Payoffs in Location Games Shuchi Chawla 1/22/2003 joint work with Amitabh Sinha, Uday Rajan & R....
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Transcript of Payoffs in Location Games Shuchi Chawla 1/22/2003 joint work with Amitabh Sinha, Uday Rajan & R....
![Page 1: Payoffs in Location Games Shuchi Chawla 1/22/2003 joint work with Amitabh Sinha, Uday Rajan & R. Ravi.](https://reader030.fdocuments.us/reader030/viewer/2022032612/56649efd5503460f94c1090d/html5/thumbnails/1.jpg)
Payoffs in Location Games
Shuchi Chawla1/22/2003
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 si2 L. Then T picks ti2 L
si = si(s1,…,si-1,t1,…,ti-1) ; ti = ti(s1,…,si,t1,…,ti-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) + ½s(v,S)=(v,T)dF(v)
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Shuchi Chawla, Carnegie Mellon Univ7
The second mover advantage
Note that if 8 i, ti = si
(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)= minL,F minn minT (M,L,F)(S)
M(S) is the worst case performance of strategy S on any metric space in M
M = maxS 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 si>1
s1 = ½ ; si = ti-1
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Shuchi Chawla, Carnegie Mellon Univ12
The 1-D case: Beaches & Icecream
(tn) · ½
(t1,…,tn-1) = (s2,…,sn)
) (S) ¸ (s2,…,sn) ¸ 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 paradoxL1
L2L3
D3
D2
D1
D1 : L1 > L3 > L2
D2 : L2 > L1 > L3
D3 : L3 > L2 > L1
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 locationsG contains edge u!v iff (u,v) 1/3Show 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 < 1/3
< 1/3
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Shuchi Chawla, Carnegie Mellon Univ19
R 2(1) ¸ 1/3
Helly’s Theorem
Given a collection {C1,C2,…,Cn} of convex sets in Euclidean space :
If every triplet of the sets has a non empty intersection, then Å1·i·n Ci ;
) 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 u1,…,uk be a cycle in G
Then,(P,ui+1) · (P,ui)
because P is in the half-space defined by the edge ui!ui+1
) (P,ui) = (P,uj) 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 Di: Li > Li+1 > Li+2 > … > Li-1
Embedding in R d+1 : Li ´ ( 0 , … , 0 , 1 , 0 , … , 0) Di ´ (d-i,d-i+1,…,1-,2, 3 , … )
Project all points down to the d dimensional plane containing {L1,…,Ld+1} – relative distances between Li and Dj 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 ii = 0 with i¸ 0
Then, p¢ii = 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 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?