Stable Invitations ‘Haden’ Hooyeon Lee Stanford University Joint work with Yoav Shoham Jan 29, 2015 ‘Haden’ Hooyeon Lee (Stanford) Stable Invitations Jan 29, 2015 1 / 11 Motivation Motivation: Group Scheduling Problems Event organizer is trying to invite agents to social event – fundraiser. Agents may have some constraints and preferences. Agent may be willing to attend only if her closest friends also attend. Another agent may be unwilling if her business competitor attends. Agents have preferences ( ) over the number of attendees (∈ [1, n]). Organizer wants to invite as many agents as possible subject to Individual Rationality (IR) Every invited agent prefers attending. No Exclusion-regret with addition (a-ER) No uninvited agent wishes to be added to the invitation list. No Exclusion-regret with replacement (r-ER) No uninvited agent wishes to replace another invited agent. ‘Haden’ Hooyeon Lee (Stanford) Stable Invitations Jan 29, 2015 2 / 11 Stable Invitation Problem Anonymous Stable Invitation Problem (ASIP) An instance of ASIP is a pair (N, P = (P1 , P2 , . . . , Pn )) where N = {a1 , a2 , . . . , an } is a set of agents Pi is preference ordering on outcomes, X = {1, . . . , n} ∪ {0} x ∈ [1, n] is an outcome of attending with x attendees (including oneself) 0 is a special outcome (an outside option) of not attending or not invited. An invitation S is a subset of N: S satisfies IR (individually rationality) if ∀ai ∈ S, |S| i 0 S satisfies a-ER (no exclusion-regret with addition) if ∀aj ∈ S, 0 j (|S| + 1) S is a-stable if it satisfies IR and a-ER. ‘Haden’ Hooyeon Lee (Stanford) Stable Invitations Jan 29, 2015 3 / 11 Stable Invitation Problem Examples of ASIP Example 1: Stable invitations may not exist. P1 : 1 0 2, P2 : 2 0 1 ∅ violates ER due to a1 (1 1 0 and thus a1 wishes to be added). {a1 } violates ER due to a2 (2 2 0 and thus a2 wishes to be added). {a2 } violates IR due to a2 (0 2 1) {a1 , a2 } violates IR due to a1 (0 1 2) ‘Haden’ Hooyeon Lee (Stanford) Stable Invitations Jan 29, 2015 4 / 11 Stable Invitation Problem Examples of ASIP Example 2: Stable invitations are not unique. P1 : 1 0 2, P2 : 1 0 2 Two stable invitations: S1 = {a1 } and S2 = {a2 } ∅ violates ER and {a1 , a2 } violates IR ‘Haden’ Hooyeon Lee (Stanford) Stable Invitations Jan 29, 2015 5 / 11 Stable Invitation Problem Technical Results for ASIP Technical Results for ASIP Given an instance (N, P) of ASIP (and non-strategic agents), Can we efficiently decide whether a stable invitation exists? Yes. If yes, can we also efficiently find a maximum stable invitation? Yes.1 Given an instance (N, P) of ASIP (and strategic agents), Can we design a non-trivial strategy-proof mechanism? No, in general. Yes, in special cases. 1 Also see ‘Group Activity Selection Problem’ by Darmann et al. WINE 2012. ‘Haden’ Hooyeon Lee (Stanford) Stable Invitations Jan 29, 2015 6 / 11 Stable Invitation Problem Proof of Impossibility Proof Sketch of Impossibility in Strategic Setting Consider two agents from Example 2: P1 : 1 0 2, P2 : 1 0 2 There are two stable invitations: S1 = {a1 } and S2 = {a2 }. If mechanism chooses S1 given (P1 , P2 ), then a2 has an incentive to misreport: If a2 reports (Pˆ2 : 1 2 0) instead of P2 Then S2 becomes the only stable invitation, given (P1 , Pˆ2 ). If mechanism chooses S2 given (P1 , P2 ), then a1 has an incentive (by symmetry). No strategy-proof mechanism is able to find a stable invitation. (Intuition?) ‘Haden’ Hooyeon Lee (Stanford) Stable Invitations Jan 29, 2015 7 / 11 Stable Invitation Problem Strategy-proof Mechanism Strategy-proof Mechanism for INC-ASIP Agent ai is said to have an increasing preference (INC): If ai prefers more attendees than fewer. In other words, there is some threshold li > 0 such that n (n − 1) · · · (li + 1) li 0 · · · . If ai happens to prefer 0 the most, we can set li = n + 1. In case of INC-ASIP (all agents have INC-preferences) Simply inviting everyone (i.e., S = N) may violate IR (Why?). Optimal, strategy-proof mechanisms exist! Idea: Check whether stable invitation of size k exists for all k in greedy manner. ‘Haden’ Hooyeon Lee (Stanford) Stable Invitations Jan 29, 2015 8 / 11 Stable Invitation Problem Recap Recap and Extension Anonymous Stable Invitation Problem (ASIP): Agents have anonymous preferences P over sizes of invitations. Objective is to find a maximum, stable invitation (if any). Agents may or may not act strategically. Technical Results: We can find a maximum stable invitation in poly-time. Strategy-proof mechanisms cannot find a stable invitation. Yet optimal, strategy-proof mechanisms exist in special cases such as INC-ASIP. General Stable Invitation Problem (GSIP) Some agent would attend only if her closest friends also attend Some agent would not attend if her business competitor attends Agents have preferences ( ) over the number of attendees (∈ [1, n]) Please see paper for definitions and technical results. ‘Haden’ Hooyeon Lee (Stanford) Stable Invitations Jan 29, 2015 9 / 11 Stable Invitation Problem Related Work Related Work Group Activity Selection Problem (GASP) [Darmann et al. (2012)] ASIP is a special subclass of GASP (with a single activity). Strategic agents vs. Truthful agents Identities vs. Anonymity (in preferences) GASP only considers a-ER (addition), but ASIP/GSIP also consider r-ER (replacement) in addition to a-ER. Related work in hedonic games Hedonic Coalitions: Optimality and Stability [DG 1980] The Stability of Hedonic Coalition Structures [BJ 2002] Computational Complexity of Hedonic Games [Ballester 2004] NP-hardness results for hedonic games and anonymous hedonic games ‘Haden’ Hooyeon Lee (Stanford) Stable Invitations Jan 29, 2015 10 / 11 Questions Questions? Presenter: ‘Haden’ Hooyeon Lee Website: Google my name with ‘Stanford’ keyword. ‘Haden’ Hooyeon Lee (Stanford) Stable Invitations Jan 29, 2015 11 / 11

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