EE595A Submodular functions, their optimization and applications Spring 2011

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1 EE595A Submodular functions, their optimization and applications Spring 2011 Prof. Jeff Bilmes University of Washington, Seattle Department of Electrical Engineering Spring Quarter, Lecture 2 - April 1st, 2011 Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 1

2 Announcements Weekly Office Hours: Wednesdays, 12:30-1:30pm, 10 minutes after class on Wednesdays. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 2

3 Scratch Paper Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 3

4 Scratch Paper Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 4

5 Scratch Paper Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 5

6 Submodular Definitions Definition (submodular) A function f : 2 V R is submodular if for any A, B V, we have that: f (A) + f (B) f (A B) + f (A B) (1) An alternate and equivalent definition is: Definition (diminishing returns) A function f : 2 V R is submodular if for any A B V, and v V \ B, we have that: f (A {v}) f (A) f (B {v}) f (B) (2) This means that 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 EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 6

7 Ground set: E or V? Submodular functions are functions defined on subsets of some finite set, called the ground set. It is common in the literature to use either E or V as the ground set. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 7

8 Ground set: E or V? Submodular functions are functions defined on subsets of some finite set, called the ground set. It is common in the literature to use either E or V as the ground set. We will follow this inconsistency in the literature and will inconsistently use either E or V as our ground set (hopefully not in the same equation, if so, please point this out). Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 7

9 Notation R E R E = {x = (x j R : j E)} (3) R E + = {x = (x j : j E) : x 0} (4) Any vector x R E can be treated as a normalized modular function, and vice versa. That is x(a) = a A x a (5) Note that x is said to be normalized since x( ) = 0. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 8

10 Other Notation: indicator vectors of sets Given an A E, define the vector 1 A R E + to be 1 A (j) = { 1 if j A; 0 if j / A (6) Sometimes this will be written as χ A 1 A. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 9

11 Other Notation: indicator vectors of sets Given an A E, define the vector 1 A R E + to be 1 A (j) = { 1 if j A; 0 if j / A (6) Sometimes this will be written as χ A 1 A. Thus, given modular function x R E, we can write x(a) in a variety of ways, i.e., x(a) = x 1 A = i A x(i) (7) Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 9

12 Other Notation: singletons and sets When A is a set and k is a singleton (i.e., a single item), the union is properly written as A {k}, but sometimes I will write just A + k. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 10

13 Summing Submodular Functions Given E, let f 1, f 2 : 2 E R be two submodular functions. Then f : 2 E R with f (A) = f 1 (A) + f 2 (A) (8) is submodular. This follows easily since f (A) + f (B) = f 1 (A) + f 2 (A) + f 1 (B) + f 2 (B) (9) f 1 (A B) + f 2 (A B) + f 1 (A B) + f 2 (A B) (10) = f (A B) + f (A B). (11) I.e., it holds for each component of f in each term in the inequality. In fact, any conic combination (i.e., non-negative linear combination) of submodular functions is submodular, as in f (A) = α 1 f 1 (A) + α 2 f 2 (A) for α 1, α 2 0. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 11

14 Summing Submodular and Modular Functions Given E, let f 1, m : 2 E R be a submodular and a modular function. Then f : 2 E R with f (A) = f 1 (A) m(a) (12) is submodular (as is f (A) = f 1 (A) + m(a)). This follows easily since f (A) + f (B) = f 1 (A) m(a) + f 1 (B) m(b) (13) f 1 (A B) m(a B) + f 1 (A B) m(a B) (14) = f (A B) + f (A B). (15) That is, the modular component with m(a) + m(b) = m(a B) + m(a B) never destroys the inequality. Note of course that if m is modular than so is m. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 12

15 Restricting Submodular Functions Given E, let f : 2 E R be a submodular functions. And let S E be an arbitrary fixed set. then is submodular. Proof. Given A B E \ v, consider f : 2 E R with f (A) = f (A S) (16) f ((A + v) S) f (A S) f ((B + v) S) f (B S) (17) If v / S, then both differences on each size are zero. If v S, then we can consider this f (A + v) f (A ) f (B + v) f (B ) (18) with A = A S and B = B S. Since A B, this holds due to submodularity of f. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 13

16 Summing Restricted Submodular Functions Given V, let f 1, f 2 : 2 V R be two submodular functions and let S 1, S 2 be two arbitrary fixed sets. Then f : 2 V R with f (A) = f 1 (A S 1 ) + f 2 (A S 2 ) (19) is submodular. This follows easily from the preceding two results. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 14

17 Summing Restricted Submodular Functions Given V, let C = {C 1, C 2,..., C k } be a set of subsets of V, and for each C C, let f C : 2 V R be a submodular function. Then f : 2 V R with f (A) = C C f C (A C) (19) is submodular. This property is critical for image processing and graphical models. For example, let C be all pairs of the form {{u, v} : u, v V }, or let it be all pairs corresponding to the edges of some undirected graphical model. We plan to revisit this topic later in the term. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 14

18 Max - normalized Given V, let c R V + be a given fixed vector. Then f : 2 V R +, where is submodular and normalized (we take f ( ) = 0). Proof. Consider which follows since we have that and f (A) = max j A c j (20) max j A c j + max j B c j max j A B c j + max j A B c j (21) max(max j A c j, max j B c j) = max j A B c j (22) min(max j A c j, max j B c j) max j A B c j (23) Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 15

19 Max Given V, let c R V be a given fixed vector (not necessarily non-negative). Then f : 2 V R, where f (A) = max j A c j (24) is submodular, where we take f ( ) min j c j (so the function is not normalized). Proof. The proof is identical to the normalized case. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 16

20 Facility Location Given V, E, let c R V E be a given V E matrix. Then f : 2 E R, where f (A) = i V max j A c ij (25) is submodular. This is a facility location function. Proof. We can write f (A) as f (A) = i V f i(a) where f i (A) = max j A c ij is submodular (max of a i th row vector), so f can be written as a sum of submodular functions. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 17

21 Facility/Plant Location (uncapacitated) Facility Location is a core problem in operations research and a strong motivation for submodular functions. Key goal is to place facilities to supply demand sites as efficiently as possible. Let E be a set of possible factory/plant locations, and V is a set of sites needing to be serviced (e.g., cities). Let c ij be the benefit (e.g., 1/c ij is the cost) of servicing city i with plant j. Let m : 2 E R E + be a plant construction modular function (vector). E.g., 1/m j is the cost to build a plant at location j. Each city needs to be serviced by only one plant but no less than one. Define f (A) as the delivery benefit plus construction benefit when the plants in set A are considered to be constructed. We can define f (A) = m(a) + i V max j A c ij. Goal is to find a set A that maximizes f (A) (the benefit) placing a bound on the number of plants A (e.g., A k). Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 18

22 Log Determinant Let Σ be an n n positive definite matrix. Let V = {1, 2,..., n} [n] be an index set, and for A V, let Σ A be the (square) submatrix of Σ obtained by including only entries in the rows/columns given by A. Then: f (A) = log det(σ A ) is submodular. (26) Proof. Suppose x R n is multivariate Gaussian, that is ( 1 x p(x) = exp 1 ) 2πΣ 2 (x µ)t Σ 1 (x µ) (27)... Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 19

23 Log Determinant...cont. f (A) = log det(σ A ) is submodular. (26) Then the (differential) entropy of the r.v. X is given by h(x ) = log 2πeΣ = log (2πe) n Σ (27) and in particular, for a variable subset A, f (A) = h(x A ) = log (2πe) A Σ A (28) Entropy is submodular (conditioning reduces entropy), and moreover where m(a) is a modular function. f (A) = h(x A ) = m(a) log Σ A (29) Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 19

24 Concave over non-negative modular Let m R E + be a modular function, and g a concave function over R. Define f : 2 E R as Proof. f(a) = g(m(a)) then f is submodular. Proof. Given A B E \ v, we have 0 a = m(a) b = m(b), and 0 c = m(v). For g concave, we have g(a + c) g(a) g(b + c) g(b), and thus g(m(a) + m(v)) g(m(a)) g(m(b) + m(v)) g(m(b)) (30) The converse is true as well. Exercise: prove this. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 20

25 Monotone difference of two functions Let f and G both be submodular functions on subsets of V and let (f g)( ) be either monotone increasing or monotone decreasing. Then h : 2 V R defined by is submodular. Proof. h(a) = min(f (A), g(a)) (31) If h(a) agrees with either f or g on both X and Y, the result follows since f (X ) + f (Y ) min(f (X Y ), g(x Y )) + min(f (X Y ), g(x Y )) g(x ) + g(y ) (32)... Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 21

26 Monotone difference of two functions Let f and G both be submodular functions on subsets of V and let (f g)( ) be either monotone increasing or monotone decreasing. Then h : 2 V R defined by is submodular....cont. h(a) = min(f (A), g(a)) (31) 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 ) (32) By monotonicity, f (X Y ) + g(y ) f (Y ) g(x Y ) giving h(x ) + h(y ) g(x Y ) + f (X Y ) h(x Y ) + h(x Y ) (33) Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 21

27 Min Let f : 2 V R be an increasing or decreasing submodular function and let k be a constant. Then the function h : 2 V R defined by is submodular. Proof. h(a) = min(k, f (A)) (34) For constant k, we have that (f k) is increasing (or decreasing) so this follows from the previous result. Note also, g(a) = min(k, a) for constant k is a concave function, so we can use the earlier result about composing a concave function with a submodular function to get this result as well. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 22

28 More on Min - saturate trick In general, the minimum of two submodular functions is not submodular. However, when wishing to maximize two monotone non-decreasing submodular functions, we can define function h : 2 V R as h(a) = 1 (min(k, f ) + min(k, g)) (35) 2 then h is submodular, and h(a) k if and only if both f (A) k and g(a) k. We plan to revisit this again later in the quarter. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 23

29 Arbitrary functions as difference between submodulars Given an arbitrary set function f, it can be expressed as a difference between two submodular functions: f = g h where both g and h are submodular. Define Proof. Let f be given and arbitrary. α = min f (X ) + f (Y ) f (X Y ) f (X Y ) (36) X,Y If α 0 then f is submodular, so by assumption α < 0. Now let h be an arbitrary strict submodular function and define Strict means β > 0. β = min h(x ) + h(y ) h(x Y ) h(x Y ) (37) X,Y... Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 24

30 Arbitrary functions as difference between submodulars Given an arbitrary set function f, it can be expressed as a difference between two submodular functions: f = g h where both g and h are submodular....cont. Define f : 2 V R as f (A) = f (A) + α h(a) (36) β Then f is submodular, and f = f (A) α β h(a), a difference between two submodular functions as desired. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 24

31 Submodular Definitions Definition (submodular) A function f : 2 V R is submodular if for any A, B V, we have that: f (A) + f (B) f (A B) + f (A B) (1) An alternate and equivalent definition is: Definition (diminishing returns) A function f : 2 V R is submodular if for any A B V, and v V \ B, we have that: f (A {v}) f (A) f (B {v}) f (B) (2) This means that 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 EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 25

32 Submodular Definitions An alternate and equivalent definition is: Definition (group diminishing returns) A function f : 2 V R is submodular if for any A B V, and C V \ B, we have that: f (A C) f (A) f (B C) f (B) (37) This means that the incremental value or gain of set C decreases as the context in which v is considered grows from A to B (diminishing returns) Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 26

33 Submodular Definitions Proposition group diminishing returns implies diminishing returns Proof. Obvious, set C = {v}. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 27

34 Submodular Definitions Proposition diminishing returns implies group diminishing returns Proof. Let C = {c 1, c 2,..., c k }. Then diminishing returns implies f (A C) f (A) (38) k 1 ( ) = f (A C) f (A {c 1,..., c i }) f (A {c 1,..., c i }) f (A) (39) = i=1 k f (A {c 1... c i }) f (A {c 1... c i 1 }) (40) i=1 k f (B {c 1... c i }) f (B {c 1... c i 1 }) (41) i=1 k 1 ( ) = f (B C) f (B {c 1,..., c i }) f (B {c 1,..., c i }) f (B) (42) i=1 = f (B C) f (B) (43) Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 27

35 Submodular Definitions are equivalent Proposition The two aforementioned definitions of submodularity submodular and diminishing returns are identical. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 28

36 Submodular Definitions are equivalent Proof. Assume submodular. Assume A B as otherwise trivial. Let B \ A = {v 1, v 2,..., v k } and define A i = A {v 1... v i }, so A 0 = A. Then by submodular, or f (A i + v) + f (A i + v i+1 ) f (A i + v + v i+1 ) + f (A i ) (44) f (A i + v) f (A i ) f (A i + v i+1 + v) f (A i + v i+1 ) (45) we apply this inductively, and use f (A i+1 + v) f (A i+1 ) = f (A i + v i+1 + v) f (A i + v i+1 ) (46) and that A k 1 + v k = B.... Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 28

37 Submodular Definitions are equivalent...cont. Assume group diminishing returns. Assume A B otherwise trivial. Define A = A B, C = A \ B, and B = B. Then giving or f (A + C) f (A ) f (B + C) f (B ) (47) f (A + C) + f (B ) f (B + C) + f (A ) (48) f (A B + A \ B) + f (B) f (B + A \ B) + f (A B) (49) which is the same as f (A) + f (B) f (A B) + f (A B) (50) Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 29

38 Submodular Definitions Definition (singleton) A function f : 2 V R is submodular if for any A V, and any a, b V \ A, we have that: f (A {a}) + f (A {b}) f (A {a, b}) + f (A) (51) This follows immediately from diminishing returns. To achieve diminishing returns, assume A B with B \ A = {b 1, b 2,..., b k }. Then f (A + a) f (A) f (A + b 1 + a) f (A + b 1 ) (52) f (A + b 1 + b 2 + a) f (A + b 1 + b 2 ) (53)... (54) f (A + b b k + a) f (A + b b k ) (55) = f (B + a) f (B) (56) Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 30

39 Gain It is often the case that we wish to express the gain of an item j V in some context, say A, namely f (A {j}) f (A). This is used so often, that there are equally as many ways to notate this. I.e., f (A {j}) f (A) = ρ j (A) (57) = ρ A (j) (58) = f ({j} A) (59) = f (j A) (60) We ll use either ρ j (A) or f (j A). Note, diminishing returns can now be stated as saying that ρ j (A) is a monotone non-increasing function of A, since ρ j (A) ρ j (B) whenever B A. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 31

40 Equivalent Definitions of Submodularity f (A) + f (B) f (A B) + f (A B), A, B E (61) ρ j (S) ρ j (T ), S T E, with j E \ T (62) ρ j (S) ρ j (S {k}), S E with j E \ (S {k}) (63) f (T ) f (S) + ρ j (S) ρ j (S T {j}), S, T E j T \S f (T ) f (S) + j T \S f (T ) f (S) + j S\T f (T ) f (S) j S\T (64) ρ j (S), S T E (65) ρ j (S \ {j}), T S E (66) ρ j (S \ {j}) + ρ j (S T ) S, T E j S\T T \S (67) Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 32

41 Equivalent Definitions of Submodularity We ve already seen that Eq. 61 Eq. 62 Eq. 63. We next show that Eq. 63 Eq. 64 Eq. 65 Eq. 63. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 33

42 Eq. 63 Eq. 64 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: f (S T ) f (S) = = r ( ) f (S {j 1,..., j t }) f (S {j 1,..., j t 1 }) t=1 r ρ jt (S {j 1,..., j t 1 }) t=1 = j T \S (68) r ρ jt (S) (69) t=1 ρ j (S) (70) Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 34

43 Eq. 63 Eq. 64 Let T \ S = {j 1,..., j r } and S \ T = {k 1,..., k q }. Next, lower bound S in the context of T : f (S T ) f (T ) = = q [f (T {k 1,..., k t }) f (T {k 1,..., k t 1 })] t=1 q ρ kt (T {k 1,..., k t } \ {k t }) t=1 = j S\T (68) q ρ kt (T S \ {k t }) t=1 (69) ρ j (S T \ {j}) (70) Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 34

44 Eq. 63 Eq. 64 Let T \ S = {j 1,..., j r } and S \ T = {k 1,..., k q }. So we have the upperbound f (S T ) f (S) ρ j (S) (68) and the lower bound f (S T ) f (T ) j S\T j T \S and subtracting the 2nd from the first gives the result. ρ j (S T \ {j}) (69) Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 34

45 Eq. 64 Eq. 65 This follows immediately since if S T, then S \ T =, and the last term of Eq. 64 vanishes. Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 35

46 Eq. 65 Eq. 63 Here, we set T = S {j, k}, j / S {k} into Eq. 65 to obtain f (S {j, k}) f (S) + ρ j (S) + ρ k (S) (70) = f (S) + f (S + {j}) f (S) + f (S + {k}) f (S) (71) = f (S + {j}) + f (S + {k}) f (S) (72) = ρ j (S) + f (S + {k}) (73) giving ρ j (S {k}) = f (S {j, k}) f (S {k}) (74) ρ j (S) (75) Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 36

47 Sources for Today s Lecture Lovasz-1983, Nemhauser,Wolsey,Fisher-1978, Narayanan-1997, and Narasimhan,Bilmes Prof. Jeff Bilmes EE595A/Spr 2011/Submodular Functions Lecture 2 - April 1st, 2011 page 37

Submodular Functions, Optimization, and Applications to Machine Learning

Submodular Functions, Optimization, and Applications to Machine Learning Spring Quarter, Lecture 3 http://j.ee.washington.edu/~bilmes/classes/ee596b_spring_2014/ Prof. Jeﬀ Bilmes University of Washington,