Two Algorithms for Additive and Fair Division of Mixed Manna Martin - PowerPoint PPT Presentation

Two Algorithms for Additive and Fair Division of Mixed Manna Martin Aleksandrov and Toby Walsh The 43rd KI Conference, 2020 M. Aleksandrov and T. Walsh (TU Berlin) Additive and Fair Division of Mixed Manna M. Aleksandrov and T. Walsh 1 / 12

Additive and fair division of mixed manna Fair division refers to the task of allocating resources to agents 1 . a set of agents N = { 1 , 2 , . . . , n } a set of indivisible items O = { o 1 , o 2 , . . . , o m } (real-valued) ∀ i ∈ N , o ∈ O , u i ( o ) ∈ R (additive) ∀ i ∈ N , B ⊆ O , u i ( B ) = � o ∈ B u i ( o ) 1 H. Steinhaus. The problem of fair division. Econometrica 16, 101–104 (1948). (TU Berlin) Additive and Fair Division of Mixed Manna M. Aleksandrov and T. Walsh 2 / 12

EF1 & EFX (Budish et al. 2011, Caragiannis et al. 2016) Allocation: A = ( A 1 , . . . , A n ) gives different A j to each j ∈ [ n ] Envy-freeness up to some item (EF1) A is EF1 if, for a , b ∈ [ n ] with u a ( A a ) < u a ( A b ), ∃ o ∈ A a ∪ A b : u a ( A a \ { o } ) ≥ u a ( A b \ { o } ). Envy-freeness up to any item (EFX) A is EFX if, for a , b ∈ [ n ], (1) ∀ o ∈ A a , u a ( o ) < 0: u a ( A a \ { o } ) ≥ u a ( A b ) and (2) ∀ o ∈ A b , u a ( o ) > 0: u a ( A a ) ≥ u a ( A b \ { o } ). EFX ⇒ EF1 (TU Berlin) Additive and Fair Division of Mixed Manna M. Aleksandrov and T. Walsh 3 / 12

EF1 3 & EFX 3 Good allocation: A + = ( A + 1 , . . . , A + n ) s.t. A + j = { o ∈ A | u j ( o ) > 0 } Bad allocation: A − = ( A − 1 , . . . , A − n ) s.t. A − j = { o ∈ A | u j ( o ) < 0 } Goods j = Agent j O = Bads j = EFX 3 requires that A is EFX, A + is EFX and A − is EFX EF1 3 requires that A is EF1, A + is EF1 and A − is EF1 EFX 3 ⇒ EFX EF1 3 ⇒ EF1 EFX 3 ⇒ EF1 3 (TU Berlin) Additive and Fair Division of Mixed Manna M. Aleksandrov and T. Walsh 4 / 12

Motivation Alice 0 0 0 1 1 -1 -1 Bob 1 1 1 0 0 -1 -1 Mary 0 0 0 1 1 -1 -1 For EF1 (EFX) and PO, A gives , , to Bob, to Alice and to Mary. A also maximizes the egalitarian welfare. Problem : But, Bob has to do both chores! EF1 3 (EFX 3 ) avoids this. For EF1 3 (EFX 3 ) and PO, B gives , to Bob, , to Alice and to Mary. (TU Berlin) Additive and Fair Division of Mixed Manna M. Aleksandrov and T. Walsh 5 / 12

The Modified Double Round-Robin (MDRR) Algorithm 1. DRR is not PO with − 1 / 0 / 1! 2. MDRR is PO with − α/ 0 /β ! The DRR algorithm 2 . 2 ACIW. Fair allocation of indivisible goods and chores. IJCAI-19, 53–59. (TU Berlin) Additive and Fair Division of Mixed Manna M. Aleksandrov and T. Walsh 6 / 12

General additive utilities (i.e. u a ( o ) ∈ R ) Theorem 1 MDDR returns an EF1 3 allocation in O (max { m 2 , mn } ) time. Proposition 1 There are problems where no allocation is EFX 3 . a b c agent 1 − 1 − 1 2 − 1 − 1 agent 2 2 By contradiction: To achieve EFX 3 , we must share a , b among 1,2. Say, 1 gets a , c and 2 gets b . This is not EFX: u 2 ( ∅ ) = 0 < 1 = u 2 ( { a , c } ). (TU Berlin) Additive and Fair Division of Mixed Manna M. Aleksandrov and T. Walsh 7 / 12

Additive utilities with identical magnitudes utilities with identical magnitudes a b c agent 1 10 1 − 4 − 10 − 1 agent 2 4 identical utilities (e.g. money) a b c agent 1 10 1 4 agent 2 10 1 4 (TU Berlin) Additive and Fair Division of Mixed Manna M. Aleksandrov and T. Walsh 8 / 12

The Minimax Algorithm (TU Berlin) Additive and Fair Division of Mixed Manna M. Aleksandrov and T. Walsh 9 / 12

Additive utilities with identical magnitudes Theorem 2 MDRR returns an EF1 3 and PO allocation in O (max { m 2 , mn } ) time. Theorem 3 Minimax returns an EFX and PO allocation in O (max { m log m , mn } ) time. Corollary 1 With identical utilities, Minimax returns an EFX and PO allocation. Note : With identical utilities, the “egal-sequential” algorithm 3 . 3 Aziz, H., Rey, S.: Almost group envy-free allocation of indivisible goods and chores. IJCAI-20. 3–45. (TU Berlin) Additive and Fair Division of Mixed Manna M. Aleksandrov and T. Walsh 10 / 12

Ternary additive utilities (i.e. u a ( o ) ∈ {− α, 0 , β } ) Theorem 4 MDRR returns an EF1 3 and PO allocation in O (max { m 2 , mn } ) time. Theorem 5 Minimax returns an EFX and PO allocation in O (max { m log m , mn } ) time. Theorem 6 With − α/ 0 /α utilities, MDRR returns an EFX 3 and PO allocation. Note : With − α/ 0 /α utilities, the “ternary-flow” algorithm 4 . 4 Aziz, H., Rey, S.: Almost group envy-free allocation of indivisible goods and chores. IJCAI-20. 3–45. (TU Berlin) Additive and Fair Division of Mixed Manna M. Aleksandrov and T. Walsh 11 / 12

Summary general ident.mag. − α/ 0 /β − α/ 0 /α property utilities utilities utilities utilities EF1 3 � , A1, P (T1) EF1 3 & PO open � , A1, P (T2) � , A1, P (T4) EFX & PO open � , A2, P (T3) � , A2, P (T5) EFX 3 × (P1) EFX 3 & PO � , A1, P (T6) Table: Key: � -possible, × -not possible, P-polynomial time, α, β ∈ R > 0 : α � = β , A1-The Modified Double Round-Robin Algorithm, A2-The Minimax Algorithm. Thank you! (TU Berlin) Additive and Fair Division of Mixed Manna M. Aleksandrov and T. Walsh 12 / 12