1.1 First Problem: Stable Matching. Six medical students and four hospitals. Student Preferences. s6 h3 h1 h4 h2. Stable Matching Problem

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1 //0 hapter Introduction: Some Representative Problems Slides by Kevin ayne. opyright 00 Pearson-ddison esley. ll rights reserved.. First Problem: Stable Matching Six medical students and four hospitals Student Preferences s h h h h Hospital Preferences s h h h h h s s s s s s s h h h h h s s s s s s s h h h h h s s s s s s s h h h h h s s s s s s s h h h h Matching Residents to Hospitals Goal. Given a set of preferences among hospitals and medical school students, design a self-reinforcing admissions process. Unstable pair: applicant x and hospital y are unstable if: x prefers y to x s assigned hospital. y prefers x to one of its admitted students. Stable assignment: ssignment with no unstable pairs. Natural and desirable condition. Individual self-interest will prevent any applicant/hospital deal from being made. Simplification of the Problem: ach student will find a hospital (create a no-pay hospital) ach student will find a different hospital (duplicate hospitals) It becomes a boy-girl matching problem. Perfect matching: n men and n women, everyone is matched to a unique one: ach man gets exactly one woman. ach woman gets exactly one man. Goal. Given n men and n women, find a "suitable" matching. Participants rate members of opposite sex. ach man lists women in order of preference from best to worst. ach woman lists men in order of preference from best to worst. Q. Is assignment -, -, - stable? Stability: no incentive for some pair of participants to undermine assignment. In matching M, an unmatched pair m-w is unstable if man m and woman w prefer each other to current partners. Unstable pair m-w could each improve by eloping. favorite least favorite st nd rd avier my ertha lare favorite least favorite st nd rd my ancey avier eus st nd rd avier my ertha lare st nd rd my ancey avier eus Stable matching: perfect matching with no unstable pairs. Stable matching problem. Given the preference lists of n men and n women, find a stable matching if one exists. ancey ertha my lare eus my ertha lare ertha avier ancey eus lare avier ancey eus ancey ertha my lare eus my ertha lare ertha avier ancey eus lare avier ancey eus

2 //0 Q. Is assignment -, -, - stable?. No. ertha and avier will hook up. Q. Is assignment -, -, - stable?. No. ertha and avier will hook up. Q. How about switch and? Q. Is assignment -, -, - stable?. No. ertha and avier will hook up. Q. How about switch and?. No. my and avier (or ancey) will hook up. st nd rd st nd rd avier my ertha lare ancey ertha my lare eus my ertha lare my ancey avier eus ertha avier ancey eus lare avier ancey eus st nd rd st nd rd avier my ertha lare ancey ertha my lare eus my ertha lare my ancey avier eus ertha avier ancey eus lare avier ancey eus st nd rd st nd rd avier my ertha lare ancey ertha my lare eus my ertha lare my ancey avier eus ertha avier ancey eus lare avier ancey eus 8 9 Propose-nd-Reject lgorithm Proof of orrectness: Termination Q. Is assignment -, -, - stable?. es. st nd rd st nd rd avier my ertha lare ancey ertha my lare eus my ertha lare my ancey avier eus ertha avier ancey eus lare avier ancey eus 0 Propose-and-reject algorithm. [Gale-Shapley 9] Intuitive method that guarantees to find a stable matching. Initialize each person to be free. while (some man is free and hasn't proposed to every woman) { hoose such a man m w = st woman on m's list to whom m has not yet proposed if (w is free) assign m and w to be engaged else if (w prefers m to her fiancé m') assign m and w to be engaged, and m' to be free else w rejects m st nd rd st nd rd avier my ertha lare ancey ertha my lare eus my ertha lare my ancey avier eus ertha avier ancey eus lare avier ancey eus Observation. Men propose to women in decreasing order of preference. Observation. Once a woman is matched, she never becomes unmatched; she only "trades up." laim. lgorithm terminates after at most n iterations of while loop. Pf. ach time through the while loop a man proposes to a new woman. There are only n possible proposals. ictor yatt avier ancey eus st nd rd th th my ertha lare iane rika st n(n-) + proposals required nd rd th th

3 //0 Proof of orrectness: Perfection Proof of orrectness: Stability Summary laim. ll men and women get matched. Suppose, for sake of contradiction, that eus is not matched upon termination of algorithm. Then some woman, say my, is not matched upon termination. y Observation, my was never proposed to. ut, eus proposes to everyone, since he ends up unmatched. laim. No unstable pairs. Suppose - is an unstable pair: each prefers each other to partner in Gale-Shapley matching S*. men propose in decreasing ase : never proposed to. order of preference prefers his GS partner to. - is stable. ase : proposed to. rejected (right away or later) prefers her GS partner to. - is stable. women only trade up S* my-ancey ertha-eus Stable matching problem. Given n men and n women, and their preferences, find a stable matching if one exists. Gale-Shapley algorithm. Guarantees to find a stable matching for any problem instance. Q. How to implement GS algorithm efficiently? Q. If there are multiple stable matchings, which one does GS find? In either case - is stable, a contradiction. fficient Implementation fficient implementation. e describe O(n ) time implementation. Representing men and women. ssume men are named,, n. ssume women are named ',, n'. ngagements. Maintain a list of free men, e.g., in a queue. Maintain two arrays wife[m], and husband[w]. set entry to 0 if unmatched if m matched to w then wife[m]=w and husband[w]=m Men proposing. For each man, maintain a list of women, ordered by preference. Maintain an array count[m] that counts the number of proposals made by man m. fficient Implementation omen rejecting/accepting. oes woman w prefer man m to man m'? For each woman, create inverse of preference list of men: pref[w][j] = m iff women w s j th preference is m iff inverse[w][m] = j onstant time access for each query after O(n ) preprocessing. my Pref st 8 nd rd th th th th 8 th my 8 th 8 th nd th Inverse rd th th st for w = to n for j = to n inverse[w][pref[w][j]] = j my () prefers man to since inverse[][] < inverse[][] Initialize each person to be free. Propose-nd-Reject lgorithm while (some man is free and hasn't proposed to every woman) { hoose such a man m w = st woman on m's list to whom m has not yet proposed if (w is free) assign m and w to be engaged else if (w prefers m to her fiancé m') assign m and w to be engaged, and m' to be free else w rejects m Initialize Mpref[][], inverse[][], stack, count[]; while (stack!= empty) { m = pop(stack); w = Mpref[m][++count[m]]; if (husband[w] == 0) { wife[m] = w; husband[w] = m; else if (inverse[w][m] < inverse[w][husband[w]]) { push(husband[w], stack); wife[husband[w]] = 0; wife[m] = w; husband[w] = m; else push(m, stack); 8

4 //0 Understanding the Solution Understanding the Solution Man Optimality Q. For a given problem instance, there may be several stable matchings. o all executions of Gale-Shapley yield the same stable matching? If so, which one? n instance with two stable matchings. -, -, -. -, -, -. avier ancey eus st nd rd my ertha lare st nd rd Q. For a given problem instance, there may be several stable matchings. o all executions of Gale-Shapley yield the same stable matching? If so, which one? ef. Man m is a valid partner of woman w if there exists some stable matching in which they are matched. Man-optimal assignment. ach man receives best valid partner. laim. ll executions of GS yield man-optimal assignment, which is a stable matching! No reason a priori to believe that man-optimal assignment is perfect, let alone stable. Simultaneously best for each and every man. laim. GS matching S* is man-optimal. Suppose some man is paired with someone other than best partner. Men propose in decreasing order of preference some man is rejected by valid partner. S Let be first such man, and let be first valid my-ancey woman that rejects him. Let S be a stable matching where and are matched. ertha-eus hen is rejected, forms (or reaffirms) engagement with a man, say, whom she prefers to. Let be 's partner in S. not rejected by any valid partner at the point when is rejected by. Thus, prefers to. since this is first rejection ut prefers to. by a valid partner Thus - is unstable in S. 9 0 Stable Matching Summary oman Pessimality xtensions: Matching Residents to Hospitals Stable matching problem. Given preference profiles of n men and n women, find a stable matching. no man and woman prefer to be with each other than assigned partner Gale-Shapley algorithm. Finds a stable matching in O(n ) time. Man-optimality. In the version of GS where men propose, each man receives best valid partner. w is a valid partner of m if there exist some stable matching where m and w are paired Q. oes man-optimality come at the expense of the women? oman-pessimal assignment. ach woman receives worst valid partner. laim. GS finds woman-pessimal stable matching S*. Pf. Suppose - matched in S*, but is not worst valid partner for. There exists stable matching S in which is paired with a man, say, whom she likes less than. Let be 's partner in S. S prefers to. man-optimality my-ancey Thus, - is an unstable in S. ertha-eus x: Men hospitals, omen med school residents. ariant. Some participants declare others as unacceptable. resident unwilling to ariant. Unequal number of men and women. work in leveland ariant. Limited polygamy. hospital wants to hire residents ef. Matching S unstable if there is a hospital h and resident r such that: h and r are acceptable to each other; and either r is unmatched, or r prefers h to her assigned hospital; and either h does not have all its places filled, or h prefers r to at least one of its assigned residents.

5 //0 xtension: Stable Roommate Problem Independent Set Q. o stable matchings always exist?. Not obvious a priori.. Five Representative Problems Input. Graph. Goal. Find maximum cardinality independent set. Stable roommate problem. n people; each person ranks others from to n-. ssign roommate pairs so that no unstable pairs. subset of nodes such that no two joined by an edge dam ob hris st nd rd -, - - unstable -, - - unstable -, - - unstable oofus Observation. Stable matchings do not always exist for stable roommate problem. One of NP-complete problems Independent Set xponential (ch. 8) Polynomial : O(n k ) Input. Graph. Goal. Find maximum cardinality independent set. subset of nodes such that no two joined by an edge Independent set. Given a graph, what is maximum size of an independent set? O(n n ) solution. numerate all subsets. Independent set of size k. Given a graph, are there k nodes such that no two are joined by an edge? k is a constant O(n k ) solution. numerate all subsets of k nodes. One of NP-complete problems xhaustive Search: G = (, ), n = for k = n downto lookforindependentset(g, k) lookforindependentset(g, k) for all subset S of, S = k, checkindepedent(g, S) omplexity of lookforindependentset: O((n, k)*k ), where (n, k) = n!/((n-k)!k!) = O(n k ) S* foreach subset S of nodes { check whether S in an independent set if (S is largest independent set seen so far) update S* S foreach subset S of k nodes { check whether S in an independent set if (S is an independent set) report S is an independent set heck whether S is an independent set = O(k ). Number of k element subsets = O(k n k / k!) = O(n k ). poly-time for k=, but not practical 8 9 0

6 //0 Interval Scheduling Interval Scheduling: Greedy lgorithm (ch..) Interval Scheduling: nalysis Input. Set of jobs with start times and finish times. Goal. Find maximum cardinality subset of mutually compatible jobs. b a c d e f jobs don't overlap Interval Scheduling can be reduced to Independent Set: reate G = (, ), where is the set of jobs and = { (j, j ) j and j overlap. Greedy algorithm. onsider jobs in increasing order of finish time. Take each job provided it's compatible with the ones already taken. Sort jobs by finish times so that f f... f n. set of jobs selected for j = to n { if (job j compatible with ) {j return Theorem. Greedy algorithm is optimal. ssume greedy is not optimal, and let's see what happens. Let i, i,... i k denote set of jobs selected by greedy. Let j, j,... j m denote set of jobs in the optimal solution with i = j, i = j,..., i r = j r for the largest possible value of r. Greedy: i i i r i r+ job i r+ finishes before j r+ g h Implementation. O(n log n). Remember job j* that was added last to. Job j is compatible with if s j f j*. OPT: j j j r j r+ why not replace job j r+ with job i r+? Interval Scheduling: nalysis eighted Interval Scheduling. eighted Interval Scheduling Theorem. Greedy algorithm is optimal. ssume greedy is not optimal, and let's see what happens. Let i, i,... i k denote set of jobs selected by greedy. Let j, j,... j m denote set of jobs in the optimal solution with i = j, i = j,..., i r = j r for the largest possible value of r. Greedy: OPT: i i i r i r+ j j j r job i r+ finishes before j r+ i r+ solution still feasible and optimal, but contradicts maximality of r. Input. Set of jobs with start times, finish times, and weights. Goal. Find maximum weight subset of mutually compatible jobs an be reduced to Independent Set of Maximum eight. 0 eighted interval scheduling problem. Job j starts at s j, finishes at f j, and has weight or value v j. Two jobs compatible if they don't overlap. Goal: find maximum weight subset of mutually compatible jobs. a b c d e f g h

7 //0 Unweighted Interval Scheduling Review eighted Interval Scheduling (ch.) ynamic Programming: inary hoice Recall. Greedy algorithm works if all weights are. onsider jobs in ascending order of finish time. dd job to subset if it is compatible with previously chosen jobs. Notation. Label jobs by finishing time: f f f n. ef. p(j) = largest index i < j such that job i is compatible with j. x: p(8) =, p() =, p() = 0. Notation. OPT(j) = value of optimal solution to the problem consisting of job requests,,..., j. ase : OPT selects job j. collect profit v j can't use incompatible jobs { p(j) +, p(j) +,..., j - Observation. Greedy algorithm can fail spectacularly if arbitrary weights are allowed. must include optimal solution to problem consisting of remaining compatible jobs,,..., p(j) optimal substructure ase : OPT does not select job j. must include optimal solution to problem consisting of remaining compatible jobs,,..., j- weight = 999 b weight = a eighted Interval Scheduling: rute Force rute force algorithm. Input: n, s,,s n, f,,f n, v,,v n Sort jobs by finish times so that f f... f n. ompute p(), p(),, p(n) ompute-opt(j) { if (j = 0) return 0 else return max(v j + ompute-opt(p(j)), ompute-opt(j-)) eighted Interval Scheduling: rute Force Observation. Recursive algorithm fails spectacularly because of redundant sub-problems exponential algorithms. x. Number of recursive calls for family of "layered" instances grows like Fibonacci sequence. p() = 0, p(j) = j eighted Interval Scheduling: Memoization Memoization. Store results of each sub-problem in a cache; lookup as needed. Input: n, s,,s n, f,,f n, v,,v n Sort jobs by finish times so that f f... f n. ompute p(), p(),, p(n) for j = to n M[j] = empty M[0] = 0 M-ompute-Opt(j) { if (M[j] is empty) M[j] = max(v j + M-ompute-Opt(p(j)), M-ompute-Opt(j-)) return M[j] 0

8 //0 eighted Interval Scheduling: Running eighted Interval Scheduling: Finding a Solution eighted Interval Scheduling: ottom-up laim. Memorized version of algorithm takes O(n log n) time. Sort by finish time: O(n log n). omputing p( ) : O(n log n) via sorting by start time. M-ompute-Opt(j): each invocation takes O() time and either (i) returns an existing value M[j] (ii) fills in one new entry M[j] and makes two recursive calls Progress measure = # nonempty entries of M[]. initially = 0, throughout n. (ii) increases by at most n recursive calls. Overall running time of M-ompute-Opt(n) is O(n). Remark. O(n) if jobs are pre-sorted by finish times. Q. ynamic programming algorithms computes optimal value. hat if we want the solution itself?. o some post-processing. Run M-ompute-Opt(n) Run Find-Solution(n) Find-Solution(j) { if (j = 0) output nothing else if (v j + M[p(j)] > M[j-]) print j Find-Solution(p(j)) else Find-Solution(j-) # of recursive calls n O(n). ottom-up dynamic programming. Unwind recursion. Input: n, s,,s n, f,,f n, v,,v n Sort jobs by finish times so that f f... f n. ompute p(), p(),, p(n) Iterative-ompute-Opt { M[0] = 0 for j = to n M[j] = max(v j + M[p(j)], M[j-]) ipartite Matching ipartite Matching. ompetitive Facility Location Input. ipartite graph. Goal. Find maximum cardinality matching. Input. ipartite graph. Goal. Find maximum cardinality matching. Input. Graph with weight on each each node. Game. Two competing players alternate in selecting nodes. Not allowed to select a node if any of its neighbors have been selected. Matching can be reduced to Independent Set: Given a graph G(, ), create another graph G (, ), where = { (e, e ) e and e share one endpoint in G, then M is maximum matching of G iff M is a maximum independent set of G. onventional Method: Use the Max-Flow lgorithm (ch..) O(mn) Research Problem: an we use the Stable Matching lgorithm? O(n ) Goal. Select a maximum weight subset of nodes. (ch. 9) 0 0 Second player can guarantee 0, but not. 8 8

9 //0 Five Representative Problems ariations on a theme: independent set. Interval scheduling: n log n greedy algorithm. eighted interval scheduling: n log n dynamic programming algorithm. ipartite matching: mn max-flow based algorithm. Independent set: NP-complete. ompetitive facility location: PSP-complete. 9 9