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Payoffs in Location Games Shuchi Chawla 1/22/2003 joint work with Amitabh Sinha, Uday Rajan & R. Ravi
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Shuchi Chawla, Carnegie Mellon Univ2 Caffeine wars in Manhattan Sam owns Starcups; Trudy owns Tazzo Every month both chains open a new outlet – Sam chooses a location first, Trudy follows Indifferent customers go to the nearest coffee shop At the end of n months, how much market share can Sam have? Trudy knows n, Sam doesnt
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Shuchi Chawla, Carnegie Mellon Univ3 An artist’s rendering of Manhattan $$ T T Sam’s Starcups T Trudy’s Tazzo
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Shuchi Chawla, Carnegie Mellon Univ4 Why bother? Product Placement many features to choose from – high dimension high cost of recall – cannot modify earlier decisions Service Location cannot move service once located
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Shuchi Chawla, Carnegie Mellon Univ5 Some history… The problem was first introduced by Harold Hotelling in 1929 Acquired the name “Hotelling Game” Originally studied on the line with n players moving simultaneously Extensions to price selection
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Shuchi Chawla, Carnegie Mellon Univ6 Formally… Given: (M,L,F) – Metric space, Location set, Distribution of demands At step i, S first picks s i 2 L. Then T picks t i 2 L s i = s i (s 1,…,s i-1,t 1,…,t i-1 ) ; t i = t i (s 1,…,s i,t 1,…,t i-1 ) S is an online player: does not know n Payoff for S at the end of n moves is: (M,L,F) (T) = 1 - (M,L,F) (S) (M,L,F) (S) = s (v,S)< (v,T) dF(v) + ½ s (v,S)= (v,T) dF(v)
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Shuchi Chawla, Carnegie Mellon Univ7 The second mover advantage Note that if 8 i, t i = s i (M,L,F) (S) = (M,L,F) (T) = ½ T can always guarantee a payoff of ½ Can S do the same? We will show that S cannot guarantee ½ but at least some constant fraction depending on M
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Shuchi Chawla, Carnegie Mellon Univ8 Some more notation M (S)= min L,F min n min T (M,L,F) (S) M (S) is the worst case performance of strategy S on any metric space in M M = max S M (S) M (1) – defined analogously when n=1
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Shuchi Chawla, Carnegie Mellon Univ9 Our Results R d (1) = 1/(d+1) ½ 1/(d+1) · R d · 1/(d+1)
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Shuchi Chawla, Carnegie Mellon Univ10 The 1-D case: Beaches & Icecream Assume a uniform demand distribution for simplicity S moves at ½, no move of T can get more than ½ ) R (1) = ½
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Shuchi Chawla, Carnegie Mellon Univ11 The 1-D case: Beaches & Icecream No subsequent move of T can get > ½ Recall T’s strategy to obtain ½ : repicate S’s moves S can use the same strategy for moves s i>1 s 1 = ½ ; s i = t i-1
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Shuchi Chawla, Carnegie Mellon Univ12 The 1-D case: Beaches & Icecream (t n ) · ½ (t 1,…,t n-1 ) = (s 2,…,s n ) ) (S) ¸ (s 2,…,s n ) ¸ ¼
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Shuchi Chawla, Carnegie Mellon Univ13 Median and Replicate Given a 1-move strategy with payoff obtain an n-move strategy with payoff /2 Use 1-move strategy for the first move, Replicate all other moves of player 2 Last move of player 2 gets at most 1- , the rest get at most half of the remaining : /2 “MEDIAN” “REPLICATE”
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Shuchi Chawla, Carnegie Mellon Univ14 Locn Game in the Euclidean plane Thm 1: R 2 (1) = 1/3 Thm 2: 1/6 · R 2 · 1/3 Proof of Thm 2: Use Median and Replicate with Thm 1
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Shuchi Chawla, Carnegie Mellon Univ15 R 2 (1) · 1/3 Condorcét voting paradox L1L1 L2L2 L3L3 D3D3 D2D2 D1D1 D 1 : L 1 > L 3 > L 2 D 2 : L 2 > L 1 > L 3 D 3 : L 3 > L 2 > L 1 S gets only 1/3 of the demand The vote is inconclusive $$ T
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Shuchi Chawla, Carnegie Mellon Univ16 R 2 (1) ¸ 1/3 Our goal: 9 a location s such that 8 t, (s,t) ¸ 1/3 Outline: Construct a digraph on locations G contains edge u ! v iff (u,v) 1/3 Show that G contains no cycles ) G has a sink s
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Shuchi Chawla, Carnegie Mellon Univ17 > 2/3 demand R 2 (1) ¸ 1/3 Each edge defines a half- space containing at least 2/3 of the demand A cycle defines an intersection of half- spaces
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Shuchi Chawla, Carnegie Mellon Univ18 If not: R 2 (1) ¸ 1/3 All triplets of half spaces must intersect! Contradiction!! < 1/3
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Shuchi Chawla, Carnegie Mellon Univ19 R 2 (1) ¸ 1/3 Helly’s Theorem Given a collection { C 1,C 2,…,C n } of convex sets in Euclidean space : If every triplet of the sets has a non empty intersection, then Å 1 · i · n C i ; ) all half-spaces defined by the graph G contain a common point P
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Shuchi Chawla, Carnegie Mellon Univ20 R 2 (1) ¸ 1/3 Let u 1,…,u k be a cycle in G Then, ( P,u i+1 ) · ( P,u i ) because P is in the half-space defined by the edge u i ! u i+1 ) ( P,u i ) = ( P,u j ) 8 i,j P P is the intersection of hyperplanes bisecting the edges
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Shuchi Chawla, Carnegie Mellon Univ21 R 2 (1) ¸ 1/3 P Let demand at P be Then each half-space has a total demand of at least 2/3 + /2 >1/3 > 2/3 - Contradiction!!
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Shuchi Chawla, Carnegie Mellon Univ22 The d-dimensional case Results on R 2 extend nicely to R d : Thm 3 : R d (1) = 1/(d+1) Thm 4 : 1/2(d+1) · R d · 1/(d+1)
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Shuchi Chawla, Carnegie Mellon Univ23 Condorcét instance in d-dimensions As before we should have D i : L i > L i+1 > L i+2 > … > L i-1 Embedding in R d+1 : L i ´ ( 0, …, 0, 1, 0, …, 0) D i ´ (d-i,d-i+1,…,1- ,2, 3, … ) Project all points down to the d dimensional plane containing {L 1,…,L d+1 } – relative distances between L i and D j are preserved
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Shuchi Chawla, Carnegie Mellon Univ24 Lower bound in d-dimensions As before, define a directed graph on locations with each half-space containing d/(d+1) demand Every set of d half-spaces must intersect By Helly’s Theorem all half-spaces must have a non empty intersection. Assume WLOG that the origin lies in this intersection. (Skip)
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Shuchi Chawla, Carnegie Mellon Univ25 Lower bound in d-dimensions Assume for the sake of contradiction that a cycle exists. Each point in the intersection is equi-distant from all vertices in the cycle We want this to hold for at most some d+1 half- spaces Arrive at a contradiction just as before.
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Shuchi Chawla, Carnegie Mellon Univ26 Lower bound in d-dimensions Let i be a vector representing the i-th edge in the cycle; Let p represent some point in the intersection Then, p ¢ i ¸ 0 8 i; i i = 0 9 a collection of · d+1 vectors i such that i i = 0 with i ¸ 0 Then, p ¢ i i = 0. But p ¢ i ¸ 0 8 i, so, p ¢ i = 0 for the d+1 vectors. Thus every point in the intersection of these half- spaces is equi-distant from all vertices in the cycle.
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Shuchi Chawla, Carnegie Mellon Univ27 Lower bound: gritty detail For any n vectors i in d dimensions with some positive linear combination summing to zero, 9 a positive linear combination of some d+1 of them Take some linear combination and eliminate the most negative term; iterate
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Shuchi Chawla, Carnegie Mellon Univ28 Concluding Remarks Results hold even when demands lie in some high dimensional space We can obtain tighter results in the line when n is bounded.
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Shuchi Chawla, Carnegie Mellon Univ29 Open Problems Closing the factor-of-2 gap for Convergence with n If S knows a lower/upper bound on n, is there a better strategy? Can he do better as n gets larger – we believe so Brand loyalty What about demand in the intermediate steps? fraction of demand at every time step becomes loyal to the already opened locations. The rest carries on to the next step.
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Shuchi Chawla, Carnegie Mellon Univ30 Questions?
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