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Incidentor coloring: Incidentor coloring: methods and resultsmethods and results
A.VA.V. . PyatkinPyatkin"Graph Theory and Interactions""Graph Theory and Interactions"
Durham, 2013Durham, 2013
4-color problem4-color problem
Reduction to the vertex Reduction to the vertex coloringcoloring
Vertex coloring problemVertex coloring problem
Edge coloring problemEdge coloring problem
Incidentor coloringIncidentor coloring
• IncidentorIncidentor is a pair is a pair ((vv,,ee) ) of a vertex of a vertex vv and and an arc (edge) an arc (edge) ee, incident with it., incident with it.
• It is a half of an arc (edge) adjoining to a It is a half of an arc (edge) adjoining to a given vertexgiven vertex..
•
• initialinitial finalfinal
• ((mated incidentorsmated incidentors))
Incidentor coloring problemIncidentor coloring problem
• Color all incidentors of a given Color all incidentors of a given multigraph by the minimum number multigraph by the minimum number of colors in such a way that the given of colors in such a way that the given restrictions on the colors of restrictions on the colors of adjacent adjacent ((having a joint vertexhaving a joint vertex) ) and and mated mated ((having a joint archaving a joint arc) ) incidentors would incidentors would be stisfiedbe stisfied
An example of incidentor An example of incidentor coloringcoloring
Incidentor coloring Incidentor coloring generalizes both vertex and generalizes both vertex and
edge coloringedge coloring
Incidentor coloring Incidentor coloring generalizes both vertex and generalizes both vertex and edge coloringedge coloring
MotivationMotivation
Central computer
… … …
1 i j ni-th object must send to j-th one dij
units of information
All links capacities are equal to 1
There are two ways of There are two ways of information transmissioninformation transmission::
• 1) 1) Directly fromDirectly from ii-th-th object to object to jj-th one-th one ((during one time unitduring one time unit););
• 2) 2) With memorizing in the central With memorizing in the central computercomputer ( (receive the message from receive the message from ii--thth object, memorize it, and transmit object, memorize it, and transmit to to jj-th one later-th one later).).
Reduction to the incidentor Reduction to the incidentor coloringcoloring
• Each object corresponds to a vertex of Each object corresponds to a vertex of the multigraph the multigraph ((nn vertices vertices).).
• Each unit of information to transmit from Each unit of information to transmit from ii-th object to -th object to jj-th one corresponds to the -th one corresponds to the arcarc ijij of the multigraph (there are of the multigraph (there are ddijij arcs arcs going from a vertex going from a vertex ii to the a vertex to the a vertex jj))..
• The maximum degree The maximum degree equals the equals the maximum load of the linkmaximum load of the link..
SchedulingScheduling
• To each information unit two time To each information unit two time moments should be assigned moments should be assigned – – when it when it goes via goes via ii-th and -th and jj-th links-th links. .
• These moments could be interpreted as These moments could be interpreted as the colors of the incidentors of the arcthe colors of the incidentors of the arc ijij..
a bi j
RestrictionsRestrictions
• The colors of adjacent incidentors The colors of adjacent incidentors must be distinctmust be distinct..
• For every arc, the color of its initial For every arc, the color of its initial incidentor is at most the color of incidentor is at most the color of the final incidentorthe final incidentor, , ii..ee. . a a b.b.
a bi j
• It is required to color all incidentors It is required to color all incidentors by the minimum number of colors by the minimum number of colors satisfying all the restrictions satisfying all the restrictions ((the the length of the schedule islength of the schedule is ).).
• For this problemFor this problem == . . Such coloring Such coloring can be found in can be found in OO((nn2222) ) timetime..
• (P., 1995)(P., 1995)
Sketch of the algorithmSketch of the algorithm
• Consider an arc that is not colored Consider an arc that is not colored yetyet
• Try to color its incidentorsTry to color its incidentors::
• 1. 1. In such a way that In such a way that a=ba=b
• 2. I2. In such a way that n such a way that a<ba<b
• Otherwise, modify the coloring Otherwise, modify the coloring (consider bicolored chains)(consider bicolored chains)
=3
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Construct a(1,3)-chain
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Construct a(2,3)-chain
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Construct a(1,2)-chain
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Further investigationsFurther investigations
• 1) 1) Modifications of initial problemModifications of initial problem
• 2) 2) Investigation of the incidentor Investigation of the incidentor coloring itselfcoloring itself
Modifications of initial problemModifications of initial problem
• 1) 1) Arbitrary capacitiesArbitrary capacities• 2) 2) Two sessions of message Two sessions of message
transmissiontransmission• 3)3) Memory restrictions Memory restrictions• 4) 4) Problem of Melnikov & VizingProblem of Melnikov & Vizing• 5)5) Bilevel network Bilevel network
Memory restrictionMemory restriction
• The memory of the central computer The memory of the central computer is at most is at most Q Q
• If If QQ=0=0 then second way of then second way of transmission is impossible and we transmission is impossible and we have the edge coloring problemhave the edge coloring problem
• IfIf Q Q nn then we can store each then we can store each message in the central computer message in the central computer during during 11 unit of time. Incidentor unit of time. Incidentor coloring problem with the following coloring problem with the following restriction on mated incidentors restriction on mated incidentors colors appears:colors appears:
• b – b – 11 a a bb
• In this caseIn this case ==
• (Melnikov, Vizing, P.; 2000)(Melnikov, Vizing, P.; 2000)..
((kk,,ll)-)-coloring of incidentorscoloring of incidentors
• LetLet 0 0 k k l l . . RestrictionsRestrictions::• 1) 1) adjacent incidentors have distinct adjacent incidentors have distinct
colorscolors;;• 2) 2) mated incidentors colors satisfymated incidentors colors satisfy• kk bb – – aa ll..• Denote the minimum number of Denote the minimum number of
colors by colors by kk,,ll((GG).).
• Case Case kk=0 =0 is solved:is solved:0,00,0((GG) ) is an edge chromatic numberis an edge chromatic number
0,10,1((GG) = ) = 0,0,((GG) = ) = (Melnikov, P., Vizing, (Melnikov, P., Vizing, 2000)2000)
• Another solved case is Another solved case is ll==: :
• kk,,((GG) = ) = max{max{, , k k ++++, , k k ++––} } (P.,1999)(P.,1999)k+1
k+–
1+ k++
Vizing’s proofVizing’s proof
• Let Let tt = = max{ max{, , k k ++++, , k k ++––}}
• 1. Construct a bipartite interpretation 1. Construct a bipartite interpretation HH of the graph of the graph GG::
• vvVV((GG) corresponds to ) corresponds to vv++,,vv––VV((HH))
• vuvuEE((GG) corresponds to ) corresponds to vv++uu––EE((HH))
Vizing’s proofVizing’s proof
• 2. Color the edges of 2. Color the edges of HH by by ((HH)) colors. Clearly, colors. Clearly, ((HH)=)= max{ max{++((GG)),,––((GG)})}
• 3. If 3. If vv++uu––EE((HH) ) is colored is colored aa, color , color aa the the initial incidentor of the arc initial incidentor of the arc vuvuEE((GG) ) and color and color aa++kk its final incidentorits final incidentor
Vizing’s proofVizing’s proof
• 4. Shift colors at every vertex4. Shift colors at every vertex
• Initial: turn Initial: turn aa11<<aa22<...<<...<aap p into into 11,,22,...,,...,pp
• Final: turn Final: turn bb11>>bb22>...>>...>bbq q into into tt,,t–t–11,...,,...,t –q+t –q+11
• We get a required incidentor coloring of We get a required incidentor coloring of GG by by tt colorscolors
ExampleExample
k=1=3+=– =2 t=3
Bipartite interpretationBipartite interpretation
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Edge coloringEdge coloring
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Shifting the colorsShifting the colors
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Equivalent problem in Equivalent problem in scheduling theoryscheduling theory
• Job Shop withJob Shop with nn machinesmachines andand mm jobs jobs, , each of which has two unit each of which has two unit operations (at different machines)operations (at different machines), , and there must be a delay at least and there must be a delay at least kk and at mostand at most ll between the end of between the end of the first operation and the beginning the first operation and the beginning of the second one.of the second one.
• It is NP-complete to find out whether It is NP-complete to find out whether there is a there is a (1,1)-(1,1)-coloring of a coloring of a multigraph by multigraph by colors even for colors even for =7=7 (Bansal, Mahdian, Sviridenko, 2006).(Bansal, Mahdian, Sviridenko, 2006).
• Reduction from Reduction from 33-edge-coloring of a -edge-coloring of a 33-regular graph-regular graph
Reduction from Reduction from 33-edge--edge-coloringcoloring
• Substitute each edge by the Substitute each edge by the following gadget: following gadget:
u u vv
Reduction from Reduction from 33-edge--edge-coloringcoloring
• It can be verified that in any It can be verified that in any (1,1)-(1,1)-coloring by coloring by 77 colors the incidentors of colors the incidentors of the initial incidentors of the red the initial incidentors of the red edges must be colored by the same edges must be colored by the same even coloreven color
Results on Results on ((kk,,ll)-)-coloringcoloring
kk,,kk((GG) = ) = kk,,∞∞((GG)) for for kk ≥ ≥ ((GG)–1 )–1
kk,,((GG)–1)–1((GG) =) = kk,,∞∞((GG)) ( (VizingVizing, 2003), 2003)
• Let Let kk,,ll(() = max{) = max{kk,,ll((GG) | deg() | deg(GG) = ) = }}
kk,,(() = ) = kk + +
kk,,kk(() ) kk,,ll(() ) kk + +
0,10,1(() = ) =
Results on Results on ((kk,,ll)-)-coloringcoloring
kk,,ll(2) = (2) = kk+2 +2 exceptexcept kk = = ll = 0 = 0
• ((MelnikovMelnikov,, P.,Vizing, P.,Vizing, 2000) 2000)kk,,ll(3) = (3) = kk+3 +3 exceptexcept kk = = ll = 0 = 0 andand k k = = l l = 1= 1
((P.P., 2003), 2003)kk,,ll(4) = (4) = kk+4 +4 exceptexcept kk = = ll = 0 = 0
• ForFor ll /2/2, , kk,,ll(() ) = = kk ++
• ((P.P., 2004), 2004)
Results on Results on (1,1)-(1,1)-coloringcoloring
• For oddFor odd ,, 11,1,1(() > ) > +1 +1 (P., 2004)(P., 2004)
Results on Results on (1,1)-(1,1)-coloringcoloring
• For oddFor odd ,, 11,1,1(() > ) > +1 +1 (P., 2004)(P., 2004)
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Results on Results on (1,1)-(1,1)-coloringcoloring
• For oddFor odd ,, 11,1,1(() > ) > +1 +1 (P., 2004)(P., 2004)
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Results on Results on (1,1)-(1,1)-coloringcoloring
• For oddFor odd ,, 11,1,1(() > ) > +1 +1 (P., 2004)(P., 2004)
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Results on Results on (1,1)-(1,1)-coloringcoloring
• For evenFor even , , it is unknown whether it is unknown whether there is a graph there is a graph GG of degree of degree such such that that 11,1,1((GG) > ) > +1. +1. If such If such GG exists, exists, then it has degree at least then it has degree at least 66..
• Theorem.Theorem. 11,1,1((44))=5=5 ((P.P., 2004), 2004)
ProofProof
• Consider an Eulerian route in a given Consider an Eulerian route in a given 44-regular multigraph-regular multigraph
• Say that an edge is red, if its Say that an edge is red, if its orientation is the same as in the orientation is the same as in the route and blue otherwiseroute and blue otherwise
• Construct a bipartite interpretation Construct a bipartite interpretation according to this route (it consists of according to this route (it consists of the even cycles)the even cycles)
• Find an edge coloring Find an edge coloring ff::EE{1,2} {1,2} such such that:that:
• 1) any two edges adjacent at the 1) any two edges adjacent at the right side have distinct colors;right side have distinct colors;
• 2) any two blue or red edges 2) any two blue or red edges adjacent at the left side have distinct adjacent at the left side have distinct colors;colors;
• 3) If a red edge 3) If a red edge ee meets a blue one meets a blue one e’e’ at the left side, then at the left side, then ff((ee))ff((e’e’)+1)+1
• Construct an incidentor coloring Construct an incidentor coloring gg in in the following way:the following way:
• 1) For the right incidentor let 1) For the right incidentor let gg=2=2ff
• 2)2) For the left red incidentor let For the left red incidentor let gg=2=2f –f –11
• 3) 3) For the left blue incidentor let For the left blue incidentor let gg=2=2f+f+11
• We obtain an incidentor We obtain an incidentor (1,1)-(1,1)-coloring coloring of the initial multigraph by of the initial multigraph by 55 colors colors
ExampleExample
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ExampleExample
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ExampleExample
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ExampleExample
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Incidentor coloring of weighted Incidentor coloring of weighted multigraphmultigraph
• Each arc Each arc ee has weighthas weight ww((ee))
• Coloring restrictionsColoring restrictions::
• 1) 1) adjacent incidentors have distinct adjacent incidentors have distinct colorscolors;;
• 2) 2) For every arc For every arc ee, , ww((ee) ) bb – – aa
Results on weighted Results on weighted coloringcoloring• Problem isProblem is NP-hardNP-hard in a strong sense forin a strong sense for
== ( (P., VizingP., Vizing; 2005); 2005)
• ForFor >> the problem is the problem is NP-hardNP-hard in a in a strong sense even for multigraphs on two strong sense even for multigraphs on two vertices vertices ((Lenstra, Hoogevan,Yu; 200Lenstra, Hoogevan,Yu; 2004)4)
• It can be solved approximately with a It can be solved approximately with a relative error relative error 3/23/2 ( (VizingVizing, 2006), 2006)
List incidentor coloringList incidentor coloring
• A weighted incidentor coloring where each A weighted incidentor coloring where each arc arc ee has a list has a list LL((ee) ) of allowed colors for its of allowed colors for its incidentorsincidentors
• Conjecture.Conjecture. If If ||LL((ee)|)|ww((ee)+)+ for every arc for every arc ee then an incidentor coloring existsthen an incidentor coloring exists
• True for True for ||LL((ee)|≥)|≥ww((ee)+)++1+1. Proved for even . Proved for even ((VizingVizing, 2001), 2001) and for and for ==33 (P. (P., 2007, 2007))
Total incidentor coloringTotal incidentor coloring
• Color incidentors and vertices in such a Color incidentors and vertices in such a way that vertex coloring is correct and a way that vertex coloring is correct and a color of each vertex is distinct from the color of each vertex is distinct from the color of all incidentors adjoining this vertexcolor of all incidentors adjoining this vertex
TTkk,,((GG)) kk+1,+1,((GG))++1 1 kk,,((GG))++2;2;
TT0,0,((GG)= )= +1+1 ( (VizingVizing, 200, 20000) )
• Conjecture. Conjecture. TTkk,,((GG)) kk,,((GG))++1 1
Interval incidentor coloringInterval incidentor coloring
• The colors of adjacent incidentors must form The colors of adjacent incidentors must form an interval an interval
II0,0,((GG)) max{ max{, , ++++–––1}–1}
II1,1,((GG)) ++++––
• For For kk ≥ 2 ≥ 2 there could be no interval incidentor there could be no interval incidentor ((kk,,))-coloring (e.g. directed cycle) -coloring (e.g. directed cycle) ((VizingVizing, , 20020011))
Undirected caseUndirected case
• Instead of Instead of bb––aa use use ||bb––a|a| for colors of for colors of mated incidentorsmated incidentors
• Undirected incidentor chromatic Undirected incidentor chromatic number is equal to the best directed number is equal to the best directed ones taken among all orientationsones taken among all orientations
Undirected caseUndirected case
kk,,((GG)= max{)= max{,,/2/2++kk}}
TTkk,,((GG)) kk,,((GG))++1 1 (Vizing,Toft, 2001)(Vizing,Toft, 2001)
• If If kk/2/2 then then kk,,kk ((GG)= )= /2/2++kk
• If If =2=2krkr then then kk,,kk ((GG)=)=
• If If =2=2kr+skr+s then then kk,,kk ((GG))++kk – – ss/2/2
• (Vizing, 2005)(Vizing, 2005)
Undirected caseUndirected case
• For every regular multigraph For every regular multigraph G G with with 22k k kk,,ll ((GG)){{,, kk,,kk ((GG)} )} depending only depending only on on ll;; in particular, in particular, kk,,ll((GG)=)=kk,,ll((HH)) for for every two regular multigraphs every two regular multigraphs GG and and HH of degree of degree (Vizing,2005)(Vizing,2005)
Undirected caseUndirected case
• Interval incidentor coloring of Interval incidentor coloring of undirected multigraphs always existsundirected multigraphs always exists
II0,0,((GG)) = = II
1,1,((GG)) = =
• For For kk≥2,≥2, IIkk,,((GG))max{max{,min{2,min{2kk,,++kk}} }} andand
IIkk,,((GG))22++kk((kk ––1)/21)/2
• (Vizing,2003)(Vizing,2003)
Undirected caseUndirected case
• The incidentor coloring of weighted The incidentor coloring of weighted undirected multigraph is NP-hard in a undirected multigraph is NP-hard in a strong sense even for strong sense even for ==
• It can be solved approximately with a It can be solved approximately with a relative error relative error 55//44
• (Vizing, P., 2008)(Vizing, P., 2008)
Open problemsOpen problems
• 1. Is it true that for every 1. Is it true that for every kk there is there is ll such that such that kk,,ll(() =) = kk,,(()=)=kk++? ?
• Proved for Proved for kk=0=0 ( (ll=1)=1). Incorrect for . Incorrect for kk,,ll((GG))
• 2. What are the values of 2. What are the values of 1,21,2(5) (5) and and 2,22,2(5)?(5)?
Open problemsOpen problems
• 3. Given a 3. Given a -regular bipartite graph with -regular bipartite graph with red and blue edges is there an edge red and blue edges is there an edge coloring coloring ff::EE{1,2,...,{1,2,...,} } such that:such that:
• 1) any two edges adjacent at the right side 1) any two edges adjacent at the right side have distinct colors;have distinct colors;
• 2) any two blue or red edges adjacent at 2) any two blue or red edges adjacent at the left side have distinct colors;the left side have distinct colors;
• 3) If a red edge 3) If a red edge ee meets a blue one meets a blue one e’e’ at at the left side, then the left side, then ff((ee))ff((e’e’))++1?1?
Open problemsOpen problems
• 4. Is it true that if 4. Is it true that if ||LL((ee)|)|≥≥ww((ee)+)+ for for every arc every arc ee then a list incidentor then a list incidentor coloring exists?coloring exists?
• 5. Is it true that 5. Is it true that TTkk,,((GG)) kk,,((GG))++1?1?
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