Branching Algorithms

Transcription

1 Lecture Branching Algorithms Author: Arkadiusz Soca la 1 Motivation Definition 1. Decision problem L Σ is called NP problem if there exists a predicate ψ P computable in polynomial time such that: 1. x L y y poly x x, y ψ such y is called a witness for x.. x L y x, y ψ. An common example of NP problem is Hamiltonian Cycle Input: A graph G Question: Does there exists a simple cycle of length V G? Suppose we know that y p x for some polynomial p. Then we can check every possible witness y in the total time of p x i poly x O p x poly x. i=0 The general question is how fast we can check entire space of witnesses and how well we can improve it. Also whether we can omit some part of that space. Suppose that we have an algorithm which works in O n time and that on our computer we can solve this problem up to n = 6. Note that if we improve our computer to be four times faster then we will be able to compute the problem up to n = 6 +. But if we improve algorithm to O n/ then we can solve the problem up to n = 5. So to sum up, our motivations are: theoretical question:,,how much we can speed up our algorithm?, solving real problems: it is better to improve algorithm than buy a new hardware, a new different view on algorithms for hard problems. Vertex Cover We study the following problem. Vertex Cover Input: A graph G and an integer k Question: Does there exist a set X V G of size at most k, such that every edge of G has at least one endpoint in G?

2 Theorem. Maximal vertex cover can be found in O n m time where n = V G and m = EG. Proof. For chosen subset of V G we can test if every of EG is covered. Testing every subset of V G can be done in total O n m time. Let s introduce O notation Definition 3. O fn = Ofn polyinput size. We decide to ignore polynomial factors almost always in the analysis, as they are theoretically irrelevant: O n n 100 O.0001 n..1 Fixed parameter tractability. Suppose we want to find a Vertex Cover, but of size not greater than some k. Let s say k = 0. Then we can use different approach than checking any subset not greater than k. Intuition: we want to create a recursion with depth bounded by k. Theorem 4. Vertex Cover can be solved in O k time and polynomial space. Proof. For any edge at least one end must be covered. So if we have any uncovered edge we must add to the cover one of the ends of this edge. We try both possibilities recursively. When we already have maximal number of vertices in the cover i.e. when k = 0 and we still have some edges uncovered then there is no solution in the current branch. The pseudocode of the algorithm: function VCG, k if EG = then return YES if k = 0 then return NO uv any ofeg return VCG u, k 1 or VCG v, k 1 end function Definition 5. Language L Σ is fixed parameter tractable FPT for parameter k : Σ N if it can be solved in Ofkx poly x time, for some computable function f. Parameter k in the definition above can be any function of input. For example it can be size of solution or some measure of the graph. The idea is to find parameters responsible for exponent in the complexity of the problem and separate them by multiplication from size of the input in polynomial. When the parameters are bounded then we can solve our problem fast in polynomial time for fixed parameters. Corollary 6. Vertex Cover is FPT for the size of the cover parameter.

3 Let s compare two algorithms for solving Vertex Cover problem: above algorithm which works in O k n time and checking all k-element possibilities in O n k time. For example if we put k = 10 then first solution works fine even for big values of n, e.g. n = 104. But the second solution even for n = 4 performs over one million operations, and for n = 8 over one billion. We can choose one of them depending on the values of k, n and some constant α : if k α n then VCG, k else Check all k-subsets. Let s analyse this algorithm for α = 3 4. For the first case we have O k O 3 n 4. For the second O n k So we have n 3 4 n = O n3 n! 3 4 n! 1 4 n. To estimate the last expression we can use Stirling s approximation: s s s! = Θ s. e 4 n! = O 1 4 n e n e n 1 4 n 3 4 n e 3 4 n = O Which is better than O n. For α = we can get even O 1.71 n.. Branching. n O 1.76 n. Now let s try another approach. We forget for a while about parameter k. Note that solving our problem for any graph with vertex degree not greater than is simple. Such graphs consist of paths and cycles only. We will use the following notation: Definition 7. v V G Nv = {u V G \ {v} : vu EV } Suppose that we have a vertex of a large degree in G. There are two possibilities: it is in Vertex Cover or all its neighbours are. This leads us to the following algorithm of finding minimal vertex cover: function VCG v vertex of maximal degree in G if degv then solve in polynomial time return min1+vcg v, degv+vcg v Nv end function Let s calculate time complexity of the above algorithm. We know that in the expression VCG v Nv we are deleting at least four vertices. So to obtain an estimation we can use a recursion

4 T n = T n 1 + T n 4. We want to find such x that T n x n. From recursion we have x n = x n 1 + x n 4. We can solve 1 = 1 x + 1 x 4 and get x Currently Oct 01 the best known algorithm works in O 0.88n O 1.3 n [3]. We can also estimate time complexity of VC in terms of parameter k. We have T k = T k 1 + T k 3, which leads us similarly as above to O k. Currently Oct 01 the best known FPT algorithm works in O1.738 k + kn [1]..3 Measure and Conquer. Let s consider slightly modified version of VC function VC3G if v V G degv = 0 then VC3G v else if v V G degv = 1 then 1+VC3G v Nv else v vertex of maximal degree in G if degv then solve in polynomial time else min1+vc3g v, degv+vc3g u Nu end function The idea is to introduce new measure for,,hardness of the instance. Note that vertices of degrees in {0, 1} are easy so we can assign them weight 0. Vertices of degree are a little harder so we assign them weight 1. Finally degree of greater weights are hard, so we assign them weight 1. For v chosen in algorithm we denote d = {u Nv, degu = }, d 3 = {u Nv, degu = 3}, d 4 = {u Nv, degu 4}. Then the recursion has the form T µ = T µ d d T We explain it in details: Expression T µ d d 3 1 µ d d 3 d 4 1. d neighbours of degree have now degree 1 change from weight 1 to weight 0 d 3 neighbours of degree 3 have now degree change from weight 1 to weight 1 1 deleting vertex v from G degv 3 so the weight is 1

5 Expression T µ d d 3 d 4 1 d deleting neighbours of degree with weights 1 d 3 deleting neighbours of degree 3 with weights 1 d 4 deleting neighbours of degree 4 with weights 1 1 deleting vertex v from G degv 3 so the weight is 1 Possibly other vertices the neighbours of the neighbours lose their degrees and weights but this expression is only upper bound so it is OK. For degv 4 in the worst case we have T µ = T µ 1 + T µ 5 when degv = 4 and all neighbours of v are of degree 4. And for degv = 3 the worst case is T µ T µ d 1 + T µ d 1 which is T µ T µ.5 when all neighbours of v are of degree. This leads us to O n. Note that we have chosen our weights basing on our intuition. We can choose them better. For example if we assign to vertices of degree 3 weight 0.99 then we obtain a slightly better complexity. So still we have the same algorithm, but we are improving our analysis of it. The above example and analysis can be found in the book of Fomin and Kratsch [4]. 3 Feedback Vertex Set Feedback Vertex Set Input: A graph G Question: What is a minimal power of X V G such that G \ X is a forest. We will be solving a dual problem of the above: Maximum Induced Forest Input: A graph G Question: What is a maximal power of Y V G such that G[X] is a forest. We will create recursive function MIFG, Y, Ŷ enlarging forest Y by any vertex v Y connected to the connected component of Y Ŷ and by an edge between v and Ŷ. To measure,,hardness of the instance we will take the following function: every v NŶ gets weight 1 every v NŶ gets weight Let s introduce notion of,,generalized neighbourhood N v = {u V G \ Y : in G[Y + v + u] exist path from v to u}. We call MIFG, Y, Ŷ recursively. Depending on N v we have the following cases: 1. For N v = 0 we take Y + = v.. For N v = 1 we also take Y + = v. 3. For N v = we do a branch:

6 Y + = v G = v, Y + = N v. If there is a solution Y which takes u 1 but no v nor u then Y u 1 + v is also a solution. N v = {u 1, u } so v is a leaf in Y u 1 + v. 4. For N v 3 we do a naive branch: Y + = v G = v Cases 1 and are simple. Let s analyse case 3. For simplicity we will be writing 3 instead of We are assuming N v =. Let a of them are in NŶ. Then we can estimate time T N T N 1 a a 1 + T N 1 a a 3. To analyse case 4 we use T N T N 1 degv 1 + T N 1 a a 3. If the current,,active component Ŷ has no neighbours then we select another connected component of G[Y ] as,,active. Every neighbour of the new,,active component lose on weight. If there is no connected component of Y with any neighbour then we take a vertex v of a maximal degree in V \ Y : If its degree is, then there are only cycles and paths in G Y so we can solve the problem in polynomial time. If its degree is 3, then we do a branch: Y + = v and Ŷ = {v} then we lose , because v has at least 3 neighbours, G = v then we lose from µ. By estimating above inequalities we can obtain O n. But this is also estimation for n n 3. The explanation of this fact is still open question Oct 01. The presented algorithm is due to Razgon [5]. Currently the fastest known algorithm for Feedback Vertex Set runs in O n []. References [1] Jianer Chen, Iyad A. Kanj, and Ge Xia. Improved upper bounds for vertex cover. Theor. Comput. Sci., : , 010. [] Fedor V. Fomin, Serge Gaspers, Artem V. Pyatkin, and Igor Razgon. On the minimum feedback vertex set problem: Exact and enumeration algorithms. Algorithmica, 5:93 307, 008. [3] Fedor V. Fomin, Fabrizio Grandoni, and Dieter Kratsch. A measure & conquer approach for the analysis of exact algorithms. J. ACM, 565, 009. [4] Fedor V. Fomin and Dieter Kratsch. Exact Exponential Algorithms. Springer, 010. [5] Igor Razgon. Exact computation of maximum induced forest. In Lars Arge and Rusins Freivalds, editors, SWAT, volume 4059 of Lecture Notes in Computer Science, pages Springer, 006.