Algorithms for multiflows

Transcription

1 Department of Computer Science, Cornell University September 29, 2008, Ithaca

2 Definition of a multiflow We are given an undirected graph G = (V, E) and a set of terminals A V. A multiflow, denoted by x = (x ab : a b, a, b A), is a set of ( ) A 2 flows xab between all pairs of terminals a b, a, b A. size(x) := size(x ab ) Problem: maximize the size of a multiflow subject to certain capacity and/or integrality constraints.

3 Special Cases A = V : max (fractional) matching A = 2: max s t flow /2 /2 /2

4 Definition of a multiflow We are given an undirected graph G = (V, E) and a set of terminals A V. A multiflow, denoted by x = (x ab : a b, a, b A), is a set of ( ) A 2 flows xab between all pairs of terminals a b, a, b A. size(x) := size(x ab ) Problem: maximize the size of a multiflow subject to certain capacity and/or integrality constraints.

5 Outline Research done at the Egerváry Research Group, Eötvös University, Budapest, Hungary, Edge-Capacitated Multiflows: Lovász-Cherkassky 972, half-integrality of the LP, Ibaraki-Karzanov-Nagamochi 998 Example: LP may have a very fractional extreme optimum. Node-Capacitated Multiflows: half-integrality of the LP, a strongly polynomial time algorithm Useful in the construction of a strongly polynomial time algorithm: Frank and Tardos 986 paper showing that with the help of the ellipsoid method a 0, LP is solvable in strongly polynomial time

6 All-one edge-capacities, Eulerian graph Lovász, Cherkassky, 972 Let G = (V, E) be eulerian, and let A V. Then the maximum size of an integral multiflow subject to all-one edge-capacities is equal to 2 λ(a, A a) a A (Essence: cut condition is necessary and sufficient.) Proof: Split-off. Implies a polynomial time algorithm.

7 All-one edge-capacities, Eulerian graph Lovász, Cherkassky, 972 Let G = (V, E) be eulerian, and let A V. Then the maximum size of an integral multiflow subject to all-one edge-capacities is equal to 2 λ(a, A a) a A (Essence: cut condition is necessary and sufficient.) Proof: Split-off. Implies a polynomial time algorithm.

8 All-one edge-capacities, Eulerian graph Lovász, Cherkassky, 972 Let G = (V, E) be eulerian, and let A V. Then the maximum size of an integral multiflow subject to all-one edge-capacities is equal to 2 λ(a, A a) a A (Essence: cut condition is necessary and sufficient.) Proof: Split-off. Implies a polynomial time algorithm.

9 All-one edge-capacities, Eulerian graph Lovász, Cherkassky, 972 Let G = (V, E) be eulerian, and let A V. Then the maximum size of an integral multiflow subject to all-one edge-capacities is equal to 2 λ(a, A a) a A (Essence: cut condition is necessary and sufficient.) Proof: Split-off. Implies a polynomial time algorithm.

10 All-one edge-capacities, Eulerian graph Lovász, Cherkassky, 972 Let G = (V, E) be eulerian, and let A V. Then the maximum size of an integral multiflow subject to all-one edge-capacities is equal to 2 λ(a, A a) a A (Essence: cut condition is necessary and sufficient.) Proof: Split-off. Implies a polynomial time algorithm.

11 All-one edge-capacities, Eulerian graph Lovász, Cherkassky, 972 Let G = (V, E) be eulerian, and let A V. Then the maximum size of an integral multiflow subject to all-one edge-capacities is equal to 2 λ(a, A a) a A (Essence: cut condition is necessary and sufficient.) Proof: Split-off. Implies a polynomial time algorithm.

12 All-one edge-capacities, Eulerian graph Lovász, Cherkassky, 972 Let G = (V, E) be eulerian, and let A V. Then the maximum size of an integral multiflow subject to all-one edge-capacities is equal to 2 λ(a, A a) a A (Essence: cut condition is necessary and sufficient.) Proof: Split-off. Implies a polynomial time algorithm.

13 All-one edge-capacities, Eulerian graph Lovász, Cherkassky, 972 Let G = (V, E) be eulerian, and let A V. Then the maximum size of an integral multiflow subject to all-one edge-capacities is equal to 2 λ(a, A a) a A (Essence: cut condition is necessary and sufficient.) Proof: Split-off. Implies a polynomial time algorithm.

14 All-one edge-capacities, Eulerian graph Lovász, Cherkassky, 972 Let G = (V, E) be eulerian, and let A V. Then the maximum size of an integral multiflow subject to all-one edge-capacities is equal to 2 λ(a, A a) a A (Essence: cut condition is necessary and sufficient.) Proof: Split-off. Implies a polynomial time algorithm.

15 All-one edge-capacities, Eulerian graph Lovász, Cherkassky, 972 Let G = (V, E) be eulerian, and let A V. Then the maximum size of an integral multiflow subject to all-one edge-capacities is equal to 2 λ(a, A a) a A (Essence: cut condition is necessary and sufficient.) Proof: Split-off. Implies a polynomial time algorithm.

16 LP of Edge-Capacitated Multiflows (condensed) Let G = (V, E) be a graph, A V be the set of terminals, and let c : E N denote a vector of integral edge-capacities. max size(x) subject to { x is a multiflow (non-negativity, flow-conservation) total.load x (e) c(e) for all e E (capacity constraint)

17 LP of Edge-Capacitated Multiflows G = (V, E), c : E N, and denote A = {a, a 2,, a k } V. max i<j x ij (δ in (a j )) subject to x 0 x ij (va l ) = 0 for i < j, l j va l E av l E x ij (zv) x ij (vz) = 0 zv E vz E x ij (a l v) = 0 for i < j, l i for i < j, v V A (x ij (pq) + x ij (qp)) c(pq) for pq E. i<j

18 Half-Integral Edge-Capacitated Multiflows Corollary Half-Integrality For any G = (V, E), A V, c : E N, the above LP has a half-integral optimum. Proof: Construct eulerian graph G c by replacing edge pq by a number of 2c(pq) paralel edges pq. Lovász-Cherkassky for G c, A implies a half-integral multiflow of size equal to the min cut upper bound. The min cut upper bound transforms into an LP dual of same value. (Same as for s t flows.)

19 Half-Integral Edge-Capacitated Multiflows Question: How to find a half-integral optimum? Possible approach: Solve LP and then play with the fractions until half-integral. Implies a polynomial time algorithm by applying the uncapacitated algorithm for the fractional part. Is this strongly polynomial? You can solve the LP in strongly polynomial time by a result of Frank and Tardos. But, by definition, you mustn t take the fractional part of numbers in a strongly polynomial time algorithm!

20 What is a strongly polynomial time algorithm? An algorithm with integer/rational numbers in the input is called strongly polynomial if it performs a polynomial number of operations +,,, /, 2 size of numbers throughout the computation stays within a polynomial factor of the size of the input numbers. Certain natural algorithms are NOT strongly polynomial: rounding a rational number (euclidian algorithm) 2 determining 2 k in decimals, 3 (most) scaling algorithms. Strongly polynomial: maximum flow, min cost circulation, min cost perfect matching, maximum weight capacitated b-matching, submodular function minimization 2 0, ± LP (Frank, Tardos, using ellipsoid method)

21 Determining 2 k (in decimals) = implying that = = requiring a total of 375 arithmetic operations.

22 Capacity Scaling for Half-Integral Edge-Capacitated Multiflows 2 2

23 Capacity Scaling for Half-Integral Edge-Capacitated Multiflows 0

24 Capacity Scaling for Half-Integral Edge-Capacitated Multiflows 0

25 Capacity Scaling for Half-Integral Edge-Capacitated Multiflows

26 Capacity Scaling for Half-Integral Edge-Capacitated Multiflows 2 2 2

27 Capacity Scaling for Half-Integral Edge-Capacitated Multiflows

28 Capacity Scaling for Half-Integral Edge-Capacitated Multiflows Capacity scaling + Lovász-Cherkassky weakly polynomial time algorithm to find a maximum half-integral multiflow. Naive running time using Goldberg-Rao for min cut: O(m 3 n min{n 2/3, m /2 } log(n 2 /m) log 2 n log U) O(n 8 log U) Remark: Babenko, Karzanov (ESA 08) constructed a faster scaling algorithm, achieving a running time of O(mn 2 min{n 2/3, m /2 } log(n 2 /m) log 2 n log 2 U) O(n 5 log 2 U), where U denotes the maximum capacity of a node.

29 Half-Integral Edge-Capacitated Multiflows Possible approach to achieve strongly polynomial time: Find an extreme optimum of the LP. Maybe it is guaranteed to be half-integral. This would imply a strongly polynomial time algorithm by Frank-Tardos. Unluckily, the following example show that the polytope of edge-capacitated multiflows may have very fractional extreme points.

30 The polytope of multiflows can have a fractional extreme point:

31 The polytope of multiflows can have a fractional extreme point: /3

32 The polytope of maximum multiflows can have a fractional extreme point:

33 The polytope of maximum multiflows can have a fractional extreme point: /3

34 The polytope of maximum multiflows can have a fractional extreme point: /3

35 The polytope of maximum multiflows can have a fractional extreme point: /3

36 Half-Integral Edge-Capacitated Multiflows A strongly polynomial time algorithm motivated by an idea of Karzanov, Ibaraki, Nagamochi (998): Decompose the graph by minimum {a, a 2 } - {a 3,.., a k } cuts into instances with three terminals. 2 Solve the three terminal case by combining two circulations.

37 Decompose by {a, a 2 } - {a 3,.., a k } cuts a a 2 a 3 a 4 a 5

38 Decompose by {a, a 2 } - {a 3,.., a k } cuts a a 2 a 3 a 4 a 5

39 Decompose by {a, a 2 } - {a 3,.., a k } cuts a a 2 a 3 a 4 a 5

40 Half-Integral Edge-Capacitated Multiflows Corollary Reduction to 3 Terminals Enough to solve for three terminals.

41 3 Terminals a β a 2 α + β = λ c ({a }, {a 2, a 3 }) β + γ = λ c ({a 2 }, {a 3, a }) γ + α = λ c ({a 3 }, {a 2, a }) We can compute α, β, γ! α γ a 3

42 3 Terminals There is a circulation z satisfying excess z (a ) = α + β excess z (a 2 ) = γ β excess z (a 3 ) = α γ excess z (v) = 0 for v V A. a β a 2 α γ We construct an integral z by a flow algorithm. a 3

43 3 Terminals And there is a circulation y satisfying excess y (a ) = α + β excess y (a 2 ) = γ β excess y (a 3 ) = γ α excess y (v) = 0 for v V A. a β a 2 α γ We construct an integral y by a flow algorithm. a 3

44 3 Terminals Then 2 (z + y) satisfies excess 2 (z+y)(a ) = α + β excess 2 (z+y)(a 2) = β excess 2 (z+y)(a 3) = α excess 2 (z+y)(v) = 0 for v V A, β a a 2 α thus 2 (z +y) is a flow of value α + β from a to a 2, a 3. a 3

45 3 Terminals Then 2 (z y) satisfies excess 2 (z y)(a ) = 0 excess 2 (z y)(a 2) = γ excess 2 (z y)(a 3) = γ excess 2 (z y)(v) = 0 for v V A, a a 2 γ thus 2 (z y) is a flow of value γ from a 3 to a 2. a 3

46 3 Terminals To show that the multiflow constructed from 2 (z + y) and 2 (z y) satisfies the capacity constraint for an edge pq, we need that 2 (z + y)(e) + 2 (z y)(e) c(e). Proof: using the elemetary inequality for all n, m R we get that n + m + n m max{2 n, 2 m } 2 (z + y)(e) + 2 (z y)(e) max{ y(e), z(e) } c(e).

47 Theorem (based on Ibaraki-Karzanov-Nagamochi) There is a strongly polynomial time algorithm to find a maximum half-integral multiflow subject to edge-capacities.

48 LP for Node-Capacitated Multiflows (condensed) Let G = (V, E) be a graph, A V be the set of terminals, and let c : V N denote a vector of integral node-capacities. max size(x) subject to { x is a multiflow total.load x (p) c(p) for all p V.

49 LP for Node-Capacitated Multiflows Let G = (V, E) be a graph, A V be the set of terminals, and let c : V N denote a vector of integral node-capacities. Moreover, let A = {a, a 2,, a k }. i<j,i l max x ij (δ in (a j )) subject to i<j x ij (δ in (a l )) = 0 for i < j, l j x ij (δ out (a l )) = 0 for i < j, l i x ij (δ in (v)) x ij (δ out (v)) = 0 for i < j, v V A x ij (δ in (v)) c(v) for v V A i<j x ij (δ in (a l )) + x lj (δ out (a l )) c(a l ) for l k. l<j

50 LP Dual for Node-Capacitated Multiflows The LP dual, on variables w : V R is equivalent with: min v V c(v) w(v) subject to w 0 w(v) v P for every path P connecting two distinct a, b A

51 Node-Capacitated Multiflows Theorem Primal and Dual Half-Integrality LP for maximum node-capacitated multiflow, and its dual both have a half-integral optimum.

52 Example Node-Capacitated Multiflows

53 Example Node-Capacitated Multiflows

54 Example Node-Capacitated Multiflows 0 0 0

55 Example 2 Node-Capacitated Multiflows

56 Example 2 Node-Capacitated Multiflows 2 2 2

57 Example 2 Node-Capacitated Multiflows /2 /2 /2

58 Example 3 Node-Capacitated Multiflows

59 Example 3 Node-Capacitated Multiflows

60 Example 3 Node-Capacitated Multiflows /2 /2 /2 /2 /2

61 Theorem Dual Half-Integrality Dual of the LP for maximum node-capacitated multiflow has a half-integral optimum. Sketch Proof (Vazirani). Let w : V R + be an LP dual optimum. For a node v V we define w as follows: If w(v) =, and there is a path P traversing v, connecting two nodes of A, such that w(p) = 0 for all p P v, then let w (v) :=. Otherwise, if w(v) > 0, and there is a path R from a node in A to v such that w(p) = 0 for all p R, then let w (v) := /2. Otherwise, let w (p) := 0. Consider a maximum multiflow x. Then x, w satisfy certain complementary slackness conditions: w(p) = for every path P in the support of a flow in x. w(v) > 0 implies that the capacity of v is saturated. Then one can show that x, w also satisfy those complementary slackness conditions. This implies w is a dual optimum.

62 Node-Capacitated Multiflows Theorem Primal Half-Integrality LP for maximum node-capacitated multiflow has a half-integral optimum. Lemma Vertices of the shortest maximum multiflow polytope are half-integral. This Lemma also implies a strongly polynomial time algorithm!

63 Node-Capacitated Multiflows LP for maximum multiflows: M := max size(x) { x total.load x (p) c(p) subject to is a multiflow for all p V LP for shortest maximum multiflows: L := min x subject to x is a multiflow total.load x (p) c(p) for all p V size(x) = M

64 Node-Capacitated Multiflows Polytope of shortest maximum multiflows: x is a multiflow total.load x (p) c(p) for all p V Lemma size(x) = M x = L Vertices of the shortest maximum multiflow polytope are half-integral. Corollary Strongly polynomial time algorithm to find a maximum half-integral node-capacitated multiflow.

65 Example with all-one edge-capacities to show that the polytope of shortest maximum multiflows is half-integral, but the polytope of maximum multiflows is not:

66 Lemma Vertices of the shortest maximum multiflow polytope are half-integral. Proof Sketch. Let y be a vertex of the shortest maximum multiflow polytope. One can show that for all a A there are certain disjoint regions R a such that a R a, and so that, with the notation R := R a and Q := V R, both of the following conditions hold: y ab (pq) = 0 for all a, b A, pq E, p V R a, q V R b 2 total.load y (q) = c(q) for all q Q.

67 This implies that y is a vertex of the following polytope: x is a multiflow total.load x (p) c(p) for all p V size(x) = M x = L x ab (pq) = 0 for a, b A, pq E, p V R a, q V R b total.load x (q) = c(q) for all q Q. Claim This polytope is the projection of the fractional b-matching polytope of an auxiliary graph.

68 Lemma Vertices of the shortest maximum multiflow polytope are half-integral.

69 Concluding Remarks Integral multiflow problems solved by Mader s min-max (978). A polynomial time algorithm to find a maximum integral multiflow subject to all-one node-capacities follows from the linear matroid matching of Lovász (98) and Schrijver s representation (200). Weakly polynomial time algorithm to find a maximum integral multiflow subject to edge-capacities by Keijsper, Pendavingh, Stougie (2006). Strongly polynomial time algorithm to find a maximum integral multiflow with node-capacities by combining P (2007) and P (2008). Open Questions: Improve running time, simplify, get rid of LP? Group-Labeled Graphs (Chudnovsky, Geelen, Gerards, Goddyn, Lohman, Seymour, 2005)