1. The Submodular Welfare Problem Lecturer: Moran Feldman Based on “Optimal Approximation for the Submodular Welfare Problem in the Value Oracle Model” By Jan Vondrák

2. 2 Talk Outline Preliminaries and the problems Reducing to continuous problems Approximating the continuous problems Constructing an integral solution Summary

3. 3 Combinatorial Auctions Instance A set P of n players A set Q of m items Utility function w j : 2 Q   + for each player. Objective Let Q j  Q denote the set of items the j th player gets. The utility of the j th player is w j (Q j ). Distribute the items among the players, maximizing the sum of utilities.

4. 4 Combinatorial Auction - Example Players –TAs TA with a presentationTA without a presentation UtilityItems 0No classroom 10Classroom and multimedia 3Classroom and markers 1Only classroom UtilityItems 0No classroom 1Classroom and multimedia 7Classroom and markers 1Only classroom Items –Classrooms –Markers –Multimedia keys

5. 5 Oracles Solutions Considering special class of utility functions with polynomial-size representation. Accessing the utility functions through oracles: –Value Oracle - Given a set Q j of items, returns the value w j (Q j ). –Demand oracle - Given an assignment p:Q   of prices to items, finds a set Q j maximizing. More powerful oracle. Obstacle The utility functions w j have exponential size in the number of items. We assume this solution

6. 6 Utility Functions Assuming nothing None possible Assuming monotone submodular 1–1/e approximation at best Assumptions about the utility functions: * Under the value oracle model Approximation with polynomial number of oracle queries: **

7. 7 Set Function Properties Given a set function  : 2 S   +, we say that  is: Monotone – if  (A) ≤  (B) for any A  B  S. Submodular – if  (A  B) +  (A  B) ≤  (A) +  (B) for any A,B  S. Say that w j is submodular, so what? Consider a player j receiving a set Q j of items. Let v be the additional value j would get if we assigned item q to j. Formally: v = w j (Q j  q) – w j (Q j ) Assign to j additional set of items Q’ j. If we assigned item q to j now, j would get no more additional value than before. Formally: v ≥ w j (Q j  Q’ j  q) – w j (Q j  Q’ j ) The utility of an item diminishes as the player gets more items!

8. 8 The Submodular Welfare Problem (SWP) Instance A set P of n players A set Q of m items Monotone submodular utility function w j : 2 Q   + for each player, accessed via value oracles. Objective Find a partition Q 1, Q 2, …, Q n of Q maximizing the total utility:

9. 9 Matroid Definition An ordered pair (X,  ), with   2 X (the sets in  are called “independent sets”), such that: 1.There is an independent set:    2.Monotonicity: A  B, B    A   3.Augmentation: If A,B   and |A| < |B|, then there exists b  B - A such that A  {b}  . Motivation: Generalization of many well known concepts: Given a vectors space, the sets of independent vectors form a matroid. Given a graph, the set of forest sub-graphs form a matroid.

10. 10 Example – Forest Sub Graphs Matroid a b c d e f X = {ab, bd, df, ef, ce, ca, bc, be, de}  = {S  X | S is a forest} Independent set Example S = {ab, bc, df, ef}  Property 1:   

11. 11 Example – Forest Sub Graphs Matroid a b c d e f Property 2: Monotonicity Removing edges from a forest leaves it a forest. Property 3: Augmentation Given two forests A,B with |A| < |B|, there is an edge e  B, such that A  {e} is also a forest. Why? No forest can have more than |A| edges inside connected components of A.

12. 12 Submodular Maximization Subject to a Matroid Constraint (SMSMC) Instance A groundset X. A monotone submodular function  : 2 X   +, accessed via a value oracle. Matroid M = (X,  ) accessed via a membership oracle. Objective Find an independent set of value: Motivation Generalize known problems, for example: SWP – (will be proven in a moment) Max k-Cover – Find in a group S 1, S 2, …, S n, k sets of maximal union. Multiple Knapsack (exponential reduction) – Same as Knapsack, but multiple Knapsacks are available.

13. 13 Reduction The Reduction Groundset: X = P  Q Given a set S  X, let S j = {q  Q | (j, q)  S}, define  : 2 X   + as: The matroid M = (X,  ) enforces the restriction that an item may be assigned to at most one player: Theorem SWP can be reduced to SMSMC. Corollary We can focus from now on SMSMC. SMSMC SWP SMSMC

15. 15 Continues Set Functions Motivation Let X be a groundset, and consider y  [0, 1] X Intuitively, we can think of y as selecting each item j  X to the extent y j. We want extend the properties of set functions to the continues case. Notation (x  y) i = max {x i,y i }  x  y is a generalization of union (x  y) i = min {x i,y i }  x  y is a generalization of intersection Definitions Let F: [0, 1] X  , F is: monotone if x  y  F(x)  F(y) submodular if F(x  y) + F(x  y)  F(x) + F(y) x = (0.4, 0, 0.3, 0.6, 1) y = (0.6, 1, 0.2, 0.2, 0.7) x  y = (0.6, 1, 0.3, 0.6, 1) x  y = (0.4, 0, 0.2, 0.2, 0.7)

16. 16 Smooth Monotone Submodularity Definition A function F: [0, 1] X   is smooth monotone submodular if: F has a second derivation throughout its domain. For each j  X, everywhere.  F is monotone. For any i,j  X (possibly i=j), everywhere.  F is submodular ( is non-increasing with respect to y i ).  F is concave in all non-negative directions.

17. 17 Extension by Expectation Extension For y  [0, 1] X, let us denote by ỹ the random set obtained from y by selecting item j  X with probability y j. Let  : 2 X   be a monotone submodular function. The canonical extension of  into a continuous function is: Extension: If y is integral, ỹ contains exactly the items selected by y. Properties Preservation: We need to prove that F is a smooth monotone submodular function. Objective Given a set function  we want to extend it into a continuous function F. Extension: F should coincide with  for integral values. Properties Preservation: F should be smooth monotone submodular, assuming  is monotone and submodular.

18. 18 Properties of F The monotonicity requirement : Follows from the monotonicity of  Follows from the submodularity of  The submodularity requirement for i  j: The submodularity requirement for i = j:

19. 19 Matroid Polytopes Definition For any set S  X we let 1 S represent the characteristic vector of S in [0, 1] X. Given a matroid M = (X,  ), its matroid polytopes P(M) is the set of convex combinations of vectors from: Example X = {a, b, c}  = { {a,b}, {b,c}, {a}, {b}, {c},  } Characteristic vectors: {a, b} (1, 1, 0) {c} (0, 0, 1)  (0, 0, 0) {b}{b} {a, b} {a}{a} {c}{c} {b, c}

20. 20 Matroid Polytopes - Properties Definition A polytopes P   + X is called down-monotone if for any 0  x  y, y  P  x  P. Lemma For any matroid M, P(M) is down-monotone. Proof Given 0  x  y  P(M), the following procedure gets us from x to y without leaving P(M). Procedure: 1.While y j > x j need to be decreased: 2. Find a set S   in the convex combination of y containing j. 3. Since M is matroid, S’ = S – { j }  . 4. Replace S by S’ in the convex combination continuously, until x j = y j or S is completely removed. {b}{b} {a, b} {a}{a} {c}{c} {b, c}

22. 22 Marginal Value Definition Let  : 2 X   be a set function. Given a set S of items, the marginal value of item j is  S (j) =  (S  j) –  (S).  is submodular  the marginal value of every item j diminishes as more items are added to S. What is it good for? Continuous optimization algorithms often use  F as guidance. Implicitly, the derivative  F/  y j is used to estimate the importance of increasing y j. Analogously, we use E [  ỹ (j)] to estimate the importance of increasing y j. E [  ỹ (1)] is low E [  ỹ (2)] is high

23. 23 The Continuous Greedy Algorithm Input Matroid M = (X,  ), monotone submodular function  : 2 X   +. Pseudo Code 1.Set   1/n 2 (n = |X|), t  0 and y(0)  0. 2.While t < 1 do 3. For each j let ω j (t) be an estimation of E [  ỹ(t) (j)] to an error of no more than OPT n 2. 4. Let s(t) be a maximal-weight independent set in M according to the weights ω j (t). 5. Set y(t +  ) = y(t) +  ∙ 1 s(t), t  t +  6.Return y(1)

24. 24 Continuous Greedy Algorithm - Demonstration Next steps Analyzing the algorithm’s performance. Surveying the implementation the algorithm. y(0) y(0.01) y(0.02) y(0.03) y(0.04)

25. 25 Lemma 1 Let OPT = max s   (S). Consider any y  [0, 1] X, then: Proof Consider a specific optimal solution O  , and any given value of ỹ. By the submodularity of F we know that: By taking the expectation over ỹ we get:

26. 26 Lemma 2 Let y be the fractional solution found by the Continuous Greedy Algorithm, then: Proof Consider a specific time t, and let Δ(t) = y(t +  ) – y(t) =  ∙ 1 s(t). D(t) - a random set containing each item j with probability Δ j (t). Then: Roadmap We want to lower bound the increase in F in time t. Taking small enough , we can ignore the contribution from D(t)’s which are not singletons. By monotonicity, since:  Pr[j  ỹ(t +  )] = y j (t) + Δ j (t)  Pr[j  ỹ(t)  D(t)] = 1 - (1-y j (t))(1-Δ j (t))  y j (t) + Δ j (t)

27. 27 Lemma 2 (cont.) Lower bound the increase in F at time t Taking advantage of s(t) properties We chose s(t) as the independent set maximizing: However ω j (t) is an estimation of E [f ỹ (j)] up to an error of OPT n 2. Therefore: Considering only singleton sets By the inequality: (1 -  ) k ≥ (1 - k  )

28. 28 Lemma 2 (cont.) Continuing the bound derivation Corollary The distance to diminishes by a factor of (1 -  ) at each step. Defining:By Lemma 1

29. 29 Lemma 2 (cont.) Warping up the proof After all 1/  steps has been performed we get: F is always positive, therefore: Algorithm Analysis - Summary Let y be the fractional solution returned by the Continuous Greedy Algorithm. y is within P(M) since it is the convex combination of 1/  = n 2 characteristic vectors of independent sets from M. The value of F in y is: F(y) ≥ (1 – 1/e – o(1))  OPT

30. 30 Continuous Greedy Algorithm - Implementation Problem Two steps of the Continuous Greedy Algorithm are not strait forward to implement: Implementing the first step w.h.p. Each time E [f ỹ(t) (j)] has to be estimated: –Perform n 5 independent samples. –Use the average as the estimation ω j (t). Notice that f ỹ(t) (j) ≤ f({ j }) ≤ OPT (unless j  S for any S   ). Calculating ω j (t), the estimation of E [f ỹ(t) (j)], to an error of no more than OPT n 2. Finding a maximum weight independent set in M according to the weights ω j (t). The probability of | ω j (t) – E [f ỹ (j)]| > OPT/n 2 is exponentially small Chernoff bound W.h.p. all estimations are not off by more than OPT/n 2 (n 3 estimations) Union bound

31. 31 Finding Maximal Weighted Independent Set Instance Matroid M = (X,  ) Weight function w on the elements of X The Greedy Algorithm 1.Sort the elements of X in non-decreasing weight order: j 1, j 2, …, j n. 2.Start with S =  3.For k = 1 to n do: 4. If S  {j k }  , add j k to S Objective Find an independent set S  , maximizing w(S).

32. 32 An old fellow Reconsider the forest sub-graphs matroid: –The edges of the graph are elements of the matroid. –The independent sets are forests. Rewriting the greedy algorithms in terms of this matroid yields: Translated Greedy Algorithm 1.Sort the edges of the graph in non-decreasing weight order: e 1, e 2, …, e m. 2.Start with F =  3.For k = 1 to m do: 4. If F  {e k } does not contain circles, add e k to F Kruskal’s algorithm

33. 33 Greedy Algorithm - Correctness Notation S k – The set S after the k th element is considered. S 0 – The set S before the first element is considered, i.e. . Lemma 3 For every 0 ≤ k ≤ n, S k  O k for some optimal set O k, and j 1, j 2, …, j k  O k - S k. Corollary For k = n lemma 3 implies: S n  O n j 1, j 2,…, j n  O n – S n  O n  S n Therefore the result of the algorithm (S n ) is the optimal set O n.  Now it all boils down to proving lemma 3. OkOk SkSk Elements we will consider Elements we considered

34. 34 Lemma 3 – Proof Proof Overview The proof is by induction of k. Induction base - for k = 0: -  = S 0  O 0 for any optimal set O 0 -No item must not be in O 0 -S 0 Induction step Prove the lemma for k, assuming it holds for k – 1: If j k is not inserted into S, set O k = O k-1 then: -S k = S k-1  O k-1 -j 1, j 2, …, j k-1  O k-1 - S k-1 -j k  O k-1 - S k-1 (because if j k  O k-1, then S k-1  {j k }  O k-1, contradicting S k-1  {j k }   ) If j k is inserted into S, then we need to construct a matching O k.

35. 35 Lemma 3 – Proof (cont.) Construction of O k 1.Initially O k = S k 2.While |O k | < |O k-1 | do 3. Find j  O k-1 – O k such that O k  { j }   4. Set O k  O k  { j } Construction Correctness Line 3 can be implemented because M is a matroid. O k = O k-1  {j k } – {j h } for some j h O k is an optimal set: –j h  O k-1 – S k-1  h  k  w(j h ) ≤ w(j k )  w(O k-1 ) ≤ w(O k ) –However, since we know that O k-1 is a maximum weight independent set: w(O k-1 ) = w(O k ). j 1, j 2, …, j k  O k – S k : –Holds for j 1, j 2, …, j k-1 because O k - O k-1 = {j k } –Holds for j k because j k  S k

36. 36 Milestone Achievement We showed how to find a fractional solution y such that: F(y)  (1 – 1/e – o(1))  OPT y  P(M) What’s next? We need to convert y into an integral solution. The integral solution must be the characteristic vector of an independent set of M. The conversion should not decrease the value of the solution significantly (in expectation).

38. 38 Rounding in the Submodular Welfare Case Notation Let y ij be the extent to which the j item is allocated to the i th player in y, i.e. the value of the pair (i, j) in y. Since y  P(M), we know that: SWP to SMSMC Reduction - Remainder Groundset: X = P  Q Given a set S  X, S j is the set of items allocated to player j. Function  : 2 X   + is defined as: The matroid M = (X,  ) allows only solutions assigning each item to at most one player: Each item is allocated to an extent  1

39. 39 Rounding Procedure Value Preservation Let R i be the set of items allocated to the i th player. Notice that R i has the same distribution as the set of items allocated to the i th player by ỹ. The expected utility of the i th player is: This is also the contribution of the i th player to F(y): Procedure For each item j, randomly allocate it to a player, the probability of assigning it to the i th player is y ij. This procedure is guaranteed to generate a valid integral solution, because each item is allocated to at most one player.

40. 40 Rounding in the General Case The rounding procedure presented crucially depends on: –The structure of the specific matroid. –The linearity of F (in terms of the utility functions). In the general case we need to use a stronger rounding method: Pipage rounding. Unlike randomized rounding, Pipage rounding preserves constraints. Pipage rounding was firstly proposed by Ageev and Sviridenko in 2004.

41. 41 Rank Functions and Tight Sets Definition Consider a matroid M = (X,  ). The rank function r M (A): 2 X  N induced by M is: r M (A) = max {|S| | S  A, S   } Informally, r M (A) maps each set to the size of the maximal independent set in it. Lemma The rank function induced by a matroid M is submodular. Proof Consider two sets A,B  X. Let S A  B be a maximal independent set in A  B. Extend S A  B to maximal independent sets S A and S B in A and B, respectively. Extend S A to a maximal independent set S A  B in A  B. SABSAB SASA SBSB SABSAB

42. 42 Rank Functions and Tight Sets (cont.) Observation Given a vector y  P(M), and a set A  X: Why? It holds for every independent set, therefore it must also hold for convex combination of independent sets. Definition Given a vector y  P(M), a set A  X is tight if: Proof - Continuation S A  B = S A  S B |S A  B |  |S A  S B |, otherwise: –|S A  B – (S A - S A  B )| > |S B | –|S A  B – (S A - S A  B )|  B, otherwise S A is not maximal. –Contradiction, because S B is maximal. |S A  B | + |S A  B |  |S A  S B | + |S A  S B | = |S A | + |S B | SABSAB SASA SBSB SABSAB

43. 43 Rank Functions and Tight Sets (cont.) Proof By the submodularity of r M : It can be easily checked that: Implying: Due to the observation: Lemma Let A and B be two tight sets, then: The intersection A  B is a tight set. The union A  B is a tight set.

44. 44 Algorithm Preliminaries Assumption We assume X is tight set under y. Otherwise: y  P(M) is a convex combination of independent sets  1,  2,…,  p. Replace each set  j with the maximal cardinality set containing it (its size must be r M (X)). y remains in P(M). Since F is monotone, F(y) does not decrease. Notation y ij (  ) - The vector y with y i increased by  and y j decreased by . y ij + = y ij (  ij + (y)),  ij + (y) = max{   0 | y ij (  )  P(M)} y ij - = y ij (  ij - (y)),  ij - (y) = min{  ≤ 0 | y ij (  )  P(M)} Ready. Set.

45. 45 The Pipage Rounding Algorithm Input A matroid M = (X,  ) Vector y such that X is tight. Pseudo Code 1.While y is not integral do 2. Let A be a minimal tight set containing i,j  A, such that y i,y j are fractional. 3. If F(y ij + )  F(y ij - ) 4. then y  y ij + 5. else y  y ij - 6.Output y Next steps Analyzing the algorithm’s performance. Describing how to implement the algorithm (sketch).

46. 46 Pipage Rounding is Well Defined Observation If a tight set A contains a non-integral value then it contains at least two fractional values in it because: Aim We need to show that there is always a valid set A, as long as y is fractional. Corollary All we only need to show is that there is a tight set containing fractional value. Consider the set X: –X is tight, the sum do not change throughout the algorithm. –X contains all items  It must include a fractional value.

47. 47 Pipage Rounding is a Rounding Lemma The algorithm converges to an integral solution within O(n 2 ) iterations. Proof For simplicity, assume that the algorithm chooses minimal tight set of minimal cardinality. Let A 1, A 2,…, A n be n sets chosen by the algorithm in n consecutive iterations. Assume no value of y becomes integral in the iterations corresponding to A 1, A 2,…, A n-1, then: –We will prove |A 1 | > |A 2 | > … > |A n |. –|A n |  2 –|A 1 |  n Therefore at least one additional value of y must become integral after every n-1 iterations. Contradiction 3.141592653

48. 48 Pipage Rounding is a Rounding (cont.) Proof The matroid polytopes can be equivalently defined as: Since no value of y becomes integral, there must be a set B  X which becomes tight, and prevents us from going further. Either i  B or j  B, but not both, otherwise does not change. Consider A i  B: –It is the intersection of two tight sets, therefore it is also tight. –It contains a fractional value. –|A i+1 |  |A i  B| < |A i | Aim Consider the iteration of A i for some 1  i  n-1. We want to show |A i+1 | < |A i |

49. 49 Pipage Rounding Preserve the Value Lemma The value of F(y) do not decrease during the rounding. Proof We need to show that one of the following holds: –F(y ij + )  F(y) –F(y ij - )  F(y) To do that we only need show F is concave in the direction: Let us replace y i by y i +t and y j by y j -t in the definition of F. By the submodularity of  we get:

50. 50 Pipage Rounding - Implementation The set A and the values  ij + (y),  ij - (y) can be computed in polynomial time using known methods. The values F(y ij + ) and F(y ij - ) can be approximated to an error polynomially small in n, by averaging over a polynomial number of samples: –The pipage rounding only loose a factor of (1 – o(1)) in each iteration w.h.p. –Using enough samples, the complete pipage rounding only loose a factor of (1 – o(1)) w.h.p, because it only makes O(n 2 ) iterations. Rounding - Summary Using the pipage rounding algorithm we get a valid integral solution for SMSMC. The approximation ratio is:

51. 51 Conclusion Remarks The approximation ratio can be improved to (1 – 1/e) using: –More samples in the estimation of ω j (t). –Setting  = o(1 / n 3 ): More iterations –More careful analysis The rounding step is necessary: –y  P(M) is a convex combination of independent sets  1,  2,…,  p. –Since  is non-linear, it is possible for  (  k ) to be negligible in comparison to F(y) for all 1  k  p. Consider the case of SWP with identical utility functions to all players: –The algorithm might end up assigning each item with equal probability to each player. –It can be shown that this is the best one can do in this case.

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