The Submodular Welfare Problem

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

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

4. Combinatorial Auction - Example • Items • Classrooms • Markers • Multimedia keys • Players • TAs TA with a presentation TA without a presentation

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

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

7. Set Function Properties Set Function Properties Given a set function : 2S +, 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 wj is submodular, so what? • Consider a player j receiving a set Qj of items. • Let v be the additional value j would get if we assigned item q to j. Formally: v = wj(Qj  q) – wj(Qj) • 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 ≥ wj(Qj  Q’j  q) – wj(Qj Q’j) The utility of an item diminishes as the player gets more items!

8. The Submodular Welfare Problem (SWP) Instance • A set P of n players • A set Q of m items • Monotone submodular utility function wj: 2Q  + for each player, accessed via value oracles. Objective • Find a partition Q1, Q2, …, Qn of Q maximizing the total utility:

9. Matroid Definition An ordered pair (X,  ), with   2X (the sets in  are called “independent sets”), such that: • There is an independent set:   • Monotonicity: A  B, B   A   • 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. b d a f c e Example – Forest Sub Graphs Matroid X = {ab, bd, df, ef, ce, ca, bc, be, de} Independent set Example S = {ab, bc, df, ef}  Property 1:     = {S X | S is a forest}

11. b d a f c e Example – Forest Sub Graphs Matroid 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. Submodular Maximization Subject to a Matroid Constraint (SMSMC) Instance • A groundset X. • A monotone submodular function : 2X  +, 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 S1, S2, …, Sn, k sets of maximal union. • Multiple Knapsack (exponential reduction) – Same as Knapsack, but multiple Knapsacks are available.

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

14. Switching to the Continuous World

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 yj. • We want extend the properties of set functions to the continues case. Notation • (x  y)i= max {xi,yi}  x  y is a generalization of union • (x  y)i= min {xi,yi}  x  y is a generalization of intersection x  y = (0.6, 1, 0.3, 0.6, 1) x = (0.4, 0, 0.3, 0.6, 1) y = (0.6, 1, 0.2, 0.2, 0.7) x  y = (0.4, 0, 0.2, 0.2, 0.7) 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)

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 issubmodular ( is non-increasing with respect to yi). F is concave in all non-negative directions.

17. Extension by Expectation 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. Extension • For y [0, 1]X, let us denote by ỹ the random set obtained from y by selecting item j X with probability yj. • Let :2X 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.

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

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

20. {a} {a, b} {c} {b} {b, c} 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: • While yj > xj need to be decreased: • Find a set S   in the convex combination of y containing j. • Since M is matroid, S’ = S – { j }  . • Replace S by S’ in the convex combination continuously, until xj = yj or S is completely removed.

21. Approximating the Continuous Case

22. E [ỹ(2)] is high E [ỹ(1)] is low Marginal Value Definition • Let : 2X  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/yj is used to estimate the importance of increasing yj. • Analogously, we use E [ỹ(j)] to estimate the importance of increasing yj.

23. The Continuous Greedy Algorithm Input Matroid M = (X,  ), monotone submodular function : 2X +. Pseudo Code • Set  1/n2 (n = |X|), t  0 and y(0)  0. • While t < 1 do • For each j let ωj(t) be an estimation of E [ỹ(t)(j)] to an error of no more than OPT n2. • Let s(t) be a maximal-weight independent set in M according to the weights ωj(t). • Set y(t + ) = y(t) +  ∙ 1s(t), t  t +  • Return y(1)

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

25. Lemma 1 Lemma 1 Let OPT = maxs (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. By monotonicity, since: • Pr[j  ỹ(t + )] = yj(t) + Δj(t) • Pr[j ỹ(t)  D(t)] = 1 - (1-yj(t))(1-Δj(t))  yj(t) + Δj(t) Lemma 2 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) =  ∙ 1s(t). • D(t) - a random set containing each item j with probability Δj(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.

27. Considering only singleton sets By the inequality: (1 - )k ≥ (1 - k) 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 ωj(t) is an estimation of E [fỹ(j)] up to an error of OPT n2. Therefore:

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

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/ = n2 characteristic vectors of independent sets from M. • The value of F in y is: F(y) ≥ (1 – 1/e – o(1))OPT

30. Chernoff bound Union bound (n3 estimations) Continuous Greedy Algorithm - Implementation Problem Two steps of the Continuous Greedy Algorithm are not strait forward to implement: • Calculating ωj(t), the estimation of E [fỹ(t)(j)], to an error of no more than OPT n2. • Finding a maximum weight independent set in M according to the weights ωj(t). Implementing the first step w.h.p. • Each time E [fỹ(t)(j)] has to be estimated: • Perform n5 independent samples. • Use the average as the estimation ωj(t). • Notice that fỹ(t)(j) ≤ f({ j }) ≤ OPT (unless j S for any S   ). The probability of |ωj(t) – E [fỹ(j)]| > OPT/n2 is exponentially small W.h.p. all estimations are not off by more than OPT/n2

31. Finding Maximal Weighted Independent Set Instance • Matroid M = (X,  ) • Weight function w on the elements of X Objective Find an independent set S  , maximizing w(S). The Greedy Algorithm • Sort the elements of X in non-decreasing weight order: j1, j2, …, jn. • Start with S =  • For k = 1 to n do: • If S  {jk}  , add jk to S

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 • Sort the edges of the graph in non-decreasing weight order: e1, e2, …, em. • Start with F =  • For k = 1 to m do: • If F  {ek} does not contain circles, add ek to F Kruskal’s algorithm

33. Elements we will consider Ok Sk Elements we considered Greedy Algorithm - Correctness Notation Sk – The set S after the kth element is considered. S0 – The set S before the first element is considered, i.e. . Lemma 3 For every 0 ≤ k ≤ n, Sk Ok for some optimal set Ok, and j1, j2, …, jk Ok - Sk. Corollary For k = n lemma 3 implies: • Sn On • j1, j2,…, jn  On – Sn On Sn Therefore the result of the algorithm (Sn) is the optimal set On.   Now it all boils down to proving lemma 3.

34. Lemma 3 – Proof • Proof Overview • The proof is by induction of k. • Induction base - for k = 0: •  = S0  O0 for any optimal set O0 • No item must not be in O0-S0 • Induction step • Prove the lemma for k, assuming it holds for k – 1: • If jk is not inserted into S, set Ok = Ok-1 then: • Sk = Sk-1 Ok-1 • j1, j2, …, jk-1 Ok-1 - Sk-1 • jk  Ok-1 - Sk-1(because if jk  Ok-1, then Sk-1  {jk}  Ok-1, contradicting Sk-1  {jk}   ) • If jk is inserted into S, then we need to construct a matching Ok.

35. Lemma 3 – Proof (cont.) Construction of Ok • Initially Ok = Sk • While |Ok| < |Ok-1| do • Find j Ok-1 – Oksuch that Ok  { j }   • Set Ok Ok { j } Construction Correctness • Line 3 can be implemented because M is a matroid. • Ok = Ok-1 {jk} – {jh} for some jh • Ok is an optimal set: • jh Ok-1 – Sk-1 h k w(jh) ≤ w(jk)  w(Ok-1) ≤ w(Ok) • However, since we know that Ok-1 is a maximum weight independent set: w(Ok-1) = w(Ok). • j1, j2, …, jk Ok – Sk: • Holds for j1, j2, …, jk-1 because Ok - Ok-1 = {jk} • Holds for jk because jk Sk

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).

37. Back To Integral Solutions

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

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

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. Rank Functions and Tight Sets Definition • Consider a matroid M = (X,  ). • The rank function rM(A): 2X Ninduced by M is: rM(A) = max {|S| | S  A, S   } • Informally, rM(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 SAB be a maximal independent set in AB. • Extend SAB to maximal independent sets SA and SB in A and B, respectively. • Extend SA to a maximal independent set SAB in AB. SAB SA SB SAB

42. SAB SA SB SAB Rank Functions and Tight Sets (cont.) Proof - Continuation • SAB = SA  SB • |SAB|  |SA  SB|, otherwise: • |SAB – (SA - SAB)| > |SB| • |SAB – (SA - SAB)|  B, otherwise SA is not maximal. • Contradiction, because SB is maximal. • |SAB| + |SAB|  |SA  SB| + |SA  SB| = |SA| + |SB| 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:

43. Rank Functions and Tight Sets (cont.) 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. Proof • By the submodularity of rM: • It can be easily checked that: • Implying: • Due to the observation:

44. Ready. Set. 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 rM(X)). • y remains in P(M). • Since F is monotone, F(y) does not decrease. Notation • yij() - The vector y with yi increased by  and yj decreased by . • yij+ = yij(ij+(y)),ij+(y) = max{ 0 | yij()  P(M)} • yij- = yij(ij-(y)),ij-(y) = min{≤ 0 | yij()  P(M)}

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

46. Pipage Rounding is Well Defined Aim We need to show that there is always a valid set A, as long as y is fractional. Observation If a tight set A contains a non-integral value then it contains at least two fractional values in it because: 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. Contradiction Pipage Rounding is a Rounding 3.141592653 Lemma The algorithm converges to an integral solution within O(n2) iterations. Proof • For simplicity, assume that the algorithm chooses minimal tight set of minimal cardinality. • Let A1, A2,…, An be n sets chosen by the algorithm in n consecutive iterations. • Assume no value of y becomes integral in the iterations corresponding to A1, A2,…, An-1, then: • We will prove |A1| > |A2| > … > |An|. • |An| 2 • |A1|  n • Therefore at least one additional value of y must become integral after every n-1 iterations.

48. Pipage Rounding is a Rounding (cont.) Aim • Consider the iteration of Ai for some 1  i  n-1. • We want to show |Ai+1| < |Ai| 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 Ai  B: • It is the intersection of two tight sets, therefore it is also tight. • It contains a fractional value. • |Ai+1|  |Ai  B| < |Ai|

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(yij+)  F(y) • F(yij-)  F(y) • To do that we only need show F is concave in the direction: • Let us replace yi by yi+t and yj by yj-tin the definition of F. • By the submodularity of  we get:

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(yij+) and F(yij-) 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(n2) iterations. Rounding - Summary • Using the pipage rounding algorithm we get a valid integral solution for SMSMC. • The approximation ratio is: