Combinatorial Auctions with kwise Dependent Valuations Vincent Conitzer CMU Tuomas Sandholm CMU Paol

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Combinatorial Auctions with k-wise Dependent
ValuationsVincent 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)
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Combinatorial auctions

• There is a set of items I for sale (Im)
• Each bidder j has valuation function vj 2I ? R
  that indicates the bidders valuation for each
  subset (bundle)
• Allows bidders to express
• complementarities v(a, b) gt v(a) v(b)
• substitutabilities v(a, b) lt v(a) v(b)
• Goal is to allocate nonoverlapping subsets
  Bj ? I to the bidders to maximize Sj vj(Bj)

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Two problems in combinatorial auctions

• Winner determination (or clearing) problem given
  the bidders valuations (under some
  representation), compute the optimal allocation
• NP-complete Rothkopf, Pekec, 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)

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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

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Existing research on structured valuations

• Rothkopf, Pekec, Harstad 98
• LaMura 99
• Nisan 00
• Tennenholtz 00
• Lehmann, OCallaghan, 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

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G2 2-wise dependent valuations
3
0
Node item
-2
3
1
1
2
Value of bundle sum of values of vertices/edges
in bundle
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Example Fashion valuation
80
250
silver watch
-2
rust sweater
-15
4
60
dark green trousers
13
8
dark brown shoes
70
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Optimal clearing is still NP-hard in G2

• Proof reduction from EXACT-COVER-BY-3-SETS

• Can get total value of 2m/3 if and only if an
  exact 3-cover exists

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Special case union of graphs is forest

• Thrm. Can solve clearing problem to optimality by
  dynamic programming in time O(mn)

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Gk k-wise dependent valuations
Value of bundle sum of values of
nodes/edges/multiedges in bundle For example,
k3
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Gk basic elicitation results

• Thrm. Every valuation function has a unique Gm
  representation
• Proof Suppose we have found the unique weights
  for multiedges up to size j. Then weight of
  multiedge over S (with S j1) must be v(S)
  SS?Sw(S)
• Thrm. A function in Gk can be elicited by
  querying all bundles of size k or less
• Proof Again, weight of multiedge over S (with
  S j1) must be v(S) SS?Sw(S), so can use
  dynamic programming

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Approximating valuations with G2 or Gk

• Thrm. Suppose there exists some v in Gk such
  that for any bundle S, v(S) v(S) d. Then,
  using O(mk) queries, we can construct a function
  g in Gk such that for any bundle S, v(S) g(S)
  d(1(S choose k)).
• Bound is tight for G2

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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!