Richard Anderson Winter 2009 Lecture 2

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Presentation transcript:

Richard Anderson Winter 2009 Lecture 2 CSE 421 Algorithms Richard Anderson Winter 2009 Lecture 2

Announcements Homework due Wednesdays Subscribe to the mailing list HW 1, Due January 14, 2009 Subscribe to the mailing list Office Hours Richard Anderson, CSE 582 Monday, 3:00-3:50 pm, Thursday, 11:00-11:50 am Aeron Bryce, CSE 216 Monday, 12:30-1:20 pm, Tuesday, 12:30-1:20 pm

Stable Marriage Input Output Preference lists for m1, m2, …, mn Preference lists for w1, w2, …, wn Output Perfect matching M satisfying stability property:  m’, w’’, (m’, w’’) is NOT an instability If (m’, w’)  M and (m’’, w’’)  M then (m’ prefers w’ to w’’) or (w’’ prefers m’’ to m’)

Proposal Algorithm Initially all m in M and w in W are free While there is a free m w highest on m’s list that m has not proposed to if w is free, then match (m, w) else suppose (m2, w) is matched if w prefers m to m2 unmatch (m2, w) match (m, w)

Result Simple, O(n2) algorithm to compute a stable matching Corollary A stable matching always exists

A closer look Stable matchings are not necessarily fair m1: w1 w2 w3 w1: m2 m3 m1 w2: m3 m1 m2 w3: m1 m2 m3 m2 w2 m3 w3 How many stable matchings can you find?

Does the M proposal algorithm give the same results as the W proposal algorithm? m1: w1 w2 m2: w2 w1 w1: m2 m1 w2: m1 m2

Algorithm under specified Many different ways of picking m’s to propose Surprising result All orderings of picking free m’s give the same matching Proving this type of result Reordering argument Prove algorithm is computing something mores specific Show property of the solution – so it computes a specific stable matching

Proposal Algorithm finds the best possible solution for M Formalize the notion of best possible solution: (m, w) is valid if (m, w) is in some stable matching best(m): the highest ranked w for m such that (m, w) is valid S* = {(m, best(m)} Every execution of the proposal algorithm computes S*

Proof See the text book – pages 9 – 12 Related result: Proposal algorithm is the worst case for W Algorithm is the M-optimal algorithm Proposal algorithms where w’s propose is W-Optimal

Best choices for one side may be bad for the other Design a configuration for problem of size 4: M proposal algorithm: All m’s get first choice, all w’s get last choice W proposal algorithm: All w’s get first choice, all m’s get last choice m1: m2: m3: m4: w1: w2: w3: w4: M’s: 1342 2431 3214 4123 W’s: 2431 1342 4123 3214

M-rank and W-rank of matching m-rank: position of matching w in preference list M-rank: sum of m-ranks w-rank: position of matching m in preference list W-rank: sum of w-ranks m1: w1 w2 w3 m2: w1 w3 w2 m3: w1 w2 w3 w1: m2 m3 m1 w2: m3 m1 m2 w3: m3 m1 m2 m1 w1 m2 w2 m3 w3 What is the M-rank? What is the W-rank?

Suppose there are n m’s, and n w’s What is the minimum possible M-rank? What is the maximum possible M-rank? Suppose each m is matched with a random w, what is the expected M-rank?

Random Preferences Suppose that the preferences are completely random m1: w8 w3 w1 w5 w9 w2 w4 w6 w7 w10 m2: w7 w10 w1 w9 w3 w4 w8 w2 w5 w6 … w1: m1 m4 m9 m5 m10 m3 m2 m6 m8 m7 w2: m5 m8 m1 m3 m2 m7 m9 m10 m4 m6 If there are n m’s and n w’s, what is the expected value of the M-rank and the W-rank when the proposal algorithm computes a stable matching?

Expected Ranks Expected M rank Expected W rank Guess – as a function of n

Expected M rank Expected M rank is the number of steps until all M’s are matched (Also is the expected run time of the algorithm) Each steps “selects a w at random” O(n log n) total steps Average M rank: O(log n)

Expected W-rank If a w receives k random proposals, the expected rank for w is n/(k+1). On the average, a w receives O(log n) proposals The average w rank is O(n/log n)

Probabilistic analysis Select items with replacement from a set of size n. What is the expected number of items to be selected until every item has been selected at least once. Choose k values at random from the interval [0, 1). What is the expected size of the smallest item.

What is the run time of the Stable Matching Algorithm? Initially all m in M and w in W are free While there is a free m w highest on m’s list that m has not proposed to if w is free, then match (m, w) else suppose (m2, w) is matched if w prefers m to m2 unmatch (m2, w) match (m, w) Executed at most n2 times

O(1) time per iteration Find free m Find next available w If w is matched, determine m2 Test if w prefer m to m2 Update matching

What does it mean for an algorithm to be efficient?

Key ideas Formalizing real world problem Model: graph and preference lists Mechanism: stability condition Specification of algorithm with a natural operation Proposal Establishing termination of process through invariants and progress measure Under specification of algorithm Establishing uniqueness of solution