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

Fast FASTFast FASTFast FASTFast FASTBy Noga Alon, Daniel Lokshtanov And By Noga Alon, Daniel Lokshtanov And

Saket SaurabhSaket Saurabh

Presentation by Gil EinzigerPresentation by Gil Einziger

Fast FAST?• Fast – A relative fast algorithm for

an NP- Complete problem.

• FAST – (minimal) Feedback Arc Set in Tournaments.

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?

,G V A

F A

' , \G V A F

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.

What is A Tournament?

• A tournament is a directed graph (digraph) obtained by assigning a direction for each edge in an undirected complete graph.

• A tournament is, a directed graph in which every pair of vertices is connected by a single directed edge.

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.

,U V T U

1-21-2

3

Not a tournament, 2 possible topological orders…

1

3

Tournament, only one topological order…

21-2

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?

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.

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?

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.

: | 1w A x R x

S A

e Sw e k

K-FAST

• NP-Complete • FPT – (the parameter will be k.) • Article improves a previous result

in this problem • from:

to:

– Interested?

14.7522.415 OkO k n

2log 12O k k On

Preliminaries: w*• For an arc weighted tournament

we define the weight function w*

*

*

:

,0

w V V R

w uv if uv Aw u v

otherwise

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.

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.

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}

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

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.

e F

w e

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

2( )O k8t k

cA

cA

Step 1:Kernelization

• Lemma 1: k-FAST has a kernel with vertices.

Proof: by explicit build of such kernel!

2O k

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.

2O k

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:

: | 1W A x x

Is the 2nd 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

N v or in N v

N v N v

N v N v

2nd 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 N v N V

Is the 2nd 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 2nd rule is safe.

T N v

T N v

1 2S S

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

2O k

Claim1: T’ has at most k(k+2) vertices

• 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 1st 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.

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

8q

8(2 ) qe

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 ?

1 2, ,..., sd d d

id

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: ?

id s i

1id s i

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:

1 2id s i q

id

1 deg 2i ki k qd s i d sumof all vertice ree q

The build and coloring?• Combining two observations: we get:

• Above hold for all i. • But what do these calculations have

to do with coloring?

2 1 2

2

i

i

d di s i di s i q

d q

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…

The Build and coloring.

• The probability that vertex number i is colored by a different color from all the other vertices is:

– Because: why?

18id

q

81 2

8id qid e

q

8 2 iq d

Why?

81 2

8

8

1 1-a- 2 0,

2: .

id qi

i

a

de

q

dmark a

q

we get e a

proof by drawing the functionwith function calculator

Indeed, the inequality holds…

So what is the probobility that G is properly

colored?• From previous result:

8 /8

1 1

1 2 28

is s

d q qi

i i

de e

q

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.

More Definitions• For a pair of integer vectors:

1 1

1

1 ' .

,..., , ,...,

if for every i p

1,1,1...,1 , 1,0,0...,0 , 0,0...,1,...,0

t t

i i

i

is ini th place

p p p q q q

p q q

e e e

Solving a Colored Instance 2

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

1 2... ,tT V V V A

iT V

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 , defineand

1 time and Ot tO tn n space

1x 1 ,...,i i ix xS v v

0iS

Lemma 3: proofLemma 3: Given a feasible t-colored tournament T, we can find a

minimum weight colorful feedback arc set in 1 time and Ot tO tn n space

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

.

p

in

1 1 1 1 1 12 2 2 2 2 23 3 3 3 3 3

2 2 3 1 2 3

4 4

T p

1 2

1 2 ... .t

tp p pT S S S

1 2... nv v v

p

1 21 1 2,..., ... t

x p p ptv v S S S V T p

Definition Example

32

4

1

42

3

1

What vertices does the vector: [1,2] stand for?

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.

*: 0min ,

i

ii p i piu V T p

FAS T p FAS T p e w v u

T p

i

ipv

(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.

*: 0min ,

i

ii p i piu V T p

FAS T p FAS T p e w v u

T p

ipiv

(1) • Proof:• Let i be an integer that minimizes

the right side. Taking the optimal ordering of and appending it with gives an ordering ofwith cost of at most

*: 0min ,

i

ii p i piu V T p

FAS T p FAS T p e w v u

iT p e

ipiv T p

* ,i

i piu V T pFAS T p e w v u

(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:

*: 0min ,

i

ii p i piu V T p

FAS T p FAS T p e w v u

T p

| iT p e iT p e

iV T p e

*

(, .i

piu V T pw v u

Completing the Proof!

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:

FAS T p

tn

1t t tnt n tn timeand n space

* ,

i

ipu V T p

w v u

t ttn timeand n space

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:

2O k

log log2 2O k k O k k

time and space

1

1 8 log/8 2

!

lg ( )

2 * 8 2 2

t

k O k kk

I beleivecoloring

A orithm tn

O e O k k k

Summery:• Lemma 4: k-fast (for a tournament

of size can be solved in expected time of

• Space required by algorithm is:

2O k

log log2 2O k k O k k

time and space

8 log2

!

lg ( )

* 2 2

t

k O k k

I beleive

A orithm n

O k k

More Results:• An algorithm to solve K-FAST in

polynomial space and time.

• De-Randomization

2log

2 1O k k

O

Thank you for your time…

Questions?