Discussion 1B

Induction Problem

A party has n2n \geq 2 people. A celebrity is a person who knows nobody, but everyone knows the celebrity. If you ask questions of the form Does AA know BB? prove that you can figure out who the celebrity is (or that there is no celebrity) using 3n43 n - 4 questions.

Stable Marriage

There are nn men and nn women. Each man has his own preference list of the women, and likewise for the women. The goal is to pair up the men and women according to their preferences.

Definitions

  • pairing - a set of nn pairs of men and women
  • rogue couple - given a pairing, a man and a woman form a rogue couple if they would rather be with each other than their current partners
  • stable - a pairing is stable if there are no rogue couples
  • optimal - the optimal woman for a man is the best woman the man could be paired with over all stable pairings

Stable Marriage Algorithm

The men go down their lists, proposing to women. When a woman receives multiple proposals, she keeps her favorite man on her string and rejects the other men, who continue down their lists. The algorithm ends when every woman receives one proposal.
Improvement Lemma - if a woman has a man on her string, then for every day after she will either have the same man or a better man on her string
Theorem - the algorithm always terminates with a male-optimal stable pairing

Worksheet

1. Stable Marriage

The algorithm takes four days (it terminates on the fifth) and the final pairing is (A,2),(B,1),(C,3)(A,2),(B,1),(C,3).

2. Propose-and-Reject Proofs

a
The Improvement Lemma says if a woman receives a proposal, then on each day after she will either have the same man on her string or get proposals from better men.
b
Suppose for the sake of contradiction that she receives no proposal on day ii but receives a proposal on a previous day jj. Then using part a, since she receives a proposal on day jj she receives proposals for all subsequent days, including day ii. This contradicts the assumption that she receives no proposal on day ii.
c
Let the number of days the algorithm takes be kk. Since the algorithm ends, every woman has a proposal on day kk. Since the algorithm does not end on day k1k-1, there is some woman who does not have a proposal on day k1k-1. Then using part b, she did not receive proposals on all days before k1k-1, so she only receives one proposal (on day kk).

3. Be a Judge

a
False. If this were true that would mean every man gets rejected n1n-1 times and every woman gives n1n-1 rejections. This contradicts what we proved before: that there is some woman who is proposed to exactly once.
If n=2n=2, can this be true?
b
True. Suppose they aren’t paired, so that in the final pairing we see the pairs (M,W)(M, W') and (M,W)(M', W). In this final pairing, MM and WW form a rogue couple since they each prefer the other over their current partners. Therefore this isn’t a stable pairing.
c
False. Consider the counterexample with n=2n=2 with men 11 and 22 and women AA and BB. Both men have the preference list B>AB > A and both women have the preference list 2>12 > 1. The algorithm terminates with the pairing (A,1),(B,2)(A, 1), (B, 2).
d
True. If for every pair (Mi,Wi)(M_i, W_i) they each have the other as their least favorite partner, then any other choice of MjM_j and WkW_k is a rogue couple.

Induction Problem Solution

Base case

For n=2n=2, we ask the two questions Does AA know BB? and Does BB know AA?

Inductive hypothesis

Assume that at a party with nn people we can figure out who the celebrity is using 3n43 n - 4 questions.

Inductive step

Our goal is to show that at a party with n+1n+1 people we need to use 3(n+1)43(n+1)-4 questions.
First we ask Does AA know BB? If AA knows BB, then AA is not a celebrity; otherwise, if AA does not know BB, then BB is not celebrity. In both cases, we are left with nn people who are potentially celebrities. So, we can use our inductive hypothesis and ask 3n43 n - 4 questions to learn one of the following:
  1. There’s no celebrity among the nn people, and since the person we excluded is not a celebrity, we know there is no celebrity among the n+1n+1 people. We use one additional question, meaning we ask a total of 3n33 n -3 questions.
  1. There’s a celebrity among the nn people, but we need to check whether this person is still a celebrity among the n+1n+1 people. This requires us to ask two questions: whether the potential celebrity and the n+1n+1th person know each other. We use three additional questions, meaning we ask a total of 3n13 n - 1 questions.