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A tableau on integer multiflow feasibility

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Explore the cut condition, directed and undirected graphs, and theorems related to integer multiflow feasibility. Discover new problems and reductions in the context of graph theory.

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A tableau on integer multiflow feasibility

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  1. A tableau on integermultiflow feasibility Guyslain Naves&András Sebő Univ. J. Fourier CNRS in Bonn: 85, 91 88-89, 92-93 Laboratoire G-SCOP, Grenoble

  2. Footnotes détaillés :

  3. Disjoint paths G =(V,E)(di)graph H= (T, E(H) ) (di)graph (T V) We want for all ab E(H) : (b,a) -path, pairwise edge-disj. That is, pairwise edge-disjoint cycles. Necessary: The Cut Condition, that is, For every partition of V, more edges in G than in H

  4. G

  5. G -1 -4 H 3 c(e) parallel copies r(e) parallel copies

  6. 1 1 -1 A remark on fractional flows Russian, Japanese schools: cone of paths Lomonosov (I heard it in Bonn 1991  1) : cone of Compact description ; Algorithmic consequence? LP (Farkas for feasibility, Duality thm for max): « Japanese theorem » cone of edges

  7. directed G directed acyclic undirected vertex- vertex- arc vertex- arc-disj arc-disj G H r disj disj disj gen Euler gen gen Euler Euler bin arb Euler: G+H Eulerian vertex-disjoint: no numbers Interesting new problems, but out of actual context un bin un 321,5 fix gen fix in P for bin 3 demand edges: Ibaraki, Poljak ‘91 un 2 fix 2  bin arb un 3  3  3 bin un fix plan fix bin H : |E(H)|= arbitrary, const or 2. un 2 fix 2 binary: numbers on edges (size = log) bin arb un unary: no numbers (parallel copies) bin G+H plan un fix fix fix: demands (sum of r) bounded by const bin un 2 2 : 2 demand-edges with a demand of 1. fix 2

  8. Examples of reductions (edge-disj.) Undirected  directed G acyclic  undirected, G + H Eulerian: Vygen’s Lemma « Metareduction »: undirected planar  acyclic planar NPCP P NPC

  9. G+H planar (Bonn 1985-89) 1.) Eulerian : Thm (Seymour ‘81): If G+H planar and Eulerian, then Cut Condition  disjoint paths Edmonds-Johnson,Lovász, Seymour, Frank-S.-Tardos, S. : ‘Quick proof …’ ( Bonn, 1985) . Pedagogically the following proof might be best :

  10. G -1 1 H « most negative dual path » 1 neg edge in coboundary

  11. G The cut condition is still satisfied, Because if not, there exists a tight cut st. switching on it the furthermost face has 2 neg edges. H

  12. 2.) G+H planar, general (not necessarily Eulerian) Thm (S. 1988, Budapest-Bonn) : pol if H fixed Thm (Middendorf, Pfeiffer 1989, Bonn) : NPC in general. Not yet NP = coNP : Open: G+H planar, H = (‘Easiest’ open special case of Lex’s Problem 50)

  13. Edge-disjoint acyclic or undirected G acyclic G+ H Euler (Vygen) + undirected G planar di, H 2 parallel classes (Müller) Terminals on outer boundary: Eulerian (Okamura-Seymour) Inner Eulerian(Frank) Gen (Schwärzler 2008 = Lex’s Problem 56) 3 terminals Naves: 2 demandsouter boundary or G+H planar

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