Discrete Optimization Lecture-11
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1 Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm 2011 () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
2 A scheduling problem Discussion A company owns 7 ships. During a 12 days period they are scheduled to visit 7 ports. See the schedule table in the next slide. Due to size limitations no two ships can dock on the same day in the same port. During this 12 days period every ship has to stop at some port and remain there for the remaining period for maintenance. So if ship S2 decides to stop on day 2 in port No. 3 it will not visit the remaining ports on its schedule and it will block the ship S3 which is not acceptable. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
3 We need to find a schedule to assign the ships to the ports according to the given schedule subject to the following rules: a. No two ships should be docked at the same port on the same day. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
4 We need to find a schedule to assign the ships to the ports according to the given schedule subject to the following rules: a. No two ships should be docked at the same port on the same day. b. Once a ship chooses a port to stop (according to its schedule) it stays there until the end of the period. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
5 We need to find a schedule to assign the ships to the ports according to the given schedule subject to the following rules: a. No two ships should be docked at the same port on the same day. b. Once a ship chooses a port to stop (according to its schedule) it stays there until the end of the period. c. No ship will interfere with the schedule of another ship; thus S5 cannot decide to stop at port 2 on day 1 because it will block S1 on day 2 or S3 on day 5. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
6 We need to find a schedule to assign the ships to the ports according to the given schedule subject to the following rules: a. No two ships should be docked at the same port on the same day. b. Once a ship chooses a port to stop (according to its schedule) it stays there until the end of the period. c. No ship will interfere with the schedule of another ship; thus S5 cannot decide to stop at port 2 on day 1 because it will block S1 on day 2 or S3 on day 5. Question This sounds like a matching problem, matching ships to ports with constraints. So how do we approach this problem? () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
7 We need to find a schedule to assign the ships to the ports according to the given schedule subject to the following rules: a. No two ships should be docked at the same port on the same day. b. Once a ship chooses a port to stop (according to its schedule) it stays there until the end of the period. c. No ship will interfere with the schedule of another ship; thus S5 cannot decide to stop at port 2 on day 1 because it will block S1 on day 2 or S3 on day 5. Question This sounds like a matching problem, matching ships to ports with constraints. So how do we approach this problem? () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
8 We need to find a schedule to assign the ships to the ports according to the given schedule subject to the following rules: a. No two ships should be docked at the same port on the same day. b. Once a ship chooses a port to stop (according to its schedule) it stays there until the end of the period. c. No ship will interfere with the schedule of another ship; thus S5 cannot decide to stop at port 2 on day 1 because it will block S1 on day 2 or S3 on day 5. Question This sounds like a matching problem, matching ships to ports with constraints. So how do we approach this problem? We can try a greedy selection: () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
9 A schedule sample A greedy selection will assign S6 to port 5 on day 12, S7 to port 6 on day 12, S5 to port 4 on day 11, S1 to port 7 on day 11 and so on? Day S1 S2 S3 S4 S5 S6 S7 1 * * 1 * 2 * * * 1 5 * * 3 3 * 5 * * * * * * 1 * 6 * 4 * * 3 7 * * * * 7 4 * * * * 11 7 * * * 12 * * * * * 5 6 () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
10 The stable marriage problem The stable marriage problem is a very famous combinatorial optimization problem with many applications. Google it and you will find hundreds of references. Basically, it is a matching problem with constraints. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
11 The stable marriage problem The stable marriage problem is a very famous combinatorial optimization problem with many applications. Google it and you will find hundreds of references. Basically, it is a matching problem with constraints. The origin of the problem is not clear. It seems that the first reference to the problem was an attempt to place medical interns in various departments in a hospital. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
12 The stable marriage problem The stable marriage problem is a very famous combinatorial optimization problem with many applications. Google it and you will find hundreds of references. Basically, it is a matching problem with constraints. The origin of the problem is not clear. It seems that the first reference to the problem was an attempt to place medical interns in various departments in a hospital. Here is what happened: Trung was placed in dermatology, but he prefered surgery. So he enquired with the surgery department, they checked the list of interns that were assigned to them and indeed found out that they prefered Trung on Hoang, so they dumped Hoang and offered Trung an internishp. Hoang was not terriby disappointed as he prefered gynecology. So he applied to them, they checked their list and found out that they liked Hoang better than Duy, so they dunped Duy and offered the internship to Hoang, so now Duy...you can imagine the rest of the story. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
13 What is a stable marriage? In 1962 David Gale and Lloyd Shapley asked whether it is possible to design an algorithm that will assign people to jobs that will be stable. That is if Duy will try to apply to surgery, surgery will not accept him becuase they prefer all their current recruits and if gynecology will try to recruit Hang she will not accept the invitation because she prefers pediatrics where she is currently accepted. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
14 What is a stable marriage? In 1962 David Gale and Lloyd Shapley asked whether it is possible to design an algorithm that will assign people to jobs that will be stable. That is if Duy will try to apply to surgery, surgery will not accept him becuase they prefer all their current recruits and if gynecology will try to recruit Hang she will not accept the invitation because she prefers pediatrics where she is currently accepted. A simple version of the problem gave it its name. Assume that we have n females and n men. Each person submits a list of preferences in which he/she lists all persons of opposite sex in order of preference. Definition A matching M is stable if it is perfect and there is no pair of matches f i m i and f j m j such that f i prefers m j over m i and m j prefers f i over f j. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
15 Example Let us match 4 couples. This is their list of preferences: 1 (3, 1, 2, 4) 1 (2, 4, 1, 3) 2 (1, 4, 3, 2) 2 (2, 1, 4, 3) 3 (3, 2, 1, 4) 3 (4, 1, 2, 3) 4 (3, 2, 4, 1) 4 (2, 1, 4, 3) () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
16 Example Let us match 4 couples. This is their list of preferences: 1 (3, 1, 2, 4) 1 (2, 4, 1, 3) 2 (1, 4, 3, 2) 2 (2, 1, 4, 3) 3 (3, 2, 1, 4) 3 (4, 1, 2, 3) 4 (3, 2, 4, 1) 4 (2, 1, 4, 3) Is the match (i, i) Stable? Try to find a a stable matching. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
17 The algorithm Given a list of preferences by n females and males does a stable marriage exist and can it be constructed efficiently. Comment Once again, we face a problem which is conceptually very simple. Try all permutations. For each permutation check whether it is stable. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
18 The algorithm Given a list of preferences by n females and males does a stable marriage exist and can it be constructed efficiently. Comment Once again, we face a problem which is conceptually very simple. Try all permutations. For each permutation check whether it is stable. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
19 The algorithm Given a list of preferences by n females and males does a stable marriage exist and can it be constructed efficiently. Comment Once again, we face a problem which is conceptually very simple. Try all permutations. For each permutation check whether it is stable. Easier said than done. Imagine doing it just for a small list of 50 pairs. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
20 The algorithm Given a list of preferences by n females and males does a stable marriage exist and can it be constructed efficiently. Comment Once again, we face a problem which is conceptually very simple. Try all permutations. For each permutation check whether it is stable. Easier said than done. Imagine doing it just for a small list of 50 pairs. The Gale-Shapley algorithm is an efficient implementation of an algorithm that also proves that a stable marriage exists regardless of the preference lists. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
21 The algorithm Given a list of preferences by n females and males does a stable marriage exist and can it be constructed efficiently. Comment Once again, we face a problem which is conceptually very simple. Try all permutations. For each permutation check whether it is stable. Easier said than done. Imagine doing it just for a small list of 50 pairs. The Gale-Shapley algorithm is an efficient implementation of an algorithm that also proves that a stable marriage exists regardless of the preference lists. During the algorihtm the men go thorugh three stages: a. Free. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
22 The algorithm Given a list of preferences by n females and males does a stable marriage exist and can it be constructed efficiently. Comment Once again, we face a problem which is conceptually very simple. Try all permutations. For each permutation check whether it is stable. Easier said than done. Imagine doing it just for a small list of 50 pairs. The Gale-Shapley algorithm is an efficient implementation of an algorithm that also proves that a stable marriage exists regardless of the preference lists. During the algorihtm the men go thorugh three stages: a. Free. b. engaged. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
23 The algorithm Given a list of preferences by n females and males does a stable marriage exist and can it be constructed efficiently. Comment Once again, we face a problem which is conceptually very simple. Try all permutations. For each permutation check whether it is stable. Easier said than done. Imagine doing it just for a small list of 50 pairs. The Gale-Shapley algorithm is an efficient implementation of an algorithm that also proves that a stable marriage exists regardless of the preference lists. During the algorihtm the men go thorugh three stages: a. Free. b. engaged. c. Matched. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
24 Gale-Shapley s algorithm () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
25 Gale-Shapley s algorithm 1 Initially all men and women are free. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
26 Gale-Shapley s algorithm 1 Initially all men and women are free. 2 Select a free man m i. Let him propose to a woman on top of his preference list among the women he has not proposed to yet. If she is free then the algorithm matches them as engaged. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
27 Gale-Shapley s algorithm 1 Initially all men and women are free. 2 Select a free man m i. Let him propose to a woman on top of his preference list among the women he has not proposed to yet. If she is free then the algorithm matches them as engaged. 3 If she is in a tentaive match with m k but she prefers m i then we engagedly match them and make m k free. Else m i remains free and we choose another man. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
28 Gale-Shapley s algorithm 1 Initially all men and women are free. 2 Select a free man m i. Let him propose to a woman on top of his preference list among the women he has not proposed to yet. If she is free then the algorithm matches them as engaged. 3 If she is in a tentaive match with m k but she prefers m i then we engagedly match them and make m k free. Else m i remains free and we choose another man. 4 When there are no more free men the algorihtm returns the list of engaged matches. This is a stable match. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
29 Analysis The algorithm has the following properties: 1 During the execution of the algorithm every man and female is in one of three stages: free, engaged or matched. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
30 Analysis The algorithm has the following properties: 1 During the execution of the algorithm every man and female is in one of three stages: free, engaged or matched. 2 Every person is engaged to at mots one other person. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
31 Analysis The algorithm has the following properties: 1 During the execution of the algorithm every man and female is in one of three stages: free, engaged or matched. 2 Every person is engaged to at mots one other person. 3 Once a female is engaged, she will remain engaged (not necessarily to the same male) till the end of the algorithm. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
32 Analysis The algorithm has the following properties: 1 During the execution of the algorithm every man and female is in one of three stages: free, engaged or matched. 2 Every person is engaged to at mots one other person. 3 Once a female is engaged, she will remain engaged (not necessarily to the same male) till the end of the algorithm. 4 Males can be engaged or become free. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
33 Analysis The algorithm has the following properties: 1 During the execution of the algorithm every man and female is in one of three stages: free, engaged or matched. 2 Every person is engaged to at mots one other person. 3 Once a female is engaged, she will remain engaged (not necessarily to the same male) till the end of the algorithm. 4 Males can be engaged or become free. 5 A step in the algorihtm is when m i proposes to w j (step number 2 in the G-S algorithm). () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
34 Analysis The algorithm has the following properties: 1 During the execution of the algorithm every man and female is in one of three stages: free, engaged or matched. 2 Every person is engaged to at mots one other person. 3 Once a female is engaged, she will remain engaged (not necessarily to the same male) till the end of the algorithm. 4 Males can be engaged or become free. 5 A step in the algorihtm is when m i proposes to w j (step number 2 in the G-S algorithm). 6 Every pair (m i, w j ) is considered at most once during the execution of the algorithm. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
35 Analysis The algorithm has the following properties: 1 During the execution of the algorithm every man and female is in one of three stages: free, engaged or matched. 2 Every person is engaged to at mots one other person. 3 Once a female is engaged, she will remain engaged (not necessarily to the same male) till the end of the algorithm. 4 Males can be engaged or become free. 5 A step in the algorihtm is when m i proposes to w j (step number 2 in the G-S algorithm). 6 Every pair (m i, w j ) is considered at most once during the execution of the algorithm. 7 At the end of the algorithm everyone will be engaged. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
36 Analysis Discussion () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
37 Analysis Discussion A1. Since there are n 2 distinct pairs and each pair is considered at most once, the algorithn executes at most n 2 steps, in other words, it is efficient. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
38 Analysis Discussion A1. Since there are n 2 distinct pairs and each pair is considered at most once, the algorithn executes at most n 2 steps, in other words, it is efficient. A2. By the end of the algorithm, if a free man remains, there will also be a free woman. So he will be able to propose to her. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
39 Analysis Discussion A1. Since there are n 2 distinct pairs and each pair is considered at most once, the algorithn executes at most n 2 steps, in other words, it is efficient. A2. By the end of the algorithm, if a free man remains, there will also be a free woman. So he will be able to propose to her. A3. We still need to prove that a stable matching will be produced by the algorithm. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
40 Analysis Discussion A1. Since there are n 2 distinct pairs and each pair is considered at most once, the algorithn executes at most n 2 steps, in other words, it is efficient. A2. By the end of the algorithm, if a free man remains, there will also be a free woman. So he will be able to propose to her. A3. We still need to prove that a stable matching will be produced by the algorithm. Theorem The gale-shapley algorithm produces a stable marriage. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
41 We can prove this by induction. Remark We first note that if there are k stable engagements and m k+1 proposes to w and w is free, then adding the pair (m k+1, w) cannot cause an instability because none of the men on w s preference list that she prefers over m k+1 proposed to her earlier, or she would not be free. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
42 Chứng minh. Induction hypothesis: When the algorithm engages k pairs, the engagements are stable. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
43 Chứng minh. Induction hypothesis: When the algorithm engages k pairs, the engagements are stable. 1 Obviously true for k = 1. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
44 Chứng minh. Induction hypothesis: When the algorithm engages k pairs, the engagements are stable. 1 Obviously true for k = 1. 2 Assume true for k = m and we now execute step 2 in the algorithm. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
45 Chứng minh. Induction hypothesis: When the algorithm engages k pairs, the engagements are stable. 1 Obviously true for k = 1. 2 Assume true for k = m and we now execute step 2 in the algorithm. 3 m k+1 is proposing to w. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
46 Chứng minh. Induction hypothesis: When the algorithm engages k pairs, the engagements are stable. 1 Obviously true for k = 1. 2 Assume true for k = m and we now execute step 2 in the algorithm. 3 m k+1 is proposing to w. 4 If she is free then they become engaged and clearly there is no in-stability in this engagement. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
47 Chứng minh. Induction hypothesis: When the algorithm engages k pairs, the engagements are stable. 1 Obviously true for k = 1. 2 Assume true for k = m and we now execute step 2 in the algorithm. 3 m k+1 is proposing to w. 4 If she is free then they become engaged and clearly there is no in-stability in this engagement. 5 If she is engaged and prefers m k+1 she will become engaged to him. So we still have only k stable engagements. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
48 Chứng minh. Induction hypothesis: When the algorithm engages k pairs, the engagements are stable. 1 Obviously true for k = 1. 2 Assume true for k = m and we now execute step 2 in the algorithm. 3 m k+1 is proposing to w. 4 If she is free then they become engaged and clearly there is no in-stability in this engagement. 5 If she is engaged and prefers m k+1 she will become engaged to him. So we still have only k stable engagements. 6 Since every male has all n females on his list of preferences and only k females are engaged, this process cannot cycle indefinitely. Eventually the k + 1 attempt to add a stable engagement wil succeed by finding a free female in his list. () Discrete Optimization Lecture-11 Ngày 2 tháng 11 năm / 12
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