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## Preferences

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**Preferences**Toby Walsh NICTA and UNSW www.cse.unsw.edu.au/~tw/teaching.html**Outline**• May 5,15:00-17:00 • Introduction, soft constraints • May 6, 10:00-12:00 • CP nets • May 7, 15:00-18:00 • Strategic games, CP-nets, and soft constraints • Voting theory • May 8, 15:00-18:00 • Manipulation, preference elicitation • May 9, 10:00-12:00 • Matching problems, stable marriage**Motivation**• Preferences are everywhere! • Alice prefers not to meet on Monday morning • Bob prefers bourbon to whisky • Carol likes beach vacations more than activity holidays …**Major questions**• Representing preferences • Soft CSPs, CP nets, … • Reasoning with preferences • What is the optimal outcome? Do I prefer A to B? How do we combine preferences from multiple agents? … • Eliciting preferences • Users don’t want to answer lots of questions! • Are users going to be truthful when revealing their preferences? • …**Preference formalisms**• Psychological relevance • Can it express your preferences? • Quantitative: I like wine twice as much as beer • Qualitative: I prefer wine to beer • Conditional: if we’re having meat, I prefer red wine to white • …**Preference formalisms**• Expressive power • What types of ordering over outcomes can it represent? • Total • Partial • Indifference • Incomplete • …**Preference formalisms**• Succinctness • How succinct is it compared to other formalisms? • Can it (compactly) represent all that another formalism can? • … • Complexity • How difficult is it to reason with? • What is the computationally complexity of ordering two choices? • What is the computationally complexity of finding the most preferred choice? • …**Utilities**• Map preferences onto a linear scale • Typically reals, naturals, … • Issues • Cardinal or ordinal utility? • Numbers meaningful or just ordering? • Different agents have different utility scales • Incomparability • Combinatorial domains • First course x Main dish x Sweet x Wine x …**Ordering relation**• I prefer A to B (written A > B) • Transitive or not: if A > B and B > C then is A > C? • Total or partial: is every pair ordered? • Strict or not: A > B or A ≥ B • … • Issues • Elicitation requires ranking O(m2) pairs • Combinatorial domains • …**Case study: combinatorial auction**• Auctioneer • Puts up number of items for sale • Agents • Submit bids for combinations of items • Winner determination • Decide which bids to accept • Two agents cannot get the same item • Maximize revenue!**Case study: combinatorial auction**• Why are bids not additive? • Complements • v(A & B) > v(A) + v(B) • Left shoe of no value without right shoe • Substitutes • v(A & B) < v(A) + v(B) • As you can only drive one car at a time, a second Ferrari is not worth as much as the first • Auction mechanism that simply assigns items in turn may be sub-optimal • How you value item depends on what you get later**Case study: combinatorial auction**• Winner determination problem • Deciding if there is a solution achieving a given revenue k (or more) • NP-complete in general • Even if each agent submits jut a single bid • And this bid has value 1**Case study: combinatorial auction**• Winner determination problem • Membership in NP • Polynomial certificate • Given allocation of goods, can compute revenue it generates**Case study: combinatorial auction**• Winner determination problem • NP-hard • Reduction from set packing • Given S, a collection of sets and a cardinality k, is there a subset of S of disjoint sets of size k? • Items in sets are goods for auction • One agent for each set in S, value 1 for goods in their set, 0 otherwise • One other agent who bids 0 for all goods**Case study: combinatorial auction**• Winner determination problem • NP-hard • One agent for each set in S, value 1 for goods in their set, 0 otherwise • One special agent who bids 0 for all goods • Allocation may not correspond to set packing • Agents may be allocated goods with 0 value (ie outside their desired set) • But can always move these goods over to special agent • Revenue equal to cardinality of the subset of S**Case study: combinatorial auction**• Winner determination problem • Tractable cases • Conflict graph: vertices = bids, edges = bids that cannot be accepted together • If conflict graph is tree, then winner determination takes polynomial time • Starting at leaves, accept bid if it is greater than best price achievable by best combination of its children**Case study: combinatorial auction**• Winner determination problem • Intractable cases • Integer programming • Heuristic search • States = accepted bids • Moves = accept/reject bid • Initial state = no bids accepted • Heuristics • Bid with high price & few goods • Bid that decomposes conflict graph**Case study: combinatorial auction**• Winner determination problem • Intractable cases • Integer programming • Heuristic search • States = accepted bids • Moves = accept/reject bid • Initial state = no bids accepted • Heuristics • Bid with high price & few goods • Bid that decomposes conflict graph**Case study: combinatorial auction**• Bidding languages • Used for agents to express their preferences over goods • If there are m goods, there are 2m possible bids • Many possibilities • Atomic bids • OR bids • XOR bids • OR* bids with dummy items • …**Case study: combinatorial auction**• Bidding languages: assumptions • Normalized • v({})=0 • Monotonic • v(A) ≤ v(B) iff A B • Implies valuations are non-negative!**Case study: combinatorial auction**• Atomic bids • (B,p) • “I want set of items B for price p” • v(X) = p if X B otherwise 0 • Note this valuation is monotonic • Very limited range of preferences expressible as atomic bids • Cannot express even simple additive valuations**Case study: combinatorial auction**• OR bids • Disjunction of atomic bids • (B1,p1) OR (B2,p2) • Value is max. sum of disjoint bundles • v(X) = max { v1(X1) + v2(X \ X1) | X1X} • Not complete • Can only express valuations without substitutes • v(X u Y) ≥ v(X) + v(Y) • Suppose you want just one item? • v(S) = max{ vj | j S }**Case study: combinatorial auction**• XOR bids • Disjunction of atomic bids but only one is wanted • (B1,p1) XOR (B2,p2) • Value is max. of two possible valuations • v(X) = max {v1(X), v2(X)} • Complete • Can express any monotonic valuation • Just list out all the differently valued sets of goods • Hence XORs are more expressive than ORs**Case study: combinatorial auction**• XOR bids • Disjunction of atomic bids but only one is wanted • (B1,p1) XOR (B2,p2) • Additive valuation requires O(2k) XORs • But only O(k) Ors • Thus, XORs are more expressive but less succinct than ORs**Case study: combinatorial auction**• OR/XOR bids • Arbitrary combinations of ORs and XORs • Bid := (B,p) | Bid OR Bid | Bid XOR Bid • Recursively define semantics as before • B1 OR B2 • v(X) = max { v1(X1) + v2(X \ X1) | X1X} • B1 XOR B2 • v(X) = max { v1(X), v2(X) }**Case study: combinatorial auction**• Two special cases • OR of XOR • Bid := XorBid | XorBid OR XorBid • XorBid := (B,p) | (B,p) XOR XorBid • XOR of OR • Bid := OrBid | OrBid XOR OrBid • OrBid := (B,p) | (B,p) OR OrBid**Case study: combinatorial auction**• Downward sloping symmetric valuation • Items symmetric • Only their number, k matters • Diminishing returns • v(k)-v(k-1) ≥ v(k+1)-v(k) • Using OR of XOR, such a valuation over n items is O(n2) in size • Let pk = v(k)-v(k-1) • Then v(k) is • ({x1},p1) XOR .. XOR ({xn},p1) OR ({x1},p2) XOR .. XOR ({xn},p2) OR .. OR ({x1},pn) XOR .. XOR ({xn},pn)**Case study: combinatorial auction**• Downward sloping symmetric valuation • Items symmetric • Only their number, k matters • Diminishing returns • v(k)-v(k-1) ≥ v(k+1)-v(k) • Using XOR of ORs (or OR) such a valuation is exponential in size • Need to represent all subsets of size k • OR of XORs is exponentially more succinct than XOR of ORs**Case study: combinatorial auction**• Monochromatic valuations • n/2 red and n/2 blue items • Want as many of one colour as possible • v(X) = max {|X Red|, |X Blue|} • With such a valuation • XOR of ORs is O(n) in size • ({red1,p}) OR .. OR ({redn/2 },p) XOR ({blue1,p}) OR .. OR ({bluen/2,p})**Case study: combinatorial auction**• Monochromatic valuations • n/2 red and n/2 blue items • Want as many of one colour as possible • v(X) = max {|X Red|, |X Blue|} • With such a valuation • OR of XORs is O(2n/2) in size • Atomic bids in OR of XORs only need be monochromatic • Removing non-monochromatic atomic bids will not change valuation of a monochromatic allocation • Atomic bids need to have price equal to their cardinality • Anything higher or lower will only value a monochromatic allocation incorrectly**Case study: combinatorial auction**• Monochromatic valuations • n/2 red and n/2 blue items • Want as many of one colour as possible • v(X) = max {|X Red|, |X Blue|} • With such a valuation • OR of XORs is O(2n/2) in size • There can be only a single XOR • Suppose there are two (or more) XORs • There are two cases: • One XOR is just blue, other is just red But then monochromatic valuation is not possible • One XOR is blue and red But then again monochromatic valuation is not possible**Case study: combinatorial auction**• Monochromatic valuations • n/2 red and n/2 blue items • Want as many of one colour as possible • v(X) = max {|X Red|, |X Blue|} • With such a valuation • OR of XORs is O(2n/2) in size • There can be only a single XOR • This must contain all O(2n/2) blue and O(2n/2) red subsets • XOR of ORs and OR of XORs are incomparable in succinctness**Case study: combinatorial auction**• OR* bids • Can modify OR bids so they can simulate XOR bids • Recall that OR bids are not complete • But XOR bids can be exponentially more succinct • Get best of both worlds? • Introduce dummy items (which cannot be shared) to OR bids to make them simulate XOR • (B u {dummy},p1) OR (C u {dummy},p2) is equivalent to (B,p1) XOR (C,p2) • Since XOR bids are complete, so are OR* bids**Case study: combinatorial auction**• OR* bids • Any OR/XOR bid of size O(s) can be represented as an OR* bid of size O(s) • Homework exercise: prove this! • This bidding language still has limitations • Majority valuation requires exponential sized OR* bid • Any allocation of m/2 or more of the items has value 1 • Any smaller allocation has value 0 • No non-zero atomic bid in the OR* bid can have less than m/2 items • Otherwise we could accept this set and violate majority valuation • So we must have every nCn/2 possible subset of size n/2**Conclusions**• Wide variety of formalisms for representing preferences • Several dimensions along which to analyse them • Completeness • Succinctness • Complexity of reasoning • …