Lecture 22. Lecturer: Michel X. Goemans Scribe: Alantha Newman (2004), Ankur Moitra (2009)

Transcription

1 8.438 Avance Combinatorial Optimization Lecture Lecturer: Michel X. Goemans Scribe: Alantha Newman (004), Ankur Moitra (009) MultiFlows an Disjoint Paths Here we will survey a number of variants of isjoint path problems, an give reuctions between these ifferent variants. We will also consier fractional relaxations of these problems, an in the special case of two source-sink pairs, we will prove a max-flow min-cut relation an in fact in this case, this flow can even be realize half-integrally.. Notation We will consier isjoint path problems in both irecte, an unirecte graphs. Aitionally, we will consier both ege-isjoint an vertex-isjoint paths problems. In the case of a igraph, we will let D = (V, A) be the igraph an we will enote the set of arcs by A. In the case of an unirecte graph, we will use G = (V, E) to enote the set of eges by E. Let s, t, s, t,... s k, t k V be terminals. Our goal is to fin isjoint paths between s i an t i for each i, i k. Again, the notion of path an the notion of isjoint epens on the variant of the isjoint path problem that we are consiering. We can in fact consier a more general problem, in which each pair of terminals s i, t i is also given an integer eman i, an our goal is to connect every source-sink pair s i, t i by i paths, an all paths connecting any source-sink pair must be isjoint. We can also generalize this question further, by allowing capacities on eges in the case of ege isjoint path problems. Here we are also given a capacity function c : A Z + or c : E Z + an our goal is again to connect each pair of terminals s i, t i by i paths, subject to the constraints that each (arc or) ege e is use at most c(e) times. So we can interpret the capacity on an (arc or) ege as a boun on the number of paths that are allowe to traverse this ege. Or we can even relax this further an consier a collection of (possibly non-integral) flows, which collectively satisfy the capacity constraints an such that the ith flow has i units between s i an t i. In the case of just a single source-sink pair s, t an integer eman, this problem is wellunerstoo: Let f be the maximum number of paths connecting s, t (subject to capacity constraints) in G = (V, E) (or in a igraph D = (V, A)). Then f is equal to the capacity of the minimum cut separating s from t. But for many source-sink pairs, the minimum cut is no longer equal to the maximum number of paths (subject to capacity constraints). In fact, each of the variants of the isjoint paths problem mentione above is NP -har, but the ifficulty of these problems is quite ifferent.. Directe an Unirecte Ege Disjoint Paths The unirecte an irecte ege isjoint path problems are rastically ifferent for a fixe number of source-sink pairs: Unirecte Ege Disjoint Paths. In the case of fixe k (i.e. there are k source-sink pairs), there is a polynomial time algorithm to ecie if there are ege isjoint paths connecting each s i, t i pair. This result follows from the Graph Minor Project of Robertson an Seymour (995), an is very involve. Yet for super-constant k, the unirecte ege isjoint paths problem is N P -complete. -

2 Directe Arc Disjoint Paths. Even for k = it is NP -complete to etermine if there are arc-isjoint paths P, P connecting s, t an s, t. In fact, we can even choose s = t an s = t an this problem still remains N P -complete. This rastic ifference in the range for which these problems are har, leas to very ifferent inapproximability results. For the problems of maximizing the number of arc-isjoint paths, in the case of a irecte graph, it is har to etermine the maximum number of arc-isjoint paths within any Ω(m ɛ ) for m arcs, an for any ɛ > 0. Yet for the case of maximizing the number of ege isjoint paths in an unirecte graph, the best known harness results are only poly-logarithmic. An in either case, the best known approximation ratio is O( m) (or actually O( n) for unirecte graphs on n vertices)..3 Reuctions Between Variants Here we consier all four problems, (unirecte or irecte) (ege or vertex) isjoint paths. We first note that the arc-isjoint path problem, an the vertex-isjoint path problem in a irecte graph are equivalent - i.e. we can give a polynomial time reuction in either irection. Given a vertex-isjoint path problem in a irecte graph, we can perform the reuction in Figure an get a are-isjoint path problem that is equivalent. Figure : Each vertex unergoes the illustrate transformation. In the case of an arc-isjoint path problem, we can perform a similar transformation given in Figure. Figure : Each vertex unergoes the illustrate transformation. Next we consier reuctions from unirecte variants of the isjoint path problem, an prove that both the ege isjoint path problem an the vertex-isjoint path problem (in unirecte graphs) can be reuce to either irecte path problem above. We can also reuce the ege isjoint path problem in unirecte graphs to the arc-isjoint path problem, an this reuction is given in Figure 3. This also implies that we can reuce the ege isjoint path problem in unirecte graphs to the irecte vertex-isjoint path problem. a b a b Figure 3: Each ege in the original unirecte graph is replace by the above gaget. -

3 For the case of the vertex isjoint path problem in unirecte graphs, we can just replace any ege (u, v) by two arcs (u, v) an (v, u). So this problem can also be reuce to either the arc-isjoint path problem or the vertex-isjoint path problem in irecte graphs. Lastly, the ege isjoint path problem in unirecte graphs can be reuce to the vertex-isjoint path problem in unirecte graphs by taking the line-graph of G. A path in the line-graph of G will also correspon to a path in the original graph G, but vertices in the line-graph correspon to eges in the original graph, so paths will be ege-isjoint in G iff the corresponing paths are vertex-isjoint in the line graph of G..4 Fractional Relaxations We focus on ege isjoint paths in unirecte graphs. When k =, flow is easy. We can fin integer flow using the max-flow min-cut theorem. In general, we can give a fractional relaxation (that can be solve because it can be written as a linear program): Let P i be set of all paths between s i an t i. We have a variable x p for every such path p P i. We have the following primal LP: max 0 x x p = i p P i x p c(e) i:p P i,e p x p 0. What oes ual mean in this case? We use variables l(e) for each ege e E, an variables b i for i =,..., k. min e E c(e)l(e) k b i i () i= l(e) b i 0 p P i i =,... k e p l(e) 0. To make the term ( k i= b i i ) small, we shoul make b i as large as possible. Fix the ege function l : E Q. Then b i is the (minimum) ist l (s i, t i ). The objective function of the ual () can be rewritten: c(e)l(e) e E k i ist l (s i, t i ). i= If the primal LP is feasible, then there is no solution for the ual LP with a negative objective value. So there exists a fractional multiflow if an only if l(e) 0, e E, the following hols: k c(e)l(e) i ist l (s i, t i ). () e E i= Duality shows that this is a necessary an sufficient for the existence of a fractional multiflow. -3

4 Integer Multiflows In general, the problem of etermining when there is an integer multiflow is NP-complete. However, there are special conitions that imply the existence of an integer multiflow in certain classes of graphs. Let R be a set of eges: R = {(s i, t i ) : i =,... k}. (3) The set of eges in E outgoing from vertex set U is enote by δ E (U) an the set of eges in R outgoing from vertex set U is enote by δ R (U). A necessary conition for the existence of a multiflow (an thus of an integer multiflow) is the cut conition: c(δ E (U)) (δ R (U)), U V. In general, the cut conition is not sufficient to guarantee the existence of an integer multiflow (or fractional multiflow) in a graph. However, in some cases of the multiflow problem, the cut conition is sufficient for the existence of a fractional multiflow. Furthermore, there are several cases known where the cut conition implies the existence of an integer multiflow when the Euler conition is satisfie: c(δ E (v)) + (δ R (v)) is even, for each vertex v. For example, when k =, we have the following implications: (i) Cut conition fractional multiflow. (ii) Cut conition an integer capacities half-integral multiflow. (iii) Cut conition, integer capacities, an Euler conition integral multiflow. s t s t Figure 4: When the capacity of each ege in this graph is an, =, the Euler conition is not satisfie. There exists a half-integral multiflow, but no integral multiflow. The first proof of (i) an (ii) for the case when k = was is ue to Hu. The proof of (iii) is ue to Rothschil an Winston. Note that (iii) implies (i) an (ii). For example, Consier the graph in Figure 4, let =, =. Let the capacity of each ege be. Note that the cut conition is satisfie but the Euler conition is not. However, suppose we ouble every capacity an eman, then the Euler conition is satisfie. We can convert an integer solution for this latter problem to a half-integral solution for the original problem. Some goo cases in which conitions (i), (ii) an (iii) are satisfie are: -4

5 . If there are two commoities, i.e. k =, then cut conition an Euler conition are sufficient for integer multiflow.. G + D has no K 5 minor, (a special case is when G + D is planar) where D is the eman graph, D = (V, R) (see (3)). 3. {(s, t ),..., (s k, t k )} G is planar an all (s i, t i ) are on bounary of outsie face. (Note that this oes not imply case because the eman graph coul be a clique) 5. If there are faces F, F an for each i, (s i, t i ) are both on the bounary of the same face F j for j {, }. 3 Two-Commoity Flows Theorem (Rothschil an Whinston) G = (V, E) is an unirecte graph such that c(e) Z + for e E. Terminals s, t, s, t are in V, an emans, are positive integers. Aitionally, the Euler conition is satisfie for G. Then G has an integer two-commoity flow if an only if the cut conition is satisfie. Proof: Our goal is to fin flows from s to t an from s to t with values an, respectively. We will show that if the cut conition is satisfie on G, then we can fin such flows. G : s s t s t t G : s s t t s t Figure 5: The graphs G an G are constructe base on the given graph G. First, base on the graph G, construct the graph G as shown in Figure 5. Note that the eges out of s are the only irecte eges in the graph G, an all other eges are inherite from G an hence unirecte. Let the eges (s, s ) an (t, t ) in G have capacity an the eges (s, s ) an (t, t ) in G have capacity. By the max-flow min-cut theorem, we can fin an integer s -t flow g with value +, since the min-cut of G has value +. Note that this s -t flow oes not necessarily give a two-commoity flow for the original problem (since some of the flow going through s may en up in t ). Since the Euler conition is satisfie, we will prove that we can assume that g(e) c(e) mo. To show this, first notice that the Euler conition implies that the total capacity incient to any vertex (except s or t ) of G is even. Furthermore, any integral flow will use up an even amount of capacity incient to any vertex. Now consier all the eges e E such that g(e) c(e) mo. Since it is the case that e δ(v) (g(e) c(e)) = 0 mo, it follows that an even number of eges ajacent to vertex v have g(e) c(e) mo. Thus, the eges such that g(e) c(e) mo make up an Eulerian graph (an o not contain the arcs incient to s an t that we ae to G to make up G, as these arcs are saturate). We can ecompose this Eulerian graph into cycles, an push push one unit of flow across all these cycles (either increasing or ecreasing the flow by one unit -5

6 along it epening on the orientation), changing the parity of g(e) for each such ege. Thus, for all eges e E, we have that g(e) c(e) mo. For G, we have the same argument. Thus, we fin an integer flow h in G with value + such that h(e) = c(e) mo, e E. Thus, for all eges e E, h(e) = g(e) mo. We arbitrarily orient the eges of E to obtain A. So for all a A, h(a) g(a) mo. Now we efine two flows on the graph G: f (a) = [g(a) + h(a)] f (a) = [g(a) h(a)]. The following properties are true for the flows f an f :. f (a), f (a) are integer flows (since g(a) an h(a) have the same parity).. f (a) + f (a) = g(a) + h(a) + g(a) h(a) max( g(a), h(a) ) c(a). 3. f is units of flow from s to t an f is units of flow from s to t. The last property hols because we can show that f (δ + (s )) f (δ (s )) = an f (δ (t )) f (δ + (t )) =. By conservation of flow, if we consier the vertex s in G, we have: Equations (4) an (5) imply f (δ + (s )) f (δ (s )) =. g(δ + (s )) g(δ (s )) = (4) h(δ + (s )) h(δ (s )) = (5) g(δ (t )) g(δ + (t )) = (6) h(δ (t )) h(δ + (t )) =. (7) Equations (6) an (7) imply f (δ (t )) f (δ + (t )) =. Similarly, we can show that the last property hols for flow f. If we consier vertices s an t in G, we have: g(δ + (s )) g(δ (s )) = (8) h(δ (s )) h(δ + (s )) = (9) g(δ (t )) g(δ + (t )) = (0) h(δ + (t )) h(δ (t )) =. () Equations (8) an (9) imply f (δ + (s )) f (δ (s )) = f (δ (t )) f (δ + (s )) =. an equations (0) an () imply As a final note, consier the problem of maximizing the sum of the flow between s an t an between s an t. This is the max biflow problem. A bicut is a cut separating s from t an s from t, thus it is either a cut separating s, s from t, t or a cut separating s, t from s, t. One can show that the following theorem follows from Theorem. Theorem The maximum biflow equals the minimum bicut. -6