The complement of PATH is in NL

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1 340 The complement of PATH is in NL Let c be the number of nodes in graph G that are reachable from s We assume that c is provided as an input to M Given G, s, t, and c the machine M operates as follows: One by one, M goes through all the m nodes of G and nondeterministically guesses whether each one is reachable from s Whenever a node u is guessed to be reachable, M attempts to verify this guess by guessing a path of length m or less from s to u If a computation branch fails to verify this guess, it rejects In addition, if a branch guesses that t is reachable, it rejects Machine M counts the number of nodes that have been verified to be reachable 341 When a branch has gone through all of G s nodes, it checks that the number of nodes that it verified to be reachable from s equals c, and rejects if not Otherwise, this branch accepts I.e., if M nondeterministically selects exactly c nodes reachable from s, not including t, and proves that each is reachable from s by guessing the path, M knows that the remaining nodes, including t, are not reachable, so it can accept The only problem is that this procedure relies on knowing c 1

2 342 A nondeterministic log space procedure whereby at least one computation branch has the correct value for c and all other branches reject For each i = 0,, m, define A i to be the collection of nodes that are at a distance of i or less from s So A 0 = { s }, each A i A i+1, and A m contains all nodes that are reachable from s Let c i be the number of nodes in A i Let us describe a procedure that calculates c i+1 from c i Repeated application of this procedure yields the desired value of c = c m To calculate c i+1 from c i the algorithm goes through all the nodes of G, determines whether each is a member of A i+1, and counts the members 343 To determine whether a node v A i+1, we use an inner loop to go through all the nodes of G and guess whether each node is in A i Each positive guess is verified by guessing the path of length at most i from s For each node u verified to be in A i, the algorithm tests whether (u, v) is an edge of G If it is an edge, v A i+1 Additionally the number of nodes verified to be in A i is counted At the completion of the inner loop, if the total number of nodes verified to be in A i is not c i, all A i has not been found, so this computation branch rejects If the count equals c i and v has not yet been found to be in A i+1, we conclude that it isn t in A i+1 Then we go to the next v in the outer loop 2

3 Advanced Topics in Complexity Theory What to do with a problem that is intractable and does not accept a deterministic exact solution in polynomial time Relax the problem: 1. Instead of solving it exactly, approximate the solution 2. Instead of using a deterministic algorithm, use a probabilistic (a.k.a. randomized) algorithm Approximation algorithm finds a solution that is guaranteed to be close to the optimal exact solution Probabilistic algorithm comes up with the exact solution with a high probability Sometimes it may fail to give the correct answer (Monte Carlo) or may have a high time requirement (Las Vegas) Approximation Algorithms Let us examine a problem, where we are given A ground set U with m elements A collection of subsets of the ground set S = { S 1,, S n } s.t. it is a cover of U: S = U The aim is to find a subcover S S, S = U, containing as few subsets as possible This problem is known as the Minimum Set Cover (minsc) One of the oldest and most studied combinatorial optimization problems 3

4 346 The corresponding decision problem Given: a ground set U, cover S and a natural number k Question: Does U have a subcover S S s.t. S' k? Theorem The decision version of minimum set cover problem is NP-complete. Proof. Obviously minsc NP: Let us guess from the given cover S a subcover S' containing k subsets and verify deterministically in polynomial time that we really have a subcover. 347 Polynomial time reduction VC mp minsc is easy to give. Let G, k be an instance of the vertex cover in which G = (V, E). We choose the mapping f: f( (V, E), k ) = E, V E, k, where V E is the collection of edges connected to the nodes of G. In other words, for each v V has a corresponding set { e E e = (v, w) }. Clearly f is computable in polynomial time and is a reduction. 4

5 S 4 S S S Hence, minsc is an intractable problem we do not know of a polynomial time algorithm for solving it Therefore, we attempt to find a polynomial time algorithm that does not necessarily give the best possible (optimal) solution, but can be shown always to be at most a function of the input length worse than the optimal solution Such an algorithm is called an approximation algorithm Let us denote by Opt the cost of the solution given by an optimal algorithm and App that of the solution given by an approximation algorithm 5

6 350 Since minsc is a minimization problem, App/Opt 1 The closer to 1 this ratio is, the better the solution produced approximates the optimal solution From an approximation algorithm one requires that the fraction is bounded by a function of the length n of the input App ( n) Opt (n) is the approximation ratio of the algorithm The algorithm is called an (n)-approximation algorithm At the best the approximation ratio does not depend at all on the length n of the input, but is constant 351 Let us examine the following algorithm for vertex cover We will show that it is an 2-approximation algorithm for the problem Input: An undirected graph G = (V, E) Output: Vertex cover C 1. C ; 2. E' E; 3. while E do a. Let (u, v) be any edge of the set E'; b.c C {u,v}; c. Remove from E all edges connected to nodes u and v; 4. od; 5. return C; 6

7 352 Selection of the first random edge: (b, c) b c d a e f g 353 We remove other edges connected with nodes b and c b c d a e f g 7

8 354 The next random choice : (e, f) and Removal of other edges connected with its nodes b c d a e f g 355 The only remaining choice (d, g) We end up with a cover of 6 nodes, while the optimal one has 3 nodes (e.g., b, d, e) b c d a e f g 8

9 356 Theorem 10.1 The above given algorithm is polynomial time 2- approximation algorithm for vertex cover. Proof. The time complexity of the algorithm, using adjacency list representation for the graph, is O(V + E), and thus uses a polynomial time. The set of nodes C returned by the algorithm obviously is a vertex cover for the edges of G, because nodes are inserted into C in the loop of row 3 until all edges have been covered. Let A be the set of edges chosen by algorithm in row 3a. In order to cover the edges of A any vertex cover in particular also the optimal vertex cover has to contain at least one of the ends of each edge in A. 357 Because the end points of the edges in A are distinct by the design of the algorithm, A is a lower bound for the size of any vertex cover. In particular, Opt A. The above algorithm always selects in row 3a an edge whose neither end point is yet in the set C. Hence, App = C = 2 A. Combining the above equations yields App = 2 A 2 Opt, and therefore App/Opt 2. 9

10 358 Also set cover has a simple greedy approximation algorithm Neither this nor any other polynomial time deterministic algorithm can attain a constant approximation ratio Input: Ground set U and its cover S Output: Set cover C 1. X U; C ; 2. while X do a. select S S s.t. S X is maximized; b.x X\S'; c. C C {S }; 3. od; 4. return C; 359 S 1 S 2 S 6 S 3 S 4 S 5 10

11 360 Greedy: 4 subsets S 1 S 2 S 6 S 3 S 4 S Optimal: 3 subsets S3 S 4 S 5 11

12 362 The greedy algorithm can quite easily be implemented to run in polynomial time in the length of the input U and S The loop in row 2 is executed at most min( U, S ) times and the body of the loop itself can be implemented to require time O( U S ) Altogether the time requirement thus is O( U S min( U, S )) It is also possible to give a linear time implementation for the greedy approximation algorithm for set cover The collection C returned by the algorithm is obviously a set cover, because the loop of row 2 is executed until there are no more elements to cover 12

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