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Incidentor coloring : methods and results

Incidentor coloring : methods and results. A.V . Pyatkin "Graph Theory and Interactions" Durham, 2013. 4-color problem. Reduction to the vertex coloring. Vertex coloring problem. Edge coloring problem. Incidentor coloring.

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Incidentor coloring : methods and results

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  1. Incidentorcoloring: methods and results A.V. Pyatkin "Graph Theory and Interactions" Durham, 2013

  2. 4-color problem

  3. Reduction to the vertex coloring

  4. Vertex coloring problem

  5. Edge coloring problem

  6. Incidentorcoloring • Incidentor is a pair (v,e) of a vertex vand an arc (edge) e, incident with it. • It is a half of an arc (edge) adjoining to a given vertex. • initialfinal • (mated incidentors)

  7. Incidentorcoloring problem • Color all incidentors of a given multigraph by the minimum number of colors in such a way that the given restrictions on the colors of adjacent (having a joint vertex) and mated (having a joint arc) incidentors would be stisfied

  8. An example of incidentorcoloring

  9. Incidentorcoloring generalizes both vertex and edge coloring

  10. Incidentorcoloring generalizes both vertex and edge coloring

  11. Motivation Central computer All links capacities areequal to 1 … … … j n 1 i i-thobject must send to j-th onedijunits of information

  12. There are two ways of information transmission: • 1) Directly fromi-thobject to j-th one (during one time unit); • 2) With memorizing in the central computer (receive the message from i-thobject, memorize it, and transmit to j-th one later).

  13. Reduction to the incidentorcoloring • Each object corresponds to a vertex of the multigraph(n vertices). • Each unit of information to transmit from i-th object to j-th one corresponds to the arcij of the multigraph (there are dij arcs going from a vertex i to the a vertex j). • The maximum degree  equals the maximum load of the link.

  14. Scheduling • To each information unit two time moments should be assigned – when it goes via i-th and j-th links. • These moments could be interpreted as the colors of the incidentors of the arcij. a b i j

  15. Restrictions a • The colors of adjacent incidentors must be distinct. • For every arc, the color of its initial incidentor is at most the color of the final incidentor, i.e. a  b. b i j

  16. It is required to color all incidentors by the minimum number of colorssatisfying all the restrictions (the length of the schedule is). • For this problem=. Such coloring can be found in O(n22) time. • (P., 1995)

  17. Sketch of the algorithm • Consider an arc that is not colored yet • Try to color its incidentors: • 1. In such a way that a=b • 2. In such a way that a<b • Otherwise, modify the coloring (consider bicolored chains)

  18. =3

  19. 1 1

  20. 2 2 1 1

  21. 2 2 3 1 1 3

  22. 2 2 3 1 1 3 2 2

  23. 2 Construct a (1,3)-chain 2 3 1 1 3 2 2

  24. 2 2 1 3 3 1 1 3 2 2

  25. 2 1 2 1 3 3 1 1 1 3 2 2

  26. 2 1 2 1 3 3 1 1 1 3 2 2 3 3

  27. 2 1 Construct a (2,3)-chain 2 1 3 3 1 1 1 3 2 2 3 3

  28. 3 2 1 3 1 2 2 1 1 3 1 3 3 3 2 2

  29. 3 2 1 Construct a (1,2)-chain 3 1 2 2 1 1 3 1 3 3 3 2 2

  30. 3 2 1 3 2 1 1 2 1 3 1 3 1 3 2 2

  31. 3 2 1 3 2 1 1 2 1 3 2 3 1 3 1 3 2 2

  32. Further investigations • 1) Modifications of initial problem • 2) Investigation of the incidentorcoloring itself

  33. Modifications of initial problem • 1) Arbitrary capacities • 2) Two sessions of message transmission • 3) Memory restrictions • 4) Problem of Melnikov & Vizing • 5)Bilevel network

  34. Memory restriction • The memory of the central computer is at most Q • If Q=0 then second way of transmission is impossible and we have the edge coloring problem

  35. IfQ  nthen we can store each message in the central computer during 1 unit of time. Incidentorcoloring problem with the following restriction on mated incidentorscolors appears: • b – 1 a  b • In this case= • (Melnikov, Vizing, P.; 2000).

  36. (k,l)-coloring of incidentors • Let0  k  l  . Restrictions: • 1) adjacent incidentors have distinct colors; • 2) mated incidentorscolors satisfy • k  b – a  l. • Denote the minimum number of colors by k,l(G).

  37. Case k=0 is solved: • 0,0(G) is an edge chromatic number • 0,1(G) = 0,(G) =  (Melnikov, P., Vizing, 2000) • Another solved case is l=: • k,(G) = max{, k ++, k +–} (P.,1999) k+1 1 + k++ k+–

  38. Vizing’s proof • Let t = max{, k ++, k +–} • 1. Construct a bipartite interpretation H of the graph G: • vV(G) corresponds to v+,v–V(H) • vuE(G) corresponds to v+u–E(H)

  39. Vizing’s proof • 2. Color the edges of H by (H)colors. Clearly, (H)= max{+(G),–(G)} • 3. If v+u–E(H) is coloreda, colora the initial incidentor of the arc vuE(G) and colora+kits final incidentor

  40. Vizing’s proof • 4. Shift colors at every vertex • Initial: turn a1<a2<...<apinto 1,2,...,p • Final: turn b1>b2>...>bqinto t,t–1,...,t –q+1 • We get a required incidentorcoloring of G by tcolors

  41. Example k=1 =3 +=–=2 t=3

  42. Bipartite interpretation 1 1 2 3 2 4 3 4 6 5 5 6

  43. Edge coloring 1 1 2 2 2 2 2 2 2 3 1 2 3 1 3 1 1 2 1 1 3 2 2 2 1 1 2

  44. Shifting the colors 1 1 3 2 2 2 2 2 1 2 3 3 2 2 3 1 3 3 1 3 1 2 1 2 1 1 3 3 2 2 2 2 1 1 3 2

  45. Equivalent problem in scheduling theory • Job Shop withnmachinesandm jobs, each of which has two unit operations (at different machines), and there must be a delay at least kand at mostlbetween the end of the first operation and the beginning of the second one.

  46. It is NP-complete to find out whether there is a (1,1)-coloring of a multigraph by  colors even for =7 (Bansal, Mahdian, Sviridenko, 2006). • Reduction from 3-edge-coloring of a 3-regular graph

  47. Reduction from 3-edge-coloring • Substitute each edge by the following gadget: v u u v

  48. Reduction from 3-edge-coloring • It can be verified that in any (1,1)-coloring by 7colors the incidentors of the initial incidentors of the red edges must be colored by the same even color

  49. Results on (k,l)-coloring • k,k(G) = k,∞(G) fork≥ (G)–1 • k,(G)–1(G) =k,∞(G) (Vizing, 2003) • Let k,l() = max{k,l(G) | deg(G) = } • k,() = k +  • k,k()  k,l()  k +  • 0,1() = 

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