The Finite Matroid-based Valuation Conjecture is False Ngoc M Tran - PowerPoint PPT Presentation

The Finite Matroid-based Valuation Conjecture is False Ngoc M Tran arXiv:1905.02287 (v2 coming Wednesday night) Department of Mathematics The University of Texas at Austin May 11, 2020

A generative description for gross substitute valuations. Valuation : u : 2 [ n ] → R , u ( ∅ ) = 0 , S ⊆ T ⇒ u ( S ) ≤ u ( T ) . Take conv { ( a, u ( a )) : a ∈ { 0 , 1 } n } and project down: get the regular subdivision ∆ u of [0 , 1] n . Definition. a u is GS iff edges of ∆ u have directions in { e i − e j , e i } . Important subclass: OXS valuations . (max-bipartite matching) u ( I ) = max { match of I to RHS } a u (1) = a , u (2) = max( a ′ , b ) 1 A a ′ b u ( { 1 , 2 } ) = max( a + b, a ′ ) 2 B a Equivalent to Kelso-Crawford via theorems of Murota and Tomizawa / Fujishige / Danilov-Koshevoy-Lang / Gelfand-Goresky- MacPherson-Serganova

A generative description for gross substitute valuations. Valuation : u : 2 [ n ] → R , u ( ∅ ) = 0 , S ⊆ T ⇒ u ( S ) ≤ u ( T ) . Take conv { ( a, u ( a )) : a ∈ { 0 , 1 } n } and project down: get the regular subdivision ∆ u of [0 , 1] n . Definition. a u is GS iff edges of ∆ u have directions in { e i − e j , e i } . Theorem. (Lehmann-Lehmann-Nissan, 2006) a Important subclass: OXS valuations . (max-bipartite matching) OXS � GS � submodular valuations . u ( I ) = max { match of I to RHS } OXS and submodulars have generative descriptions: a u (1) = a , u (2) = max( a ′ , b ) 1 A OXS = merging of unit demands: a ′ b a a u ( { 1 , 2 } ) = max( a + b, a ′ ) 2 B 1 1 1 A A ′ ′ a a b = b ∗ 2 2 2 B B a Equivalent to Kelso-Crawford via theorems of Murota and Tomizawa / Fujishige / Danilov-Koshevoy-Lang / Gelfand-Goresky- Generative descriptions are useful! MacPherson-Serganova Is there a generative description for GS? a Lehmann, Lehmann and Nissan: Combinatorial auctions with de- creasing marginal utilities

Progress: the EAV (2005) and MBV (2015) conjectures Recap: Theorem. (Lehmann-Lehmann-Nissan 2006) OXS � GS � SM. OXS = (unit demand, merging) a a 1 1 1 A A ′ ′ a a b = b ∗ 2 2 2 B B Can start from OXS and make it bigger. Hatfield-Milgrom idea . Add endowment , another operation that preserves OXS. EAV = (unit demand, (merging, endowment)) EAV Conjecture. a EAV = GS . a Hatfield and Milgrom, Matching with contracts, 2005

Progress: the EAV (2005) and MBV (2015) conjectures Theorem. (Ostrovsky and Paes Leme) a Recap: Theorem. (Lehmann-Lehmann-Nissan 2006) EAVs are strongly exchangeable; GS may not. OXS � GS � SM. In particular, EAV � GS . OXS = (unit demand, merging) Ostrovsky-Leme idea . Start with a bigger a a generating set 1 1 1 A A ′ ′ a a b = b ∗ Weighted matroid rank ρ w : 2 [ n ] → R : 2 2 2 B B Can start from OXS and make it bigger. ρ w ( S ) = max � w i . I ∈I ,I ⊆ S Hatfield-Milgrom idea . Add endowment , i ∈ I another operation that preserves OXS. MBV m,n = ( { ρ w : ground set at most [ m ] } , EAV = (unit demand, (merging, endowment)) (merging, endowment)) EAV Conjecture. a EAV = GS . ρ w ∈ GS n ⇒ MBV m,n ⊆ GS n . a Hatfield and Milgrom, Matching with contracts, 2005 Best possible case: m = n . The only case where { ρ w } ⊂ GS n . Finite MBV conjecture. a For each n , there exists m ( n ) s.t. MBV m ( n ) ,n = GS n . a Gross substitutes and endowed assignment valuations, 2015

Could the MBV conjecture be true? Recap: OXS = (unit demand, merging) Why it could be true. a a 1 1 1 A A ′ ′ a a b b = ∗ 2 2 2 B B EAV = (unit demand, (merging, endowment)). MBV m,n = ( { ρ w : ground set at most [ m ] } , (merging, endowment)) Finite MBV conjecture. a For each n , ∃ m ( n ) s.t. MBV m ( n ) ,n = GS n .

Could the MBV conjecture be true? Recap: OXS = (unit demand, merging) Why it could be true. a a Gross substitutes = generalization of matroid 1 1 1 A A ′ ′ a a ranks b b = ∗ 2 2 2 B B 1. GS n contains all matroid ranks on ground set EAV = (unit demand, (merging, endowment)). S ⊆ [ n ] . MBV m,n = ( { ρ w : ground set at most [ m ] } , 2. Operations that preserve GS are (merging, endowment)) generalizations of matroid operations. Finite MBV conjecture. a For each n , ∃ m ( n ) s.t. merging = matroid union; endowment = MBV m ( n ) ,n = GS n . contraction 3. Good intuition on why EAV fails. OXS generalizes transversals. EAV generalizes gammoids. Not all matroids are gammoids ⇒ EAV fails. 4. MBV contains all weighted matroid ranks ⇒ big enough...?

Unfortunately... Best possible case m ( n ) = n is not possible. Theorem. (T. 2019). For m ( n ) = n , n ≥ 4 ⇐ ⇒ MBV n,n � GS n .

Why the MBV fails for m ( n ) = n MBV m,n = ( { ρ w : ground set at most [ m ] } , Why the MBV fails for m ( n ) = n (merging, endowment)) 1. Endowment+merging cannot create irreducible valuations u : 2 [ n ] → R , Finite MBV conjecture. a For each n , ∃ m ( n ) s.t. MBV m ( n ) ,n = GS n . u = v ∗ w, v, w ∈ GS n ⇒ u = v or u = w. Theorem. (T. 2019). For m ( n ) = n , n ≥ 4 ⇐ ⇒ MBV n,n � GS n Analogue of irreducible matroid ranks. 2. Let G n = { v : 2 [ n ] → R , v ∈ GS n , v irreducible } . Then finite MBV conjecture is true ⇒ G n ⊆ { ρ w } . 3. Counter-example: construct a family of irreducibles C n outside of weighted matroid ranks: C n ⊂ G n but C n ∩ { ρ w } = ∅ .

Recipe 1 for irreducibles Recipe 1. C n = partition valuations . Simple family indexed by set partitions of [ n ] . (+) Works for n ≥ 4 (+) direct proof (-) little insight.

Recipe 2: geometric construction of irreducibles Geometry of merging. If F ∈ ∆ u ∗ v , then F = ( F u + F v ) ∩ [0 , 1] n for some F u ∈ ∆ u , F v ∈ ∆ v . M -irreducible polytopes are obstructions to reducibility: P = ( P 1 + P 2 ) ∩ [0 , 1] n ⇒ P = P 1 or P = P 2 . Lemma. If ∆ u has a full-dimensional M -irreducible face F , then u is irreducible. Proposition. For P ⊂ [0 , 1] n is M ♮ , define ρ P : 2 [ n ] → R ρ P ( I ) = max { x I : x ∈ P } . If ρ P is the rank function of an irreducible matroid, then P is M -irreducible. Bonus. ρ w is irreducible in MBV n,n iff the matroid is irreducible. So MBV n,n is generated by weighted rank of irreducible matroids.

Recipe 2: geometric construction of irreducibles Geometry of merging. If F ∈ ∆ u ∗ v , then Recipe for irreducible valuations . F = ( F u + F v ) ∩ [0 , 1] n for some F u ∈ ∆ u , F v ∈ ∆ v . M -irreducible polytopes are obstructions to reducibility: P = ( P 1 + P 2 ) ∩ [0 , 1] n ⇒ P = P 1 or P = P 2 . Lemma. If ∆ u has a full-dimensional M -irreducible face F , then u is irreducible. a. ρ w = weighted rank of irreducible matroid of Proposition. For P ⊂ [0 , 1] n is M ♮ , define rank ≥ 2 (smallest one is M ( K 4 ) , n = 6 ) ρ P : 2 [ n ] → R b. Modify ∆ ρ w : split the independence polytope v = ρ w + c · ( 1 − 1 ∅ ) ρ P ( I ) = max { x I : x ∈ P } . If ρ P is the rank function of an irreducible c. v ∈ GS n , ∆ v has an M -irreducible face, ⇒ v matroid, then P is M -irreducible. is irreducible. Bonus. ρ w is irreducible in MBV n,n iff the matroid is irreducible. So MBV n,n is generated by weighted rank of irreducible matroids.

What’s next? Summary. What’s next? • More operations? OXS = (unit demand, merging) • Consider m ( n ) > n ? a a 1 1 1 A A • Characterize all irreducibles? ′ ′ a a b b = ∗ • A different approach: matroid rank sums? 2 2 2 B B EAV = (unit demand, (merging, endowment)). arXiv:1905.02287 (v2 coming Wednesday MBV m,n = ( { ρ w : ground set at most [ m ] } , night) (merging, endowment)) Thanks to: Rakesh Vohra, Renato Paes Leme, Kazuo Murota, and two anonymous referees. We have OXS n � EAV n � MBV n,n � GS n � SM n .

What’s next? Summary. What’s next? • More operations? OXS = (unit demand, merging) • Consider m ( n ) > n ? a a 1 1 1 A A • Characterize all irreducibles? ′ ′ a a b b = ∗ • A different approach: matroid rank sums? 2 2 2 B B EAV = (unit demand, (merging, endowment)). arXiv:1905.02287 (v2 coming Wednesday MBV m,n = ( { ρ w : ground set at most [ m ] } , night) (merging, endowment)) Thanks to: Rakesh Vohra, Renato Paes Leme, Kazuo Murota, and two anonymous referees. We have OXS n � EAV n � MBV n,n � GS n � SM n . Thank you!