Payoffs in Location Games Shuchi Chawla 1/22/2003 joint work with Amitabh Sinha, Uday Rajan & R. Ravi

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

Shuchi Chawla, Carnegie Mellon Univ3 An artist’s rendering of Manhattan $$ T T Sam’s Starcups T Trudy’s Tazzo

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

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

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)

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

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

Shuchi Chawla, Carnegie Mellon Univ9 Our Results  R d (1) = 1/(d+1) ½ 1/(d+1) ·  R d · 1/(d+1)

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

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

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

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”

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

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

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

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

Shuchi Chawla, Carnegie Mellon Univ18 If not:  R 2 (1) ¸ 1/3 All triplets of half spaces must intersect! Contradiction!! < 1/3

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

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

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!!

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)

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

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)

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.

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.

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

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.

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.

Shuchi Chawla, Carnegie Mellon Univ30 Questions?