#### DMCA

## Criticality for multicommodity flows

Venue: | J. Combinatorial Theory, Ser. B |

Citations: | 36 - 1 self |

### Citations

901 |
Combinatorial Optimization - Polyhedra and Efficiency
- Schrijver
- 2003
(Show Context)
Citation Context ...f p (and indeed the solution is unique), as follows: • take 2p + 1 disjoint sets of vertices A1, . . . , A2p+1, each of cardinality p + 2, and let Ai = {a1i , . . . , a p i , ui, vi} for 1 ≤ i ≤ 2p + 1; • let aji be adjacent to a j i+1 for all i, j (where a j 2p+2 means a j 1), and let ui, vi be adjacent to a1i , . . . , a p i for all i; • let k = p(2p + 1), and let the list (s1, t1), . . . , (sk, tk) consist of (ui, vi+p) for 1 ≤ i ≤ 2p + 1 (reading subscripts modulo 2p + 1), together with p − 1 copies of (ui, vi) for 1 ≤ i ≤ 2p + 1. A more complicated construction due to Lomonosov [10] (see [19] for a simplified version) shows more-or-less the same thing, with k0 = 3 in addition. Let us return to 1.1, and dispose of the case p = ∞ before we start on the serious proofs. 1.7 If (G, s1, t1, . . . , sk, tk) is R–critical, then there exists an integer p > 1 (depending on G) such that (G, s1, t1, . . . , sk, tk) is Z/p–critical. Consequently, to prove 1.1 when p = ∞ it suffices to prove 1.1 for all finite p > 1. Proof. Since (G, s1, t1, . . . , sk, tk) is R–critical, it follows that for every edge e of G, there are k flows in G/e satisfying the requirements of the multicommodity flow probl... |

312 |
Multi-Terminal Network Flows.
- Gomory, Hu
- 1961
(Show Context)
Citation Context ...si, ti. This is always necessary for Z/p–feasibility (and indeed for R–feasibility); and it is sufficient when all the pairs (si, ti) are the same, and it was heavily investigated in early work on the multicommodity flow problem. It turns out that the cut condition is not sufficient for Z–feasibility for any k > 1, and not sufficient for R–feasibility for any k ≥ 3, and yet in some interesting circumstances the cut condition is necessary and sufficient for Z/2–feasibility. For instance, this is the case when • there are only two distinct pairs (si, ti) (this is Hu’s two-commodity flow theorem [6]), or more generally, the set {s1, t1, . . . , sk, tk} has at most four elements [21] • the graph obtained from G by adding edges to make all the pairs si, ti adjacent is planar [22], or • G is planar, and can be drawn so that s1, t1, . . . , sk, tk all belong to the infinite face [14], and more (see [9]). Our result 1.3 is a sort of generalization of these, and has the same complexity consequences. For instance, to check if (G, s1, t1, . . . , sk, tk) is Z/p–feasible (where p > 1), it suffices to check if there is some partition of V (G) into a bounded number of sets with the property that af... |

298 | On the complexity of timetable and multicommodity flow problems - Even, Itai, et al. - 1976 |

269 |
A new polynomial time algorithm for linear programming
- Karmarkar
- 1984
(Show Context)
Citation Context ...ertex v of G, φi(δ +(v)) − φi(δ −(v)) = 0 if v 6= si, ti 1 if v = si 6= ti −1 if v = ti 6= si 0 if v = si = ti, where δ+(v), δ−(v) denote the sets of non-loop edges with tail v and head v respectively, and for X ⊆ E(G), φi(X) denotes ∑ e∈X φi(e), and • for every edge e, ∑ 1≤i≤k |φi(e) |≤ 1. (We allow si = ti, for convenience when we contract edges, although then φi might as well be identically zero.) If there is a solution we say the k-commodity flow problem is R–feasible. This is a linear programme of polynomial size, and so we can check R–feasibility in polynomial time [4, 8], independent of k. In this paper we are interested in restricting the values φi(e). Let p ≥ 1 be an integer, and let Z/p denote the set of all rationals q/p where q is an integer. If φ1, . . . , φk can be chosen as above so that in addition • φi(e) ∈ Z/p for 1 ≤ i ≤ k and for each edge e ∈ E(G) we say the problem is Z/p–feasible (or Z–feasible if p = 1) and we call deciding Z/p–feasibility the (k, p)-commodity flow problem. To include the original R–feasibility problem in this language, let us set p = ∞; thus, Z/∞–feasibility means R–feasibility. From the algorithmic point of view, checking Z... |

150 |
Quickly excluding a planar graph,
- Robertson, Seymour, et al.
- 1994
(Show Context)
Citation Context ... ∩ B) |≥ w1 + min(|W |/3, w2) ≥ |W |/3 since w1 + w2 ≥ |W |/2. It follows that |V (A ∩ B) |≥ |W |/(6k) ≥ K/(6k), a contradiction. This proves our claim that if (A,B) ∈ T then |E(A) ∩W |< |W |/3, and the second tangle axiom follows. For the third, let (A,B) ∈ T , and suppose that V (A) = V (G). Since |E(B) ∩ W |≥ |W |/2, it follows that at least |W |/2 edges have both ends in V (A∩B); but there are at most k|V (A∩B) |such edges since every vertex has degree at most 2k, and so k|V (A∩B) |≥ |W |/2 ≥ K/2, a contradiction. This proves (3). Since K/(6k) = 2064(b+s+1) 5 , (3) and the main theorem of [18] imply that there is a wall M of height b + s + 1 in G, such that for every (A,B) ∈ T of order at most b + s + 1, B includes a row of the wall. (The theorem of [23] gives a grid minor rather than a wall subgraph, so we adjusted the numbers to get a 2(b + s + 3)× 2(b + s + 3) grid minor, which gives a wall of height b + s + 1.) This wall has b + s diagonal vertices. (4) There are at least b diagonal vertices of M that are pairwise four-edge-connected. We recall that by hypothesis, either there are at most s edges of G that are not parallel to other 19 edges, or there are at most s vertices in G... |

117 | Graph searching, and a min-max theorem for treewidth,
- Seymour, Thomas
- 1993
(Show Context)
Citation Context ...) ∈ T then |E(A) ∩W |< |W |/3, and the second tangle axiom follows. For the third, let (A,B) ∈ T , and suppose that V (A) = V (G). Since |E(B) ∩ W |≥ |W |/2, it follows that at least |W |/2 edges have both ends in V (A∩B); but there are at most k|V (A∩B) |such edges since every vertex has degree at most 2k, and so k|V (A∩B) |≥ |W |/2 ≥ K/2, a contradiction. This proves (3). Since K/(6k) = 2064(b+s+1) 5 , (3) and the main theorem of [18] imply that there is a wall M of height b + s + 1 in G, such that for every (A,B) ∈ T of order at most b + s + 1, B includes a row of the wall. (The theorem of [23] gives a grid minor rather than a wall subgraph, so we adjusted the numbers to get a 2(b + s + 3)× 2(b + s + 3) grid minor, which gives a wall of height b + s + 1.) This wall has b + s diagonal vertices. (4) There are at least b diagonal vertices of M that are pairwise four-edge-connected. We recall that by hypothesis, either there are at most s edges of G that are not parallel to other 19 edges, or there are at most s vertices in G that have odd degree. Let us say a branch of M is a path P of M with distinct ends, both of degree three in M , and such that every internal vertex of P has degree... |

92 |
Multicommodity flows in planar graphs
- Okamura, Seymour
- 1981
(Show Context)
Citation Context ...or Z–feasibility for any k > 1, and not sufficient for R–feasibility for any k ≥ 3, and yet in some interesting circumstances the cut condition is necessary and sufficient for Z/2–feasibility. For instance, this is the case when • there are only two distinct pairs (si, ti) (this is Hu’s two-commodity flow theorem [6]), or more generally, the set {s1, t1, . . . , sk, tk} has at most four elements [21] • the graph obtained from G by adding edges to make all the pairs si, ti adjacent is planar [22], or • G is planar, and can be drawn so that s1, t1, . . . , sk, tk all belong to the infinite face [14], and more (see [9]). Our result 1.3 is a sort of generalization of these, and has the same complexity consequences. For instance, to check if (G, s1, t1, . . . , sk, tk) is Z/p–feasible (where p > 1), it suffices to check if there is some partition of V (G) into a bounded number of sets with the property that after identifying the vertices in each set, the problem is not Z/p–feasible. We can do this in polynomial time; for let P be the set of edges with ends in different sets of this partition. We may assume that |P |is bounded, because if there are more than k edges between any two of the se... |

55 |
Disjoint paths in graphs”,
- Seymour
- 1980
(Show Context)
Citation Context ...rge Z/3–critical instances with |W |= 3. For instance, let n ≥ 0 be an integer, let G be the complete bipartite graph K3,9n+2, and let u, v,w be its three vertices of degree 9n + 2. Let k = 12n + 3. For 1 ≤ i ≤ k, let (si, ti) = (u, v) if 1 ≤ i ≤ k/3 (v,w) if k/3 < i ≤ 2k/3 (w, u) if 2k/3 < i ≤ k; then (G, s1, t1, . . . , sk, tk) is Z/3–critical. These can be converted to unbounded Z/3–critical instances with k0 = 2, by adding a new vertex adjacent to u, v. There are similar examples for all odd p > 1. (Incidentally, when |W |= 3, checking Z/p-feasibility is polynomial-time solvable [20] with no bound on k.) Let us try p = 2 instead; and in that case setting |W |≤ 4 gives us no problem, as all critical instances have two vertices [21]. But we cannot go much further. If we hope for boundedness of the critical instances in the class, it is necessary that deciding Z/2-feasibility in the class is polynomiallysolvable; and Middendorf and Pfeiffer (published in [15]) showed that checking Z/2-feasibility is NP-complete when k0 = 3. Incidentally, a recent result of Moitra [13] implies that there are R–critical instances with an unbounded number of vertices when k0 = 3, although R-fea... |

43 |
On the complexity of the disjoint path problem
- Middendorff, Pfeiffer
- 1993
(Show Context)
Citation Context .... If φ1, . . . , φk can be chosen as above so that in addition • φi(e) ∈ Z/p for 1 ≤ i ≤ k and for each edge e ∈ E(G) we say the problem is Z/p–feasible (or Z–feasible if p = 1) and we call deciding Z/p–feasibility the (k, p)-commodity flow problem. To include the original R–feasibility problem in this language, let us set p = ∞; thus, Z/∞–feasibility means R–feasibility. From the algorithmic point of view, checking Z–feasibility and checking Z/2–feasibility behave similarly, in that both problems are solvable in polynomial time when k is fixed [17], and both are NP-complete if k is not fixed [3, 12]. Nevertheless, there is a significant difference between the two problems, as we shall see. If a k-commodity flow problem is feasible, and we contract some edge, this results in a new kcommodity flow problem that is also feasible. (Contraction may make loops or parallel edges.) Thus, given G and s1, t1, . . . , sk, tk as before, let us say (G, s1, t1, . . . , sk, tk) is Z/p–critical if 1 • the corresponding k-commodity flow problem is not Z/p–feasible; • for every edge e, if we contract e then the k-commodity flow problem becomes Z/p–feasible; and • no vertex different from s1, t1, . . . , sk... |

37 | obstructions to tree-decomposition, - unknown authors - 1991 |

36 |
Combinatorial Approaches to Multiflow Problems.”
- Lomonosov
- 1985
(Show Context)
Citation Context ...ultiples of p (and indeed the solution is unique), as follows: • take 2p + 1 disjoint sets of vertices A1, . . . , A2p+1, each of cardinality p + 2, and let Ai = {a1i , . . . , a p i , ui, vi} for 1 ≤ i ≤ 2p + 1; • let aji be adjacent to a j i+1 for all i, j (where a j 2p+2 means a j 1), and let ui, vi be adjacent to a1i , . . . , a p i for all i; • let k = p(2p + 1), and let the list (s1, t1), . . . , (sk, tk) consist of (ui, vi+p) for 1 ≤ i ≤ 2p + 1 (reading subscripts modulo 2p + 1), together with p − 1 copies of (ui, vi) for 1 ≤ i ≤ 2p + 1. A more complicated construction due to Lomonosov [10] (see [19] for a simplified version) shows more-or-less the same thing, with k0 = 3 in addition. Let us return to 1.1, and dispose of the case p = ∞ before we start on the serious proofs. 1.7 If (G, s1, t1, . . . , sk, tk) is R–critical, then there exists an integer p > 1 (depending on G) such that (G, s1, t1, . . . , sk, tk) is Z/p–critical. Consequently, to prove 1.1 when p = ∞ it suffices to prove 1.1 for all finite p > 1. Proof. Since (G, s1, t1, . . . , sk, tk) is R–critical, it follows that for every edge e of G, there are k flows in G/e satisfying the requirements of the multicommodity ... |

21 | odd cuts and plane multicommodity flows”,
- Seymour
- 1981
(Show Context)
Citation Context ...n early work on the multicommodity flow problem. It turns out that the cut condition is not sufficient for Z–feasibility for any k > 1, and not sufficient for R–feasibility for any k ≥ 3, and yet in some interesting circumstances the cut condition is necessary and sufficient for Z/2–feasibility. For instance, this is the case when • there are only two distinct pairs (si, ti) (this is Hu’s two-commodity flow theorem [6]), or more generally, the set {s1, t1, . . . , sk, tk} has at most four elements [21] • the graph obtained from G by adding edges to make all the pairs si, ti adjacent is planar [22], or • G is planar, and can be drawn so that s1, t1, . . . , sk, tk all belong to the infinite face [14], and more (see [9]). Our result 1.3 is a sort of generalization of these, and has the same complexity consequences. For instance, to check if (G, s1, t1, . . . , sk, tk) is Z/p–feasible (where p > 1), it suffices to check if there is some partition of V (G) into a bounded number of sets with the property that after identifying the vertices in each set, the problem is not Z/p–feasible. We can do this in polynomial time; for let P be the set of edges with ends in different sets of this partit... |

13 | and P.Ragde, “Characterizing multiterminal flow networks and computing flows in networks of small treewidth”,
- Hagerup, Nishimura
- 1998
(Show Context)
Citation Context ...for G, v1, . . . , vk and c) is a graph H, distinct vertices w1, . . . , wk of H, and a map d : E(H) → R+, satisfying the following. For every I ⊆ {1, . . . , k} and every r ∈ R+ , there exists A ⊆ V (G) satisfying • for 1 ≤ i ≤ k, vi ∈ A if and only if i ∈ I • ∑ e∈δG(A) c(e) ≤ r if and only if there exists B ⊆ V (H) satisfying • for 1 ≤ i ≤ k, wi ∈ B if and only if i ∈ I • ∑ e∈δH(B) d(e) ≤ r. 6 (δG(A) or δ(A) is the set of non-loop edges of G with exactly one end in A.) The objective is to find mimicking networks with as few vertices as possible; and Hagerup, Katajainen, Nishimura, and Ragde [5] showed that for every choice of G, v1, . . . , vk, c there is a mimicking network with at most 22 k vertices. But the flow problem that mimicking networks handle is the one-commodity multiterminal flow problem, not the multicommodity flow problem, so our result is different, and more general (except that we do not permit capacities). 2 Demand systems It is helpful to reformulate the various problems a little. First, let us add a new vertex v0 and 2k new edges, each incident with v0, and respectively incident with s1, t1, . . . , sk, tk. (Thus if two of s1, t1, . . . , sk, tk are equal we add ... |

8 |
A.Schrijver, Geometric Algorithms and Combinatorial Optimization,
- Grotschel
- 1988
(Show Context)
Citation Context ...ertex v of G, φi(δ +(v)) − φi(δ −(v)) = 0 if v 6= si, ti 1 if v = si 6= ti −1 if v = ti 6= si 0 if v = si = ti, where δ+(v), δ−(v) denote the sets of non-loop edges with tail v and head v respectively, and for X ⊆ E(G), φi(X) denotes ∑ e∈X φi(e), and • for every edge e, ∑ 1≤i≤k |φi(e) |≤ 1. (We allow si = ti, for convenience when we contract edges, although then φi might as well be identically zero.) If there is a solution we say the k-commodity flow problem is R–feasible. This is a linear programme of polynomial size, and so we can check R–feasibility in polynomial time [4, 8], independent of k. In this paper we are interested in restricting the values φi(e). Let p ≥ 1 be an integer, and let Z/p denote the set of all rationals q/p where q is an integer. If φ1, . . . , φk can be chosen as above so that in addition • φi(e) ∈ Z/p for 1 ≤ i ≤ k and for each edge e ∈ E(G) we say the problem is Z/p–feasible (or Z–feasible if p = 1) and we call deciding Z/p–feasibility the (k, p)-commodity flow problem. To include the original R–feasibility problem in this language, let us set p = ∞; thus, Z/∞–feasibility means R–feasibility. From the algorithmic point of view, checking Z... |

7 |
Half-integral five-terminus flows”,
- Karzanov
- 1987
(Show Context)
Citation Context ... any k > 1, and not sufficient for R–feasibility for any k ≥ 3, and yet in some interesting circumstances the cut condition is necessary and sufficient for Z/2–feasibility. For instance, this is the case when • there are only two distinct pairs (si, ti) (this is Hu’s two-commodity flow theorem [6]), or more generally, the set {s1, t1, . . . , sk, tk} has at most four elements [21] • the graph obtained from G by adding edges to make all the pairs si, ti adjacent is planar [22], or • G is planar, and can be drawn so that s1, t1, . . . , sk, tk all belong to the infinite face [14], and more (see [9]). Our result 1.3 is a sort of generalization of these, and has the same complexity consequences. For instance, to check if (G, s1, t1, . . . , sk, tk) is Z/p–feasible (where p > 1), it suffices to check if there is some partition of V (G) into a bounded number of sets with the property that after identifying the vertices in each set, the problem is not Z/p–feasible. We can do this in polynomial time; for let P be the set of edges with ends in different sets of this partition. We may assume that |P |is bounded, because if there are more than k edges between any two of the sets we could replace... |

4 |
and J.Wyllie, “The directed subgraph homeomorphism problem”,
- Fortune
- 1980
(Show Context)
Citation Context .... If φ1, . . . , φk can be chosen as above so that in addition • φi(e) ∈ Z/p for 1 ≤ i ≤ k and for each edge e ∈ E(G) we say the problem is Z/p–feasible (or Z–feasible if p = 1) and we call deciding Z/p–feasibility the (k, p)-commodity flow problem. To include the original R–feasibility problem in this language, let us set p = ∞; thus, Z/∞–feasibility means R–feasibility. From the algorithmic point of view, checking Z–feasibility and checking Z/2–feasibility behave similarly, in that both problems are solvable in polynomial time when k is fixed [17], and both are NP-complete if k is not fixed [3, 12]. Nevertheless, there is a significant difference between the two problems, as we shall see. If a k-commodity flow problem is feasible, and we contract some edge, this results in a new kcommodity flow problem that is also feasible. (Contraction may make loops or parallel edges.) Thus, given G and s1, t1, . . . , sk, tk as before, let us say (G, s1, t1, . . . , sk, tk) is Z/p–critical if 1 • the corresponding k-commodity flow problem is not Z/p–feasible; • for every edge e, if we contract e then the k-commodity flow problem becomes Z/p–feasible; and • no vertex different from s1, t1, . . . , sk... |

1 |
T.Klimosova, and P.Seymour, “Immersion in four-edge-connected graphs”, submitted for publication (manuscript
- Chudnovsky
- 2013
(Show Context)
Citation Context ....1, assuming 5.1. Let k ≥ 0, and let K be as in the first statement of 5.1. We claim that K satisfies 4.1. For let (G, v0,D, p) be a demand system of degree at most k, with p > 1, and let Y ⊆ V (G)\{v0}, where Y is robust and |δ(Y ) |≥ K. Choose X ⊆ V (G) with Y ⊆ X ⊆ V (G)\{v0}, with δ(X) minimum. Then X is K-supported, and hence p-porous by 5.1. But then E(G|X) is contractible by 3.1, and hence so is E(G|Y ). This proves 4.1. There are three alternative hypotheses in 5.1, and the first is handled by applying a lemma proved in [11]. The second is handled by another method, using a lemma from [1] (and the third could be done either way, and we choose to use the second method for it). We begin with the first method. A separation of order k in a graph G is a pair (A,B) of subgraphs of G such that A ∪ B = G, E(A ∩ B) = ∅, and |V (A ∩ B) |= k. If θ ≥ 1 is an integer, a tangle of order θ in a graph G is a set T of separations of G, each of order less than θ, such that • for every separation (A,B) of order less than θ, T contains at least one of (A,B), (B,A) • if (Ai, Bi) ∈ T for i = 1, 2, 3, then A1 ∪ A2 ∪ A3 6= G • if (A,B) ∈ T then V (A) 6= V (G). Let G,H be graphs, where H is simple. A ... |