1. Mechanism Design without Money Lecture 8 1

2. The Match – a success story Stability – 10 years before Gale Shapley Truthfulness “THE MOST IMPORTANT THING FOR APPLICANTS TO REMEMBER: Simply rank internship programs based on your TRUE preferences, without consideration for where you believe you might be ranked by these programs. List the program that you want most as rank #1, followed by your next most-preferred program as rank #2, and so on.” – APPIC “For over 50 years, most United States medical school seniors have chosen to use a matching program … “Before such matching programs … medical students often felt pressure, at an unreasonably early stage… accept offers from residency programs. … This situation was inefficient, chaotic, and unfair…” - Congress

3. Rural hospitals 3 The problem with allocation of doctors to rural hospitals. Theorem: When preferences are strict, the set of doctors employed and positions filled is the same at every stable matching. Theorem: When preferences are strict, any hospital that does not fill its quota at some stable matching is assigned precisely the same set of students at every stable matching.

4. Couples 4 With time and progress, couples became a problem for the NRMP. The “leading member” adjustment didn’t work very well… This was one of the main reasons the original NRMP algorithm had to be replaced.

5. stable matching with couples 5

6. Detour – complements and substitutes Complements: want both A and B (cucumbers and tomatoes to make salad) Substitutes: Given A, I want B less (apples and bananas when I need a fruit) Substitutes: greedy allocation can’t be bad Couples create complementarities 6

7. Where does DA break? Suppose we run a doctor’s proposing DA If a one of the members of the couple gets a negative answer, both leave Problem – if a couple is temporarily assigned and then gets kicked out, a vacancy is created 7

8. Roth and Peranson (1999) 8 The new algorithm works as follows: – First do doctor-proposing deferred-acceptance, with only single doctors involved. – Then add the couples one by one (in random order) and using similar “proposing” mechanism. – If any cycles are detected, start over. If the algorithm terminates, the resulting matching is stable. And while supposedly there is no good reason for that, this algorithm always terminates (when using data from previous years or when using random preferences). Furthermore, it is generally considered a success story, and it was adopted for other programs as well…

9. Roth and Peranson (1999) 9 Clearinghouses currently using the algorithm (Taken from Roth’s slides):

10. Some questions 10 When is the set of stable matchings (with couples) non-empty? – NPC Why does the Roth-Peranson algorithm works? What about truthfulness?

11. Large markets again… 11

12. Extending Roth’s result 12

13. Preferences n “strong” doctors ; many “weak” doctors 1.5n residency programs k=n 1-e couples, all strong doctors Each hospital ranks all the strong doctors uniformly at random, and then the weak doctors Each single ranks all hospitals uniformly at random Each couple ranks pairs of hospitals uniformly at random, but only considers pairs which are in the same city

14. Algorithm 1.Assign the “strong” singles, according to GS (doctors propose) 2.Insert couples one by one: a)If inserting couple c i didn’t reject any couple - continue b)If assigning c i caused couple c p to get rejected: 1. Backtrack to the time where we assigned c p. 2. Assign c i before c p and continue 3.Assign all weak doctors (doctors proposing)

15. Example Alice Bob Charlie C2C2 C1C1 1.Assign Alice Bob and Charlie 2.Assign C 1. Charlie gets unassigned 3.Assign Charlie 4.Assign C 2. Bob get unassigned 5.Assign Bob. C 1 gets unassigned 6.Need to backtrack Redo, with C 2 and then C 1 H1H1 H2H2 H3H3 H4H4 H5H5 H6H6 H7H7

16. C1C1 Example Alice Bob Charlie C2C2 1.Assign Alice Bob and Charlie 2.Assign C 2. Bob gets unassigned 3.Assign Bob 4.Try to Assign C 1. They don’t get their first choice 5.Assign C 1. They get their second choice H1H1 H2H2 H3H3 H4H4 H5H5 H6H6 H7H7

17. Termination  success G - a graph on the couples, such that there is an edge from c i to c p if c i causes c p to get unassigned Want: a topological sort on G Defining such a graph is hard: - Whether c i gets c p to be unassigned depends on the rest of the system (who else is assigned right now) - Worst case bounds are not enough: Every couple can potentially unassign many other couples – no topological sort exists If no couple ever gets unassigned – the algorithm ends successfully

18. Example: (c 2 -husband, H 2 ), (c 2 -wife, H 3 )  Blob(c 2,0) Blobs and the dependency graph From blobs to edges; if: 1.(H,d’)  Blob(c p,0) 2.(H,d)  Blob(c i,0) 3.Hospital H prefers d to d’ Then there is an edge c i  c p Alice Bob C2C2 C1C1 H1H1 H2H2 H3H3 H4H4 H5H5 H6H6 (c 1 -husband, H 5 )  Blob (c 1,0), and H 5 prefers Bob There is an edge from c 2  c 1 (Bob,H 5 )  Blob(c 2,0) Blob(c i,0) = set of hospitals doctor pairs which c i can effect, if they were inserted as the first couple

19. Not everyone can be first: controlling dependencies using high order blobs If a couple is not inserted first, they might see a different picture, e.g. because the hospital they would naturally get into is now taken by a couple who was there before them Blob(c i,r) = set of hospitals doctor pairs which c i can effect, if they were inserted as the first couple and an adversary would be allowed to close r hospitals We add an edge c i  c p for high order blobs as well: (H,d’)  B(c i,r) and (H,d)  B(c i,r) with H preferring d to d’

20. High order blob example C1C1 Alice Bob Charlie C2C2 H1H1 H2H2 H3H3 H4H4 H5H5 H6H6 H7H7 Example: (c 1 -husband, H 5 ), (c 1 -wife, H 6 ),  Blob(c 1,0) (c 1 -husband, H 1 ), (c 1 -wife, H 7 ),  Blob(c 1,1)

21. Proof intuition Build the dependency graph G based on blobs of order r=3/  Prove that G is a DAG, and can be topologically sorted Main Lemma: Blobs of size 3/  are conservative enough, such that no couple will kick someone outside of their blob Show that our algorithm implicitly generates the blob graph and approximates a topological sort of it

22. Proof intuition: Unraveling the dependencies G - a graph on the couples, such that there is an edge from c i to c p if c i causes c p to get unassigned Want: a topological sort on G Defining such a graph is hard: - Whether c i gets c p to be unassigned depends on the rest of the system (who else is assigned right now) - Worst case bounds are not enough: Every couple can potentially unassign many other couples – no topological sort exists The algorithm we run implicitly eliminates dependencies If no couple ever gets unassigned – the algorithm ends successfully

23. Truthfulness in the Match “Programs should be ranked in sequence, according to the applicant's true preferences... It is highly unlikely that either applicants or programs will be able to influence the outcome of the Match in their favor by submitting a list that differs from their true preferences.” NRMP

24. Problem: S 1 can cheat H2H2 s1s1 c2c2 s2s2 Hospital Preferences Capacity is 1 s2s2 H2H2 H4H4 c=(c 1,c 2 ) H 3,H 4 H 1,H 2 Student Preferences s1s1 H4H4 H1H1 H3H3 H2H2 H1H1 c1c1 s1s1 H4H4 s2s2 s1s1 c2c2 H3H3 c1c1 s1s1 s1s1 H1H1 H2H2 H3H3 c 1  H 1, c 2  H 2, s 1  H 3, s 2  H 4 s 1  H 1, s 2  H 2, c 1  H 3, c 2  H 4 The only stable marriage

25. A linear fraction of couples The number of married people is proportional to the number of people Simulations show decent success rates for a constant fraction of couples Is there a way to insert the couples, to get a stable matching? Thm : consider a random market, with n singles, n couples and more than 20n hospitals. With constant probability, there is no stable outcome

26. Proof idea – isolated markets 1.Find a small structure, which prevents a stable outcome A few hospitals and doctors, which (if left alone) can not form a stable outcome 2.Show that this small structure exists with constant probability 3.Show that no one outside the structure ever enters a hospital in the isolated market

27. Instable structure For a single s and a couple c, with probability O(1/n 2 ) we have the structure If the structure occurs – no “local” stable outcome There are n singles and n couples, so with constant probability this structure will occur H1H1 … s … c1c1 … H2H2 … c2c2 … s … S H2H2 H1H1 … c=(c 1,c 2 ) H 1,H 2 whatever Doctor Preferences Hospital Preferences Capacity is 1

28. Isolated market The only solution is to insert someone else to h 1,h 2 thus avoiding the problem There is an excess of positions, so if a doctor goes to h 1,h 2 there are hospitals which are left free. We need to show that the doctor prefers them A quantitative version of the Rural Hospital Theorem – Define a probabilistic process, show it’s a martingale, use Azuma’s inequality

29. Extensions Roommate problems, multi-sided matching Many-to-one with discrete money and substitutable preferences (Crawford and Knoer, 1981; Kelso and Crawford, 1982) Many-to-many with responsive preferences (Roth, 1984) Matching with contracts (Hatfield and Milgrom, 2005) Many-to-many matching with contracts (Echenique and Oveido, 2006) Matching in supply chains (Ostrovsky, 2008) Matching in networks with bilateral contracts (Hatfield, Kominers, Nichifor, Ostrovsky and Westkamp, working paper) Matching with minimum quotas, regional caps, etc. (Biro, Fleiner, Irving and Manlove, 2010, Kamada and Kojima, 2013) 29

30. Related topics Roth and Vande Vate (1990) – Random paths to stability Jackson and Watts (2002) Ausubel and Milgrom (2000) on package bidding 30

31. Questions? 31

32. Extra Slides 32

33. Chicken 33

34. Road example AB 1 hour N minutes 50 people want to get from A to B There are two roads, each one has two segments. One takes an hour, and the other one takes the number of people on it 34

35. Nash in road example In the Nash equilibrium, 25 people would take each route, for a travel time of 85 minutes AB 1 hour N minutes 35

36. Braess’ paradox Now suppose someone adds an extra road which takes no time at all. Travel time goes to 100 minutes AB 1 hour N minutes Free 36