Parameterized Algorithms for dhitting Set: the Weighted Case Henning Fernau. Univ. Trier, FB 4 Abteilung Informatik Trier, Germany


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1 Parameterize Algorithms for Hitting Set: the Weighte Case Henning Fernau Trierer Forschungsberichte; Trier: Technical Reports Informatik / Mathematik No. 086, July 2008 Univ. Trier, FB 4 Abteilung Informatik Trier, Germany Theoretische Informatik Trier
2 Parameterize Algorithms for Hitting Set: the Weighte Case Henning Fernau 1 Univ. Trier, FB 4 Abteilung Informatik, Trier, Germany Abstract. We are going to analyze simple search tree algorithms for Weighte Hitting Set. Although the algorithms are simple, their analysis is technically rather involve. However, this approach allows us to even improve on elsewhere previously publishe running time estimates for the more restricte case of (unweighte) Hitting Set. 1 1 Introuction Our approach in general. We exhibit how to systematically esign an analyze search tree algorithms within the framework of parameterize algorithmics [3]. Here, we avocate a topown approach as oppose to a rather bottomup esign, because the resulting algorithms ten to be simpler than via the opposite approach, an they sometimes pretty much resemble heuristic pruning techniques as use in branchancut algorithms for solving har problems. Moreover, this approach is quite moular in the sense that it prouces algorithms whose search tree backbone, i.e., the branching pattern of the algorithm as such, is not affecte by the optimization techniques reflecte in what we will call heuristic priorities (accoring to which the branching is performe) an the employe reuction rules. This not only moularizes correctness proofs for such algorithms, but also favors rapi prototyping of implementations. We will exemplify this approach by eveloping an analyzing simple algorithms for Weighte Hitting Set (WHS) problems. No prior research on parameterize algorithms has been reporte for these problems. Problem statement. Weighte Hitting Set (WHS) can be viewe as a weighte vertex cover problem on hypergraphs. More formally, this problem can be state as follows: Given: A weighte hypergraph G = (V, E, w) with ege size boune by, i.e., e E( e ), an a weight function w : V [1, ) Parameter: a nonnegative integer k Question: Is there a (weighte) hitting set C of total weight of at most k, i.e., C V e E(C e ) an w(c) := x C w(x) k? 1 This is a full version of the paper presente in [9]. Most of the work on this material has been one while the author was with Univ. Tübingen, Germany, as well as with Univ. Hertforshire, Hatfiel, UK. The accoring support is gratefully acknowlege.
3 2 Why Hitting Set? Hitting Set problems show up in many places; e.g., Reiter s grounbreaking research on moelbase iagnosis [12,17] relates the automatic iagnosis of systems to Hitting Set. The thrive for minimum hitting sets is in that context motivate by the parsimony principle in two ways: (a) the simplest iagnosis tens to fin the actual cause, an (b) when the iagnosis implies exchanging (possibly) faulty components (as a consequence of a selfiagnosis of an autonomous system, e.g., in space), then a minimum hitting set might also be the cheapest repair solution; in that particular scenario, however, the weighte case seems to be even more interesting than the unweighte one. As a further application, in [10], connections between a twotree rawing problem that is important in bioinformatics an 4WHS are shown, where the weights reflect further natural restrictions from biological backgroun knowlege. The algorithmics of this paper can be immeiately transferre to both applications. Previous work. For the unweighte case (which is a special case of the weighte setting if all weights are equal to one), there is one publishe paper presenting a search tree algorithm for Unweighte Hitting Set (HS), > 2, from a parameterize perspective [13]. The exponential base of the running time estimate for these algorithms tens to 1 with growing, although in the simplest case = 3, it is still relatively far off from that boun: that basis is By an intricate case analysis of a comparatively complicate algorithm, they were able to arrive at an O (2.270 k ) algorithm for the (unweighte) 3HS problem (i.e., all weights equal one). This was improve in [6] to about O (2.179 k ) by using a similar methoology as explaine here for the weighte case. In fact, this result was recently improve by Wahlström [19] by a ifferent although relate methoology to O ( k ). Notice that we are ealing with search tree algorithms an apply a parameterize analysis of the search tree size. If we then say that the algorithm has O (f(k)) running time, where k is the parameter, this means that the search tree has size (number of leaves) O(f(k)), since the work in each search tree noe will be at worst polynomial in n. In actual fact, all analysis that follows will be a clever estimate on the size of the search tree. For the special case of 2HS, likewise known as Vertex Cover, in a kin of race (using more an more intricate case analysis) an O(1.285 k +kn)algorithm [1] has been obtaine. For 2WHS, likewise known as Weighte Vertex Cover, the best that was obtaine is on O (1.396 k ), see [14]. Our approach seems not to be suitable to tackle the case = 2. The results of this paper. As in the unweighte case [6], 2 our analysis is base on the introuction of a secon auxiliary parameter that allows us to account for gains obtaine by using appropriate reuction rules an heuristic priorities. This technique can be useful in other areas of parameterize algorithms, as we believe. We get the following table for the bases c of an O (c k ) algorithm for  WHS; the bases are better than those for the unweighte case publishe in [13]: 2 A revise version of that report is going to be publishe with Algorithmica.
4 c (1) General notions an efinitions. We introuce some terminology on hypergraphs as neee for Hitting Set. A hypergraph G = (V, E) is given by its finite set of vertices V an its set of (hyper)eges E, where a hyperege is a subset of V. The carinality e of a hyperege e is also calle its size. The carinality of the set of eges which contain the vertex v is calle the egree of v, written δ(v). 2 Heuristics an reuctions for Weighte Hitting Set 2.1 A simple branching algorithm Since each hyperege must be covere an the weights are all at least one, there exists a trivial O ( k )algorithm for WHS. simplewhs(g = (V, E, w), k, S): IF k > 0 AND G has some eges THEN choose some ege e; // to be refine S = ; // solution to be constructe FOREACH x e DO // recursively branch G = (V \ {x}, {e E x / e}); S = simplewhs(g, k w(x), S {x}) IF S failure THEN break return S ELSIF E = THEN return S ELSE return failure Obviously, the base of the exponential running time of this algorithm heavily epens on the necessary amount of branching. Observe that accoring to the problem specification, in a WHS instance, there might be eges of size up to alreay in the very beginning. Small eges may also be introuce later uring the run of the algorithm. A natural heuristic woul first branch on small eges. We woul therefore refine: simplewhs(g, k, S): IF k > 0 AND G has some eges THEN choose some ege e of smallest size;... // as before Can we make use of this heuristic priority in our analysis? We therefore now efine reuction rules which we will always exhaustively apply at the beginning of each recursive call. Moreover, we switch towars a binary branching at vertices (instea of branching on eges), as can be seen in Alg. WHSST below.
5 4 2.2 Reuction rules First reuction rule: vertex omination. The vertex omination rule that was use in [6,13] for the unweighte case is invali in full generality in the weighte case, but has to be replace by the following weighte vertex omination rule: If, for all eges e, x e implies y e an if w(y) w(x), then elete x. This reuction rule implies the following one (reuction rule for egreeonevertices): If x, y e with δ(x) = 1 an w(y) w(x), then remove x. The sounness of this rule is easily seen: the only reason for taking a vertex x into the hitting set, in a situation as escribe by the reuction rule, is that it might be cheap. Conserving expensive vertices makes no sense. This reuction rule immeiately implies: Lemma 1. In a reuce instance, there is no ege with more than one vertex of egree one. The next lemma is again an easy consequence from the weighte vertex omination rule an is of particular importance when > 3. Lemma 2. In a reuce instance, for any two eges e 1 an e 2, there is at most one x e 1 e 2 with δ(x) = 2. Other rules state in [6] literally transfer to the weighte case: Secon reuction rule: ege omination. An ege e is ominate by another ege f if f e. Then, we elete e, since covering f will automatically also cover e. Thir reuction rule: small eges. Delete all eges of size one an place the corresponing vertices into the hitting set. The small ege rule, together with the vertex omination rule, proves the nonexistence of isolate eges in the following precise sense: Lemma 3. In a reuce instance, there is no ege e such that all vertices x e have egree one. Fourth reuction rule: ege cover rule. If G contains a component C that is of maximum vertex egree two, then resolve C in polynomial time. This (last) rule is justifie by the following lemma: Lemma 4. If G is a weighte hypergraph of maximum vertex egree of two, then a minimum weighte hitting set can be foun in polynomial time. Proof. To G, there correspons an egeweighte graph G whose vertices are the eges of G an whose vertexajacency relation is the egeajacency relation of G. Then, a minimum weighte hitting set of G correspons to a minimum weighte ege cover of G that can be compute in polynomial time.
6 5 2.3 Branching rules an their analysis The iea of making favorable branches first has also another bearing, this time on the way we are going to analyze the search tree algorithm, base on an auxiliary parameter l. Let T l (k), l 0 enote the size (more precisely, the number of leaves) of the search tree when assuming that at least l eges in the given instance (with parameter k) have a size of (at most) 1. The intuition is that T 3 (k) woul escribe a situation which is more like ( 1)WHS than T 2 (k). The unerlying iea is that search trees with many small eges are smaller than search trees with only a few; hence: k : T l (k) T l+1 (k). (2) Regaring an upper boun on the size T (k) of the search tree of the whole problem, we can equate T (k) = T 0 (k) by following the same intuition. Eq. (2) also shows that, upon analyzing a T l situation, we can always assume that there are exactly l eges that have a size of at most 1, an these small eges o have a size of exactly 1. Our algorithm will make choices with the bias of what we will call heuristic priorities. They can be refine if necessary along the analysis of the algorithm. The simplest list to start with might contain a single rule that shoul be intuitively clear: Choose a vertex of highest egree within an ege of smallest size. We will upate the list of priorities whenever necessary. WHSST(G = (V, E, w), k, S): exhaustively apply reuction rules; IF k > 0 THEN IF E = THEN return S; choose some vertex x accoring to the heuristic priorities S = ; // solution to be constructe E = { e E x / e }; S = WHSST((V \ {x}, E ), k w(x), S {x}); IF S ==failure THEN E = { e \ {x} e E }; S = WHSST((V \ {x}, E ), k, S); return S ELSIF G contains some eges or k < 0 return failure ELSE // G contains no eges an k is zero return S In the very beginning, given the instance (G, k), we call WHSST(G, k, ). We assume that reuction rules may also change the parameter value k an the solution S. The algorithm is quite generic: the list of reuction rules may grow an we might also change the heuristic priorities. The simple binary branching structure of WHSST enables a straightforwar inuctive proof of its correctness:
7 6 Theorem 1. If the reuction rules are correct, then WHSST(G, k, ) either returns a correct hitting set to the WHS instance (G, k) or it returns failure, if there is no solution of size at most k. Proof. The proof is by straightforwar inuction on the number of vertices of the hypergraph, similar to the unweighte case consiere in [6]. 3 A simple branching analysis We will now unertake a simple analysis, only consiering T 0, T 1 an (partially) T 2 an T 3. Lemma 5. T 0 (k) T 0 (k 1) + T 3 (k). Proof. Whenever we select an ege of size to branch on (accoring to the heuristic priorities), we can fin an ege that contains a vertex x of egree three or larger ue to the ege cover rule. One branch is that x is put into the hitting set. This reuces the amissible weight by at least one. If x is not put into the hitting set, then at least three new eges of size two are create. T 1 branching. In the next lemma, we show a first step into a strategy which will finally give us better branching behaviors. Namely, we try to exploit the effect of reuction rules triggere in ifferent subcases. This alreay necessitates a refinement in the choice of heuristic priorities: within a smallest ege e of size j <, we prefer branching at x e that maximizes the number of incient eges of size j + 1. Lemma 6. T 1 (k) max{t 0 (k 1) + T 1 (k 1) + T 2 (k 1) + ( 4)T 3 (k 1), T 0 (k 1) + T 1 (k 1) + (j 2)T 2 (k 1) + ( 2 (2j + 1) + (j 2 + j))t 0 (k 2) : j = 2, 3,..., 2} if 4; if T 0 (k) ( 1) k or if = 3, this may be simplifie: T 1 (k) T 0 (k 1) + ( 2)T 1 (k 1). Proof. The instance G has an ege e = {x 1, x 2,..., x 1 } of size ( 1). Case 1. If there is an ege f of size such that 2 j := e f 2 (this can only happen if 4), then we woul first branch at the vertices in e f; ue to weighte vertex omination, at least one of the j branches that take one of the vertices of e f into the hitting set is an T 1 (k 1)branch an j 2 are even T 2 (k 1)branches (or better). If none of the vertices from e f goes into the hitting set, then, in orer to cover e, there are 1 j many possibilities left, an in orer to cover f, there are j remaining possibilities. This explains the other (( j) 1)( j) many T 0 (k 2)branches. We can neglect these cases in our time analysis when assuming T 0 (k) ( 1) k. For reaability, we refer for this analysis to the appenix. Case 2. If the previous case oes not occur, then for all eges f e, e f 1. Assume that x 1 is the vertex of maximum egree in e, so that we branch at x 1. If δ(x 1 ) = 1, we can eterministically resolve the case with the reuction rules (apply 1 times the weighte vertex omination rule an then the small ege
8 7 rule) an get one T 0 (k 1)branch. This is obviously better than the inequality claime in the lemma. Therefore, we can now assume that δ(x 1 ) 2. If we take x 1 into the hitting set, then we get a T 0 (k 1)branch. If we o not take x 1 into the hitting set, we create one new ege e 1 of size ( 1) an we get the ege e = e \ {x 1 } of size ( 2). In the next recursive call, e is the ege of smallest size. There is no other ege of that size, since Case 1 i not apply. We therefore continue branching at the vertex (say x 2 ) of maximum egree in e. Again, δ(x 2 ) = 1 is better than the case we are going to pursue next. If δ(x 2 ) 2, then we again have two cases: either we take x 2 into the hitting set or not. If x 2 goes into the hitting set, then this is a T 1 (k 1)branch; namely, since Case 1 i not apply, x 2 / e 1, so that the small ege e 1 will be preserve. If x 2 oes not go into the hitting set, then there will be a new ege e 2 of size ( 1) ( new ue to ege omination). In the next recursive call, e = e \ {x 1, x 2 } is the ege of smallest size. The argument continues an shows that branches of type T j (k 1) will show up, for j = 2, 3,..., 2. This shows the claim, taking into account that T j (k 1) T 3 (k 1) for j 3 ue to Eq. (2). Estimating branching numbers. By using the inequality T 3 (k) T 2 (k) T 1 (k), Lemmas 5 an 6 yiel: T 0 (k) T 0 (k 1) + T 1 (k) (3) T 1 (k) T 0 (k 1) + ( 2)T 1 (k 1) With c being the largest positive real root of the characteristic polynomial x 2 x + 2, i.e., c = = + ( 2) (4) we can see that by setting T 0 (k) = c k an T 1 (k) = (c 1)c k 1, the inequalities system (3) can be solve. The larger, the closer c gets to 1. Hence: T (k) 2.62 k 3.42 k 4.31 k 5.24 k 9.13 k k Obviously, this is worse than what Nieermeier an Rossmanith got in [13] for the (general) unweighte case (ue to the lack of the vertex omination rule in full generality), but shows the same limit behavior (when is large). Can we o better? Let us give a simple trial to incorporate T 2 an T 3 into the analysis in the special case of Weighte 3Hitting Set. 4 Weighte 3Hitting Set We will use subscripts in the functions that escribe the search tree sizes to inicate this special case. We branch accoring to the following heuristic priorities. Let s be the size of the smallest ege in the instance G = (V, E, w).
9 8 Let E s be the collection of smallest size eges. (P 3 1) Let the set of (first) branching caniates B be e E s e. (P 3 2) If e is a smallest ege that is isjoint with all other e E s, refine B = e. (P 3 3) If no such isolate smallest ege exists, then upate B to collect the vertices of maximum egree in the hypergraph ( e E s e, E s ). (P 3 4) Select x B to be a vertex of maximum egree in G. It is easy to check that the analyses of Lemmas 5 an 6 are still vali uner these heuristic priorities. Lemma 7. T 2 3 (k) max{t 1 3 (k 1) + T 2 3 (k 1), T 0 3 (k 1) + T 0 3 (k 2)} Proof. We consier first the situation that the two eges e 1 an e 2 of size two are isjoint (see (P 3 2)). Then, basically the analysis of Lemma 6 applies, showing the claim. More precisely, we have T3 2 (k) T3 1 (k 1) + T3 2 (k 1). Otherwise, e 1 e 2, i.e., e 1 = {x, y} an e 2 = {x, z}. Accoring to the heuristic priority (P 3 3), we branch at x. If we take x into the hitting set, we get a T3 0 (k 1)branch. Not taking x into the hitting set enforces y an z into the hitting set, which is a T3 0 (k 2)branch. T Lemma 8. T (k 1) + T3 0 (k 2), (k) max T3 0 (k 1) + T3 0 (k 3), T3 2 (k 1) + T3 3 (k 1) Proof. If there is a ege e of size two that has nonempty intersection with any other ege of size two, ue to (P 3 2) we branch on e without estroying the at least two other eges of size two. The reasoning given in Lemma 6 therefore yiels the upper boun T 2 3 (k 1) + T 3 3 (k 1) in this case. If the first case oes not apply, the all eges of size two are connecte. Let e 1, e 2, e 3 be three connecte eges of size two. If x e 1 e 2 e 3 exists, then we branch at x ue to (P 3 3). This gives the (trivial) upper boun of T 0 3 (k 1) + T 0 3 (k 3). Otherwise, we branch at some x containe in two small eges ue to (P 3 3); w.l.o.g.: x e 1 e 2. Since x / e 3, the case that we take x into the hitting set is inee a T 1 3 (k 1)branch. This explains the upper boun T 1 3 (k 1) + T 0 3 (k 2). Theorem 2. Weighte 3Hitting Set can be solve in time O ( k ). The algebra justifying this claim can be foun in the following subsection. We only mention for the reaer that likes to skip this section in a first rea that the exact solution of the inequalities system can be escribe by the largest positive root c 3 of the polynomial x 3 2x 2 x + 1, which then gives T3 0 (k) = c k 3, T 1 3 = c k 3/(c 3 1), T 2 3 (k) = c k 3/(c 3 1) 2, an T 3 3 (k) = c k 1 3 (c 3 1). This worst case is realize when all T 3 branches are accoring to the T 1 3 (k 1) + T 0 3 (k 2) estimate. Improving on that particular case woul not help too much, however, since the other extreme cases show also branching behaviors worse than 2.2 k. Observe that this also means that a search tree in the T 3 (k)case is only about half the size of a search tree in the T 0 (k)case.
10 9 4.1 The algebra for Weighte 3Hitting Set In the following, we suppress the subscript 3, since we are only ealing with this case. Let us first show some algebra in case that we only analyze up to T 2 (k), i.e., if we put T 3 (k) = T 2 (k). What branching behavior o we observe in either case in Lemma 7? 1. If T 2 (k) T 1 (k 1) + T 2 (k 1), we get (by Lemma 5), (T 0 (k) T 0 (k 1)) = T 1 (k 1) + (T 0 (k 1) T 0 (k 2)), which gives as a recurrence By Lemma 6, T 1 (k 1) = T 0 (k) 2T 0 (k 1) + T 0 (k 2). T 0 (k + 1) 2T 0 (k) + T 0 (k 1) = T 0 (k 1) + T 0 (k) 2T 0 (k 1) + T 0 (k 2). Therefore, 0 = T 0 (k + 1) 3T 0 (k) + 2T 0 (k 1) T 0 (k 2). This is resolve by T 0 (k) k. 2. T 2 (k) T 0 (k 1)+T 0 (k 2) gives immeiately the characteristic polynomial x 2 2x 1 with largest positive real root x = ; this is hence the worst case here. It might be surprising at first glance that we treate the weak inequalities as if they were equalities. This is base on experience: taking this approach we were always able to come up with a solution to the envisage system of inequalities. However, these computations shoul be justifie by further analysis. Since this is only an (encouraging) intermeiate result, we refrain from giving such valiation here; we will however give it in the general case. For Weighte 3Hitting Set, we erive the following recurrences: T 0 (k) T 0 (k 1) + T 3 (k) T 1 (k) T 0 (k 1) + T 1 (k 1) T 2 (k) max{t 1 (k 1) + T 2 (k 1), T 0 (k 1) + T 0 (k 2)} T 1 (k 1) + T 0 (k 2), T 3 (k) max T 0 (k 1) + T 0 (k 3), T 2 (k 1) + T 3 (k 1) How o we arrive at possible solutions? Firstly, we try to solve extreme cases that are obtaine by treating the weak inequalities as if they were equalities an by iscussing all possible combinations as escribe by the maximum operator. In our case, this woul in principle result in six systems of equations.
11 10 However, since T 2 shows up only in one place in a righthan sie of a T 3 (k)  inequality, we only get four cases. It is noteworthy to see that, when assuming T 0 (k) = c k, (for all situations) the first equation gives T 3 (k) = c k 1 (c 1) an the secon equation gives T 1 (k) = c k /(c 1). Notice that finally we are looking for an overall solution for all T j, i.e., T j (k) = α j c k 3 with c 3 an the α j still to be etermine. Our consierations so far entail: α 0 = 1 α 1 = 1/(c 3 1) α 2 α 3 = (c 3 1)/c 3 T 0 (k) = T 0 (k 1) + T 3 (k) T 1 (k) = T 0 (k 1) + T 1 (k 1) (T 2 (k) = max{t 1 (k 1) + T 2 (k 1), T 0 (k 1) + T 0 (k 2)}) T 3 (k) = T 1 (k 1) + T 0 (k 2) Obviously, the function T 2 (k) oes not come into play if we are primarily intereste in looking for solutions for T 0 (k) in this case. Plugging in the expressions for T 0 (k), T 1 (k) an T 3 (k) in the last equation yiels: c k 1 (c 1) = c k 1 /(c 1) + c k 2. After multiplication with (c 1)c 2 k an some reorering, this becomes the characteristic polynomial: 0 = c(c 1) 2 c (c 1) = c 3 2c 2 c + 1. Its largest positive real root can be boune by from above. As can be seen, this is the claime worst case. T 0 (k) = T 0 (k 1) + T 3 (k) T 1 (k) = T 0 (k 1) + T 1 (k 1) (T 2 (k) = max{t 1 (k 1) + T 2 (k 1), T 0 (k 1) + T 0 (k 2)}) T 3 (k) = T 0 (k 1) + T 0 (k 3) The same solution strategy provies: c k 1 (c 1) = c k 1 + c k 3. Multiplication with c 3 k an some reorering yiels: 0 = c 3 2c 2 1;
12 11 the largest positive real root can be boune by from above. In the following two cases, we have to istinguish the two ifferent upper bouns for T 2. T 0 (k) = T 0 (k 1) + T 3 (k) T 1 (k) = T 0 (k 1) + T 1 (k 1) T 2 (k) = T 1 (k 1) + T 2 (k 1) T 3 (k) = T 2 (k 1) + T 3 (k 1) From T 1 (k) = c k /(c 1), we can euce from the thir equation: T 2 (k) = c k /((c 1) 2 ). Therefore, the last equation gives: c k (c 1) = c k 1 /((c 1) 2 ) + c k 1 (c 1). Multiplication with c 1 k (c 1) 2 results in: 0 = (c 1) 4 c = c 4 4c 3 + 6c 2 5c + 1, whose largest positive real root can be estimate by A irect plugin yiels: T 0 (k) = T 0 (k 1) + T 3 (k) T 1 (k) = T 0 (k 1) + T 1 (k 1) T 2 (k) = T 0 (k 1) + T 0 (k 2) T 3 (k) = T 2 (k 1) + T 3 (k 1) T 0 (k) = T 0 (k 1) + T 3 (k) = T 0 (k 1) + T 2 (k 1) + T 3 (k 1) This gives again = T 0 (k 1) + T 0 (k 2) + T 0 (k 3) + (T 0 (k 1) T 0 (k 2)) = 2T 0 (k 1) + T 0 (k 3) 0 = c 3 2c 2 1; the largest positive real root can be boune by from above. So, the worst scenario gives the characteristic polynomial c 3 2c 2 c + 1, whose largest positive real root c 3 can be boune from above by We haven t yet etermine α 2. From T 2 (k) T 1 (k 1)+T 2 (k 1), we woul get α 2 = 1/(c 3 1) 2, an from T 2 (k) T 0 (k 1) + T 0 (k 2), we arrive at α 2 = (c 3 +1)/c 2 3. However, c 2 3 = (c 3 +1)(c 3 1) 2 = (c 2 3 1)(c 3 1) = c 3 3 c 2 3 c 3 +1 is true, since c 3 is a root of the characteristic polynomial, so that both (seemingly ifferent) values of α 2 are in fact equal. In other wors, we foun:
13 12 α 0 = 1 α 1 = 1/(c 3 1) 0.80 α 2 = (c 3 + 1)/c α 3 = (c 3 1)/c After having obtaine these numbers, we shoul verify that inee T 0 (k) = c k 3, T 1 = c k 3/(c 3 1), T 2 (k) = c k 2 3 (c 3 +1), an T 3 (k) = c k 1 3 (c 3 1) satisfies the whole system of inequalities. This amounts in showing that the extreme case we foun is inee maximizing all righthan sie maxima functions. In fact, our reasoning with etermining α 2 in the preceing paragraph alreay shows that T 1 (k 1) + T 2 (k 1) = T 0 (k 1) + T 0 (k 2) for our functions. We still have to eal with the T 3 (k) inequalities. 1. To show: T 1 (k 1) + T 0 (k 2) T 0 (k 1) + T 0 (k 3). Substituting the functions we erive an multiplying with c 3 k (c 1) leaves us to show: c 2 + (c 2 c) (c 3 c 2 ) + (c 1) 0 c 3 3c 2 + 2c 1 which is true for c = c To show: T 1 (k 1) + T 0 (k 2) T 2 (k 1) + T 3 (k 1). Substituting the functions we erive an iviing by c k 2 leaves us to show: c c c2 + c c 2 + c2 c = 1 + c2 c c This in turn means we have to verify (again) for c = c 3 : 0 c 3 3c 2 + 2c 1. 5 Weighte Hitting Set with 4 How well o our consierations transfer to the more general case? We analyze possible T 2 branches in what follows. Since the obtaine bases are quite satisfactory, we refrain from analyzing the T 3 branches. In our analysis, we apply the following heuristic priorities to a given (reuce) instance G = (V, E, w): Let s be the size of the smallest ege in the instance G = (V, E, w). Let E s be the collection of smallest size eges. (P 1) Let the set of (first) branching caniates B be e E s e. (P 2) Define G B = (B, E s ) an upate B to be the set of vertices in G B of maximum egree. (P 3) Choose a vertex x B of maximum egree in G. One can check that Lemmas 5 an 6 are still vali when assuming these heuristic priorities. Since in our opinion solving Hitting Set for larger is of less practical importance, we will efer some etails of the following analysis to the appenix. Analyzing T 2. We will istinguish several cases in what follows: Lemma 9. Let e 1 an e 2 be two eges of size 1. If e 1 e 2 =, then we can estimate T 2(k) T 1(k 1) + ( 2)T 2 (k 1).
14 13 This can be basically inherite from Lemma 6 ue to ege omination. As we will see, this is the secon worst case branching. Being the simplest case, we give some etails. As justifie in the appenix, we solve the next set of equations: T 0 (k) = T 0 (k 1) + T 2 (k) (5) T 1 (k) = T 0 (k 1) + T 1 (k 1) + ( 3)T 2 (k 1) T 2 (k) = T 1 (k 1) + ( 2)T 2 (k 1) This yiels, after some algebra: 0 = T 0 (k + 1) T 0 (k) + ( 1)T 0 (k 1) T 0 (k 2). (6) Theorem 3. Let c enote the largest positive real root of the polynomial x 3 x 2 + ( 1)x 1. Then T 0 (k) = ck, T 1 (k) = α,1c k with α,1 = (c + 2)(c 1)/c an T 2 (k) = α,2c k with α,2 = (c 1)/c solve the system (5). The following table lists some of the exponential bases c for (5): c (7) Lemma 10. Let e 1 an e 2 be two eges of size 1. If e 1 e 2 = j {1, 2,..., 2}, then we can estimate T 2 (k) T 0 (k 1) + T 1 (k 1) + (j 2)T 2 (k 1) + ( 1 j) 2 T 0 (k 2), thereby assuming that T 0 (k) ( 1)k, i.e., c 1. Proof. The priorities (P 1) an (P 2) let us branch at a vertex x e 1 e 2. If j > 1, the weighte vertex omination rule moreover guarantees that there is a vertex of egree at least three in e 1 e 2, an (P 3) will select one such vertex x for branching. Hence, when x is not taken into the hitting set, then we gain at least one ege of size 1 if j > 1 ue to vertex omination, see Lemma 2, since we will continue selecting vertices within e 1 e 2 accoring to (P 2). The case that eges that intersect with e 1 e 2 might contain more than one vertex in this intersection turns out not to be the worst case (assuming ( 1) k as a lower boun of our approach) along the lines of Lemma 6. If e 1 e 2 is exhauste, then in the case that we take none of the vertices from e 1 e 2 into the hitting set, we are left with two very small eges e 1 = e 1 \ e 2 an e 2 = e 2 \ e 1. P 1 lets us continue branching at say e 1. Having selecte x e 1 to go into the hitting set, e 2 will be the smallest ege (of size ( 1 j)), an hence P 1 continues to branch on e 2 in the next recursion step. This explains that we get (very grossly estimate) ( 1 j) 2 many T 0 (k 2)branches. In orer to prove Theorem 4, the following technical lemma is important:
15 14 Lemma 11. If j > 1 an > 3, then T 1 (k 1) + ( 2)T 2 (k 1) T 0 (k 1) + T 1 (k 1) + (j 2)T 2 (k 1) + ( 1 j) 2 T 0 (k 2) for T 0 (k) = ck an T 2 (k) = ck c k 1 with 1 c, inepenent of T 1. We nee a somewhat stronger result (compare to Lemma 10) in the case j = 1 that escribes our worst case (for > 4): Lemma 12. Let e 1 an e 2 be two eges of size 1. If e 1 e 2 = 1, then we can estimate T 2 (k) T 0 (k 1) + ( 3)T 1 (k 2) + [( 2)( 3) + 1]T 0 (k 2). Moreover, T 1 (k 1)+( 2)T 2 (k 1) T 0 (k 1)+( 3)T 1 (k 2)+[( 2)( 3)+1]T 0 (k 2) for T l as efine in Theorem 4 below. Proof. We only explain the branching in what follows (for the algebra, see the appenix). Assume that {x} = e 1 e 2. x is selecte for branching accoring to (P 1). If x oes not go into the hitting set, then we may continue branching on e 1. The claim is that, for any y e 1 \ {x} (with one possible exception, if δ(y) = 1 for some y e 1 ; but ue to Lemma 1, there is at most one vertex of egree one in e 1 an (P 3) avois branching at that vertex), there is an ege e y e 1 with y e y such that there is a vertex z y e 2 \ e 1 with z y / e y. For, if (e 2 \{x}) e y, then the ege omination rule woul have triggere. The branch that takes y an z y into the hitting set is a T 1 (k 2)branch (possibly better). Theorem 4. WHS can be solve in time O (c k ), where c is the largest positive root of the characteristic polynomial x 4 3x 3 ( 2 5+5)x 2 +x+( 2 6+9). Some values of c are liste below: c (8) Is it worthwhile trying to further improve on the exponential bases as erive in this section? In principle, yes of course; however, one woul nee a ifferent approach for substantial improvements: (a) the seconworst case is only slightly better than the worst case that we analyze, an (b) with growing, the lower boun ( 1) assume in (some) estimates is alreay quite well approximate. The most interesting case that remains seems to be = 4, which we tackle in the following separate section.
16 Hitting Set We are claiming that Lemma 9 actually provies the worst case for 4WHS, base on a eeper analysis an (again) slightly change heuristic priorities (which we will not make explicit in this case but which will become clear from the analysis). So, we are going to show in the remainer of this section the following result: Theorem 5. Let c 4 enote the largest positive real root of the polynomial x 3 4x 2 +3x 1. Then T 0 4 (k) = c k 4, T 1 4 (k) = α 4,1 c k 4 with α 4,1 = (c 4 2)(c 4 1)/c 4 an T 2 4 (k) = α 4,2 c k 4 with α 4,2 = (c 4 1)/c 4 solve the system (12). Moreover, O (c k 4) is an upper boun on the running time of our algorithm for solving Weighte 4Hitting Set. We can boun c 4 from above by As we have alreay seen before, the worst case (from above) we have to eal with is the case of two eges e 1, e 2 with e 1 = e 2 = 3 an {x} = e 1 e 2. We will analyze two subcases: (a) e 1, e 2: e i (e i \ {x}) but e i e 3 i = for i = 1, 2. (b) e 1, e 2 with e i (e i \ {x}) : e i e 3 i as well, for i = 1, 2. In case (a), we can branch as follows: Taking x into the hitting set gives a T 0 4 (k 1)branch. Otherwise, ue to the conition, let us continue branching at {x 1 } = e 1 e 1. (Observe that ue to weighte vertex omination, not all e 1 satisfying the conition (a) may contain {x 1, x 2 } = e 1 \ {x}.) Similarly, there is some {y 1 } = e 2 e 2. So, if we take both x 1 an y 1 into the hitting set, we get a T 0 4 (k 2)branch. If we take y 1 into the hitting set but not x 1, we must select x 2. Since y 1 / e 1 by (a), this is a T 1 4 (k 2)branch. Similarly, taking x 1 into the hitting set but not y 1 is a T 1 4 (k 2)branch. If neither x 1 nor y 1 go into the hitting set, then we gain two new small eges ue to conition (a), so this is even a T 2 4 (k 2)branch. Altogether, we have erive in case (a): T 2 4 (k) T 0 4 (k 1) + T 0 4 (k 2) + 2T 1 4 (k 2) + T 2 4 (k 2). (9) In case (b), there must be an ege e with (e 1 \{x}) e an (e 2 \{x}) e, since otherwise (i.e., if the forall conition is vacuously satisfie) the weighte vertex omination rule woul trigger an result in the following branching: T 2 4 (k) T 0 4 (k 1) + 2T 0 4 (k 2). We will see that the case we are going to consier will result in a branching that is strictly worse, so that we can neglect this case. Moreover, we can also assume that (e 1 \ {x}) e = 1, for if not, the weighte vertex omination rule woul trigger in the case that x is not taken into the hitting set. This means that one of the two vertices from (e 1 \ {x}) e must go into the hitting set. Moreover, since e is now hit, two subcases arise: either both vertices from e 1 \ {x} are containe in e an there are no eges that contain vertices from e 2 \ {x} apart from e 2 an possibly e; then, the weighte vertex omination rule woul trigger once more an altogether yiel the branch we alreay observe before, namely: T 2 4 (k) T 0 4 (k 1) + 2T 0 4 (k 2);
17 16 or, say y 1 e 2 \ {x} is containe in one other ege e y besies e 2 an e, an no vertex from e 1 \ {x} is containe in e y. Branching at y 1 woul hence gain us one small ege at least in the situation that y 1 is not going into the hitting set. Altogether, we get as estimate: T 2 4 (k) T 0 4 (k 1) + T 0 4 (k 2) + T 1 4 (k 2). This is again always better than the general estimate that we erive now. So, we can assume now that {x 1 } = (e 1 \ {x}) e an that {y 1 } = (e 2 \ {x}) e. Assume we start branching at x 1. Let us call {x 2 } = e 1 \ {x, x 1 }. We istinguish two subcases regaring {y 2 } = (e 2 \ {x, y 1 }): (i) δ(y 2 ) = 1 an (ii) δ(y 2 ) 2. Case (i) is again split into two cases: (ia) {e E y 1 e, x 1 / e} = 1 an (ib) {e E y 1 e, x 1 / e} > 1. In case (ia), if x 1 is taken into the hitting set, we will elete either y 1 or y 2 ue to the weighte vertex omination rule, an then this ege is resolve by the small ege rule. This gives a T 0 4 (k 2)branch. If x 1 is not taken into the hitting set, x 2 must be in. Moreover, if we continue branching at y 2, we get a T 0 4 (k 2)branch when y 1 goes into the hitting set an two T 0 4 (k 3)branches when not y 1 but y 2 is in the hitting set, since then one of the two remaining vertices from e must be in the hitting set, too. Overall, we get in this case: T 2 4 (k) T 0 4 (k 1) + 2T 0 4 (k 2) + 2T 0 4 (k 3). (Notice that this is strictly speaking a tight analysis for δ(y 1 ) = 2.) Again, this is not the worst case to consier. In case (ib), if x 1 is taken into the hitting set, we might take y 1 into the hitting set. This gives a T 0 4 (k 2)branch. If y 1 oes not go into the hitting set, then y 2 will, giving a T 1 4 (k 2)branch (gaining a small ege by the case assumption). If x 1 is not going into the hitting set but x 2, we get one T 0 4 (k 2)branch an two T 1 4 (k 3)branches. This yiels: T 2 4 (k) T 0 4 (k 1) + 2T 0 4 (k 2) + T 1 4 (k 2) + 2T 1 4 (k 3); again, this is not the worst case to consier. In case (ii), we can assume that there is an ege e y that contains y 2 but none of the vertices from {x, y 1, x 1 }. Namely, since δ(y 2 ) 2, there is at least one ege e y besies e 2 that contains y 2. As can be seen, δ(y 2 ) = 2 is the worst case we assume henceforth. If y 1 e y, the weighte ege omination rule woul have triggere, yieling a better branching as analyze before. If x 1 e y, we have a situation analyze uner case (i) above. We branch as follows: If x 1 goes into the hitting set, we can gain a small ege in the case that y 2 oes not go into the hitting set (assume we continue branching at y 2 ); hence, this is one T 0 4 (k 2)branch an one T 1 4 (k 2)branch. If x 1 is not in the hitting set, let us continue branching at y 1. By a simple analysis, we get one T 0 4 (k 2) an two T 0 4 (k 3)branches (similar as in the previous cases). This means: T 2 4 (k) T 0 4 (k 1) + 2T 0 4 (k 2) + T 1 4 (k 2) + 2T 0 4 (k 3). (10) We will show now that the case that e 1 e 2 = is in fact the worst case that yiels the estimate T 2 4 (k) = T 1 4 (k 1) + 2T 2 4 (k 1).
18 17 We first consier Eq. (9): we have to fin an upper boun for T 2 4 (k) + T 0 4 (k 1) + T 0 4 (k 2) + 2T 1 4 (k 2) + T 2 4 (k 2). With the settings of the functions as formulate in Theorem 5, this expression means: c k 4 + 2c k c k (c 4 2)(c 4 1)c k c k 2 4 c k 3 4. After multiplication with c 3 k 4, we get the following chain: c c c (c 4 2)(c 4 1) 1 = c c c c 2 4 6c = [ c c 2 4 3c 4 + 1] c The expression in square brackets vanishes, since c 4 is a root of the characteristic polynomial mentione in Theorem 5, an the inequality follow from 3 c 4. Let us consier Eq. (10): we will fin an upper boun for T 2 4 (k) + T 0 4 (k 1) + 2T 0 4 (k 2) + T 1 4 (k 2) + 2T 0 4 (k 3) uner the settings of the functions as formulate in Theorem 5. This means we have to upperboun: c k 4 + 2c k c k (c 4 2)(c 4 1)c k c k 3 4. After multiplication with c 3 k 4, we get the following chain: c c c (c 4 2)(c 4 1) + 2 = c c 2 4 c = [ c c 2 4 3c 4 + 1] c c c c 4 = c 4 (3 c 4 ) 0 The expression in square brackets vanishes, since c 4 is a root of the characteristic polynomial mentione in Theorem 5, an the two inequalities follow from 3 c 4. 7 Conclusions Let us allow first one final remark regaring the correctness of our analysis: Remark 1. The astute reaer might have notice that there are possible situations say with two or more eges of size 1 where the reuction rules actually estroy all of them an replace them by one ege of size 2 or smaller. However, it can be easily verifie that, for all 3 an l 2 as far as analyze, T l (k) ( 2)k. This is base on the assumption c 1. Therefore, this omission in the analysis will never affect the worst case running time claime in the paper.
19 18 We have evelope an analyze a novel, topown methoology for parameterize search tree algorithms. Up to now, we have applie this methoology to Hitting Set [6], biplanarization problems [8] (thereby improving on the constants erive in [4]), linear arrangement problems (in the long version of [7]) an to Weighte Hitting Set (this paper). A further natural caniate for applying this technique woul be (Weighte) Directe Feeback Vertex / Arc Set in Tournaments, as consiere in [15], as well as variants thereof [2]. In orer to apply this metho, we nee a kin of secon auxiliary parameter in the problem which we try to improve on in case the main parameter cannot be improve upon binary branching. In the case of (Weighte) Hitting Set, the number of eges of small size is such an auxiliary parameter. Our results show that this methoology is a quite powerful tool of algorithm analysis. For example, while the gap between the running times of the (very sophisticate) best search tree algorithms for Weighte Vertex Cover an for Vertex Cover [1,14] o iffer significantly (both algorithms being approximately of the same complexity), this paper shows that with our analysis metho of a comparatively simple algorithm for 3WHS, we can even (slightly) improve on the previous analysis of a much more sophisticate algorithm for Unweighte 3HS [13]. It may be interesting to compare the way the analysis of the recurrences guie by the auxiliary parameter is unertaken in this paper with the analysis metho of Wahlström [18] or with Eppstein s quasiconvex metho [5]. It woul be also interesting to see this approach applie to other, ifferent problems with accoringly ifferent auxiliary parameters. More generally speaking, there seems to be a recent thrive in Exact Algorithmics towars simple algorithms. The Minimum Dominating Set algorithm of Fomin, Granoni an Kratsch is only one more example (see [11]) that incientally also uses a (special) Hitting Set algorithm. This irection of research certainly brings practical an theoretical research on attacking har problems closer together, since one coul also envisage a kin of interplay between algorithm analysis an algorithm testing in the near future. Can an appropriate analysis then explain certain observe phenomena of the implementation? The moular ecomposition of such an algorithm into the actual recursive search tree backbone an the reuction rules an (in particular) the heuristic priorities also opens up a whole area of experimental algorithmics: uner which circumstances (or, in a more theoretical formulation: for which classes of hypergraphs) is a certain set of rules the most successful? Can this be prove? Due to the simple overall structure of the algorithms, also an analysis of expecte running times (possibly aing coin tossing into the heuristic priorities) might be possible. Incientally, improvements in parameterize algorithms for Hitting Set also entail improvements in exact algorithms for Minimum Hitting Set, measure in terms of number of vertices: in the case of 3Hitting Set, the use of the algorithm exhibite in [6] improve Wahlström s algorithm [18] from
20 19 O ( n ) own to O ( n ) (personal communication by Wahlström ating from 2005). The results of this paper will immeiately entail new running time bouns for exact algorithms for Minimum Weighte Hitting Set. For example, along the lines sketche by Raman, Saurabh an Sikar, in [16], we get an exact algorithm for Minimum Weighte 4Hitting Set that runs in time O (1.97 n ), using our parameterize Weighte 4Hitting Set algorithm. References 1. J. Chen, I. A. Kanj, an W. Jia. Vertex cover: further observations an further improvements. Journal of Algorithms, 41: , M. Dom, J. Guo, F. Hüffner, R. Nieermeier, an A. Truß. Fixeparameter tractability results for feeback set problems in tournaments. In T. Calamoneri, I. Finocchi, an G. F. Italiano, eitors, Conference on Algorithms an Complexity CIAC, volume 3998 of LNCS, pages Springer, R. G. Downey an M. R. Fellows. Parameterize Complexity. Springer, V. Dujmović, M. R. Fellows, M. Hallett, M. Kitching, G. Liotta, C. McCartin, N. Nishimura, P. Rage, F. A. Rosamon, M. Suerman, S. Whitesies, an D. R. Woo. A fixeparameter approach to 2layer planarization. Algorithmica, 45: , D. Eppstein. Quasiconvex analysis of backtracking algorithms. In Proc. 15th Symp. Discrete Algorithms SODA, pages ACM an SIAM, January H. Fernau. A topown approach to searchtrees: Improve algorithmics for 3 Hitting Set. Technical Report TR04073, Electronic Colloquium on Computational Complexity ECCC, An upate version has been accepte to Algorithmica. 7. H. Fernau. Parameterize algorithmics for linear arrangement problems. In U. Faigle, eitor, CTW 2005: Workshop on Graphs an Combinatorial Optimization, pages University of Cologne, Germany, Long version to appear in Discrete Applie Mathematics. 8. H. Fernau. Twolayer planarization: improving on parameterize algorithmics. In P. Vojtáš, M. Bieliková, B. CharronBost, an O. Sýkora, eitors, SOFSEM, volume 3381 of LNCS, pages Springer, H. Fernau. Parameterize algorithms for hitting set: the weighte case. In T. Calamoneri, I. Finocchi, an G. F. Italiano, eitors, Conference on Algorithms an Complexity CIAC, volume 3998 of LNCS, pages Springer, H. Fernau, M. Kaufmann, an M. Poths. Comparing trees via crossing minimization. In R. Ramanujam an Saneep Sen, eitors, Founations of Software Technology an Theoretical Computer Science FSTTCS 2005, volume 3821 of LNCS, pages Springer, F. V. Fomin, F. Granoni, an D. Kratsch. Measure an conquer: omination a case stuy. In L. Caires, G. F. Italiano, L. Monteiro, C. Palamiessi, an M. Yung, eitors, Automata, Languages an Programming, 32n International Colloquium, ICALP, volume 3580 of LNCS, pages Springer, J. e Kleer, A. K. Mackworth, an R. Reiter. Characterizing iagnoses an systems. Artificial Intelligence, 56: , R. Nieermeier an P. Rossmanith. An efficient fixeparameter algorithm for 3Hitting Set. Journal of Discrete Algorithms, 1:89 102, R. Nieermeier an P. Rossmanith. On efficient fixe parameter algorithms for weighte vertex cover. Journal of Algorithms, 47:63 77, 2003.
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