42/66 (pg.99/187)

102 Submodularity and Concave This submodular/concave relationship is more simply done with singletons. Recall four points definition: A function is submodular if for all X V and j, k 2 V \ X f(x + j)+f(x + k) f(x + j + k)+f(x) (5.90) This gives us a simpler notion corresponding to concavity. Define gain as r j (X) =f(x + j) f(x), aformofdiscretegradient. Trivially becomes a second-order condition, akin to concave functions: AfunctionissubmodularifforallX V and j, k 2 V,wehave: r j r k f(x) apple 0 (5.91) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F42/66 (pg.100/187)

103 Example: Rank function of a matrix Consider the following 4 8 matrix, so V = {1, 2, 3, 4, 5, 6, 7, 8} = x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 Let A = {1, 2, 3}, B = {3, 4, 5}, C = {6, 7}, A r = {1}, B r = {5}. Then r(a) =3, r(b) =3, r(c) =2. r(a [ C) =3, r(b [ C) =3. r(a [ A r )=3, r(b [ B r )=3, r(a [ B r )=4, r(b [ A r )=4. r(a [ B) =4, r(a \ B) =1 <r(c) =2. 6= r(a)+r(b) =r(a [ B)+r(C) >r(a [ B)+r(A \ B) = 5 Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F43/66 (pg.101/187)

104 On Rank Let rank :2 V! Z + be the rank function. 0 Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F44/66 (pg.102/187)

105 On Rank Let rank :2 V! Z + be the rank function. In general, rank(a) apple A, and vectors in A are linearly independent if and only if rank(a) = A. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F44/66 (pg.103/187)

106 On Rank Let rank :2 V! Z + be the rank function. In general, rank(a) apple A, and vectors in A are linearly independent if and only if rank(a) = A. If A, B are such that rank(a) = A and rank(b) = B, with A < B, thenthespacespannedbyb is greater, and we can find a vector in B that is linearly independent of the space spanned by vectors in A. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F44/66 (pg.104/187)

107 On Rank Let rank :2 V! Z + be the rank function. In general, rank(a) apple A, and vectors in A are linearly independent if and only if rank(a) = A. If A, B are such that rank(a) = A and rank(b) = B, with A < B, thenthespacespannedbyb is greater, and we can find a vector in B that is linearly independent of the space spanned by vectors in A. To stress this point, note that the above condition is A < B, not A B which is sufficient (to be able to find an independent vector) but not required. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F44/66 (pg.105/187)

108 On Rank Let rank :2 V! Z + be the rank function. In general, rank(a) apple A, and vectors in A are linearly independent if and only if rank(a) = A. If A, B are such that rank(a) = A and rank(b) = B, with A < B, thenthespacespannedbyb is greater, and we can find a vector in B that is linearly independent of the space spanned by vectors in A. To stress this point, note that the above condition is A < B, not A B which is sufficient (to be able to find an independent vector) but not required. In other words, given A, B with rank(a) = A &rank(b) = B, then A < B,9an b 2 B such that rank(a [{b}) = A +1. Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F44/66 (pg.106/187)

109 Spanning trees/forests 1 1 o 2 We are given a graph G =(V,E), andconsidertheedgese = E(G) as an index set. Consider the V E incidence matrix of undirected graph G, which is the matrix X G =(x v,e ) v2v (G),e2E(G) where ( 1 if v 2 e x v,e = (5.92) 0 if v/2 e B C 7@ A (5.93) Prof. Jeff Bilmes EE563/Spring 2018/Submodularity - Lecture 5 - April 9th, 2018 F45/66 (pg.107/187)