1. Fast FAST By Noga Alon, Daniel Lokshtanov And Saket Saurabh Presentation by Gil Einziger

2. Fast FAST? Fast – A relative fast algorithm for an NP- Complete problem. FAST – (minimal) Feedback Arc Set in Tournaments.

3. Feedback Arc Set Given a Directed Graph We want to find a set of arcs such as: is a DAG. We want F to be minimal, how?

4. Feedback Arc Set – The Problem: 1.NP Complete – but we can expect it. 2.In a general Directed graph, un-weighted Feedback arc set is APX- Hard – meaning that there exist a constant k such as there is no polynomial time k approximation algorithm for this problem.

5. What is A Tournament? A tournament is a directed graph (digraph) obtained by assigning a direction for each edge in an undirected complete graph.directed graph undirectedcomplete graph A tournament is, a directed graph in which every pair of vertices is connected by a single directed edge.

6. Tournament Important Observations Let T(V,A) be a Tournament 1. for is a Tournament 2. if a Tournament is DAG, it have a unique topological order. 1-2 3 Not a tournament, 2 possible topological orders… 1 3 Tournament, only one topological order… 2 1-2

7. Tournament and FAS Assume we have n tennis players. each tennis player is playing 1 game against all other tennis player. How can we decide who the best tennis player is? How can we rank the players?

8. Tournament and FAS 1 If the results of the tournament are acyclic, we can use topological ordering to determine both the winner and the full rank of the players. No player have any reason to complain since all the players I won, are always ranked lower then me.

9. Tournament and FAS 2 If the results aren’t acyclic, we can’t satisfy ALL the players. So we want a solution satisfying as many players as possible. Given a minimal feedback arc set, we have such solution. »Why?

10. K-Weighted Feedback Arc Set On Tournaments Given a tournament T=(V,A). A weight function And an Integer k. Question: Is there an arc set such that and T’=(V,A\S) is a DAG.

11. K-FAST NP-Complete FPT – (the parameter will be k.) Article improves a previous result in this problem from: to: – Interested?

12. Preliminaries: w* For an arc weighted tournament we define the weight function w*

13. Preliminaries: D{F} Let D=(V,A) a directed graph. And a set F of arcs in A. We define D{F} to be a directed graph obtained from D by reversing all arcs of F.

14. D{F} And FAS 1 Claim: let T=(V,A) be a tournament. F is a minimal FAS of T=(V,A) if and only if F is a minimal set of arcs such that T{F} is a DAG. In other words, you don’t have to remove FAS arcs in a minimal solution, you can simply REVERSE them.

15. D{F} And FAS 2 Explanation/Prove: Given a minimal feedback arc set F of a tournament T, the ordering corresponding to F is the unique topological ordering of T{F}

16. D{F} And FAS 3 Conversely, given an ordering of the vertices of T, the feedback arc set F corresponding to is the set of arcs whose endpoint appears before their start point in

17. D{F} And FAS conclusion We showed that every vertex ordering define a FAS and that every FAS define a vertex ordering. The cost of an arc set F is: And the cost of a vertex ordering is the cost of the corresponding FAS.

18. The Algorithm 1 Perform a data reduction to obtain a tournament T’ of size 2. Let. Color the vertices of T’ uniformly at random with colors from {1,…,t} 3.Let be the set of arcs whose endpoints have different color, find a minimum FAS contained in, or conclude that no such FAS exist

19. Step 1: Kernelization Lemma 1: k-FAST has a kernel with vertices. Proof: by explicit build of such kernel!

20. Lemma 1: k-FAST has a kernel with vertices We use these reduction rules: 1.If an arc (e) is contained in at least k+1 directed triangles reverse the arc and reduce k by w(e). 2.If a vertex (v) is not contained in any triangle delete v from T.

21. Is the 1st rule safe ? First rule is safe because any feedback arc set that does not contain the arc (e) must contain at least one arc from each of the k+1 triangles containing e, and thus must have weight of at least k+1. (remember why?) Refresh the definition of W in the start:

22. Is the 2 nd rule safe ? Looking at a vertex (v), and assume v is not contained on any triangle. Observation 1: Since we are in a tournament each vertex in the graph is either in Observation 2: Any arc connecting And is in the direction from to

23. 2 nd rule drawing/intuitive explanation Since all the arcs go only from N-(v) to N+(v)… if both sub-graphs are DAG, adding v and all the arcs associated with v won’t add a cycle to the graph. V

24. Is the 2 nd rule safe ? From Observation 1 + Observation 2 one can deduce that any optimal FAS S1 on and an optimal FAS S2 on satisfy: Is an optimal FAS on T Therefore the 2 nd rule is safe.

25. Lemma 1: k-FAST has a kernel with vertices We showed the build to be legal build. We still need to count now how many vertices are there in the reduced graph (T’). Claim1: T’ has at most k(k+2) vertices

26. Let S be a feedback arc set of T’ with weight of at most k’. The Set S contains at most k arcs. For every arc (e) in S, aside from the two endpoints of e, there are at most k vertices that are contained in a triangle containing e, (otherwise 1 st rule will apply) Since every triangle in T’ contains an arc of S and every vertex of T’ is in a triangle T’ has at most k(k+2) vertices.

27. Step 2: color the vertices of T’ uniformly at random with colors from {1,…,t} What is the probability of a a good coloring? Lemma 2: if a graph on q edges is colored randomly with colors then the probability that G is properly colored is at least

28. Calculating the probability of a good coloring. To prove the lemma we will make use of the following build: Arrange the vertices of the graph by repeatedly removing a vertex of lowest degree. Let be the degrees of the vertices when they have been removed. What can we say about ?

29. Analyzing the build: First we notice that for all i, since the degree of the vertex removed can not exceed the number of remaining vertices. Now what can we say about: ?

30. Analyzing the build 2 since when a vertex i is removed each vertex had degree of at least Why is that? Sum of all vertex degrees in a graph is 2q. Therefore for the i’th step:

31. The build and coloring? Combining two observations: we get: Above hold for all i. But what do these calculations have to do with coloring?

32. The Build and coloring. Consider the colors of each vertex one by one starting from the last one, that is vertex number s. When vertex number i is colored, the probability that it will be colored by a color that differs from all those di neighbors following it is at least…

33. The Build and coloring. The probability that vertex number i is colored by a different color from all the other vertices is: –Because:why?

34. Why? Indeed, the inequality holds…

35. So what is the probobility that G is properly colored? From previous result:

36. Solving a Colored Instance Definitions: Given a t-colored tournament T, We will say that an arc set F is colorful if no arc of F is monochromatic. An ordering of T is colorful if the feed back arc set corresponding to it is colorful. An optimal colorful ordering of T is a color ordering of T with the minimum cost among all other colorful orderings.

37. More Definitions For a pair of integer vectors:

38. Solving a Colored Instance 2 Let be a t-colored tournament. There exists a colorful feedback arc set of T if and only if induces an acyclic tournament for every i. (and we’ll call such T feasible)

39. Solving a Colored Instance 3 Lemma 3: Given a feasible t-colored tournament T, we can find a minimum weight colorful feedback arc set in For an integer, define and

40. Lemma 3: proof Lemma 3: Given a feasible t-colored tournament T, we can find a minimum weight colorful feedback arc set in Given an integer vector of length t in witch the i’th entry is between 0 and and let be observe that for any ordering of V corresponding to a colorful FAS F of T and any integer x there exist a such that. 111111 222 222 3 33 333 223123 44

41. Definition Example 32 4 1 4 2 3 1 What vertices does the vector: [1,2] stand for?

42. The Algorithm-idea The Idea is to try all possible candidates for the last vertex v of an optimal ordering of For every i the vertex is the only candidate for v with color i.

43. (1) The idea behind (1) is to try all possible candidates for the last vertex v of an optimal ordering of For every i the vertex is the only candidate with color i.

44. (1) Proof: Let i be an integer that minimizes the right side. Taking the optimal ordering of and appending it with gives an ordering of with cost of at most

45. (1) Proof: Let be an optimal colorful ordering of and let v be the last vertex in this ordering. There is an i such as. Thus is a colorful ordering of and the total weight of arcs with start points in v and end points in is exactly: Completing the Proof!

46. Dynamic Programming: Implementation Table: containing for every p. There are table entries. For each entry it takes us: nt time Thus we can get: Working a bit harder and calculating: will yield the result of:

47. Summery: Lemma 4: k-fast (for a tournament of size can be solved in expected time of Combining lemma 1, 2 and 3 yields expected running time of:

48. Summery: Lemma 4: k-fast (for a tournament of size can be solved in expected time of Space required by algorithm is:

49. More Results: An algorithm to solve K-FAST in polynomial space and time. De-Randomization

50. Thank you for your time… Questions?