Lecture Note: Rounding From pseudo-polynomial to FPTAS

Transcription

1 Lecture Note: Rounding From pseudo-polynomial to FPTAS for Approximation Algorithms course in CCU Bang Ye Wu CSIE, Chung Cheng University, Taiwan

2 Rounding Rounding is a kind of techniques used to design a fully polynomial time approximation scheme (FPTAS) from a pseudo-polynomial time exact algorithm. It has been shown for many problems. In this slides we show two rounding techniques by two typical examples. Suppose we have an algorithm for finding an exact solution for an (NP-hard) optimization problem P. T : set of elements in an instance w(t): non-negative integral weight of an element t T time complexity: f(n, W ) (f is a polynomial in n and W, n = T and W = t T w(t)

3 Since the time complexity is a polynomial in W, it is a pseudo-polynomial. Remarks: in the standard definition, W should be the maximum weight of the elements but it does not matter since the sum is bounded by n max{t }. Let g be the measurement function, i.e., g maps a feasible solution S to Z +0 and the problem looks for a feasible solution S with minimum g(s) amongst all feasible solutions. Intuitively, for an instance T, we may scale down the weights of all elements by a factor ρ > 1 and round the results to integers to construct another instance T of P. The exact algorithm takes f(n, W/ρ ) time to compute an exact solution S for T. Due to the rounding error, we cannot expect that S is also an optimal for T. But, hopefully, g(i ) does not differ g(opt(t ))

4 too much. To show that it is an FPTAS, what we need to show is that g(i )/g(opt(t )) can be infinitely close to 1 for suitable ρ. Usually ρ = W h(n) to make the time complexity polynomial.

5 Two types The rounding technique can be divided into two types. Direct rounding: bound. when the optimal has a One-side error Binary search: when the optimal is unbounded. We shall take examples to illustrate the two techniques.

6 Direct Rounding we use the next simple example. Problem: Balance partition Instance: n positive integers Goal: partition the integers into two subsets (S 1, S 2 ) such that min{w(s 1 ), w(s 2 )} is maximized, in which w(s 1 ) is the sum of elements in S 1. The problem is a maximization version of the Partition problem and is therefore NP-hard.

7 Balance partition There exists an O(nW ) time exact algorithm (using the dynamic programming approach to generate all possible weights of subsets). Algorithm 1 DP-Partition y[i] = 0, for 1 i W/2; y[0] = 1; for i 1 to n do for j 0 to W/2 x i do if y[j] = 1 then y[j + x i ] 1; end if end for end for output the maximum j W/2 such that y[j] = 1.

8 Direct Rounding For an instance T = {x i 1 i n} with weight w, we construct another weight w, in which w (x i ) = w(x i )/ρ. In O(nW/ρ) time we find an optimal solution S for T. By the optimality, x S w (x) x S w (x) Since ρw (x i ) w(x i ) < ρw (x i ) + ρ, w(x) ρ w (x) ρ x S x S > x S w(x) ρ S > x S w(x) ρn x S w (x) Since S is an optimal w.r.t. w, OPT x S w(x) < 1 + ρn x S w(x) 1 + ρn OPT ρn

9 Bound of the optimal If x i such that w(x i ) W/2, the optimal is trivial. For otherwise we can show that OPT W/3. Then OPT x S w(x) < 1 + ρn W/3 ρn By choosing ρ = (1 + 1/ε) W 3n, the ratio is 1 + ε and the time complexity is O((1 + 1/ε)n 2 ). It is an FPTAS.

10 One-side error Binary search An important reason that the above technique works is the bound of the optimal. But for some problems there is no such bound, for example, the path problems in graphs. There are some path problems which are NPhard. One of them is to find a minimum cost path with restriction to time delay. In this problem, we are given a graph G = (V, E) and each edge e is associated with a cost w(e) and a time delay t(e). The goal is to find a minimum cost path such that the total time delay is at most a given threshold. Anyway, it is not necessary to know the detail. All we need to know is that there is an exact algorithm with time complexity f(n, W ) and the feasible solution is a simple path.

11 The key difference from the balance partition problem is that this is a minimization problem and the optimal path weight has no lower bound (it is possible the optimal is very small).

12 Test procedure Constructing an optimal in O(f(n, W )) (usually) determining whether opt W in the same time complexity. For any r 1, let w /r (e) = w(e)/r and G /r = (V, E, w /r ). Algorithm 2 /* test if OPT(G) > M or OPT(G) (1 + δ)m*/ procedure Test(M, δ) r Mδ/n; construct G /r ; if OPT(G /r ) n/δ then return Yes ; else return No ; end if end procedure Note: condition OPT(G /r ) n/δ can be computed by the exact algorithm in O(f(n, n/δ)) time.

13 Lemma 1: If the test procedure returns Yes, OPT(G) (1+δ)M; and otherwise OPT(G) > M. Proof: Let Q be an optimal path, i.e., OPT(G) = w(q). If OPT(G /r ) n/δ, a path P such that w /r (P ) n/δ. Since rw /r (e) w(e) < r(w /r (e) + 1), e E and any simple path has at most n 1 edges, w(p ) rn/δ + r(n 1). By r = Mδ/n, w(p ) M(1 + δ), which completes the proof of Yes-part. On the other hand, if OPT(G /r ) > n/δ, OPT(G) > rn/δ = M.

14 Note: the test procedure determines whether the optimal is larger or smaller than a given value M. If it says larger, it is sure that the optimal > M. But it is not precise when it says smaller. It only ensures that the optimal (1 + δ)m. That is, there is an uncertain range. The time complexity doest not depend on M but only on n and δ. We can iteratively narrow down the uncertain range by giving suitable M and δ, similar to the standard binary search.

15 How to search Suppose: a lower bound L and an upper bound U of the optimal. To find a (1 + ε)-approximation, the search process should be performed until U (1+ε)L. Now we have two questions: What are the initial lower and upper bounds? What is the best query strategy? That is, in each iteration, how should we choose M and δ in each iteration?

16 First try traditional binary search Let L 0 and U 0 be the initial bounds. The first thought: choosing M = (L + U)/2 and δ = ε. Time: (roughly) f(n, n/ε) log(u 0 L 0 ) (lucky enough, the range is cut in half every time.) Already an FPTAS because trivially L 0 = 0 and U 0 = W. (convince yourself that log W is polynomial in the length of an instance.) We can do it better:

17 Balancing the two cases Since a test divides the range into 3-subranges ([L, M], [M, (1+δ)M], and [(1+δ)M], U]), balancing the yes-case and the no-case may be a good idea. At this thought we choose M = (U + 2L)/3 and (1 + δ)m = (2U + L)/3 and the range will be reduced by a factor 2/3 every iteration. δ = U L ( U/L 1 U + 2L = L U/L + 2 ( ) 3 = L 1 > 1/4 U/L + 2 as long as U 2L > 0. That is, each test takes O(f(n, 4n)) time. What a magic! The parameter ε disappears. But two things need to be overcome. ) It works only when U 2L; and

18 we need ensure L > 0.

19 Two phases search Stopping binary search when range is small. When U 2L: we use the exact algorithm to find OPT(G /r ) with r = Lε/n and this is a (1 + ε)-approximation. Let P be an optimal path w.r.t. G /r. w /r (P ) = OPT(G /r ). Because w(p ) r OPT(G /r ) + r(n 1) and OPT(G /r ) OPT(G)/r and r = Lε/n w(p ) OPT(G) + Lε(n 1)/n < (1 + ε)opt(g) Remember OPT(G) L. The time complexity: The optimal is at most U in G. the weight of the optimal in G /r is at most U Lε/n = 2n/ε.

20 It takes f(n, 2n/ε) time but only once. The total time complexity will be log(u 0 L 0 )f(n, 4n) + f(n, 2n/ε) if we can ensure L 0 > 0. Another magic! we separate the factors log W and ε.

21 Better initial bounds Let w 1 < w 2 <... < w p, p E, be the distinct edge weights in G. Let E t = {e E w(e) t} and G t = (V, E t ). By binary search we can find the minimum w {w i 1 i p} such that the graph G w is feasible. By the definition of w, any optimal solution uses at least an edge of weight w, and therefore we have w OPT(G). Furthermore, since a simple path has at most n 1 edges, we have OPT(G) (n 1)w. Note that if w = 0, we find the optimal 0; and if G wp is infeasible, there is no feasible solution. We can exclude both the cases.

22 The time complexity of this step: We solve the 0-weighted version of the problem O(log n) times. For the time delay path problem, the feasibility can be tested by finding a shortest path. For other more general problems, it usually takes less time than the search stage. By a preprocessing stage, we can determine an upper bound U 0 and a lower bound L 0 of OPT(G) such that U 0 nl 0 and L 0 > 0.

23 Another killer binary search in logarithmic scale Temporarily forget the error. In a binary search, we want to reduce the range U L. The best way is to choose M = (U + L)/2 (arithmetic mean). It takes log(u 0 L 0 ) iterations. Could we reduce it if we only want to make U/L a constant? Yes! By choosing M = UL (geometric mean), both U/M and M/L are the square root of U/L. The number of iterations is log log(u 0 /L 0 ). ( ) (1/2) x U0 L 0 2 x = log log(u 0 /L 0 ) Remember we have U 0 nl 0 and L 0 > 0. We need only log log n iterations.

24 Final killer Idea: two-phases search+logarithmic binary search +balancing two cases ((1 + δ)m/l = U/M) Algorithm 3 Rounding-and-Searching find U = U 0 and L = L 0 ; while U > 2L do δ U/L 1; M if UL/(1 + δ); Test(M, δ) returns Yes then U M(1 + δ); else L M; end if end while r Lε/n; find and output OPT(G /r ). Should we expect a time complexity like (log log n)f(n, n) + f(n, n/ε)? It is the time to see the real magic.

25 Time complexity The time complexity is only f(n, n/ε). The factor log log n disappears. The last step takes O(f(n, n/ε)) time (shown previously). Let U i, L i, δ i and M i, 0 i q, be the parameters used in the i-th iteration of the while loop. (ending at (q + 1)-th iteration.) It is sufficient to show that q i=0 f(n, n/δ i ) = f(n, n). Usually q i=0 f(n, n/δ i) f(n, n)( q i=0 1/δ i). (E.g. f(n, n/δ) = f(n, n)(1/δ) k, k 1.) What we need to show: q i=0 (1/δ i) = O(1).

26 At each iteration, we set either U i+1 = M i (1 + δ i ) or L i+1 = M i. Since δ i +1 = U i /L i and M i = U i L i /(1 + δ i ) = U 1/4 i L 3/4 i, for both cases we have U i+1 /L i+1 = (U i /L i ) 3/4. (1) By definition, 1/δ i = 1 U i /L i 1 = L i /U i < c 1 L i /U i L i /U i since U i > 2L i (c is a constant). Hence, By (1), q i=0 (1/δ i ) = O q i=0 L i /U i L i /U i = (L q /U q ) (1/2)(4/3)q i, and then q i=0 L i /U i = q j=0 since U q > 2L q. (L q /U q ) (1/2)(4/3)j < q 2 (4/3)j j=0

27 Using 2 (4/3)j+1 = ( /3 ) (4/3)j (2 1/3 ) 2 (4/3)j repeatedly, we have 2 (4/3)j (2 1/3 ) j (2 1 ). Therefore q j=0 2 (4/3)j < 2 1 q j=0 (2 1/3 ) j < 2 1 O(1) 1 2 1/3

28 References Lenstra, J. K., Shmoys, D. B. and Tardos, E. (1987). Approximate Algorithms for Scheduling Unrelated Parallel Machines. Proc. 28th FOCS, Hochbaum, D. and Shmoys, D. (1988). A Polynomial Approximation Scheme for Scheduling on Uniform Processors: Using the Dual Approach. SIAM J. Comput R. Hassin, Approximation schemes for the restricted shortest path problem, Math. Oper. Res. 17 (1) (1992) D.H. Lorenz and D. Raz, A simple efficient approximation scheme for the restricted shortest path problem, Oper. Res. Lett. 28 (2001)

29 F. Ergun, R. Sinha and L. Zhang, An improved FPTAS for restricted shortest path, Inform. Process. Lett. 83 (2002) B.Y. Wu (2010) A note on approximating the min-max vertex disjoint paths on directed acyclic graphs. Journal of Computer and System Sciences, in press, doi: /j.jcss