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courses:cs211:winter2014:journals:colin:chapter1 [2014/01/09 02:02] – created mohnacsccourses:cs211:winter2014:journals:colin:chapter1 [2014/01/16 07:02] (current) – Initial commit mohnacsc
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 ===== Chapter 1: Introduction ===== ===== Chapter 1: Introduction =====
  
-=== 1.1 - Stable Matching ===+==== 1.1 - Stable Matching ===
 + 
 +The Stable Matching Problem was created by David Gale and Lloyd Shapley in 1962 based on the question: //Could one design a college admissions process, or a job recruiting process, that was self-enforcing?// 
 + 
 +Concerns with the process: 
 +Better offers come along and change the matching of applicants, causing a domino effect 
 + 
 +Both applicants and employers must be happy to maintain a stable state.  One of the two must hold true: 
 +  * E prefers every one of its accepted applicants to A; or 
 +  * A prefers her current situation over working for employer E 
 +  
 + 
 +=== The Problem === 
 + 
 +Consider matching men and women where set M = {m(1),m(2),…,m(n)} represents n men and set W = {w(1),w(2),…w(n)} represents n women.   
 + 
 +M X W denotes the set of all ordered pairs of the form (m, w). 
 + 
 +A matching S is a set of ordered pairs with the property that each member of M and W appears in at most one pair in S.  A perfect matching S’ is a matching with the property that each member of M and W appears in exactly one pair in S’. 
 + 
 +Preferences must also be accounted for - in this problem, men rank all women.  For example: m  of (m,w) and w’ of (m’,w’) may prefer each other to their matching in S.  In this case, the matching is unstable. 
 + 
 +Matching S is stable if: 
 +  * It is perfect 
 +  * There is no instability with respect to S 
 + 
 + 
 +=== Designing the Algorithm === 
 + 
 +Basic ideas: 
 +  * Initially, every person is free.  During the matching process, pairs may enter a state of engagement. 
 +  * If an engaged women w of pair (m,w) receives a more preferred offer from m’, she becomes engaged to m’ and m becomes free. 
 +  * The algorithm terminates when no one is free. 
 + 
 +Concrete description of Gale-Shapley algorithm: 
 +<code> 
 +Initially all m (in M) and w (in W) are free 
 +While there is a man m who is free and hasn’t proposed to every woman 
 +        Choose such a man m 
 +        Let w be the highest-ranked woman in m’s preference list to whom m has not yet proposed 
 + If w is free then 
 + (m,w) become engaged 
 + Else w is currently engaged to m’ 
 + If w prefers m’ to m then 
 + m remains free 
 + Else w prefers m to m’ 
 + (m,w) become engaged 
 + m’ becomes free 
 + Endif 
 + Endif 
 +Endwhile 
 +Return the set S of engaged pairs 
 +</code> 
 + 
 +=== Analyzing the Algorithm === 
 + 
 +w remains engaged from first proposal, and her partners increase in terms of preference. 
 + 
 +The sequence of women to whom m proposes gets worse in terms of preference. 
 + 
 +The G-S algorithm terminates after at most n^2 iterations of the while loop. 
 +  * Proof:  Each iteration is a man proposing to a new woman.  There are only n^2 possible pairs of men and women and so only n^2 iterations at most. 
 + 
 +If m is free at some point, then there remains a free w 
 + 
 +The set S returned at termination is a perfect matching 
 +  * Proof:  At termination, m must have proposed to every woman, otherwise the while loop would not have exited.  This contradicts the statement that there can’t be a free man who has proposed to every woman. 
 + 
 +Consider and execution of the G-S algorithm that returns a set of pairs S.  The set S is a stable matching. 
 +  * Proof:  Because S is a perfect matching, an instability is assumed to create a contradiction.  We must prove that (i) m prefers w’ to w or (ii) w’ prefers m to m’.  Following the proven statements above, it concludes that S is a stable matching. 
 + 
 +All executions yield the same matching, set S*. 
 +  * Proof:  Suppose that an execution resulted in a match with a woman who is not his valid partner.  Since men propose in descending order of preference, this means the man is rejected because the woman has a preferred partner or he was left for a preferred partner.  Either way, w continues an engagement to a man.  Because the matching must adhere to the code of descending preference, this statement becomes a contradiction. 
 + 
 +In the stable matching S*, each woman is paired with her worst valid partner. 
 +  * Proof:  Suppose there were a pair (m, w) in S* where m is not the worst valid partner.  Then there is a stable matching with a man m’ who she likes less than m.  In this situation, m is paired with a woman w’; since w is the best valid partner of m, and w’ is a valid partner of m, we see that m prefers w to w’.  This shows an instability in the stable matching, leading to a contradiction. 
 + 
 +From this we can conclude that the gender doing the proposing ends up with a more desirable stable matching. 
  
-=== 1.2 - Five Representative Problems === 
courses/cs211/winter2014/journals/colin/chapter1.1389232929.txt.gz · Last modified: 2014/01/09 02:02 by mohnacsc
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