1. Combinatorial Auctions with k-wise Dependent Valuations Vincent Conitzer (CMU) Tuomas Sandholm (CMU) Paolo Santi (Pisa) (Some of the results in this paper (or weaker versions thereof) were simultaneously and independently obtained by Chevaleyre, Endriss, Estivie, and Maudet, “Multiagent resource allocation with k-additive utility functions”)

2. Combinatorial auctions There is a set of items I for sale (|I|=m) Each bidder j has valuation function v j : 2 I  R that indicates the bidder’s valuation for each subset (bundle) –Allows bidders to express complementarities: v({a, b}) > v({a}) + v({b}) substitutabilities: v({a, b}) < v({a}) + v({b}) Goal is to allocate nonoverlapping subsets B j  I to the bidders to maximize Σ j v j (B j )

3. Two problems in combinatorial auctions Winner determination (or clearing) problem: given the bidders’ valuations (under some representation), compute the optimal allocation –NP-complete [Rothkopf, Pekeč, Harstad 98], even to approximate [Sandholm 02] –Often solved fast in practice Preference elicitation problem: elicit enough information from the bidders about their valuation functions to determine the optimal allocation –[Nisan and Segal 05] exponential lower bound –arguably the bigger problem in practice –in this paper, we focus on value queries (how much is a given bundle worth to a given agent?) (Mechanism design problems not considered here)

4. Structure in valuation functions In practice, valuation functions display various kinds of structure Such structure can help by –making the winner determination problem easier to solve –making the preference elicitation problem easier to solve Even if the bidders’ valuations do not display a particular type of structure exactly, there may still be a structured valuation function that is close to the true valuation We may also force bidders to submit only valuations satisfying the particular structure –Not a good idea unless the real valuations have approximately this structure

5. Existing research on structured valuations Rothkopf, Pekeč, Harstad 98 LaMura 99 Nisan 00 Tennenholtz 00 Lehmann, O’Callaghan, Shoham 02 Sandholm 02 Sandholm, Suri, Gilpin, Levine 02 Chang, Li, Smith 03 Sandholm & Suri 03 Zinkevich, Blum, Sandholm 03 Blum, Jackson, Sandholm, Zinkevich 04 Conitzer, Derryberry, Sandholm 04 Lahaie & Parkes 04 Santi, Conitzer, Sandholm 04 ……

6. 0+1+2 = 3 G 2 = 2-wise dependent valuations 1 3 3 -2 0 2 1 Node = item Value of bundle = sum of values of vertices/edges in bundle

7. Example: Fashion valuation rust sweater silver watch dark green trousers dark brown shoes 80 250 60 70 13 -2 4 -15 8

8. Optimal clearing is still NP-hard in G 2 Proof: reduction from EXACT-COVER-BY-3- SETS 1 1 1 1 1 1 Can get total value of 2m/3 if and only if an exact 3-cover exists

9. Special case: union of graphs is forest Thrm. Can solve clearing problem to optimality by dynamic programming in time O(mn) 3 6 7 1 1 9 2

10. G k = k-wise dependent valuations Value of bundle = sum of values of nodes/edges/multiedges in bundle For example, k=3: 1 3 3 -2 0 2 1 Node = item 1 3-edge

11. G k basic elicitation results Thrm. Every valuation function has a unique G m representation –Proof: Suppose we have found the unique weights for multiedges up to size j. Then weight of multiedge over S (with |S| = j+1) must be v(S) – Σ S’  S w(S’) Thrm. A function in G k can be elicited by querying all bundles of size k or less –Proof: Again, weight of multiedge over S (with |S| = j+1) must be v(S) – Σ S’  S w(S’), so can use dynamic programming

12. Approximating valuations with G 2 or G k Thrm. Suppose there exists some v’ in G k such that for any bundle S, |v(S) – v’(S)| ≤ δ. Then, using O(m k ) queries, we can construct a function g in G k such that for any bundle S, |v(S) – g(S)| ≤ δ(1+(|S| choose k)). –Bound is tight for G 2

13. Conclusion k-wise dependency gives a natural family of easy-to-elicit classes of valuations –k = m gives fully general valuations –also can approximate valuations Although 2-wise dependency already makes the winner determination problem hard, additional structure (joint graph is a forest) makes the problem easy again Thank you for your attention!