1. IOI/ACM ICPC Training
4 June 2005

2. Euler Tour
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3. Can we draw the following figure in one line?

4. Can we draw the following figure in one line?
Fail!

5. How to determine if a graph can be drawn in one line?
Theorem:
A graph G has an Euler path if and only if
Exactly two vertices u satisfies: |indegree(u)-outdegree(u)| = 1
All other vertices v satisfies: indegree(v)=outdegree(v) 3 2 3 3 3 3

6. Time analysis
We can compute the degrees of all vertices in O(|E|) time.
Then, it requires O(|V|) time to check if the vertices satisfies the previous theorem.
In total, O(|V|+|E|) time.

7. Topological sort
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8. Topological order b d a c e
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

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

10. 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;

11. Example b d e c a b e c a b c a b a a d has no successor! Choose d!
Both b and e have no successor! Choose e! b e c a
Both b and c have no successor! Choose c! b c a
Only b has no successor! Choose b! b a a
Choose a!
The topological order is a,b,c,e,d

12. 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|).

13. 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;

14. Spanning Tree
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15. Spanning Tree
Given a connected undirected graph G, a spanning tree of G is a subgraph of G that contains all of G’s nodes and enough of its edges to form a tree. v2 v3 v1
Tell the students some application of spanning tree. v5 v4
Spanning tree
Spanning tree is not unique!

16. DFS spanning tree
Generate the spanning tree edge during the DFS traversal.
Algorithm dfsSpanningTree(v)
mark v as visited;
for (each unvisited node u adjacent to v) {
mark the edge from u to v;
dfsSpanningTree(u); }

17. Example of generating spanning tree based on DFS
stack v3 v2
v3, v2 v1
v3, v2, v1
backtrack v4
v3, v2, v4 v5
v3, v2, v4 , v5
empty v2 v3 v1 x x x x x v5 v4 G

18. BFS spanning tree
Similar to DFS, the spanning tree edges can be generated based on BFS traversal.

19. Minimum Spanning Tree
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20. Telephone line placing problem
Consider a connected undirected graph where
Each node x represents a country x
Each edge (x, y) has a number which measures the cost of placing telephone line between country x and country y
Problem: connecting all countries while minimizing the total cost
Solution: find a spanning tree with minimum total weight, that is, minimum spanning tree

21. Formal definition of minimum spanning tree
Given a connected undirected graph G.
Let T be a spanning tree of G.
cost(T) = eTweight(e)
The minimum spanning tree is a spanning tree T which minimizes cost(T) v1 v4 v3 v5 v2 5 2 3 7 8 4
Minimum
spanning
tree

22. Prim’s algorithm v2 v1 v4 v3 v5 5 2 3 7 8 4 v2 v1 v4 v3 v5 5 2 3 7 8 4
Start from v5, find the minimum edge attach to v5 v2 v1 v4 v3 v5 5 2 3 7 8 4
Find the minimum edge attach to v3 and v5 v2 v1 v4 v3 v5 5 2 3 7 8 4
Find the minimum edge attach to v2, v3 and v5 v2 v1 v4 v3 v5 5 2 3 7 8 4 v2 v1 v4 v3 v5 5 2 3 7 8 4 v2 v1 v4 v3 v5 5 2 3 7 8 4
Find the minimum edge attach to v2, v3 , v4 and v5

23. Prim’s algorithm (II) Algorithm PrimAlgorithm(v)
Mark node v as visited and include it in the minimum spanning tree;
while (there are unvisited nodes) {
find the minimum edge (v, u) between a visited node v and an unvisited node u;
mark u as visited;
add both v and (v, u) to the minimum spanning tree; }

24. Bipartite Matching
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25. Unweighted Bipartite Matching

26. Definitions
Matching
Free Vertex

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

28. Definition
Note that maximum matching is not unique.

29. 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!

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

31. Alternating Path Alternating between matching and non-matching edges.f g h i j
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

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

33. 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!)

34. 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.

35. 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

36. Labeling Algorithm
Start with arbitrary matching

37. Labeling Algorithm
Pick a free vertex in the bottom

38. Labeling Algorithm
Run BFS

39. Labeling Algorithm
Alternate unmatched/matched edges

40. Labeling Algorithm
Until a augmenting path is found

41. Augmenting Tree

42. Flip!

43. Repeat
Pick another free vertex in the bottom

44. Repeat
Run BFS

45. Repeat
Flip

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

47. 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.

48. 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|)

49. 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|)

50. Weighted Bipartite Graph3 4 6 6

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

52. Maximum Weighted Matching
Score: =13 3 4 6 6

53. 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 3 4 6 6
Note: augmenting path need not start and end at free vertices!

54. 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.

55. 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.

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

57. Stable Marriage Problem
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58. 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.

59. Stable? A set of marriages is unstable if
two persons who are not married, but both of them 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 A B C D E 2 5 1 3 4

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

61. Naïve solution (2) Does not work! E.g. A1 B3 C2 D4 E5 A2 B3 C1 D4 E5

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

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

64. Time analysis
If there are N men and N women,
O(N2) time

65. Planar Graph
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66. Planar graph
A graph is planar if we can draw the graph without crossing. K4 K4

67. Non-planar
Below shows two non-planar graphs
K3,3 K5

68. Theorem for planar graph
A graph G is non-planar
if and only if
G contains K5 or K3,3.

69. Question How to check if a graph contains cycle or not?
Consider a set S of fixed length sequences, says, {ACG, ATC, CAT, CGC, GCA}. S represents the string ACGCATC.
Given any set S, can you give an algorith to determine if it represents a string or not?

70. Questions
Please give an O(|V|+|E|) time algorithm to recover the Euler path
Implement the spanning tree algorithm
Please implement the labeling algorithm for maximum matching.
Please give the pseudo-code for the stable marriage problem. Implement it.