1. All Pair Shortest Path IOI/ACM ICPC Training June 2004

2. All Pair Shortest Path Note: Dijkstra ’ s Algorithm takes O((V+E)logV) time All Pair Shortest Path Problem can be solved by executing Dijkstra ’ s Algorithm |V| times Running Time: O(V(V+E)log V) Floyd-Warshall Algorithm: O(V 3 )

3. Idea Label the vertices with integers 1..n Restrict the shortest paths from i to j to consist of vertices 1..k only (except i and j) Iteratively relax k from 1 to n. i k j

4. Definition Find shortest distance from i to j using vertices 1.. k only i k j

5. Example 1 2 4 35 2 4 1 1 5 3 1 1 3

6. i = 4, j = 5, k = 0 1 2 4 35 2 4 1 1 5 3 1 1 3

7. i = 4, j = 5, k = 1 1 2 4 35 2 4 1 1 5 3 1 1 3

8. i = 4, j = 5, k = 2 1 2 4 35 2 4 1 1 5 3 1 1 3

9. i = 4, j = 5, k = 3 1 2 4 35 2 4 1 1 5 1 1 1 3

10. Idea k j i

11. The tables i j k=4 k=5 i j k k

12. The code for i = 1 to |V| for j = 1 to |V| a[i][j][0] = cost(i,j) for k = 1 to |V| for i = 1 to |V| for j = 1 to |V| a[i][j][k] = min( a[i][j][k-1], a[i][k][k-1] + a[k][j][k-1])

13. Topological sort IOI/ACM ICPC Training June 2004

14. Topological order Consider the prerequisite structure for courses: Each node x represents a course x (x, y) represents that course x is a prerequisite to course y Note that this graph should be a directed graph without cycles. A linear order to take all 5 courses while satisfying all prerequisites is called a topological order. E.g.  a, c, b, e, d  c, a, b, e, d b d e c a

15. Topological sort Arranging all nodes in the graph in a topological order Applications:  Schedule tasks associated with a project

16. Topological sort algorithm Algorithm topSort1 n = |V|; Let R[0..n-1] be the result array; for i = 1 to n { select a node v that has no successor; R[n-i] = v; delete node v and its edges from the graph; } return R;

17. Example b d e c a 1.d has no successor! Choose d! a 5.Choose a! The topological order is a,b,c,e,d 2.Both b and e have no successor! Choose e! b e c a 3.Both b and c have no successor! Choose c! b c a 4.Only b has no successor! Choose b! b a

18. Time analysis Finding a node with no successor takes O(|V|+|E|) time. We need to repeat this process |V| times. Total time = O(|V| 2 + |V| |E|). We can implement the above process using DFS. The time can be improved to O(|V| + |E|).

19. Algorithm based on DFS Algorithm topSort2 s.createStack(); for (all nodes v in the graph) { if (v has no predecessors) { s.push(v); mark v as visited; } while (s is not empty) { let v be the node on the top of the stack s; if (no unvisited nodes are children to v) { // i.e. v has no unvisited successor aList.add(1, v); s.pop(); // blacktrack } else { select an unvisited child u of v; s.push(u); mark u as visited; } return aList;

20. Bipartite Matching IOI/ACM ICPC Training June 2004

21. Unweighted Bipartite Matching

22. Definitions Matching Free Vertex

23. Definitions Maximum Matching: matching with the largest number of edges

24. Definition Note that maximum matching is not unique.

25. Intuition Let the top set of vertices be men Let the bottom set of vertices be women Suppose each edge represents a pair of man and woman who like each other Maximum matching tries to maximize the number of couples!

26. Applications Matching has many applications. For examples,  Comparing Evolutionary Trees  Finding RNA structure  … This lecture lets you know how to find maximum matching.

27. Alternating Path Alternating between matching and non-matching edges. a b cde f g hij d-h-e: alternating path a-f-b-h-d-i: alternating path starts and ends with free vertices f-b-h-e: not alternating path e-j: alternating path starts and ends with free vertices

28. Idea “Flip” augmenting path to get better matching Note: After flipping, the number of matched edges will increase by 1! 

29. Idea Theorem (Berge 1975): A matching M in G is maximum iff There is no augmenting path Proof: (  ) If there is an augmenting path, clearly not maximum. (Flip matching and non-matching edges in that path to get a “better” matching!)

30. Proof for the other direction (  ) Suppose M is not maximum. Let M’ be a maximum matching such that |M’|>|M|. Consider H = M  M’ = (M  M’)-(M  M’) i.e. a set of edges in M or M’ but not both H has two properties:  Within H, number of edges belong to M’ > number of edges belong to M.  H can be decomposed into a set of paths. All paths should be alternating between edges in M and M’. There should exist a path with more edges from M’. Also, it is alternating.

31. Idea of Algorithm Start with an arbitrary matching While we still can find an augmenting path  Find the augmenting path P  Flip the edges in P

32. Labelling Algorithm Start with arbitrary matching

33. Labelling Algorithm Pick a free vertex in the bottom

34. Labelling Algorithm Run BFS

35. Labelling Algorithm Alternate unmatched/matched edges

36. Labelling Algorithm Until a augmenting path is found

37. Augmenting Tree

38. Flip!

39. Repeat Pick another free vertex in the bottom

40. Repeat Run BFS

41. Repeat Flip

42. Answer Since we cannot find any augmenting path, stop!

43. Overall algorithm Start with an arbitrary matching (e.g., empty matching) Repeat forever  For all free vertices in the bottom, do bfs to find augmenting paths  If found, then flip the edges  If fail to find, stop and report the maximum matching.

44. Time analysis We can find at most |V| augmenting paths (why?) To find an augmenting path, we use bfs! Time required = O( |V| + |E| ) Total time: O(|V| 2 + |V| |E|)

45. Improvement We can try to find augmenting paths in parallel for all free nodes in every iteration. Using such approach, the time complexity is improved to O(|V| 0.5 |E|)

46. Weighted Bipartite Graph 4 6 6 3

47. Weighted Matching 4 6 6 3 Score: 6+3+1=10

48. Maximum Weighted Matching 4 6 6 3 Score: 6+1+1+1+4=13

49. Augmenting Path (change of definition) Any alternating path such that total score of unmatched edges > that of matched edges The score of the augmenting path is  Score of unmatched edges – that of matched edges 4 6 6 3 Note: augmenting path need not start and end at free vertices!

50. Idea for finding maximum weight matching Theorem: Let M be a matching of maximum weight among matchings of size |M|.  If P is an augmenting path for M of maximum weight,  Then, the matching formed by augmenting M by P is a matching of maximum weight among matchings of size |M|+1.

51. Overall Algorithm Start with an empty matching Repeat forever  Find an augmenting path P with maximum score  If the score > 0, then flip the edges  Otherwise, stop and report the maximum weight matching.

52. Time analysis The same! Time required = O(|V| 2 + |V| |E|)

53. Stable Marriage Problem IOI/ACM ICPC Training June 2004

54. Stable Marriage Problem Given N men and N women, each person list in order of preference all the people of the opposite sex who would like to marry. Problem:  Engage all the women to all the men in such a way as to respect all their preferences as much as possible.

55. Stable? A set of marriages is unstable if  two people who are not married both prefer each other than their spouses E.g. Suppose we have A1 B3 C2 D4 E5. This is unstable since  A prefer 2 more than 1  2 prefer A more than C ABCDE 2 5 1 3 4 1 2 3 4 5 2 3 5 4 1 1 3 2 4 5 5 3 2 1 4 1 E A D B C 2 D E B A C 3 A D B C E 4 C B D A E 5 D B C E A

56. Naïve solution Starting from a feasible solution. Check if it is stable.  If yes, done!  If not, remove an unstable couple. Is this work?

57. Naïve solution (2) Does not work! E.g.  A1 B3 C2 D4 E5  A2 B3 C1 D4 E5  A3 B2 C1 D4 E5  A3 B1 C2 D4 E5 ABCDE 2 5 1 3 4 1 2 3 4 5 2 3 5 4 1 1 3 2 4 5 5 3 2 1 4 1 E A D B C 2 D E B A C 3 A D B C E 4 C B D A E 5 D B C E A

58. Solution 1. Let X be the first man. 2. X proposes to the best woman in the remaining on his list. (Initially, the first woman on his list!) 3. If α is not engaged  Pair up (X, α). Then, set X=next man and goto 1. 4. If α prefers X more than her fiancee Y,  Pair up (X, α). Then, set X=Y and goto 1. 5. Goto 1

59. Example ABCDE 2 5 1 3 4 1 2 3 4 5 2 3 5 4 1 1 3 2 4 5 5 3 2 1 4 1 E A D B C 2 D E B A C 3 A D B C E 4 C B D A E 5 D B C E A ABCDE 2 5 1 1 2 3 4 2 3 5 1 3 5 3 2

60. Time analysis If there are N men and N women,  O(N 2 ) time

61. Algorithm prefer[m][s]=w means the woman w is on the s- th position in the preference list of the man m Let next[m] be the current best woman in his remaining list. (Initially, next[m]=0) fiancee[w]=m means the man m engaged to woman w. (Initially, fiancee[w]=0) Let rank[w][m] is the ranking of the man m in the preference list of the woman w. For(m=1;m<=N;m++) { For(s=m;s!=0; }