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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


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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

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Additive and fair division of mixed manna

Fair division refers to the task of allocating resources to agents1. a set of agents N = {1, 2, . . . , n} a set of indivisible items O = {o1, o2, . . . , om} (real-valued) ∀i ∈ N, o ∈ O, ui(o) ∈ R (additive) ∀i ∈ N, B ⊆ O, ui(B) =

  • ∈B ui(o)
  • 1H. Steinhaus. The problem of fair division. Econometrica 16, 101–104 (1948).

(TU Berlin) Additive and Fair Division of Mixed Manna

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EF1 & EFX (Budish et al. 2011, Caragiannis et al. 2016)

Allocation: A = (A1, . . . , An) gives different Aj to each j ∈ [n]

Envy-freeness up to some item (EF1)

A is EF1 if, for a, b ∈ [n] with ua(Aa) < ua(Ab), ∃o ∈ Aa ∪ Ab: ua(Aa \ {o}) ≥ ua(Ab \ {o}).

Envy-freeness up to any item (EFX)

A is EFX if, for a, b ∈ [n], (1) ∀o ∈ Aa, ua(o) < 0: ua(Aa \ {o}) ≥ ua(Ab) and (2) ∀o ∈ Ab, ua(o) > 0: ua(Aa) ≥ ua(Ab \ {o}). EFX ⇒ EF1

(TU Berlin) Additive and Fair Division of Mixed Manna

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EF13 & EFX3

Good allocation: A+ = (A+

1 , . . . , A+ n ) s.t. A+ j = {o ∈ A|uj(o) > 0}

Bad allocation: A− = (A−

1 , . . . , A− n ) s.t. A− j = {o ∈ A|uj(o) < 0}

Goodsj = Agent j O = Badsj = EFX3 requires that A is EFX, A+ is EFX and A− is EFX EF13 requires that A is EF1, A+ is EF1 and A− is EF1 EFX3 ⇒ EFX EF13 ⇒ EF1 EFX3 ⇒ EF13

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Motivation

Alice 1 1

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Bob 1 1 1

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Mary 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! EF13 (EFX3) avoids this. For EF13 (EFX3) and PO, B gives , to Bob, , to Alice and to Mary.

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The Modified Double Round-Robin (MDRR) Algorithm

The DRR algorithm2.

  • 1. DRR is not PO with −1/0/1!
  • 2. MDRR is PO with −α/0/β!
  • 2ACIW. Fair allocation of indivisible goods and chores. IJCAI-19, 53–59.

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General additive utilities (i.e. ua(o) ∈ R)

Theorem 1

MDDR returns an EF13 allocation in O(max{m2, mn}) time.

Proposition 1

There are problems where no allocation is EFX3. a b c agent 1 −1 −1 2 agent 2 −1 −1 2 By contradiction: To achieve EFX3, we must share a, b among 1,2. Say, 1 gets a, c and 2 gets b. This is not EFX: u2(∅) = 0 < 1 = u2({a, c}).

(TU Berlin) Additive and Fair Division of Mixed Manna

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Additive utilities with identical magnitudes

utilities with identical magnitudes a b c agent 1 10 1 −4 agent 2 −10 −1 4 identical utilities (e.g. money) a b c agent 1 10 1 4 agent 2 10 1 4

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The Minimax Algorithm

(TU Berlin) Additive and Fair Division of Mixed Manna

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Additive utilities with identical magnitudes

Theorem 2

MDRR returns an EF13 and PO allocation in O(max{m2, 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” algorithm3.

3Aziz, H., Rey, S.: Almost group envy-free allocation of indivisible goods and chores.

IJCAI-20. 3–45.

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Ternary additive utilities (i.e. ua(o) ∈ {−α, 0, β})

Theorem 4

MDRR returns an EF13 and PO allocation in O(max{m2, 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 EFX3 and PO allocation. Note: With −α/0/α utilities, the “ternary-flow” algorithm4.

4Aziz, H., Rey, S.: Almost group envy-free allocation of indivisible goods and chores.

IJCAI-20. 3–45.

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Summary

property general ident.mag. −α/0/β −α/0/α utilities utilities utilities utilities EF13 , A1, P (T1) EF13 & PO

  • pen

, A1, P (T2) , A1, P (T4) EFX & PO

  • pen

, A2, P (T3) , A2, P (T5) EFX3 × (P1) EFX3 & 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!

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