Download to read offline
![Many complex Business rules
of admissions
• Affirmative action rules
• Freeze/Float/Slide/withdraw
• Supernumerary Seats for Female Candidates [’18 - ]
– No increase in male seats
– No adverse affect on chances for the male candidates
– Create supernumerary seats (not reservation at all !)](https://image.slidesharecdn.com/lecture6-stable-matching-and-applications-251113112634-8678ddaf/75/OPERATING-SYSTEMS-AND-LINUX1OPERATING-SYSTEMS-AND-LINUX1-30-2048.jpg)
![Many complex Business rules
of admissions
• Affirmative action rules
• Freeze/Float/Slide/withdraw
• Supernumerary Seats for Female Candidates [’18 - ]
– No increase in male seats
– No adverse affect on chances for the male candidates
– Create supernumerary seats (not reservation at all !)](https://crownmelresort.com/image.slidesharecdn.com/lecture6-stable-matching-and-applications-251113112634-8678ddaf/75/OPERATING-SYSTEMS-AND-LINUX1OPERATING-SYSTEMS-AND-LINUX1-30-2048.jpg)
More Related Content
OPERATING SYSTEMS AND LINUX1OPERATING SYSTEMS AND LINUX1
- 1. Design and Analysisof Algorithms Lecture 6 Stable Matching Algorithm and Applications 1 CS345
- 2. Recap from thelast Lecture 2
- 3. Interpolation There is apolynomial with ’s unknown. Given: the following (point, value) representation of : Aim: To compute ’s . 3 {}
- 4. 4 Interpolation Polynomial Interpolation =Polynomial evaluation Can you guess the polynomial whose evaluation will help solve polynomial interpolation ? Make an obvious guess.
- 5. Interpolation Solution: Define a polynomial ,where . Evaluate on . What is ? 5 Homework: Can you guess what is relation between and ?
- 6. • It wasvery disheartening to see that only a handful of students attempted the homework. • Please attempt each homework sincerely. It is for your own benefit. • If you are not able to do the homework, you may send email to the instructor 1 day in advance for a short meeting (avoid holidays and week- ends). 6
- 7. Real challenge of multiplemerit lists 7
- 8. Fairness 8 Merit list q p qp Ram Mohan Choice list of Mohan of programme q Ram Mohan
- 9. How to tackleMultiple Merit Lists using “Single Merit list algorithm” 9
- 10. A common man’salgorithm Allocation_IIT(); Allocation_NIT(); ; While ( ) { For each do { is asked to choose between the two choices; The preference list of is trimmed accordingly; } Allocation_IIT(); Allocation_NIT(); ; } 10 Round Round Round + Round and ) Homework (not for exam): Prove that the algorithm outputs a fair seat allocation.
- 11. Multiple merit lists IIT NIT ARCH 11 MeritLists A B C Choice Lists IIT NIT ARCH IIT NIT ARCH IIT NIT ARCH Allocation Allocation Allocation IIT NIT ARCH NIT NIT IIT ARCH ARCH IIT C B A A C B B A C C A B A B C B C A A B C Candidate Optimal Candidate Pessimal
- 12. The Stable MarriageProblem 12
- 13. Stable Marriage Problem •prefers to • prefers to Definition: A marriage is said to be stable if there is no unstable pair in the society. 13 𝑤𝑙 𝑤𝑖 𝑚𝑘 𝑚𝑗 divorce divorce (, ) : an unstable pair Meaning of unstability
- 14. Stable Marriage Problem •: set of men • : set of women For each man , : a preference list of all women For each woman , : a preference list of all men Aim: To compute a stable marriage. 14
- 15. Algorithm for StableMarriage Man proposes, God disposes 15 1962, David Gale and Lloyd Shapley Woman Nobel Prize in Economic Sciences, 2012
- 16. We designed aniterative algorithm in the last class based on these rules. • A man proposes in the decreasing order of preference. • Once rejected by a woman, the man never proposes to her again. • A woman, if single, accepts a proposal. • A woman, if engaged, accepts a proposal from a man if she prefers him to her current partner. 16 Rule for Men Rule for Women
- 17. GaleShapley(, ) (Proof ofstability) Lemma : If prefers to prefers to . 17 𝑤𝑙 𝑤𝑖 𝑚𝑘 𝑚𝑗 Each new engagement gives a woman a better partner (mate). There is no unstable pair in the marriage. The key observation : How to restate it so that we can prove it easily ? Consider the pair (, ) If prefers to , nothing to be done
- 18. Lemma : If prefersto prefers to . Proof: (a sketch (the details emerged from the interaction in the class)) must have proposed to . At the moment of the proposal, was either engaged or single. If was single, would have accepted the offer at that time but rejected later. But a woman rejects her present mate only when she gets a better mate. So would have surely got a better mate than . Since the mate of a woman only improves in future rounds, would indeed be preferred to . If was engaged, …fill in the details along similar lines as stated above… 18 𝑤𝑙 𝑤𝑖 𝑚𝑘 𝑚𝑗 GaleShapley(, ) (Proof of stability) The following slide gives the steps of the analysis.
- 19.
- 20. GaleShapley(, ) (Proof ofstability) Lemma : If prefers to prefers to . 20 𝑤𝑙 𝑤𝑖 𝑚𝑘 𝑚𝑗
- 21. GaleShapley(, ) Theorem: GaleShapley(, )indeed computes a stable marriage. Question : Does there exist a unique stable marriage ? Answer: No. Homework: Give an example graph with multiple stable matchings.
- 22. Gale Shapley Algorithm Theorem: Therecannot exist any other stable matching 22 Man Optimal Woman pessimal Man proposing that matches a man to a better woman.
- 23. Gale Shapley Algorithm 23 ManPessimal Woman Optimal Woman proposing
- 24. Gale Shapley Algorithm 24 WomenMen 𝑤1 𝑤2 𝑤3 𝑤𝑛 𝑤𝑖 𝑚1 𝑚2 𝑚3 𝑚𝑗 𝑚𝑛 Can you think of using this algorithm to solve the problem of joint seat allocation ?
- 25. Gale Shapley Algorithm 25 ProgramsMen 𝑃1 𝑃2 𝑃3 𝑃𝑛 𝑃𝑖 𝑚1 𝑚2 𝑚3 𝑚𝑗 𝑚𝑛
- 26. Gale Shapley Algorithm 26 ProgramsCandidates 𝑃1 𝑃2 𝑃3 𝑃𝑛 𝑃𝑖 𝑐1 𝑐2 𝑐3 𝑐 𝑗 𝑐𝑚 Deferred Acceptance Algorithm Preference lists Merit lists Unstability Unfairness
- 27. 27 Programs Candidates 𝑃1 𝑃2 𝑃3 𝑃𝑛 𝑃𝑖 𝑐1 𝑐2 𝑐3 𝑐 𝑗 𝑐𝑚 DeferredAcceptance Algorithm
- 28. Algorithms 28 Common man’s algorithmSoftware Industry’s algorithm Deferred Acceptance algorithm Gale Shapley algorithm Candidate proposing Programme proposing Programme proposing Limited rounds Candidate Optimal Candidate Pessimal Candidate Pessimal
- 29. Merits of Deferred AcceptanceAlgorithm • Any number of merit lists • Candidate optimality guaranteed • Efficient Algorithm 29 Joint Seat Allocation Deferred Acceptance Algorithm
- 30. Many complex Businessrules of admissions • Affirmative action rules • Freeze/Float/Slide/withdraw • Supernumerary Seats for Female Candidates [’18 - ] – No increase in male seats – No adverse affect on chances for the male candidates – Create supernumerary seats (not reservation at all !)
- 31. Impact of JointSeat Allocation No. of Seats Vacancies Vacancies in IITs 𝟓𝟖𝟕 𝟑𝟎𝟖 𝟏𝟗𝟎 𝟏𝟗𝟖
- 32. Impact of JointSeat Allocation • A single vacancy removed many candidates getting better program (due to cascading effect) No. of Candidates getting better Seats 𝟏𝟖𝟗𝟎 𝟏𝟕𝟔𝟕 𝟑𝟔𝟕𝟐 𝟏↔𝟑𝟎𝟎
- 33. Algorithms do havegreat impact in real world … 33
- 34. Impact of JointSeat Allocation • INFORMS J. Appl. Anal., 49(5) 338-354 (2019) The above research paper is available on the homepage of the instructor. Read it to see the role of algorithms in your journey to IITK Not meant for exam or quizzes Centralized Admissions for Engineering Colleges in India S. Baswana, P. P. Chakrabarti, S. Chandran, Y. Kanoria, U. Patange
- 35. Gayle Shapley Algorithm Proofof Man optimality 35 Man-Proposing version
- 36. Gale Shapley Algorithm 36 WomenMen 𝑤1 𝑤2 𝑤3 𝑤𝑛 𝑤𝑖 𝑚1 𝑚2 𝑚3 𝑚𝑗 𝑚𝑛
- 37. Let be thematching produced by GS algorithm. Let be any other stable matching. Theorem (Man Optimality): There does not exist any man such that precedes in the preference list of . 37
- 38. At the endof the th iteration, the set of men matched by GS algorithm. 38 Theorem (Man Optimality): There does not exist any man such that precedes in the preference list of . Theorem (Man Optimality): At the end of the th iteration, There does not exist any man such that precedes in the preference list of .
- 39. 39 Theorem (Man Optimality):At the end of the th iteration, There does not exist any man such that precedes in the preference list of . 𝒈 𝒇 𝒊
- 40. 1st Iteration 40 Theorem (ManOptimality): At the end of the th iteration, There does not exist any man such that precedes in the preference list of . 𝒈 𝒇 𝟏
- 41. th iteration: 41 Theorem (ManOptimality): At the end of the th iteration, There does not exist any man such that precedes in the preference list of . 𝒇 𝒊 −𝟏 𝑚 𝑗 Execution of th iteration: 𝒈
- 42. 42 Theorem (Man Optimality):At the end of the th iteration, There does not exist any man such that precedes in the preference list of . 𝒇 𝒊 𝑚 𝑗 Execution of th iteration: 𝑤 𝑗 𝑤 𝑗′ 𝑚𝑗 𝒈 Homework: Prove that if prefers to , must be unstable.