16.1 Min-Cut as an LP

Transcription

1 / Approximation Algorithms Lecturer: Michael Dinitz Topic: LPs as Metrics: Min Cut and Multiway Cut Date: 4//5 Scribe: Gabriel Kaptchuk 6. Min-Cut as an LP We recall the basic definition of the Min Cut Problem Input: Graph G = (V, E) Feasible: Objective: min e A Source s V and Terminal t T A E s.t. G A has no s-t path We note that this definition of Min Cut Problem can be written in an equivalent form Input: Graph G = (V, E) Definition 6.. P s,t = {all S T paths} Source s V and Terminal t T Feasible: S V s.t. s S and t S Objective: min (S, S) We define the following LP, whose integer solutions are solutions to the Min Cut Problem min subject to x e x e e p p P s,t 0 x e e E Intuitively, this states that along each path at least one edge must be in the cut. Theorem 6.. This LP can be solved in polytime, even though it has an exponential number of constraints

2 Proof: Suppose x is not a feasible solution to the LP. There must thus be a path p P s,t s.t. e p x e <. We now think of x e as a length assigned to edge e E. We can easily find a shortest S T path. Because it is the shortest path, and p P s,t with e p x e = e p length(e) <, we can use shortest path as a separation oracle for the ellipsoid method. Theorem 6..3 If x is a feasible solution to the LP with cost( x) = x e = Z, then we can find an integral solution with cost Z in polynomial time. Proof: We begin by defining a few variables. Definition 6..4 Let d(u) denote the shortest path distance from s to u under the edge lengths x e x. Definition 6..5 Let B(s, r) = {v V d(v) r} Definition 6..6 If S V, let δ(s) = E(S, S) = set of edges with one endpoint in S and one endpoint in S We note that using the shortest path metric for all given nodes v V turns the graph into a metric space. Algorithm LP rounding for Min Cut Input: Graph G = (V, E) and solution to the LP x Output: S E Choose r uniformly at random in [0, ] S B(s, r) return A δ(s) Claim 6..7 Let e = {u, v} E. P r[e A] x e Proof: WLOG let d(u) d(v). For e A the radius of the ball B = (s, r) must be enough to contain u, but not v. So, P r[e A] = P r[r [d(u), d(v)]] d(v) d(u) d(u, v) x e, where d(u, v) denotes the length of the shortest u v path. Hence by linearity of expectation, By linearity of expectations, E[c(A)] = P r[e A] = x e = Z. We can take this seemingly randomized algorithm and make it deterministic by simply trying all possibilities. Suppose node w such that d(u) d(w) d(v). Thus any random r [d(u), d(v)] will yield the same cut. Thus there are a linear number of possible cuts, which can each be tested in linear time the entire algorithm can be run in polynomial time. 6. Multiway Cut We define the Multiway cut problem.

3 Input: Graph G = (V, E) T = {s, s,..., s k } V Feasible: A E s.t. G A has no s i s j path i, j {,,..., k} Objective: min e A Definition 6.. Let P u,v = {all simple u v paths} We define the following LP, whose integer solutions are solutions to the Multiway cut problem min subject to x e x e e p i, j {,,..., k}, p P si,s j 0 x e e E Note that the LP solution x induces a metric d on the nodes through the shortest-path distances, and the constraints guarantee that d(s i, s j ) for all i, j [k]. We use the same separation oracle as in the above Min Cut problem example, but simply use it to check for each pair of {i, j}. Algorithm LP rounding for Multiway Cut Input: Graph G = (V, E) and solution to the LP x Output: S E A for all i {,,.., k} do Randomly choose r [0, ] A i δ(b(s i, r)) A A A i end for return A Theorem 6.. If x is a feasible solution to the LP with cost( x) = x e = Z, then there is a polynomial time algorithm to find an integral solution x such that cost( x ) Z which can be trivially reduced to ( k )Z. Proof: We begin by proving a claim Claim 6..3 e = {u, v} E, P r[e A] x e Proof: Let w V. By the triangle inequality, d(s i, w) + d(w, s j ) d(s i, s j ). Definition 6..4 Let C i = {v V d(s i, v) }. Clearly C i C j = i, j 3

4 Case : u, v C i for some i,,..., k. WLOG we assume d(s i, u) d(s i, v). P r[e A] = P r[e A i ] = P r[r [d(s i, u), d(s i, v)]] = d(s i, v) d(s i, u) d(u, v) x e Case : u C i, v C i and v C j for i j P r[e A] P r[e A i ] + P r[e A j ] P r[r [d(s i, u), ]] + P r[r [d(s j, v), ]] = d(s i, u) + So in all cases P r[e A] x e. d(s j, u) = ( d(s i, u) d(s j, v)) (d(s i, s j ) d(s i, u) d(s j, v)) d(u, v) x e By linearity of expectations E[c(A)] = P r[e A] (x e) x e Z. Thus this is a polynomial time algorithm to find a integral solution given a fractional LP solution x. In the same way as the previous example, we can test polynomial possibilities in polynomial time. 6.. Integrality Gap Is our analysis tight? We consider the star with k nodes around the outside connected by a single node v. We choose as our k terminal nodes the outside nodes. The optimal solution OP T is clearly given by cutting all but one edge. So cost is given by k. The worst case LP solution is given by assigning all edges. So cost is given by k Thus the gap is given by OP T LP 6.. A better LP = k k = ( k ). So our analysis is tight. We consider a better solution to the Multiway cut problem. The problem itself stays the same. We create pieces C, C,...C k s.t. s i C i i,,..., k and C i C j = for i j. We define X i u = Z i e = { u Ci 0 else { e δ(ci ) 0 else We now define an LP using these indicator variables 4

5 min i= k subject to x i u = i= k Ze i u V Ze i Xu i Xv i e = u, v E, i,,..., k Ze i Xv i Xu i e = u, v E, i,,..., k Xs i i = i,,..., k 0 Xu i 0 Ze i It is straightforward to verify that this is a valid relaxation of the multiway cut problem. We will give a more compact way of writing this LP which makes the connection to metrics clear. Definition 6..5 Let x, y R k then their l -distance is x y = k i= xi y i where x i is the i th coordinate of the vector x. Definition 6..6 Let k = {x R k k i= xi =, x i 0 i} where x i is the i th coordinate of the vector x. Definition 6..7 Let e i be a vector with a in the i th coordinate and zeros elsewhere Let X u = (X u, X u,..., X k u) and X v = (X v, X v,..., X k v ). Note that Z i e = X i u X i v in any optimal LP solution because of the constraints on Z i e and because we are minimizing the objective function. So, X u X v = k i= Zi e. Using all these definitions we can rewrite the new LP as follows min X u X v e={u,v} E subject to X si = e i i,,..., k X u k Using this LP, there exists a rounding that yields a 3 approximation of the problem 5