CSC2556 Lecture 5 Facility Location Stable Matching CSC2556 - - PowerPoint PPT Presentation

CSC2556 Lecture 5 Facility Location Stable Matching CSC2556 - Nisarg Shah 1

Facility Location CSC2556 - Nisarg Shah 2

Apprx Mechanism Design 1. Define the problem: agents, outcomes, values 2. Fix an objective function (e.g., maximizing sum of values) 3. Check if the objective function is maximized through a strategyproof mechanism 4. If not, find the strategyproof mechanism that provides the best worst-case approximation ratio of the objective function CSC304 - Nisarg Shah 3

Facility Location • Set of agents 𝑂 • Each agent 𝑗 has a true location 𝑦 𝑗 ∈ ℝ • Mechanism 𝑔 ➢ Takes as input reports ෤ 𝑦 = (෤ 𝑦 1 , ෤ 𝑦 2 , … , ෤ 𝑦 𝑜 ) ➢ Returns a location 𝑧 ∈ ℝ for the new facility • Cost to agent 𝑗 : 𝑑 𝑗 𝑧 = 𝑧 − 𝑦 𝑗 • Social cost 𝐷 𝑧 = σ 𝑗 𝑑 𝑗 𝑧 = σ 𝑗 𝑧 − 𝑦 𝑗 CSC2556 - Nisarg Shah 4

Facility Location • Social cost 𝐷 𝑧 = σ 𝑗 𝑑 𝑗 𝑧 = σ 𝑗 𝑧 − 𝑦 𝑗 • Q: Ignoring incentives, what choice of 𝑧 would minimize the social cost? • A: The median location med(𝑦 1 , … , 𝑦 𝑜 ) ➢ 𝑜 is odd → the unique “ (n+1)/2 ” th smallest value ➢ 𝑜 is even → “ n/2 ” th or “ (n/2)+1 ” st smallest value ➢ Why? CSC2556 - Nisarg Shah 5

Facility Location • Social cost 𝐷 𝑧 = σ 𝑗 𝑑 𝑗 𝑧 = σ 𝑗 𝑧 − 𝑦 𝑗 • Median is optimal (i.e., 1 -approximation) • What about incentives? ➢ Median is also strategyproof (SP)! ➢ Irrespective of the reports of other agents, agent 𝑗 is best off reporting 𝑦 𝑗 CSC2556 - Nisarg Shah 6

Median is SP No manipulation can help CSC2556 - Nisarg Shah 7

Max Cost • A different objective function 𝐷 𝑧 = max 𝑧 − 𝑦 𝑗 𝑗 • Q: Again ignoring incentives, what value of 𝑧 minimizes the maximum cost? • A: The midpoint of the leftmost ( min 𝑦 𝑗 ) and the 𝑗 rightmost ( max 𝑦 𝑗 ) locations 𝑗 • Q: Is this optimal rule strategyproof? • A: No! CSC2556 - Nisarg Shah 8

Max Cost • 𝐷 𝑧 = max 𝑗 𝑧 − 𝑦 𝑗 • We want to use a strategyproof mechanism. • Question: What is the approximation ratio of median for maximum cost? 1. ∈ 1,2 2. ∈ 2,3 3. ∈ 3,4 4. ∈ 4, ∞ CSC2556 - Nisarg Shah 9

Max Cost • Answer: 2 -approximation • Other SP mechanisms that are 2 -approximation ➢ Leftmost: Choose the leftmost reported location ➢ Rightmost: Choose the rightmost reported location ➢ Dictatorship: Choose the location reported by agent 1 ➢ … CSC2556 - Nisarg Shah 10

Max Cost • Theorem [Procaccia & Tennenholtz, ‘ 09] No deterministic SP mechanism has approximation ratio < 2 for maximum cost. • Proof: CSC2556 - Nisarg Shah 11

Max Cost + Randomized • The Left-Right-Middle (LRM) Mechanism ➢ Choose min 𝑦 𝑗 with probability ¼ 𝑗 ➢ Choose max 𝑦 𝑗 with probability ¼ 𝑗 ➢ Choose (min 𝑦 𝑗 + max 𝑦 𝑗 )/2 with probability ½ 𝑗 𝑗 • Question: What is the approximation ratio of LRM for maximum cost? (1/4)∗2𝐷+(1/4)∗2𝐷+(1/2)∗𝐷 3 • At most = 𝐷 2 CSC2556 - Nisarg Shah 12

Max Cost + Randomized • Theorem [Procaccia & Tennenholtz, ‘ 09]: The LRM mechanism is strategyproof. • Proof: 1/4 1/2 1/4 2𝜀 𝜀 1/4 1/2 1/4 CSC2556 - Nisarg Shah 13

Max Cost + Randomized • Exercise for you! Try showing that no randomized SP mechanism can achieve approximation ratio < 3/2 . CSC2556 - Nisarg Shah 14

Stable Matching CSC2556 - Nisarg Shah 15

Stable Matching • Recap Graph Theory: • In graph 𝐻 = (𝑊, 𝐹) , a matching 𝑁 ⊆ 𝐹 is a set of edges with no common vertices ➢ That is, each vertex should have at most one incident edge ➢ A matching is perfect if no vertex is left unmatched. • 𝐻 is a bipartite graph if there exist 𝑊 1 , 𝑊 2 such that 𝑊 = 𝑊 1 ∪ 𝑊 2 and 𝐹 ⊆ 𝑊 1 × 𝑊 2 CSC2556 - Nisarg Shah 16

Stable Marriage Problem • Bipartite graph, two sides with equal vertices ➢ 𝑜 men and 𝑜 women (old school terminology  ) • Each man has a ranking over women & vice versa ➢ E.g., Eden might prefer Alice ≻ Tina ≻ Maya ➢ And Tina might prefer Tony ≻ Alan ≻ Eden • Want: a perfect, stable matching ➢ Match each man to a unique woman such that no pair of man 𝑛 and woman 𝑥 prefer each other to their current matches (such a pair is called a “ blocking pair ” ) CSC2556 - Nisarg Shah 17

Example: Preferences Albert Diane Emily Fergie Bradley Emily Diane Fergie Charles Diane Emily Fergie Diane Bradley Albert Charles Emily Albert Bradley Charles Fergie Albert Bradley Charles ≻ ≻ CSC2556 - Nisarg Shah 18

Example: Matching 1 Albert Diane Emily Fergie Bradley Emily Diane Fergie Charles Diane Emily Fergie Diane Bradley Albert Charles Emily Albert Bradley Charles Fergie Albert Bradley Charles Question: Is this a stable matching? CSC2556 - Nisarg Shah 19

Example: Matching 1 Albert Diane Emily Fergie Bradley Emily Diane Fergie Charles Diane Emily Fergie Diane Bradley Albert Charles Emily Albert Bradley Charles Fergie Albert Bradley Charles No, Albert and Emily form a blocking pair . CSC2556 - Nisarg Shah 20

Example: Matching 2 Albert Diane Emily Fergie Bradley Emily Diane Fergie Charles Diane Emily Fergie Diane Bradley Albert Charles Emily Albert Bradley Charles Fergie Albert Bradley Charles Question: How about this matching? CSC2556 - Nisarg Shah 21

Example: Matching 2 Albert Diane Emily Fergie Bradley Emily Diane Fergie Charles Diane Emily Fergie Diane Bradley Albert Charles Emily Albert Bradley Charles Fergie Albert Bradley Charles Yes! (Charles and Fergie are unhappy, but helpless.) CSC2556 - Nisarg Shah 22

Does a stable matching always exist in the marriage problem? Can we compute it in a strategyproof way? Can we compute it efficiently? CSC2556 - Nisarg Shah 23

Gale-Shapley 1962 • Men-Proposing Deferred Acceptance (MPDA): 1. Initially, no proposals, engagements, or matches are made. 2. While some man 𝑛 is unengaged: ➢ 𝑥 ← 𝑛 ’s most preferred woman to whom 𝑛 has not proposed yet ➢ 𝑛 proposes to 𝑥 ➢ If 𝑥 is unengaged: o 𝑛 and 𝑥 are engaged ➢ Else if 𝑥 prefers 𝑛 to her current partner 𝑛′ o 𝑛 and 𝑥 are engaged, 𝑛′ becomes unengaged ➢ Else: 𝑥 rejects 𝑛 3. Match all engaged pairs. CSC2556 - Nisarg Shah 24

Example: MPDA Albert Diane Emily Fergie Bradley Emily Diane Fergie Charles Diane Emily Fergie Diane Bradley Albert Charles Emily Albert Bradley Charles Fergie Albert Bradley Charles = engaged = proposed = rejected CSC2556 - Nisarg Shah 25

Running Time • Theorem: DA terminates in polynomial time (at most 𝑜 2 iterations of the outer loop) • Proof: ➢ In each iteration, a man proposes to someone to whom he has never proposed before. ➢ 𝑜 men, 𝑜 women → 𝑜 × 𝑜 possible proposals ➢ Can actually tighten a bit to 𝑜 𝑜 − 1 + 1 iterations • At termination, it must return a perfect matching. CSC2556 - Nisarg Shah 26

Stable Matching • Theorem: DA always returns a stable matching. • Proof by contradiction: ➢ Assume (𝑛, 𝑥) is a blocking pair. ➢ Case 1: 𝑛 never proposed to 𝑥 o 𝑛 cannot be unmatched o/w algorithm would not terminate. o Men propose in the order of preference. o Hence, 𝑛 must be matched with a woman he prefers to 𝑥 o (𝑛, 𝑥) is not a blocking pair CSC2556 - Nisarg Shah 27

Stable Matching • Theorem: DA always returns a stable matching. • Proof by contradiction: ➢ Assume (𝑛, 𝑥) is a blocking pair. ➢ Case 2: 𝑛 proposed to 𝑥 o 𝑥 must have rejected 𝑛 at some point o Women only reject to get better partners o 𝑥 must be matched at the end, with a partner she prefers to 𝑛 o (𝑛, 𝑥) is not a blocking pair CSC2556 - Nisarg Shah 28

Men-Optimal Stable Matching • The stable matching found by MPDA is special. • Valid partner: For a man 𝑛 , call a woman 𝑥 a valid partner if (𝑛, 𝑥) is in some stable matching. • Best valid partner: For a man 𝑛 , a woman 𝑥 is the best valid partner if she is a valid partner, and 𝑛 prefers her to every other valid partner. ➢ Denote the best valid partner of 𝑛 by 𝑐𝑓𝑡𝑢(𝑛) . CSC2556 - Nisarg Shah 29

Men-Optimal Stable Matching • Theorem: Every execution of MPDA returns the “ men- optimal ” stable matching: every man is matched to his best valid partner. ➢ Surprising that this is a matching. E.g., it means two men cannot have the same best valid partner! • Theorem: Every execution of MPDA produces the “ women- pessimal ” stable matching: every woman is matched to her worst valid partner. CSC2556 - Nisarg Shah 30