The Acyclic Chromatic Index is Less than the Double of the Max Degree
TThe Acyclic Chromatic Index is Less than theDouble of the Max Degree
Lefteris Kirousis a, ∗ , John Livieratos a a Department of Mathematics, National & Kapodistrian University of Athens, Greece
Abstract
The acyclic chromatic index of a graph G is the least number of colors neededto properly color its edges so that none of its cycles is bichromatic. In this work,we show that 2 ∆ − ∆ is the maximum degree of the graph. In contrast with most extant randomizedalgorithmic approaches to the chromatic index, where the algorithms presupposeenough colors to guarantee properness deterministically and use randomnessonly to deal with the bichromatic cycles, our randomized, Moser-type algorithmproduces a not necessarily proper random coloring, in a structured way, tryingto avoid cycles whose edges of the same parity are homochromatic, and onlywhen this goal is reached it checks for properness. It repeats until propernessis attained. Keywords:
Acyclic chromatic index, Acyclic edge coloring, Lov´asz LocalLemma
1. Introduction
Let G = ( V, E ) be a (simple) graph with l vertices and m edges. Thechromatic index of G , often denoted by χ (cid:48) ( G ), is the least number of colorsneeded to properly color its edges, i.e., to color them so that no adjacent edgesget the same color. If ∆ is the maximum degree of G , it is known that itschromatic index is either ∆ or ∆ + 1 (Vizing [15]).The acyclic chromatic index of G , often denoted by χ (cid:48) a ( G ), is the leastnumber of colors needed to properly color the edges of G so that no cycle ofeven length is bichromatic , i.e. no even length cycle’s edges are colored by onlytwo colors. Notice that in any properly colored graph, no cycle of odd lengthcan be bichromatic. It has been conjectured (J. Fiamˇcik [7] and Alon et al. [2])that the acyclic chromatic index of any graph with maximum degree ∆ is atmost ∆ + 2.There is an extensive literature on the acyclic chromatic index. The resultsinclude: ∗ Corresponding author
Email addresses: [email protected] (Lefteris Kirousis), [email protected] (John Livieratos) a r X i v : . [ m a t h . C O ] A ug For planar graphs χ (cid:48) α ( G ) ≤ ∆ + 6 (Wang and Zhang [16]). Also, forsome constant M , all planar graphs with ∆ ≥ M have χ (cid:48) α ( G ) = ∆ (Cranston [4]). • For every (cid:15) > g such that if the girth of G is at least g , then χ (cid:48) α ( G ) = (1 + (cid:15) ) ∆ + O (1) (Cai et al. [3]). • For a random d -regular graph, χ (cid:48) α ( G ) = d + 1, asymptotically almostsurely (Ne˘set˘ril and Wormald [13]). • For d -degenerate graphs, i.e. graphs whose vertices can be ordered sothat every vertex has at most d neighbors greater than itself, χ (cid:48) α ( G ) ≤(cid:100) (2 + (cid:15) ) ∆ (cid:101) , for (cid:15) = 16 (cid:112) ( d/∆ ) (Achlioptas and Iliopoulos [1]).For general graphs, the known upper bounds for the chromatic index are O ( ∆ ). Specifically, Esperet and Parreau [5] proved that χ (cid:48) α ( G ) ≤ ∆ − (cid:100) . ∆ − (cid:101) + 1 by Giotis et al. [9]. Also, animprovement of the 4( ∆ −
1) bound was announced by Gutowski et al. [10](the specific coefficient for ∆ is not given in the abstract of the announcement).The best bound until now was recently given by Fialho et al. [6], where it isproved that χ (cid:48) α ( G ) ≤ (cid:100) . ∆ − (cid:101) + 1. These results are proved by designingrandomized algorithms that with high probability halt and produce a properacyclic edge coloring. Such algorithms, whose main idea is based on the algo-rithmic proofs of the L´ovasz Local Lemma by Moser [11] and Moser and Tardos[12], assign at each step a randomly chosen color in a way that properness is notdestroyed. This approach, unfortunately, necessitates a palette of more than2( ∆ −
1) colors to deterministically guarantee properness at each step, and thenanother O ( ∆ ) colors to probabilistically guarantee acyclicity.In this work we get rid of the necessity to initially have a separate palette of2( ∆ −
1) colors that guarantee properness by a simple idea: ignore propernessaltogether when choosing a color and design the Moser-type algorithm in a wayto guarantee that no even-length cycle exists whose edges of the same parityhave the same color (same parity edges are edges that are one edge apart ina cycle traversal —cycles in a non-proper coloring whose edges of the sameparity have the same color could have all their edges with the same color); thenrepeat this process anew until a coloring that is proper is obtained. We showthat with high probability the algorithm halts within a polynomial number ofsteps. Thus we show that 2 ∆ − ∆ −
2. Acyclic Edge Coloring
Let G = ( V, E ) be a (simple) graph with l vertices and m edges (both l and m are considered constants). The degree of a vertex u of G , denoted by deg( u ),is the number of edges incident on u . The maximum degree of G is denotedby ∆ and we assume, to avoid trivialities, that it is >
1. A cycle of length k ( k - cycle ) is a set of k ≥ u . . . , u k such that { u i .u i +1 } , { u k , u } ∈ E , i = 1 , . . . , k −
1. An edge-coloring of G is proper if no adjacent edges have thesame color. A proper edge-coloring is k - acyclic if there are no bichromatic k -cycles, k ≥ acyclic chromatic index of G , denoted by χ (cid:48) a ( G ), is the least number ofcolors needed to produce a proper, acyclic edge-coloring of G .In the algorithms of this paper, not necessarily proper edge-colorings areconstructed by independently selecting one color for each edge from a palette of K > e ∈ E and anycolor i ∈ { , . . . , K } , Pr[ e receives color i ] = 1 K . (1)We assume that we have K = (cid:100) (2 + (cid:15) )( ∆ − (cid:101) colors at our disposal, where (cid:15) > G . Therefore, since for any ∆ , there exists a constant (cid:15) > (cid:100) (2 + (cid:15) )( ∆ − (cid:101) ≤ ∆ −
1) + 1 = 2 ∆ −
1, it follows that χ (cid:48) a ( G ) ≤ ∆ − G . In all that follows, we assume the existence of some arbitraryordering of the edges and the cycles of G . Edges that in some traversal of aneven-length cycle are one edge apart are said to be of the same parity. Amongthe two consecutive traversals of the edges of a cycle, we arbitrarily select oneand call it positive . Given an edge e and a 2 k -cycle C containing it, we define C ( e ) := { e = e C , . . . , e C k } to be the set of edges of C , in the positive traversalstarting from e . The two disjoint and equal cardinality subsets of C ( e ) comprisedof edges of the same parity that are at even (odd, respectively) distance from e are to be denoted by C ( e ) ( C ( e ), respectively).We now give a cornerstone result proven by Esperet and Parreau [5]: Lemma 1 (Esperet and Parreau [5]) . At any step of any successive coloring ofthe edges of a graph, there are at most ∆ − colors that should be avoided inorder to produce a proper 4-acyclic coloring.Proof Sketch. Notice that for each edge e , one has to avoid the colors of alledges adjacent to e , and moreover for each pair of homochromatic (of the samecolor) edges e , e adjacent to e at different endpoints (which contribute one tothe count of colors to be avoided), one has also to avoid the color of the at mostone edge e that together with e, e , e define a cycle of length 4. Thus, easily,the total count of colors to be avoided does not exceed the number of adjacentedges of e , which is at most 2( ∆ − .2. EdgeColor
We first present below algorithm
EdgeColor . Notice that
EdgeColor
Algorithm 1
EdgeColor for each e ∈ E do Choose a color for e from the palette, independently for each e , and u.a.r.(not caring for properness) end for while there is an edge contained in a cycle of even length ≥ e be the least suchedge and C be the least such cycle and do Recolor ( e, C ) end while return the current coloring Recolor ( e, C ), where C = C ( e ) = { e = e C , . . . , e C k } , k ≥ for i = 1 , . . . , k − do Choose a color for each e Ci independently and u.a.r. (not caring forproperness) end for while there is an edge in { e C , . . . , e C k − } contained in a cycle of evenlength ≥ e (cid:48) be the least such edge and C (cid:48) the least such cycle and do Recolor ( e (cid:48) , C (cid:48) ) end while may not halt, and perhaps worse, even if it stops, it may generate a non-propercoloring or a coloring that is proper but has bichromatic 4-cycles. However, it isobvious, because of the while -loops in the main part of EdgeColor and in theprocedure
Recolor , that if the algorithm halts, then it outputs a coloring withno cycles of even length ≥ MainAlgorithm that follows, we repeat
EdgeColor until the desired coloring is obtained.
Algorithm 2
MainAlgorithm while the color generated by EdgeColor is not proper or is proper butcontains a bichromatic 4-cycle do Execute
EdgeColor anew end while Obviously
MainAlgorithm , if and when it stops, generates a proper acycliccoloring. The rest of the paper is devoted to compute the probability distribu-tion of the number of steps it takes.A call of the Recolor procedure from line 5 of the algorithm
EdgeColor is a root call of
Recolor , while one made from within the execution of another
Recolor procedure is called a recursive call. Each iteration of
Recolor iscalled a phase . We also call phase, the initial one, the for -loop of
EdgeColor .In the sequel, we count phases rather than steps of color assignments. Because4he number m of the edges of the graph is constant, this does not affect theform of the asymptotics of the number of steps.We prove the following progression lemma, which shows that at every timea Recolor ( e, C ) procedure terminates, some progress has indeed been made,which is then preserved in subsequent phases. Lemma 2.
Consider an arbitrary call of
Recolor ( e, C ) and let E be the setof edges that at the beginning of the call are not contained in a cycle of evenlength ≥ and having homochromatic edges of the same parity. Then, if andwhen that call terminates, no such edge in E ∪ { e } exists.Proof. Suppose that
Recolor ( e, C ) terminates and there is an edge e (cid:48) ∈ E ∪{ e } contained in a cycle of even length ≥ e (cid:48) = e , then by line 4, Recolor ( e, C ) could not have terminated.Thus, e (cid:48) ∈ E .Since e (cid:48) is not contained in a cycle as described at the beginning of Re-color ( e, C ), it must be the case that at some point during this call, somecycle, with e (cid:48) among its edges, turned into one having homochromatic edges ofthe same parity because of some call of Recolor . Consider the last time thishappened and let
Recolor ( e ∗ , C ∗ ) be the causing call. Then, there is somecycle C (cid:48) of even length ≥ e (cid:48) ∈ C (cid:48) , such that the recoloring of theedges of C ∗ resulted in C (cid:48) having homochromatic edges of the same parity andstaying such until the end of the Recolor ( e, C ) call. Then there is at least oneedge e ∗ contained in both C ∗ and C (cid:48) that was recolored Recolor ( e ∗ , C ∗ ). Byline 4 of Recolor ( e ∗ , C ∗ ), this procedure could not have terminated, and thusneither could Recolor ( e, C ), a contradiction.By Lemma 2, we get: Lemma 3.
There are at most m , the number of edges of G , i.e. a constant,repetitions of the while -loop of the main part of EdgeColor . However, a while -loop of
Recolor or MainAlgorithm could last in-finitely long. In the next section we analyze the distribution of the numberof steps they take.
3. Analysis of the Algorithm
In this section we will prove two things: • The probability that
EdgeColor lasts for at least n phases is inverseexponential in n . • The probability that the while -loop of
MainAlgorithm is repeated atleast n times is inverse exponetial in n .From the above two facts, yet to be proved , Lemma 3, and because (cid:15) in thenumber of colors (cid:100) (2 + (cid:15) )( ∆ − (cid:101) of the palette is an arbitrary positive constant,we get the Theorem below and its corollary, our main results. The proof ofboth the above will be possible by coupling the EdgeColor algorithm with avalidation algorithm that does not have the stochastic dependencies of the colorassignments of
EdgeColor and thus is amenable to probabilistic analysis.5 heorem.
Assuming ∆ − colors are available, the probability that MainAl-gorithm , which if and when it stops produces a proper acyclic edge coloring,lasts for at least n steps is inverse exponential in n . Therefore:
Corollary. ∆ − colors suffice to properly and acyclically color a graph. We will proceed as follows: in Subsection 3.1, we use a graph structure todepict the action of an execution of
EdgeColor organized in phases. Thegoal is to present in a structured way any undesirable behavior of
EdgeColor (and the corresponding validation algorithm), and thus get a bound on theprobability of it happening. Finally, in Subsection 3.2, we will give still anotheralgorithm, to show that can be coupled with
EdgeColor whose probabilitywe will bound.
We will depict the action of execution of
EdgeColor organized in phaseswith a rooted forest , that is an acyclic graph whose connected components ( trees )all have a designated vertex as their root. We label the vertices of such forestswith pairs ( e, C ), where e is an edge and C a 2 k -cycle containing e , for some k ≥
3. If a vertex u of F is labeled by ( e, C ), we will sometimes say that e isthe edge-label and C the cycle-label of u . The number of nodes of a forest isdenoted by |F| . Definition 1.
A labeled rooted forest F is called feasible , if the following twoconditions hold:i. Let e and e (cid:48) be the edge-labels of two distinct vertices u and v of G . Then,if u, v are both either roots of F or siblings (i.e. they have a commonparent) in F , then e and e (cid:48) are distinct.ii. If ( e, C ) is the label of a vertex u that is not a leaf, where C has half-length k ≥ , and e (cid:48) is the edge-label of a child v of u , then e (cid:48) ∈ { e C , . . . , e C k − } . Given a feasible forest, we order its trees and the siblings of each nodeaccording to their edge-labels. By traversing F in a depth-first fashion, re-specting the ordering of trees and siblings, we obtain the label sequence L ( F ) =( e , C ) , . . . , ( e |F| , C |F| ) of F .Given an execution of EdgeColor with at least n phases, we constructa feasible forest with n nodes by creating one node u labeled by ( e, C ) foreach phase, corresponding to a call (root or recursive) of Recolor ( e, C ). Westructure these nodes according to the order their labels appear in the recursivestack implementing EdgeColor : the children of a node u labeled by ( e, C )correspond to the recursive calls of Recolor made by line 5 of
Recolor ( e, C ),with the leftmost child corresponding to the first such call and so on.Let now P n be the probability that EdgeColor lasts for at least n phases,and Q n be the probability that EdgeColor (i) lasts for strictly less than n phases, and (ii) the coloring generated when it halts is either not proper or,alternatively, it is proper and has a bichromatic 4-cycle. In the next subsection,we will compute upper bounds for these probabilities.6 .2. Validation Algorithm We now give the validation algorithm:
Algorithm 3
ColorVal ( F )Input: L ( F ) = ( e , C ) , . . . , ( e |F| , C |F| ) : C i ( e i ) = { e i = e C i , . . . , e C i k i } . Color the edges of G , independently and selecting for each a color u.a.r.from { , . . . , K } . for i = 1 , . . . , |F| do if C i ( e i ) and C i ( e i ) are both monochromatic (have one color) then Recolor e C i , . . . , e C i k i − by selecting independently colors u.a.r.from { , . . . , K } else Recolor e C i , . . . , e C i k i − by selecting independently colors u.a.r.from { , . . . , K } return failure and exit end if end for return success We call each iteration of the for -loop of lines 2–9 a phase of
ColorVal ( F ). Lemma 4.
At the end of each phase of
ColorVal the colors are distributedas if they were assigned for a first and single time to each edge, selecting inde-pendently for each edge a color u.a.r. from the palette.Proof.
This is because at each phase, the edges e C i , . . . , e C i k i − , the only edgesfor which information is obtained during the phase, are recolored selecting in-dependently a color u.a.r.For a feasible forest F , let V F be the event that ColorVal ( F ) reports success . First, we compute the probability of V F . Lemma 5.
Let F be a feasible forest, with label sequence L ( F ) = ( e , C ) , . . . , ( e |F| , C |F| ) . Assume that C i has half-length k i ≥ , i = 1 , . . . , n . Then: Pr[ V F ] = |F| (cid:89) i =1 (cid:32) K (2 k i − (cid:33) . Proof.
For each i , whatever the colors of e C i k i − and e C i k i are, we need the other2 k i − C to have the same color with one of them. The probability ofthis being true for each edge is, by Equation (1) and Lemma 4, K .We now let ColorVal ( F ) be executed independently for all feasible forestswith at most n nodes. We let ˆ P n be the probability that ColorVal ( F ) succeedsfor at least one F with exactly n nodes, and let ˆ Q n be the probability that ColorVal ( F ) succeeds for at least one F such that (i) F has strictly less than n nodes, and (ii) the coloring generated when ColorVal ( F ) comes to an endis either not proper or, alternatively, it is proper and has a bichromatic 4-cycle.7 emma 6. We have that P n ≤ ˆ P n and Q n ≤ ˆ Q n . Proof.
Consider an execution of
EdgeColor and let F be the feasible forestwith n nodes generated in case the execution lasts for at least n phases, or bethe forest generated until EdgeColor halts, in case it lasts for strictly lessthan n phases. Execute now ColorVal ( F ) making the random choices of EdgeColor . Lemma 7.
We have that ˆP n ≤ (cid:80) |F| = n Pr[ V F ] and ˆ Q n ≤ − (cid:16) − (cid:16) (cid:15) (cid:17)(cid:17) m . Proof.
The first inequality is obvious. Now, just for the sake of notationalconvenience, let us call below a coloring strongly proper if it is proper and4-acyclic. Also by just random we mean a coloring generated by independentlycoloring all edges once, choosing for each a color u.a.r from the palette. For thesecond inequality, first observe that by Lemma 4 we have that ˆ Q n ˆ Q n = Pr[ ColorVal succeeds for at least one F with |F| < n | the generated color is not strongly proper] × Pr[a random coloring is not strongly proper] ≤ Pr[a random coloring is not strongly proper] . The given bound then follows from the cornerstone result of Esperet and Parreaugiven in Lemma 1.The second inequality of the Lemma above has as corollary, by Lemma 6,that the probability that the number of repetitions of the while -loop of
MainAl-gorithm is n is inverse exponential in n . So all that remains to be proved tocomplete the proof of the main Theorem is to show that (cid:80) |F| = n Pr[ V F ] is in-verse exponential in n . We do this in the next subsection, expressing the sumas a recurrence. We will estimate (cid:80) |F| = n Pr[ V F ] by purely combinatorial arguments. To-wards this end, we first define the weight of a forest, denoted by (cid:107)F(cid:107) , to be thenumber |F| (cid:89) i =1 (cid:32) K (2 k i − (cid:33) , (recall that |F| denotes the number of nodes of F ), and observe that by Lemma 5 (cid:88) |F| = n Pr[ V F ] = (cid:88) |F| = n (cid:107)F(cid:107) . (2)Assume that the empty tree is a feasible forest with number of nodes 0 andweight 1. From the definition of a feasible forest, we have that such a forest iscomprised of m possibly empty trees, in one to one correspondence to the edges,the root of each nonempty one having as edge-label the corresponding edge. For j = 1 , . . . , m , let T j be the set of all possible feasible trees corresponding to theedge e j and let T be the collection of all m -ary sequences ( T , . . . , T m ) with T j ∈ T j . 8ow, obviously: (cid:88) |F| = n (cid:107)F(cid:107) = (cid:88) ( T ,...,T m ) ∈T| T | + ···| T m | = n (cid:107) T (cid:107) · · · (cid:107) T m (cid:107) = (cid:88) n + ··· + n m = nn ,...,n m ≥ (cid:32)(cid:16) (cid:88) T ∈T : | T | = n (cid:107) T (cid:107) (cid:17) · · · (cid:16) (cid:88) T m ∈T m : | T m | = n m (cid:107) T m (cid:107) (cid:17)(cid:33) (3)We will now obtain a recurrence for each factor of the rhs of (3). Let: q = ∆ − K = ∆ − (cid:100) (2 + (cid:15) )( ∆ − (cid:101) . (4) Lemma 8.
Let T e be anyone of the T j . Then: (cid:88) T ∈T e | T | = n (cid:107) T (cid:107) ≤ R n , (5) where R n is defined as follows: R n := (cid:88) k ≥ q k − (cid:32) (cid:88) n + ··· + n k − = n − n ,...,n k − Q n . . . Q n k − (cid:33) (6) and R = 1 .Proof. Indeed, the result is obvious if n = 0, because the only possible T isthe empty tree, which has weight 1. Now if n >
0, observe that there are atmost ∆ k − possible cycles with 2 k edges, for some k ≥
3, that can be thecycle-edge of the root of a tree T ∈ T e with | T | >
0. Since the probability ofeach such cycle having homochromatic equal parity sets is (cid:0) K (cid:1) k − , the lemmafollows.We will now asymptotically analyze the coefficients of the OGF R ( z ) of R n .Multiply both sides of (6) by z n and sum for n = 1 , . . . , ∞ to get R ( z ) − (cid:88) k ≥ (cid:32) q k − zR ( z ) k − (cid:33) , (7)with R (0) = 1. Setting W ( z ) = R ( z ) − W ( z ) = (cid:88) k ≥ (cid:32) q k − z ( W ( z ) + 1) k − (cid:33) , (8)with W (0) = 0. For notational convenience, set W = W ( z ). Then from (8) weget: W = z (cid:88) k ≥ (cid:32) q k ( W + 1) k (cid:33) = z ( q ( W + 1)) − ( q ( W + 1)) . (9)9ow, set: φ ( x ) = ( q ( x + 1)) − ( q ( x + 1)) , (10)to get from (9): W = zφ ( W ) . (11)By [8, Proposition IV.5, p. 278], we have that if φ is a function analytic at 0 hav-ing non-negative Taylor coefficients and such that φ (0) (cid:54) = 0 and if r is the radiusof convergence of the series representing φ at 0 and finally if lim x → r − φ ( x ) = + ∞ we get that [ z n ] R (cid:46)(cid:47) (1 /ρ ) n , i.e. lim sup ([ z n ] R ) /n = 1 /ρ , where ρ = τφ ( τ ) , and τ is the (necessarily unique) solution of the characteristic equation: τ φ (cid:48) ( τ ) φ ( τ ) = 1 (12)within (0 , r ) (for the asymptotic notation “ (cid:46)(cid:47) ” see [8, IV.3.2, p. 243]).In our case, with φ given in (10). The above hypotheses are easily satisfiedfor each q ∈ (0 ,
1) with r = (1 /q ) −
1. Also, if q = , then r = 1, and the(necessarily unique) solution of the characteristic equation τφ (cid:48) ( τ ) φ ( τ ) = 1 within(0 ,
1) is τ = − √
5, and therefore ρ = τφ ( τ ) = 1 . Now, as it is shown in the the proof of [8, Proposition IV.5, p. 278], thefunction xφ (cid:48) ( x ) φ ( x ) is increasing with x ∈ (0 , r ). Therefore, for each x ∈ (0 , r ), itis also increasing with q ∈ (0 , / ( x + 1)), because q appears in the argument of φ as q ( x + 1). So, if q > /
2, the unique solution of τφ (cid:48) ( τ ) φ ( τ ) = 1 within (0 , r )becomes smaller than − √
5. Now φ (cid:48) ( x ) is increasing with x ∈ (0 , r ) ( φ ’sTaylor series has positive coefficients), and so [ z n ] R (cid:46)(cid:47) (1 /ρ ) n , for some ρ > ρ = τφ ( τ ) = φ (cid:48) ( τ ) .By the above, and since there are at most n m sequences n , . . . , n m of integersthat add up to n , we get by Lemma 7, equations (2) and (3), and Lemma 8that: ˆP n ≤ n m (cid:16) ρ (cid:17) n . (13)Thus, by Equation (13) and Lemma 6, we get that: Lemma 9.
For any (cid:15) > , and for any graph G for which l , m and ∆ (resp.the number of vertices, the number of edges and the maximum degree of G ) areconsidered constant, and given the availability of at least (2 + (cid:15) )( ∆ − colors,there exists a constant c ∈ (0 , , depending on l, m, ∆ and (cid:15) , such that theprobability that EdgeColor executes at least n steps is ≤ c n . This completes the proof of our two probabilistic claims at the introductoryparagraph of Section 3, and the proofs of the Theorem and Corollary, our mainresults, given there.
Acknowledgement
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