CSC304 Lecture 13 Finishing Facility Location: Randomized - PowerPoint PPT Presentation

CSC304 Lecture 13 Finishing Facility Location: Randomized Left-Right-Middle Mechanism Begin Stable Matching: Gale-Shapley Algorithm CSC304 - Nisarg Shah 1

Recap: Facility Location • Set of agents 𝑂 • Each agent 𝑗 has a true location 𝑦 𝑗 ∈ ℝ • Mechanism 𝑔 takes as input reported locations ෤ 𝑦 = (෤ 𝑦 1 , ෤ 𝑦 2 , … , ෤ 𝑦 𝑜 ) , and places the facility at 𝑧 ∈ ℝ • Cost to agent 𝑗 : 𝑑 𝑗 𝑧 = 𝑧 − 𝑦 𝑗 CSC304 - Nisarg Shah 2

Recap: Facility Location • Social cost 𝐷 𝑧 = σ 𝑗 𝑑 𝑗 𝑧 = σ 𝑗 𝑧 − 𝑦 𝑗 ➢ Median is strategyproof (SP) and optimal (i.e., provides a 1 -approximation) • Maximum cost 𝐷 𝑧 = max |𝑧 − 𝑦 𝑗 | 𝑗 ➢ Median, Leftmost, Rightmost, Dictatorship, etc are strategyproof and provide a 2 -approximation ➢ No deterministic SP mechanism can provide < 2 approximation CSC304 - Nisarg Shah 3

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 CSC304 - Nisarg Shah 4

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 CSC304 - Nisarg Shah 5

Max Cost + Randomized • Exercise for you! Try showing that no randomized SP mechanism can achieve approximation ratio < 3/2 • Suggested outline ➢ Consider two agents with 𝑦 1 = 0 and 𝑦 2 = 1 ➢ Show that one of them has expected cost at least ½ ➢ What happens if that agent moves 1 unit farther from the other agent? CSC304 - Nisarg Shah 6

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 CSC304 - Nisarg Shah 7

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”) CSC304 - Nisarg Shah 8

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 ≻ ≻ CSC304 - Nisarg Shah 9

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? CSC304 - Nisarg Shah 10

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 . CSC304 - Nisarg Shah 11

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: What about this matching? CSC304 - Nisarg Shah 12

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.) CSC304 - Nisarg Shah 13

Does a stable matching always exist in the marriage problem? Can we compute it in a strategyproof way? Can we compute it efficiently? CSC304 - Nisarg Shah 14

Gale-Shapley 1962 • Men-Proposing Deferred Acceptance (MPDA): 1. Initially, no one has proposed, no one is matched. 2. While some man 𝑛 is unmatched: ➢ 𝑥 ← 𝑛 ’s most preferred woman to whom 𝑛 has not proposed yet ➢ 𝑛 proposes to 𝑥 ➢ If 𝑥 is unmatched: 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. CSC304 - Nisarg Shah 15

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 CSC304 - Nisarg Shah 16

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. CSC304 - Nisarg Shah 17

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 GS: 𝑛 cannot be unmatched o/w algorithm would not terminate. o GS: 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 CSC304 - Nisarg Shah 18

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 GS: 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 CSC304 - Nisarg Shah 19

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) . CSC304 - Nisarg Shah 20

Men-Optimal Stable Matching • Theorem: Every execution of MPDA returns the men- optimal stable matching in which every man is matched to his best valid partner 𝑐𝑓𝑡𝑢 𝑛 . ➢ Surprising that this is even a matching. E.g., why can’t two men have the same best valid partner? ➢ Every man is simultaneously matched with his best possible partner across all stable matchings • Theorem: Every execution of MPDA produces the women- pessimal stable matching in which every woman is matched to her worst valid partner. CSC304 - Nisarg Shah 21

Men-Optimal Stable Matching • Theorem: Every execution of MPDA returns the men- optimal stable matching. • Proof by contradiction: ➢ Let 𝑇 = matching returned by MPDA. ➢ 𝑛 ← first man rejected by 𝑐𝑓𝑡𝑢 𝑛 = 𝑥 ➢ 𝑛 ′ ← the more preferred man due to which 𝑥 rejected 𝑛 ➢ 𝑥 is valid for 𝑛 , so (𝑛, 𝑥) part of stable matching 𝑇′ ➢ 𝑥′ ← woman 𝑛′ is matched to in 𝑇′ ➢ We show that 𝑇′ cannot be stable because (𝑛 ′ , 𝑥) is a blocking pair. CSC304 - Nisarg Shah 22

Men-Optimal Stable Matching • Theorem: Every execution of MPDA returns the men- optimal stable matching. • Proof by contradiction: Blocking pair 𝑛′ 𝑛′ Not yet rejected by a valid partner ⇒ X hasn’t proposed to 𝑥′ 𝑛 𝑥 𝑛 𝑥 ⇒ prefers 𝑥 to 𝑥′ 𝑥′ 𝑇 𝑇′ First to be rejected by Rejects 𝑛 because best valid partner ( 𝑥 ) prefers 𝑛′ to 𝑛 CSC304 - Nisarg Shah 23

Strategyproofness • Theorem: MPDA is strategyproof for men, i.e., reporting the true ranking is a weakly dominant strategy for every man. ➢ We’ll skip the proof of this. ➢ Actually, it is group-strategyproof. • But the women might want to misreport. • Theorem: No algorithm for the stable matching problem is strategyproof for both men and women. CSC304 - Nisarg Shah 24

Women-Proposing Version • Women-Proposing Deferred Acceptance (WPDA) ➢ Just flip the roles of men and women ➢ Strategyproof for women, not strategyproof for men ➢ Returns the women-optimal and men-pessimal stable matching CSC304 - Nisarg Shah 25