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Neighbour sum distinguishing total colourings via the Combinatorial Nullstellensatz
Discrete Applied Mathematics 202 (2016) 163–173
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Neighbour sum distinguishing total colourings via the Combinatorial Nullstellensatz Jakub Przybyło ∗ AGH University of Science and Technology, al. A. Mickiewicza 30, 30-059 Krakow, Poland
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Article history: Received 19 January 2014 Received in revised form 21 August 2015 Accepted 26 August 2015 Available online 16 September 2015 Keywords: Total neighbour sum distinguishing number List total colouring Combinatorial Nullstellensatz Adjacent strong chromatic index Neighbour sum distinguishing index 1–2-Conjecture Graph labelling
Consider a simple graph G = (V , E ) and its (proper) total colouring c with elements of the set {1, 2, . . . , k}. We say that c is neighbour sum distinguishing if for every edge uv ∈ E, the sums of colours met by u and v differ, i.e., c (u) + e∋u c (e) ̸= c (v) + e∋v c (e). The least k guaranteeing the existence of such a colouring is denoted χ′′ (G). We investigate a daring conjecture presuming that χ′′ (G) ≤ ∆(G) + 3 for every graph G, a seemingly demanding problem if confronted with up-to-date progress in research on the Total Colouring Conjecture itself. We note that χ′′ (G) ≤ ∆(G) + 2col(G) − 1 and apply Combinatorial Nullstellensatz to prove a stronger bound: χ′′ (G) ≤ ∆(G) + ⌈ 35 col(G)⌉. This imply an upper bound of the form χ ′′ (G) ≤ ∆(G) + const. for many classes of graphs with unbounded maximum degree. In particular we obtain χ′′ (G) ≤ ∆(G) + 10 for planar graphs. In fact we show that identical bounds also hold if we use any set of k real numbers instead of {1, 2, . . . , k} as edge colours, and moreover the same is true in list versions of the both concepts. © 2015 Elsevier B.V. All rights reserved.
1. Introduction Let c : V ∪ E → R be a (proper) total colouring of a simple graph G = (V , E ). For every vertex v ∈ V denote by sc (v) := c (v) +
c (uv)
(1)
u∈N (v)
the sum of colours met by v , i.e., the sum of colours of the edges incident with v and the colour of v itself. We shall also write s(v) instead of sc (v) if the colouring c is unambiguous. The least k such that there is a (proper) total colouring c : V ∪ E → {1, 2, . . . , k} with s(u) ̸= s(v) for every edge uv ∈ E – so-called neighbour sum distinguishing total colouring – ′′ is called the total neighbour sum distinguishing number of G, and denoted by χ (G). The least number of colours in a proper ′′ total colouring of a graph, the total colouring number, is denoted by χ ′′ (G). Obviously, χ ′′ (G) ≤ χ (G). The famous Total
Colouring Conjecture, posed independently by Vizing [40] and Behzad [6], presumes that χ ′′ (G) ≤ ∆(G) + 2 for every graph G, where ∆(G) is the maximum degree of G. Thus far the best upper bound is due to Molloy and Reed, who proved that χ ′′ (G) ≤ ∆(G) + C , where C is some (large) constant, via a complex probabilistic argument, see [28]. Regardless, the following buoyant conjecture was posed in [30].
∗
Tel.: +48 12 617 46 38; fax: +48 12 617 31 65. E-mail address: [email protected].
http://dx.doi.org/10.1016/j.dam.2015.08.028 0166-218X/© 2015 Elsevier B.V. All rights reserved.
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′′ Conjecture 1. For any graph G, χ (G) ≤ ∆(G) + 3.
This holds e.g. for the complete graphs, cycles, bipartite graphs, subcubic graphs [30], K4 -minor free graphs [26], and planar graphs of maximum degree at least 13 [25]. The roots of the general field of additive graph colourings date back to the paper of Chartrand, Erdős and Oellermann [9], discussing various potential definitions of an irregular graph. The problem originated from the well known fact that there exists no antonym of a regular graph, understood as a graph whose all vertices have pairwise distinct degrees, except for the trivial 1-vertex case. This limitation does not concern multigraphs though. Consequently, the following extension of research from [9] was developed as an attempt of designing a graph invariant measuring the level of ‘irregularity’ of a graph. Suppose that given a simple graph G = (V , E ) we are allowed to multiply some of its edges. How small can be the largest necessary multiplicity of an edge so that we are able to construct an irregular multigraph of G, i.e., a multigraph with pairwise distinct vertex degrees? This value was named the irregularity strength of G, see [10] for details. Alternatively one may investigate (not necessarily proper) edge colourings c : E → {1, 2, . . . , k} such that for every pair of (distinct) vertices u, w ∈ V , sc (u) ̸= sc (w), where sc (v), v ∈ V , is understood the same as in (1) but with c (v) = 0. Then the irregularity strength of G is simply the least integer k admitting such edge colouring c of G. The concept of the irregularity strength has been investigated in numerous papers, see e.g. [3,7,11,12,15,17,21,24,27,29, 32,33] and many others, but it also gave rise to now very rich and fast developing branch of graph theory, which might be referred to as additive graph colourings or more generally — vertex distinguishing graph colourings. This includes now well known 1–2–3-Conjecture, see e.g. [1,2,22,23], and 1–2-Conjecture, see e.g. [20,36]. The latter of this is devoted to almost the same concept as discussed in this paper, but with one significant difference, that total colourings investigated within it do not have to be proper. The 1–2–3-Conjecture in turn constitutes its protoplast whose domain are edge colourings, i.e., it investigates a graph invariant similar to the irregularity strength, where only neighbours are required to receive distinct sums. Our research are also very closely related to the concept of the neighbour sum distinguishing index of a graph, denoted ′ by χ (G), and defined as the least integer k so that a proper edge colouring c : E → {1, 2, . . . , k} exists with sc (u) ̸= sc (v) for every edge uv of G. See [13,16,31,34,35,41] for some results concerning this problem, which was also motivated by an another central problem of this discipline called the Zhang’s Conjecture, see e.g. [19,44], and also [18,24,38] for interesting surveys of the field. List colourings of graphs were first introduced by Vizing [39] and independently by Erdős, Rubin and Taylor [14]. Most of the concepts mentioned above have been investigated also in a frame where colours for respective edges or vertices have to be chosen from arbitrary lists of a given length prescribed to them, see [5,34,35,37,42,43] for details. In fact we shall prove our results in this more general list setting. Suppose that we are given a set (La )a∈V ∪E of lists of real numbers. We say that G can be totally coloured from these lists if there exists a (proper) total colouring c : V ∪ E → R such that c (a) ∈ La for every a ∈ V ∪ E. The least integer k such that for every set of lists of reals (each) of length k there exists a total colouring c of G from these lists so that sc (u) ̸= sc (v) for every uv ∈ E is called the list total neighbour sum distinguishing number, and denoted by ch′′ (G). Note that by this definition, ′′ χ (G) ≤ ch′′ (G)
(2)
for every graph G. In this paper we prove a general result which implies an upper bound of the form ch′′ (G) ≤ ∆(G) + const . for many classes of graphs with unbounded maximum degree, including e.g. the planar graphs. The detailed results follow in the next section. Entire Section 3 is devoted to the proof of our main result -Theorem 4. Some comments follow next in Section 4. 2. General approach and results Recall that the colouring number of a graph G, denoted by col(G), is defined as the least integer k such that G has a vertex enumeration in which each vertex is preceded by fewer than k of its neighbours, hence in particular 1 ≤ col(G) − 1 ≤ ∆(G) for every graph with at least one edge. (The colouring number of a graph also exceeds its degeneracy exactly by one.) Given any enumeration v1 , v2 , . . . , vn of the vertices of G, the neighbours vj of a vertex vi shall be called backward if j < i, or forward if j > i. Analogously, an edge vi vj shall be referred to as a backward or forward edge of vi if j < i or j > i, respectively. First we observe the following. Observation 2. For every graph G, ch′′ (G) ≤ ∆(G) + 2col(G) − 1. Proof. Let G = (V , E ), and let (La )a∈V ∪E be a set of arbitrary (∆(G) + 2col(G) − 1)-element lists of real numbers. Choose any enumeration v1 , . . . , vn of the vertices of G in which every vertex is preceded by at most col(G) − 1 neighbours. We first fix a proper edge colouring from the given lists. For that purpose we analyse the consecutive vertices in the sequence and for every subsequent vertex vi one after another we choose colours for all its backward edges. Since for every consecutive such edge e at most (∆(G) − 1) + (col(G) − 2) edges incident with e might have fixed colours prior to e, then there is always some available colour in Le , |Le | = ∆(G) + 2col(G) − 1.
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Now one after another analyse the consecutive vertices in the sequence and fix their colours so that every vertex v obtains a sum s(v) distinct from the sums of all its backward neighbours. Since there are at most col(G) − 1 such neighbours, at most col(G) − 1 colours if chosen for v might cause a conflict between the sums at v and some of its preceding neighbours, thus we must forbid these (at most col(G) − 1) colours for v . Moreover, since the constructed total colouring is supposed to be proper, v additionally cannot get the colours of its (at most ∆(G)) incident edges nor the ones assigned to its (at most col(G) − 1 preceding) neighbours. Thus v still has at least one available colour in Lv , |Lv | = ∆(G) + 2col(G) − 1. Suppose now that we change our strategy and colour some part of the edges and vertices simultaneously in stages, and additionally allow some modifications of vertex colours during the process. Then we may improve the upper bound above. Surprisingly this might be achieved by colouring the forward edges (rather than backward ones, what due to their limited number seemed potentially more beneficial to us at first glance) of every consecutive vertex in the sequence. Unfortunately, the analysis of this procedure gets much more complex and requires introducing multivariable polynomials. Some of these shall be handled via an algebraic theorem Combinatorial Nullstellensatz, popularised by Alon [4]. Theorem 3 (Combinatorial Nullstellensatz). Let F be an arbitrary field, and let P = P (x1 , . . . , xn ) be a polynomial in F[x1 , k1 kn . . . , xn ]. Suppose the coefficient n of a monomial x1 . . . xn , where each ki is a non-negative integer, is non-zero in P and the degree deg(P ) of P equals i=1 ki . If moreover S1 , . . . , Sn are any subsets of F with |Si | > ki for i = 1, . . . , n, then there are s1 ∈ S1 , . . . , sn ∈ Sn so that P (s1 , . . . , sn ) ̸= 0. Our main result is the following general upper bound for ch′′ (G). Theorem 4. For every graph G, ch′′ (G) ≤ ∆(G) + ⌈ 35 col(G)⌉. Since by Euler’s Formula, col(G) ≤ 6 for the planar graphs, we immediately obtain the following corollary, slightly better than its correspondent deriving from Observation 2. Corollary 5. For every planar graph G, ch′′ (G) ≤ ∆(G) + 10. ′′ ′′ Note that by (2) these also imply that χ (G) ≤ ∆(G) + ⌈ 35 col(G)⌉ (and χ (G) ≤ ∆(G) + 10 for the planar graphs).
3. Proof of Theorem 4 Let G = (V , E ) be a graph of maximum degree ∆ ≥ 1, and suppose we are given a set (La )a∈V ∪E of arbitrary (∆ + ⌈ 35 col(G)⌉)-element lists of real numbers. We shall denote the colouring number of G, col(G), by ‘col’ for short, hence 1 ≤ col − 1 ≤ ∆. As previously, choose any enumeration v1 , . . . , vn of the vertices of G in which every vertex is preceded by at most col − 1 neighbours. We begin with the empty partial total colouring of G, i.e., initially no colours are assigned to the edges or vertices. This shall be gradually extended to a desired one in n steps via a series of partial total colourings, each corresponding to a consecutive vertex vi , i = 1, . . . , n, in the sequence. Given any partial total colouring of G, by the sum of a vertex v we shall mean s(v), where every uncoloured edge (and vertex) contributes 0 to this sum. We shall obey the following simple rules within our construction.
• Once we reach vi , i ∈ {1, . . . , n}, we shall colour all its uncoloured (forward) edges and colour or recolour vi and all its forward neighbours so that
• the sum of vi is distinct from the sums of all its neighbours and • the sums of all forward neighbours of vi are distinct from the sums of their respective backward neighbours vj with j ≤ i. (In fact we do not have to require this for all forward neighbours, only for those whose edges joining them with vi are the last yet uncoloured ones incident with these neighbours, but the adopted unification of the rules for all vertices eliminates possible confusion in the algorithm description.) Moreover,
• our choices from lists must guarantee that every coloured edge has a different colour than its (coloured) incident edges and its two endvertices,
• the vertex vi has a colour distinct from the colours of all its neighbours and • each forward neighbour of vi has a colour distinct from the colour of its every neighbour vj with j ≤ i (since these will not change colours after step i). Note that as a result, we shall obtain a neighbour sum distinguishing total colouring of G after the nth step of such procedure. Suppose that we are about to analyse a vertex v , and thus far all our requirements have been fulfilled. Let d− (v) and d+ (v) be the numbers of backward and forward, resp., neighbours of v in the sequence. Denote the consecutive forward neighbours of v by u1 , u2 , . . . , ud+ (v) (if d+ (v) > 0), and note that all backward edges of v are already (properly) coloured. Consider the following cases.
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3.1. Case 1 Suppose first that d+ (v) = 0. Then the set of backward neighbours of v must be nonempty (or v is an isolated vertex). Let v ′ be the last element of this set in the sequence. As all the construction’s requirements have been fulfilled thus far, all its features expected for v at the current stage of the construction were guaranteed priory (in particular) while considering the vertex v ′ . 3.2. Case 2 Assume that d(v) ≤ col + 1 and d+ (v) ≥ 1. Then we first choose (or modify) colours for all forward edges and forward neighbours of v except for v u1 and u1 , consistent with the specified rules. For this aim, we temporarily disregard the colour and the sum of v (this shall be adjusted at the end via appropriate choices for v , v u1 and u1 ), and for every j = 2, . . . , d+ (v) successively, we choose colours for v uj and uj together so as to distinguish uj from its neighbours preceding v . For this purpose we cannot use the colours of at most d− (v) + (d+ (v) − 2) = d(v) − 2 ≤ ∆ − 2 (where d(v) − 2 ≤ col − 1) already coloured edges incident with v and at most col − 2 edges incident with uj to colour v uj , thus we are left with at least (∆ +⌈ 35 col⌉)−(∆ − 2)−(col − 2) = ⌈ 23 col⌉+ 4 available colours in the list Lvuj . Analogously, we cannot use the colours of at most col − 2 already coloured edges incident with uj and the colours of at most col − 2 colours of (backward) neighbours of uj preceding v to colour uj , thus we are left with at least (∆ +⌈ 35 col⌉)− 2(col − 2) = ∆ −(col − 1)+⌈ 23 col⌉+ 3 ≥ ⌈ 32 col⌉+ 3 available colours in the list Luj . Out of these possibilities, we wish to choose colours for v uj and uj so that the obtained sum of uj is distinct from the sums of its at most col − 2 backward neighbours preceding v . This is however easily feasible via the following simple lemma, which might be found e.g. in [16]. Below we include a slight reformulation of its proof, as a generalisation of such interpretation shall be very useful later on. Lemma 6. If X and Y are two finite nonempty sets of real numbers, then there are at least |X | + |Y | − 3 distinct sums in the set
{x + y : x ∈ X , y ∈ Y , x ̸= y}. Proof. If |X |, |Y | ≥ 2, remove the minimal element of X from Y , and then remove the maximal remaining element of Y from X , i.e., let Y ′ = Y r {min X } and let X ′ = X r {max Y ′ }. Then for x′ = min X ′ and y′′ = max Y ′ (where x′ ̸∈ Y ′ and y′′ ̸∈ X ′ ), the sums of the forms x′ + y with y ∈ Y ′ and x + y′′ with x ∈ X ′ r {x′ } are all pairwise distinct (increasing) and prove the thesis. If |X | = 1 or |Y | = 1, the thesis is straightforward. Finally, to choose colours for v , u1 and v u1 , we shall use the following lemma. Prior to the proof, we first specify the details of its application below. 2t +2 2t +2 in the expansion of the polynomial x2 Lemma 7. Let t be a positive integer, and let at be the coefficient of the monomial x2t 0 x1
Pt (x0 , x1 , x2 ) = (x0 − x1 )(x0 − x2 )2 (x1 − x2 )(x0 + x1 )3t +1 (x1 + x2 )3t −1 .
(3)
Then at ̸= 0. Note that the colours of at most d(v) − 1 ≤ ∆ − 1 (already coloured) edges incident with v are forbidden for v and v u1 . The colours of at most col − 2 backward edges of u1 cannot be used on u1 and v u1 . Moreover, the colours of at most d(v)− 1 ≤ col neighbours of v and at most col − 2 backward neighbours of u1 might be forbidden for v and u1 , respectively. Denote the lists of the remaining colours by L′v , L′v u1 , L′u1 , respectively. Then
′ 2 L ≥ ∆ + 5 col − (∆ − 1) − col = col + 1, v 3 3 5 2 |L′vu1 | ≥ ∆ + col − (∆ − 1) − (col − 2) = col + 3, 3 3 5 2 |L′u1 | ≥ ∆ + col − 2(col − 2) ≥ col + 3. 3
3
Let s1 , s2 , . . . , sp be the sums of all the neighbours of v except for u1 , hence p = d(v) − 1 ≤ col, and let s′1 , s′2 , . . . , s′q be the sums of all the backward neighbours of u1 preceding v , thus q ≤ col − 2. Denote additionally by s0 and s′0 , resp., the sums of v and u1 . Associate variables x0 , x1 , x2 with v , v u1 , u1 , respectively. These will stand for the colours which shall be assigned to these vertices and edge. Consider the following polynomial: P (x0 , x1 , x2 ) = (x0 − x1 )(x0 − x2 )(x1 − x2 )(s0 + x0 + x1 − s′0 − x1 − x2 )
×
p i=1
(s0 + x0 + x1 − si )
q i =1
(s′0 + x1 + x2 − s′i ),
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and suppose that there exist x′0 ∈ L′v , x′1 ∈ L′v u1 and x′2 ∈ L′u1 for which P (x′0 , x′1 , x′2 ) ̸= 0. Then by setting the colours of v, v u1 , u1 as x′0 , x′1 , x′2 , resp., the construction of P guarantees that these three colours are pairwise distinct, the obtained sums of v and u1 differ, and moreover the obtained sum of v is distinct also from the sums of all its other neighbours, while the sum of u1 is different from the sums of all its backward neighbours preceding v . Such colouring is consistent with all our requirements. Note that for
t :=
2 col 3
2
,
2t ≤ ⌈ 23 col⌉ < |L′v | and 2t + 2 ≤ ⌈ 23 col⌉+ 2 < |L′v u1 |, |L′u1 |. On the other hand, p ≤ col ≤ 3t + 1 and thus q ≤ col − 2 ≤ 3t − 1. Now let us consider the polynomial: Q (x0 , x1 , x2 ) := P (x0 , x1 , x2 ) × (x0 + x1 )3t +1−p (x1 + x2 )3t −1−q . Observe that if Q (x′0 , x′1 , x′2 ) ̸= 0, then also P (x′0 , x′1 , x′2 ) ̸= 0. Note that deg Q = 6t + 4 = 2t + (2t + 2) + (2t + 2), hence 2t +2 2t +2 x2t x2 is a maximal (with respect to the sum of its powers) monomial of Q . In order to prove the existence of desired 0 x1 ′ x1 , x′2 , x′3 via Combinatorial Nullstellensatz it is thus sufficient to prove that the coefficient of this monomial is nonzero in Q . However, by the maximality of this monomial (which must ‘take’ a variable, not a constant, from every bracket), its coefficient is the same as the coefficient at of the same monomial in the polynomial Pt defined in (3). The thesis thus follows by Lemma 7 and Theorem 3. Proof of Lemma 7. Note that Pt (x0 , x1 , x2 ) = (x0 − x1 )(x0 − x2 )(x1 − x2 )(x0 − x2 )(x0 + x1 )3t +1 (x1 + x2 )3t −1
= (x30 x1 − x20 x1 x2 − x30 x2 + x20 x22 − x20 x1 x2 + x0 x1 x22 + x20 x22 − x0 x32 − x20 x21 + x0 x21 x2 + x20 x1 x2 − x0 x1 x22 + x0 x21 x2 − x21 x22 − x0 x1 x22 + x1 x32 ) × (x0 + x1 )3t +1 (x1 + x2 )3t −1 . (We deliberately do not reduce monomials of degree four above, so as not to obscure the general idea of our approach, and thus to make accessible the analysis of its more complex correspondent which shall follow in the third case of our main 2t +2 2t +2 proof.) We now may write the formula for the coefficient at of x2t x2 in Pt via analysing every consecutive monomial 0 x1 2t +2 2t +2 of degree four above and the number of ways we may choose its ‘complement’ to x2t c c i c0−xi 1 x2 from the two factors in the c last line of the formula above (or exploiting the expansion (a + b) = valid for every positive integer c). In i=0 i a b
= 0 if d < 0 or a < d: 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 at = − − + 2t − 3 2t + 2 2t − 2 2t + 1 2t − 3 2t + 1 2t − 2 2t 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 − + + − 2t − 2 2t + 1 2t − 1 2t 2t − 2 2t 2t − 1 2t − 1 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 − + + − 2t − 2 2t + 2 2t − 1 2t + 1 2t − 2 2t + 1 2t − 1 2t 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 + − − + 2t − 1 2t + 1 2t 2t 2t − 1 2t 2t 2t − 1 (3t − 1)! (3t + 1)! · × [2t (2t − 1)(2t − 2)t (t − 1)(t − 2) − 2t (2t − 1)(t + 4)(2t + 2)t (t − 1) = (2t )!(t + 4)! (2t + 2)!t ! − 2t (2t − 1)(2t − 2)(2t + 2)t (t − 1) + 2t (2t − 1)(t + 4)(2t + 2)(2t + 1)t − 2t (2t − 1)(t + 4)(2t + 2)t (t − 1) + 2t (t + 4)(t + 3)(2t + 2)(2t + 1)t + 2t (2t − 1)(t + 4)(2t + 2)(2t + 1)t − 2t (t + 4)(t + 3)(2t + 2)(2t + 1)2t − 2t (2t − 1)(t + 4)t (t − 1)(t − 2) + 2t (t + 4)(t + 3)(2t + 2)t (t − 1) + 2t (2t − 1)(t + 4)(2t + 2)t (t − 1) − 2t (t + 4)(t + 3)(2t + 2)(2t + 1)t + 2t (t + 4)(t + 3)(2t + 2)t (t − 1) − (t + 4)(t + 3)(t + 2)(2t + 2)(2t + 1)t − 2t (t + 4)(t + 3)(2t + 2)(2t + 1)t + (t + 4)(t + 3)(t + 2)(2t + 2)(2t + 1)2t ] (3t + 1)! (3t − 1)! · · 6t (8 − 10t − 71t 2 − 9t 3 + 2t 4 ). = (2t )!(t + 4)! (2t + 2)!t !
the following,
a d
(The last equality above might be verified via tedious by hand calculations, or via computer program,1 as it was done in our case.) By the Rational Root Theorem, every possible integer-valued root of a polynomial 8 − 10t − 71t 2 − 9t 3 + 2t 4 must 1 One might use e.g. Mathematica.
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divide 8. By substituting 1, 2, 4 and 8, we then obtain that this polynomial has no roots which are positive integers, and thus at ̸= 0 for every integer t ≥ 1. 3.3. Case 3 Suppose now that col + 2 ≤ d(v) ≤ ⌈ 34 col⌉ (and thus d+ (v) ≥ 3). Then, analogously as in the previous case, we first choose colours for all forward edges and forward neighbours of v except for v u1 , v u2 , v u3 and u1 , u2 , u3 , consistent with the specified rules. This is again feasible via Lemma 6. To colour the remaining vertices and edges incident with v (and v itself), we shall apply the following more complex correspondent of Lemma 7. +1 2t +2 2t +2 2t +2 2t +3 2t +2 Lemma 8. Let t be a positive integer, and let bt be the coefficient of the monomial xt0+3 x2t x2 x3 x4 x5 x6 in the 1 expansion of the polynomial
Qt (x0 , x1 , x2 , x3 , x4 , x5 , x6 ) = (x0 − x1 )(x0 − x2 )(x0 − x3 )(x0 − x4 )(x0 − x5 )(x0 − x6 )
× (x1 − x2 )(x1 − x3 )(x1 − x5 )(x3 − x4 )(x3 − x5 )(x5 − x6 ) × (x0 + x3 + x5 − x2 )(x0 + x1 + x5 − x4 )(x0 + x1 + x3 − x6 ) × (x1 + x2 )3t (x3 + x4 )3t (x5 + x6 )3t (x0 + x1 + x3 + x5 )4t .
(4)
Then bt ̸= 0. As previously we first discuss an application of Lemma 8, and only just then — the proof itself. Analogously as above, we denote by L′v u1 , L′v u2 , L′v u3 the lists of colours available for v u1 , v u2 , v u3 formed of the original lists Lv u1 , Lv u2 , Lv u3 , resp., by removing the colours of (already coloured) edges incident with the respective out of these three edges. Let also L′v , L′u1 , L′u2 , L′u3 be the lists of available colours for v, u1 , u2 , u3 after excluding (from Lv , Lu1 , Lu2 , Lu3 ) the colours used by their respective incident edges and neighbours preceding v in the sequence (and the colours of u4 , . . . , ud+ (v) in the case of v ). Observe that then:
2 5 col − (∆ − 3) − (col − 2) = col + 5, |L′vu1 |, |L′vu2 |, |L′vu3 | ≥ ∆ + 3 3 5 4 1 5 ′ col − 2(d(v) − 3) ≥ col − col + 6 ≥ col + 5, |Lv | ≥ ∆ + 3 3 3 3 2 2 5 col − 2(col − 2) ≥ d(v) − col + col + 4 ≥ col + 6. |L′u1 |, |L′u2 |, |L′u3 | ≥ ∆ + 3
3
3
Associate variables x1 , x3 , x5 and x0 , x2 , x4 , x6 with v u1 , v u2 , v u3 and v, u1 , u2 , u3 , respectively. Let s0,1 , s0,2 , . . . , s0,p denote the sums of all the neighbours of v except for u1 , u2 and u3 , hence p = d(v) − 3 ≤ ⌈ 43 col⌉ − 3. For every i ∈ {1, 2, 3}, let also si,1 , si,2 , . . . , si,qi be the sums of all the backward neighbours of ui preceding v , thus qi ≤ col − 2. Denote additionally by s0 , s1 , s2 , s3 the sums of v, u1 , u2 , u3 , resp., and consider the following polynomial:
P (x0 , x1 , x2 , x3 , x4 , x5 , x6 ) = (x0 − x1 )(x0 − x2 )(x0 − x3 )(x0 − x4 )(x0 − x5 )(x0 − x6 ) × (x1 − x2 )(x1 − x3 )(x1 − x5 )(x3 − x4 )(x3 − x5 )(x5 − x6 ) × (s0 + x0 + x1 + x3 + x5 − s1 − x1 − x2 ) × (s0 + x0 + x1 + x3 + x5 − s2 − x3 − x4 ) × ( s0 + x0 + x1 + x3 + x5 − s3 − x5 − x6 ) q1 q2 × (s1 + x1 + x2 − s1,i ) (s2 + x3 + x4 − s2,i ) ×
i =1
i =1
q3
p
i =1
(s3 + x5 + x6 − s3,i )
(s0 + x0 + x1 + x3 + x5 − s0,i ).
i =1
Analogously as in the corresponding application of Lemma 7, if we find a nonzero substitution for P from the lists defined above, then the respective colours of v, v u1 , u1 , v u2 , u2 , v u3 and u3 shall be consistent with all our requirements. Note that for 1 col , t := 3 t + 3 < |L′v |, 2t + 1 < 2t + 2 < 2t + 3 ≤ ⌈ 23 col⌉ + 4 < |L′v u1 |, |L′v u2 |, |L′v u3 | and analogously 2t + 2 ≤ ⌈ 23 col⌉ + 3 < |L′u1 |, |L′u2 |, |L′u3 |. On the other hand, p ≤ ⌈ 34 col⌉ − 3 ≤ 4t and qi ≤ col − 2 ≤ 3t for i = 1, 2, 3. Let us consider the polynomial:
P (x0 , x1 , x2 , x3 , x4 , x5 , x6 ) × (x1 + x2 )3t −q1 Q (x0 , x1 , x2 , x3 , x4 , x5 , x6 ) := × (x3 + x4 )3t −q2 (x5 + x6 )3t −q3 (x0 + x1 + x3 + x5 )4t −p .
J. Przybyło / Discrete Applied Mathematics 202 (2016) 163–173
169
+1 2t +2 2t +2 2t +2 2t +3 2t +2 Note that deg Q = 13t + 15, hence xt0+3 x2t x2 x3 x4 x5 x6 is a maximal monomial of Q . Moreover, its coefficient 1 in the expansion of Q is the same as in Qt , given by (4), and thus equal to bt , which is nonnegative by Lemma 8. By Combinatorial Nullstellensatz we thus obtain that there is a nonzero substitution for Q (and hence also for P) from the lists L′v , L′v u1 , L′u1 , L′v u2 , L′u2 , L′v u3 , L′u3 . To conclude this case it thus remains to prove Lemma 8.
Proof of Lemma 8. Let us define a matrix W = [wij ] by setting:
1 2 3 4 5 6 2 W = [wij ] = 3 5 4 5 6 2 4 6
∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅
0 0 0 0 0 0 1 1 1 3 3 5 5 5 3
3 1 1
∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅ . ∅ ∅ ∅ ∅ 0 0 0
(Note that the consecutive rows of W correspond to the first 15 linear factors of Qt from (4) and include the variables’ indicators arranged in the reversed order, so that the indicator of a unique variable preceded by a minus sign from a given bracket is always in the first column; the entries ‘∅’ shall be insignificant for the forthcoming formulas, and thus could as well be replaced by any other symbol.) Also define a function O : Z → Z such that
O(k) =
1 0
if k = 0, if k = ̸ 0.
Analogously as in the proof of Lemma 7 we may thus write that
15 2 4 4 4 3t 3t O(ji −1) 15 15 × bt = ... (−1)i=1 2t + 2 − O(wiji − 2) 2t + 2 − O(wiji − 4) j 1 =1 j 2 =1 j12 =1 j13 =1 j14 =1 j15 =1 2 2
i=1
3t
×
2t + 2 −
15
4t
O(wiji − 6)
t +3−
i =1
4t −
t +3−
O(wiji )
15
O(wiji )
15 15 2t + 1 − O(wiji − 1) − 3t − 2t + 2 − O(wiji − 2) i =1
i =1
15 i =1
×
i=1
i=1
15 15 O (w ) − 2t + 1 − O (w − 1 ) − 3t − 2t + 2 − O (w − 2 ) 4t − t + 3 − iji iji iji i =1 i =1 i=1 × . 15 15 2t + 2 − O(wiji − 3) − 3t − 2t + 2 − O(wiji − 4)
15
i =1
i =1
The summation above runs through all ordered 15-tuples within every of which each subsequent entry corresponds to 15 the index of one of the variables appearing in the appropriate consecutive bracket. The sum i=1 O(ji − 1) within the formula above counts the number of variables corresponding to a given 15-tuple which are preceded by a minus symbol in the respective brackets. Further, the first binomial coefficient counts the number of ways we may complement a given (obtained) monomial of degree 15 to obtain (altogether) 2t + 2 ‘copies’ of x2 , where these remaining copies (i.e, except for 15 the ones appearing already in the monomial of degree of 15, and counted by the formula i=1 O(wiji − 2)) must be chosen from the factor (x1 + x2 )3t . The following three binomial coefficients count the analogous numbers of ways we may obtain all together 2t + 2, 2t + 2 and t + 3 copies of x4 , x6 and x0 by complementing those from a given monomial of degree 15, while +1 choosing from (x3 + x4 )3t , (x5 + x6 )3t and (x0 + x1 + x3 + x5 )4t , respectively. Then we count in how many ways x2t might 1
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J. Przybyło / Discrete Applied Mathematics 202 (2016) 163–173
be achieved by choosing the necessary number of copies of x1 from the remaining (unused) factors of (x0 + x1 + x3 + x5 )4t , +2 and afterwards — in how many ways x2t might be achieved by choosing the necessary number of copies of x3 from the 3 4t remaining factors of (x0 + x1 + x3 + x5 ) (x5 shall be automatically chosen from the remaining factors). We thus obtain:
bt =
2 2
2 4 4 4
...
j1 =1 j2 =1
15
(−1)i=1
j12 =1 j13 =1 j14 =1 j15 =1
O(ji −1)
3t
3t
×
2t + 2 −
15
O(wiji − 2)
2t + 2 −
4t
3t
×
15
2t + 2 −
O(wiji − 6)
t +3−
3t − 3 +
× O(wiji )
i=1
i=1
15
O(wiji − 4)
i =1
i =1
15
15
O(wiji )
i =1
t +3−
15
(O(wiji − 1) + O(wiji − 2))
i=1
15
2t − 6 + (O(wiji ) + O(wiji − 1) + O(wiji − 2)) i =1 × 15 t + 4 − (O(wiji − 3) + O(wiji − 4)) i=1
=
2 2
2 4 4 4
...
j 1 =1 j 2 =1
15
(−1)
O(ji −1)
i=1
j12 =1 j13 =1 j14 =1 j15 =1
(3t )!
× (2t + 2 −
15
O(wiji − 2))!(t − 2 +
i=1
O(wiji − 2))!
i=1
(3t )!
× (2t + 2 −
15
O(wiji − 4))!(t − 2 +
i=1
15
O(wiji − 4))!
i=1
(3t )!
× (2t + 2 −
15
O(wiji − 6))!(t − 2 +
i=1
×
15
15
O(wiji − 6))!
i=1
(4t )! 15 (t + 3 − O(wiji ))!(t + 3 − (O(wiji − 1) + O(wiji − 2)))! 15
i =1
i=1
1
× (t + 4 −
15
(O(wiji − 3) + O(wiji − 4)))!
i =1
1
× (t − 10 +
15
,
(O(wiji ) + O(wiji − 1) + O(wiji − 2) + O(wiji − 3) + O(wiji − 4)))!
i=1
where we assume every ingredient containing a factor a! with a < 0 (in a denominator) to be equal to 0 in the sum above. Similarly as in the proof of Lemma 7 we may thus now extract the common factors of all the elements under the sum. In order to do it possibly most efficiently, we analyse the first 15 brackets of the polynomial Qt (or the rows of matrix W ) and note that x2 appears in just 3 of these, and so does x4 and x6 . Moreover, (at least) one bracket is formed exclusively of x1 and x2 , and one of x3 and x4 . Similarly, there is (at least) one bracket which contains none of x0 , x1 , x2 , x3 , x4 . We thus finally a−1 have (assuming that k=a f (k) = 1 for any function f and integer a): bt =
(3t )! (3t )! (3t )! (4t )! · · · (2t + 2)!(t + 1)! (2t + 2)!(t + 1)! (2t + 2)!(t + 1)! (t + 3)!(t + 2)!(t + 3)!(t + 4)! 2
×
2
j1 =1 j2 =1
2
...
4
4
4
15
j12 =1 j13 =1 j14 =1 j15 =1
(−1)
i=1
15
O(ji −1)
i=1
×
O(wij −2)−1 i
3
(2t + 2 − k)
k=0 k=
15 i=1
(t − 2 + k)
O(wij −2)+1 i
J. Przybyło / Discrete Applied Mathematics 202 (2016) 163–173 15
15
O(wij −4)−1 i
i=1
×
3
(2t + 2 − k)
k=0 k=
15 i=1
15
15
O(wij )−1 i
i=1
15 i=1
15 i=1
15 i=1
(t − 2 + k)
O(wij −6)+1 i
(O(wiji −3)+O(wiji −4))−1
(t + 3 − k) ×
14
k=
k=
i
(t + 4 − k)
k =1
k=1
×
3
(2t + 2 − k)
k=0
O(wij −4)+1
k=0
=
(t − 2 + k) ×
(O(wiji −1)+O(wiji −2))−1
(t + 3 − k)
i
i=1
×
O(wij −6)−1
i=1
171
(t − 10 + k)
(O(wiji )+O(wiji −1)+O(wiji −2)+O(wiji −3)+O(wiji −4))+1
(3t )! (2t + 2)!(t + 1)!
3
(4t )! · 12 · (1152 + 18336t + 127472t 2 (t + 2)!((t + 3)!)2 (t + 4)!
+513920t 3 + 1347344t 4 + 2444534t 5 + 3188541t 6 + 3062765t 7 + 2197996t 8 +1186064t 9 + 479902t 10 + 143110t 11 + 30108t 12 + 4066t 13 + 285t 14 + 5t 15 ), where the last equality above is accessible only via application of a computer program.2 By the Rational Root Theorem, every possible integer-valued root of the polynomial of degree 15 in t above must divide 1152 = 27 · 32 . By substituting all (altogether 24) positive divisors of 1152, we then obtain that this polynomial has no roots which are positive integers, and thus bt ̸= 0 for every integer t ≥ 1. 3.4. Case 4 Assume eventually that d(v) ≥ ⌈ 43 col⌉ + 1 (hence d+ (v) ≥ 3 and ∆ ≥ ⌈ 43 col⌉ + 1). We then first fix a colour of v chosen from Lv so that it is distinct from the colours of (at most ∆) neighbours of v and from the coloured (at most col − 1) edges incident with v . Each forward edge v ui of v has then at most d− (v)+(col − 2) colours blocked by its coloured incident edges, and at most one by v (we disregard the colour of ui , which shall be changed later if necessary). Thus every v ui is left with the list L′v ui of at least ∆ +⌈ 35 col⌉− d− (v)−(col − 2)− 1 ≥ (d− (v)+ d+ (v))+⌈ 32 col⌉− d− (v)+ 1 ≥ d+ (v)+ 3 available colours. We shall simultaneously choose colours from these lists for all forward edges of v so that all these edges’ colours are pairwise distinct and the obtained sum of v is different from the sums of its backward neighbours. This might be achieved using Combinatorial Nullstellensatz via analysis of a properly defined polynomial. Such approach would however require a quite complex argument, see the last section for further comments. We thus use a simpler method inspired by [8], applicable in this particular case, which might be viewed at as a generalisation of Lemma 6. For this purpose, we first dynamically modify the lists L′v u1 , . . . , L′v u + . Subsequently, for i = 1, 2, . . . , d+ (v) − 1, we take min L′v ui (where L′v ui always refers to the up-tod
(v)
date remainder of this list on a given stage of our modifying procedure) and remove it from all current lists L′v uj with j > i.
Then, subsequently, for i = d+ (v), d+ (v) − 1, . . . , 2, we find max L′v ui and remove it from all up-to-date lists L′v uj with j < i,
and denote the finally constructed respective lists by L′′v u1 , . . . , L′′v u
d+ (v)
. This way at most d+ (v) − 1 element were removed
from every list L′v ui and for every i < j < k, L′′v uj contains neither min L′′v ui nor max L′′v uk . Thus if we denote the consecutive (increasing) elements of each of the obtained lists L′′v ui by ci,1 , ci,2 , . . . , ci,li (hence ci,1 = min L′′v ui and ci,li = max L′′v ui ), then
d+ (v) each of the following at least i=1 |L′′v ui | − d+ (v) + 1 ≥ d+ (v)((d+ (v) + 3) − (d+ (v) − 1)) − d+ (v) + 1 = 3d+ (v) + 1 distinct sums consists of d+ (v) pairwise distinct integers: c1,1 + c2,1 + · · · + cd+ (v)−2,1 + cd+ (v)−1,1 + cd+ (v),1
< c1,1 + c2,1 + · · · + cd+ (v)−2,1 + cd+ (v)−1,1 + cd+ (v),2 ... < c1,1 + c2,1 + · · · + cd+ (v)−2,1 + cd+ (v)−1,1 + cd+ (v),ld+ (v)
< c1,1 + c2,1 + · · · + cd+ (v)−2,1 + cd+ (v)−1,2 + cd+ (v),ld+ (v) ... < c1,l1 + c2,l2 + · · · + cd+ (v)−2,ld+ (v)−2 + cd+ (v)−1,ld+ (v)−1 + cd+ (v),ld+ (v) .
2 E.g., Mathematica.
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J. Przybyło / Discrete Applied Mathematics 202 (2016) 163–173
Since v has at most col − 1 ≤ 3(( 43 col + 1) − (col − 1)) ≤ 3(d(v) − d− (v)) = 3d+ (v) backward neighbours, at least one of these sums must yield s(v) distinct from the sums of all of these neighbours. We then choose colours of the forward edges of v corresponding to this sum. At the end we subsequently adjust the colour of every vertex ui , with i = 1, 2, . . . , d+ (v), so that its colour is different from the colours of its backward edges and neighbours, and so that the obtained sum of ui is distinct from the sums of its backward neighbours. Altogether at most 3(col − 1) colours might thus be forbidden for ui . However, in the analysed case, |Lui | ≥ ∆ + ⌈ 35 col⌉ ≥ (⌈ 43 col⌉ + 1) + ⌈ 35 col⌉ > 3(col − 1), thus we may always choose an appropriate colour from Lui for ui , i = 1, . . . , d+ (v). The proof of Theorem 4 is thus completed. 4. Remarks Within the general approach presented in this paper we could not obtain a better bound than roughly, i.e. up to a small additive constant, ∆ + 53 col, even with more detailed algebraic analysis applied. Our main result thus could not be significantly improved. It is due to the second case analysed in the proof above, which we could not omit. To see that this case prevents a significant reduction of the number of colours necessary for our algorithm, assume that e.g. ∆ is roughly equal to col, and a vertex v , which we are about to analyse, has only one forward neighbour u1 , where v and u1 have both roughly col neighbours preceding v . Suppose that we strive at proving a somewhat smaller upper bound than ∆ + 53 col for ′′ ch′′ (G) (or χ (G)). Then the variables x0 , x1 , x2 , representing colours sought for v, v u1 , u1 , resp., each have at most roughly
∆ + 53 col − 2col ≈ 23 col colours available (i.e., not blocked by the colours of other edges or vertices), say 1, 2, . . . , 23 col for all of these. Now within these possibilities, e.g. col smallest achievable values for x0 + x1 might be blocked by the backward neighbours of v (hence x1 would have to be larger than 31 col), while col largest achievable values for x1 + x2 might be blocked
by the backward neighbours of u1 (and thus x1 should be at most 31 col, a contradiction). Using the same approach, we might however be able to improve our main result for graphs with maximum degree (much) larger than their colouring number, i.e. decrease for them the multiplicative factor 35 preceding col(G) in the bound from Theorem 4. The main obstacle on the way towards such goal is Case 3 from the proof above. Within this, instead of just one polynomial (4), we would have to analyse a whole (possibly infinite) family of its correspondents of increasing complexity. Moreover, as already occurred while investigating polynomial (4), a monomial with a non-zero coefficient in the expansion of a given representative of such a family might presumably lack symmetry (in its variables’ powers), what might cause additional complications for our reasonings. (Our experimental analysis showed e.g. that Lemma 8 was false +1 2t +2 2t +2 2t +2 2t +3 2t +2 for ‘more symmetrical’ correspondents of the monomial xt0+3 x2t x2 x3 x4 x5 x6 .) 1 To complement our aforementioned comment, let us also note that a polynomial whose analysis would be equivalent to our reasoning in Case 4, has the following form:
(xi − xj )
1≤i
3d+ (v)
(s0 + x1 + x2 + · · · + xd+ (v) − si ),
(5)
i =1
where the investigated vertex v has a (temporary) sum s0 , variables x1 , . . . , xd+ (v) correspond to colours of consecutive forward edges of v , while we estimated in Case 4 that v may have at most 3d+ (v) backward neighbours, whose sums are denoted by s1 , . . . , s3d+ (v) above. The reason why in this case we could come up with a simpler approach than the one based on Combinatorial Nullstellensatz was the fact that, unlike in the previous cases, we could concentrate on just one sum of variables (corresponding to the sum of v ), namely x1 + x2 + · · · + xd+ (v) , in order to evade a number of potential conflicts (while in previous cases we had to control two or even four corresponding sums associated with respective vertices simultaneously). It is also worth noting that the construction used for handling Case 4 is optimal (and might also be easily generalised to ‘less symmetrical’ cases). To see that, suppose accordingly to our assumptions from that case, that every variable xi is associated with the list of d+ (v) + 3 available values, say 1, 2, . . . , d+ (v) + 3 for each of them. Then, given that the first factor of the polynomial (5) corresponds to the required feature that distinct variables take different values (colours), there might be at most (4 + 5 + · · · + (d+ (v) + 3)) − (1 + 2 + · · · + d+ (v)) + 1 = 3d+ (v) + 1 distinct values available for the sum x1 + x2 + · · · + xd+ (v) (disregarding the second factor of (5)), and exactly this many were guaranteed by our construction in Case 4. Acknowledgements The research contained in this paper were supported by the National Science Centre, Poland, grant no. 2014/13/ B/ST1/01855, and partially supported by the Polish Ministry of Science and Higher Education. References [1] L. Addario-Berry, K. Dalal, C. McDiarmid, B.A. Reed, A. Thomason, Vertex-colouring edge-weightings, Combinatorica 27 (1) (2007) 1–12. [2] L. Addario-Berry, K. Dalal, B.A. Reed, Degree constrained subgraphs, Discrete Appl. Math. 156 (7) (2008) 1168–1174. [3] M. Aigner, E. Triesch, Irregular assignments of trees and forests, SIAM J. Discrete Math. 3 (4) (1990) 439–449.
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Neighbour sum distinguishing total colourings via the Combinatorial Nullstellensatz Jakub Przybyło ∗ AGH University of Science and Technology, al. A. Mickiewicza 30, 30-059 Krakow, Poland
article
abstract
info
Article history: Received 19 January 2014 Received in revised form 21 August 2015 Accepted 26 August 2015 Available online 16 September 2015 Keywords: Total neighbour sum distinguishing number List total colouring Combinatorial Nullstellensatz Adjacent strong chromatic index Neighbour sum distinguishing index 1–2-Conjecture Graph labelling
Consider a simple graph G = (V , E ) and its (proper) total colouring c with elements of the set {1, 2, . . . , k}. We say that c is neighbour sum distinguishing if for every edge uv ∈ E, the sums of colours met by u and v differ, i.e., c (u) + e∋u c (e) ̸= c (v) + e∋v c (e). The least k guaranteeing the existence of such a colouring is denoted χ′′ (G). We investigate a daring conjecture presuming that χ′′ (G) ≤ ∆(G) + 3 for every graph G, a seemingly demanding problem if confronted with up-to-date progress in research on the Total Colouring Conjecture itself. We note that χ′′ (G) ≤ ∆(G) + 2col(G) − 1 and apply Combinatorial Nullstellensatz to prove a stronger bound: χ′′ (G) ≤ ∆(G) + ⌈ 35 col(G)⌉. This imply an upper bound of the form χ ′′ (G) ≤ ∆(G) + const. for many classes of graphs with unbounded maximum degree. In particular we obtain χ′′ (G) ≤ ∆(G) + 10 for planar graphs. In fact we show that identical bounds also hold if we use any set of k real numbers instead of {1, 2, . . . , k} as edge colours, and moreover the same is true in list versions of the both concepts. © 2015 Elsevier B.V. All rights reserved.
1. Introduction Let c : V ∪ E → R be a (proper) total colouring of a simple graph G = (V , E ). For every vertex v ∈ V denote by sc (v) := c (v) +
c (uv)
(1)
u∈N (v)
the sum of colours met by v , i.e., the sum of colours of the edges incident with v and the colour of v itself. We shall also write s(v) instead of sc (v) if the colouring c is unambiguous. The least k such that there is a (proper) total colouring c : V ∪ E → {1, 2, . . . , k} with s(u) ̸= s(v) for every edge uv ∈ E – so-called neighbour sum distinguishing total colouring – ′′ is called the total neighbour sum distinguishing number of G, and denoted by χ (G). The least number of colours in a proper ′′ total colouring of a graph, the total colouring number, is denoted by χ ′′ (G). Obviously, χ ′′ (G) ≤ χ (G). The famous Total
Colouring Conjecture, posed independently by Vizing [40] and Behzad [6], presumes that χ ′′ (G) ≤ ∆(G) + 2 for every graph G, where ∆(G) is the maximum degree of G. Thus far the best upper bound is due to Molloy and Reed, who proved that χ ′′ (G) ≤ ∆(G) + C , where C is some (large) constant, via a complex probabilistic argument, see [28]. Regardless, the following buoyant conjecture was posed in [30].
∗
Tel.: +48 12 617 46 38; fax: +48 12 617 31 65. E-mail address: [email protected].
http://dx.doi.org/10.1016/j.dam.2015.08.028 0166-218X/© 2015 Elsevier B.V. All rights reserved.
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′′ Conjecture 1. For any graph G, χ (G) ≤ ∆(G) + 3.
This holds e.g. for the complete graphs, cycles, bipartite graphs, subcubic graphs [30], K4 -minor free graphs [26], and planar graphs of maximum degree at least 13 [25]. The roots of the general field of additive graph colourings date back to the paper of Chartrand, Erdős and Oellermann [9], discussing various potential definitions of an irregular graph. The problem originated from the well known fact that there exists no antonym of a regular graph, understood as a graph whose all vertices have pairwise distinct degrees, except for the trivial 1-vertex case. This limitation does not concern multigraphs though. Consequently, the following extension of research from [9] was developed as an attempt of designing a graph invariant measuring the level of ‘irregularity’ of a graph. Suppose that given a simple graph G = (V , E ) we are allowed to multiply some of its edges. How small can be the largest necessary multiplicity of an edge so that we are able to construct an irregular multigraph of G, i.e., a multigraph with pairwise distinct vertex degrees? This value was named the irregularity strength of G, see [10] for details. Alternatively one may investigate (not necessarily proper) edge colourings c : E → {1, 2, . . . , k} such that for every pair of (distinct) vertices u, w ∈ V , sc (u) ̸= sc (w), where sc (v), v ∈ V , is understood the same as in (1) but with c (v) = 0. Then the irregularity strength of G is simply the least integer k admitting such edge colouring c of G. The concept of the irregularity strength has been investigated in numerous papers, see e.g. [3,7,11,12,15,17,21,24,27,29, 32,33] and many others, but it also gave rise to now very rich and fast developing branch of graph theory, which might be referred to as additive graph colourings or more generally — vertex distinguishing graph colourings. This includes now well known 1–2–3-Conjecture, see e.g. [1,2,22,23], and 1–2-Conjecture, see e.g. [20,36]. The latter of this is devoted to almost the same concept as discussed in this paper, but with one significant difference, that total colourings investigated within it do not have to be proper. The 1–2–3-Conjecture in turn constitutes its protoplast whose domain are edge colourings, i.e., it investigates a graph invariant similar to the irregularity strength, where only neighbours are required to receive distinct sums. Our research are also very closely related to the concept of the neighbour sum distinguishing index of a graph, denoted ′ by χ (G), and defined as the least integer k so that a proper edge colouring c : E → {1, 2, . . . , k} exists with sc (u) ̸= sc (v) for every edge uv of G. See [13,16,31,34,35,41] for some results concerning this problem, which was also motivated by an another central problem of this discipline called the Zhang’s Conjecture, see e.g. [19,44], and also [18,24,38] for interesting surveys of the field. List colourings of graphs were first introduced by Vizing [39] and independently by Erdős, Rubin and Taylor [14]. Most of the concepts mentioned above have been investigated also in a frame where colours for respective edges or vertices have to be chosen from arbitrary lists of a given length prescribed to them, see [5,34,35,37,42,43] for details. In fact we shall prove our results in this more general list setting. Suppose that we are given a set (La )a∈V ∪E of lists of real numbers. We say that G can be totally coloured from these lists if there exists a (proper) total colouring c : V ∪ E → R such that c (a) ∈ La for every a ∈ V ∪ E. The least integer k such that for every set of lists of reals (each) of length k there exists a total colouring c of G from these lists so that sc (u) ̸= sc (v) for every uv ∈ E is called the list total neighbour sum distinguishing number, and denoted by ch′′ (G). Note that by this definition, ′′ χ (G) ≤ ch′′ (G)
(2)
for every graph G. In this paper we prove a general result which implies an upper bound of the form ch′′ (G) ≤ ∆(G) + const . for many classes of graphs with unbounded maximum degree, including e.g. the planar graphs. The detailed results follow in the next section. Entire Section 3 is devoted to the proof of our main result -Theorem 4. Some comments follow next in Section 4. 2. General approach and results Recall that the colouring number of a graph G, denoted by col(G), is defined as the least integer k such that G has a vertex enumeration in which each vertex is preceded by fewer than k of its neighbours, hence in particular 1 ≤ col(G) − 1 ≤ ∆(G) for every graph with at least one edge. (The colouring number of a graph also exceeds its degeneracy exactly by one.) Given any enumeration v1 , v2 , . . . , vn of the vertices of G, the neighbours vj of a vertex vi shall be called backward if j < i, or forward if j > i. Analogously, an edge vi vj shall be referred to as a backward or forward edge of vi if j < i or j > i, respectively. First we observe the following. Observation 2. For every graph G, ch′′ (G) ≤ ∆(G) + 2col(G) − 1. Proof. Let G = (V , E ), and let (La )a∈V ∪E be a set of arbitrary (∆(G) + 2col(G) − 1)-element lists of real numbers. Choose any enumeration v1 , . . . , vn of the vertices of G in which every vertex is preceded by at most col(G) − 1 neighbours. We first fix a proper edge colouring from the given lists. For that purpose we analyse the consecutive vertices in the sequence and for every subsequent vertex vi one after another we choose colours for all its backward edges. Since for every consecutive such edge e at most (∆(G) − 1) + (col(G) − 2) edges incident with e might have fixed colours prior to e, then there is always some available colour in Le , |Le | = ∆(G) + 2col(G) − 1.
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Now one after another analyse the consecutive vertices in the sequence and fix their colours so that every vertex v obtains a sum s(v) distinct from the sums of all its backward neighbours. Since there are at most col(G) − 1 such neighbours, at most col(G) − 1 colours if chosen for v might cause a conflict between the sums at v and some of its preceding neighbours, thus we must forbid these (at most col(G) − 1) colours for v . Moreover, since the constructed total colouring is supposed to be proper, v additionally cannot get the colours of its (at most ∆(G)) incident edges nor the ones assigned to its (at most col(G) − 1 preceding) neighbours. Thus v still has at least one available colour in Lv , |Lv | = ∆(G) + 2col(G) − 1. Suppose now that we change our strategy and colour some part of the edges and vertices simultaneously in stages, and additionally allow some modifications of vertex colours during the process. Then we may improve the upper bound above. Surprisingly this might be achieved by colouring the forward edges (rather than backward ones, what due to their limited number seemed potentially more beneficial to us at first glance) of every consecutive vertex in the sequence. Unfortunately, the analysis of this procedure gets much more complex and requires introducing multivariable polynomials. Some of these shall be handled via an algebraic theorem Combinatorial Nullstellensatz, popularised by Alon [4]. Theorem 3 (Combinatorial Nullstellensatz). Let F be an arbitrary field, and let P = P (x1 , . . . , xn ) be a polynomial in F[x1 , k1 kn . . . , xn ]. Suppose the coefficient n of a monomial x1 . . . xn , where each ki is a non-negative integer, is non-zero in P and the degree deg(P ) of P equals i=1 ki . If moreover S1 , . . . , Sn are any subsets of F with |Si | > ki for i = 1, . . . , n, then there are s1 ∈ S1 , . . . , sn ∈ Sn so that P (s1 , . . . , sn ) ̸= 0. Our main result is the following general upper bound for ch′′ (G). Theorem 4. For every graph G, ch′′ (G) ≤ ∆(G) + ⌈ 35 col(G)⌉. Since by Euler’s Formula, col(G) ≤ 6 for the planar graphs, we immediately obtain the following corollary, slightly better than its correspondent deriving from Observation 2. Corollary 5. For every planar graph G, ch′′ (G) ≤ ∆(G) + 10. ′′ ′′ Note that by (2) these also imply that χ (G) ≤ ∆(G) + ⌈ 35 col(G)⌉ (and χ (G) ≤ ∆(G) + 10 for the planar graphs).
3. Proof of Theorem 4 Let G = (V , E ) be a graph of maximum degree ∆ ≥ 1, and suppose we are given a set (La )a∈V ∪E of arbitrary (∆ + ⌈ 35 col(G)⌉)-element lists of real numbers. We shall denote the colouring number of G, col(G), by ‘col’ for short, hence 1 ≤ col − 1 ≤ ∆. As previously, choose any enumeration v1 , . . . , vn of the vertices of G in which every vertex is preceded by at most col − 1 neighbours. We begin with the empty partial total colouring of G, i.e., initially no colours are assigned to the edges or vertices. This shall be gradually extended to a desired one in n steps via a series of partial total colourings, each corresponding to a consecutive vertex vi , i = 1, . . . , n, in the sequence. Given any partial total colouring of G, by the sum of a vertex v we shall mean s(v), where every uncoloured edge (and vertex) contributes 0 to this sum. We shall obey the following simple rules within our construction.
• Once we reach vi , i ∈ {1, . . . , n}, we shall colour all its uncoloured (forward) edges and colour or recolour vi and all its forward neighbours so that
• the sum of vi is distinct from the sums of all its neighbours and • the sums of all forward neighbours of vi are distinct from the sums of their respective backward neighbours vj with j ≤ i. (In fact we do not have to require this for all forward neighbours, only for those whose edges joining them with vi are the last yet uncoloured ones incident with these neighbours, but the adopted unification of the rules for all vertices eliminates possible confusion in the algorithm description.) Moreover,
• our choices from lists must guarantee that every coloured edge has a different colour than its (coloured) incident edges and its two endvertices,
• the vertex vi has a colour distinct from the colours of all its neighbours and • each forward neighbour of vi has a colour distinct from the colour of its every neighbour vj with j ≤ i (since these will not change colours after step i). Note that as a result, we shall obtain a neighbour sum distinguishing total colouring of G after the nth step of such procedure. Suppose that we are about to analyse a vertex v , and thus far all our requirements have been fulfilled. Let d− (v) and d+ (v) be the numbers of backward and forward, resp., neighbours of v in the sequence. Denote the consecutive forward neighbours of v by u1 , u2 , . . . , ud+ (v) (if d+ (v) > 0), and note that all backward edges of v are already (properly) coloured. Consider the following cases.
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3.1. Case 1 Suppose first that d+ (v) = 0. Then the set of backward neighbours of v must be nonempty (or v is an isolated vertex). Let v ′ be the last element of this set in the sequence. As all the construction’s requirements have been fulfilled thus far, all its features expected for v at the current stage of the construction were guaranteed priory (in particular) while considering the vertex v ′ . 3.2. Case 2 Assume that d(v) ≤ col + 1 and d+ (v) ≥ 1. Then we first choose (or modify) colours for all forward edges and forward neighbours of v except for v u1 and u1 , consistent with the specified rules. For this aim, we temporarily disregard the colour and the sum of v (this shall be adjusted at the end via appropriate choices for v , v u1 and u1 ), and for every j = 2, . . . , d+ (v) successively, we choose colours for v uj and uj together so as to distinguish uj from its neighbours preceding v . For this purpose we cannot use the colours of at most d− (v) + (d+ (v) − 2) = d(v) − 2 ≤ ∆ − 2 (where d(v) − 2 ≤ col − 1) already coloured edges incident with v and at most col − 2 edges incident with uj to colour v uj , thus we are left with at least (∆ +⌈ 35 col⌉)−(∆ − 2)−(col − 2) = ⌈ 23 col⌉+ 4 available colours in the list Lvuj . Analogously, we cannot use the colours of at most col − 2 already coloured edges incident with uj and the colours of at most col − 2 colours of (backward) neighbours of uj preceding v to colour uj , thus we are left with at least (∆ +⌈ 35 col⌉)− 2(col − 2) = ∆ −(col − 1)+⌈ 23 col⌉+ 3 ≥ ⌈ 32 col⌉+ 3 available colours in the list Luj . Out of these possibilities, we wish to choose colours for v uj and uj so that the obtained sum of uj is distinct from the sums of its at most col − 2 backward neighbours preceding v . This is however easily feasible via the following simple lemma, which might be found e.g. in [16]. Below we include a slight reformulation of its proof, as a generalisation of such interpretation shall be very useful later on. Lemma 6. If X and Y are two finite nonempty sets of real numbers, then there are at least |X | + |Y | − 3 distinct sums in the set
{x + y : x ∈ X , y ∈ Y , x ̸= y}. Proof. If |X |, |Y | ≥ 2, remove the minimal element of X from Y , and then remove the maximal remaining element of Y from X , i.e., let Y ′ = Y r {min X } and let X ′ = X r {max Y ′ }. Then for x′ = min X ′ and y′′ = max Y ′ (where x′ ̸∈ Y ′ and y′′ ̸∈ X ′ ), the sums of the forms x′ + y with y ∈ Y ′ and x + y′′ with x ∈ X ′ r {x′ } are all pairwise distinct (increasing) and prove the thesis. If |X | = 1 or |Y | = 1, the thesis is straightforward. Finally, to choose colours for v , u1 and v u1 , we shall use the following lemma. Prior to the proof, we first specify the details of its application below. 2t +2 2t +2 in the expansion of the polynomial x2 Lemma 7. Let t be a positive integer, and let at be the coefficient of the monomial x2t 0 x1
Pt (x0 , x1 , x2 ) = (x0 − x1 )(x0 − x2 )2 (x1 − x2 )(x0 + x1 )3t +1 (x1 + x2 )3t −1 .
(3)
Then at ̸= 0. Note that the colours of at most d(v) − 1 ≤ ∆ − 1 (already coloured) edges incident with v are forbidden for v and v u1 . The colours of at most col − 2 backward edges of u1 cannot be used on u1 and v u1 . Moreover, the colours of at most d(v)− 1 ≤ col neighbours of v and at most col − 2 backward neighbours of u1 might be forbidden for v and u1 , respectively. Denote the lists of the remaining colours by L′v , L′v u1 , L′u1 , respectively. Then
′ 2 L ≥ ∆ + 5 col − (∆ − 1) − col = col + 1, v 3 3 5 2 |L′vu1 | ≥ ∆ + col − (∆ − 1) − (col − 2) = col + 3, 3 3 5 2 |L′u1 | ≥ ∆ + col − 2(col − 2) ≥ col + 3. 3
3
Let s1 , s2 , . . . , sp be the sums of all the neighbours of v except for u1 , hence p = d(v) − 1 ≤ col, and let s′1 , s′2 , . . . , s′q be the sums of all the backward neighbours of u1 preceding v , thus q ≤ col − 2. Denote additionally by s0 and s′0 , resp., the sums of v and u1 . Associate variables x0 , x1 , x2 with v , v u1 , u1 , respectively. These will stand for the colours which shall be assigned to these vertices and edge. Consider the following polynomial: P (x0 , x1 , x2 ) = (x0 − x1 )(x0 − x2 )(x1 − x2 )(s0 + x0 + x1 − s′0 − x1 − x2 )
×
p i=1
(s0 + x0 + x1 − si )
q i =1
(s′0 + x1 + x2 − s′i ),
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and suppose that there exist x′0 ∈ L′v , x′1 ∈ L′v u1 and x′2 ∈ L′u1 for which P (x′0 , x′1 , x′2 ) ̸= 0. Then by setting the colours of v, v u1 , u1 as x′0 , x′1 , x′2 , resp., the construction of P guarantees that these three colours are pairwise distinct, the obtained sums of v and u1 differ, and moreover the obtained sum of v is distinct also from the sums of all its other neighbours, while the sum of u1 is different from the sums of all its backward neighbours preceding v . Such colouring is consistent with all our requirements. Note that for
t :=
2 col 3
2
,
2t ≤ ⌈ 23 col⌉ < |L′v | and 2t + 2 ≤ ⌈ 23 col⌉+ 2 < |L′v u1 |, |L′u1 |. On the other hand, p ≤ col ≤ 3t + 1 and thus q ≤ col − 2 ≤ 3t − 1. Now let us consider the polynomial: Q (x0 , x1 , x2 ) := P (x0 , x1 , x2 ) × (x0 + x1 )3t +1−p (x1 + x2 )3t −1−q . Observe that if Q (x′0 , x′1 , x′2 ) ̸= 0, then also P (x′0 , x′1 , x′2 ) ̸= 0. Note that deg Q = 6t + 4 = 2t + (2t + 2) + (2t + 2), hence 2t +2 2t +2 x2t x2 is a maximal (with respect to the sum of its powers) monomial of Q . In order to prove the existence of desired 0 x1 ′ x1 , x′2 , x′3 via Combinatorial Nullstellensatz it is thus sufficient to prove that the coefficient of this monomial is nonzero in Q . However, by the maximality of this monomial (which must ‘take’ a variable, not a constant, from every bracket), its coefficient is the same as the coefficient at of the same monomial in the polynomial Pt defined in (3). The thesis thus follows by Lemma 7 and Theorem 3. Proof of Lemma 7. Note that Pt (x0 , x1 , x2 ) = (x0 − x1 )(x0 − x2 )(x1 − x2 )(x0 − x2 )(x0 + x1 )3t +1 (x1 + x2 )3t −1
= (x30 x1 − x20 x1 x2 − x30 x2 + x20 x22 − x20 x1 x2 + x0 x1 x22 + x20 x22 − x0 x32 − x20 x21 + x0 x21 x2 + x20 x1 x2 − x0 x1 x22 + x0 x21 x2 − x21 x22 − x0 x1 x22 + x1 x32 ) × (x0 + x1 )3t +1 (x1 + x2 )3t −1 . (We deliberately do not reduce monomials of degree four above, so as not to obscure the general idea of our approach, and thus to make accessible the analysis of its more complex correspondent which shall follow in the third case of our main 2t +2 2t +2 proof.) We now may write the formula for the coefficient at of x2t x2 in Pt via analysing every consecutive monomial 0 x1 2t +2 2t +2 of degree four above and the number of ways we may choose its ‘complement’ to x2t c c i c0−xi 1 x2 from the two factors in the c last line of the formula above (or exploiting the expansion (a + b) = valid for every positive integer c). In i=0 i a b
= 0 if d < 0 or a < d: 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 at = − − + 2t − 3 2t + 2 2t − 2 2t + 1 2t − 3 2t + 1 2t − 2 2t 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 − + + − 2t − 2 2t + 1 2t − 1 2t 2t − 2 2t 2t − 1 2t − 1 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 − + + − 2t − 2 2t + 2 2t − 1 2t + 1 2t − 2 2t + 1 2t − 1 2t 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 3t + 1 3t − 1 + − − + 2t − 1 2t + 1 2t 2t 2t − 1 2t 2t 2t − 1 (3t − 1)! (3t + 1)! · × [2t (2t − 1)(2t − 2)t (t − 1)(t − 2) − 2t (2t − 1)(t + 4)(2t + 2)t (t − 1) = (2t )!(t + 4)! (2t + 2)!t ! − 2t (2t − 1)(2t − 2)(2t + 2)t (t − 1) + 2t (2t − 1)(t + 4)(2t + 2)(2t + 1)t − 2t (2t − 1)(t + 4)(2t + 2)t (t − 1) + 2t (t + 4)(t + 3)(2t + 2)(2t + 1)t + 2t (2t − 1)(t + 4)(2t + 2)(2t + 1)t − 2t (t + 4)(t + 3)(2t + 2)(2t + 1)2t − 2t (2t − 1)(t + 4)t (t − 1)(t − 2) + 2t (t + 4)(t + 3)(2t + 2)t (t − 1) + 2t (2t − 1)(t + 4)(2t + 2)t (t − 1) − 2t (t + 4)(t + 3)(2t + 2)(2t + 1)t + 2t (t + 4)(t + 3)(2t + 2)t (t − 1) − (t + 4)(t + 3)(t + 2)(2t + 2)(2t + 1)t − 2t (t + 4)(t + 3)(2t + 2)(2t + 1)t + (t + 4)(t + 3)(t + 2)(2t + 2)(2t + 1)2t ] (3t + 1)! (3t − 1)! · · 6t (8 − 10t − 71t 2 − 9t 3 + 2t 4 ). = (2t )!(t + 4)! (2t + 2)!t !
the following,
a d
(The last equality above might be verified via tedious by hand calculations, or via computer program,1 as it was done in our case.) By the Rational Root Theorem, every possible integer-valued root of a polynomial 8 − 10t − 71t 2 − 9t 3 + 2t 4 must 1 One might use e.g. Mathematica.
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divide 8. By substituting 1, 2, 4 and 8, we then obtain that this polynomial has no roots which are positive integers, and thus at ̸= 0 for every integer t ≥ 1. 3.3. Case 3 Suppose now that col + 2 ≤ d(v) ≤ ⌈ 34 col⌉ (and thus d+ (v) ≥ 3). Then, analogously as in the previous case, we first choose colours for all forward edges and forward neighbours of v except for v u1 , v u2 , v u3 and u1 , u2 , u3 , consistent with the specified rules. This is again feasible via Lemma 6. To colour the remaining vertices and edges incident with v (and v itself), we shall apply the following more complex correspondent of Lemma 7. +1 2t +2 2t +2 2t +2 2t +3 2t +2 Lemma 8. Let t be a positive integer, and let bt be the coefficient of the monomial xt0+3 x2t x2 x3 x4 x5 x6 in the 1 expansion of the polynomial
Qt (x0 , x1 , x2 , x3 , x4 , x5 , x6 ) = (x0 − x1 )(x0 − x2 )(x0 − x3 )(x0 − x4 )(x0 − x5 )(x0 − x6 )
× (x1 − x2 )(x1 − x3 )(x1 − x5 )(x3 − x4 )(x3 − x5 )(x5 − x6 ) × (x0 + x3 + x5 − x2 )(x0 + x1 + x5 − x4 )(x0 + x1 + x3 − x6 ) × (x1 + x2 )3t (x3 + x4 )3t (x5 + x6 )3t (x0 + x1 + x3 + x5 )4t .
(4)
Then bt ̸= 0. As previously we first discuss an application of Lemma 8, and only just then — the proof itself. Analogously as above, we denote by L′v u1 , L′v u2 , L′v u3 the lists of colours available for v u1 , v u2 , v u3 formed of the original lists Lv u1 , Lv u2 , Lv u3 , resp., by removing the colours of (already coloured) edges incident with the respective out of these three edges. Let also L′v , L′u1 , L′u2 , L′u3 be the lists of available colours for v, u1 , u2 , u3 after excluding (from Lv , Lu1 , Lu2 , Lu3 ) the colours used by their respective incident edges and neighbours preceding v in the sequence (and the colours of u4 , . . . , ud+ (v) in the case of v ). Observe that then:
2 5 col − (∆ − 3) − (col − 2) = col + 5, |L′vu1 |, |L′vu2 |, |L′vu3 | ≥ ∆ + 3 3 5 4 1 5 ′ col − 2(d(v) − 3) ≥ col − col + 6 ≥ col + 5, |Lv | ≥ ∆ + 3 3 3 3 2 2 5 col − 2(col − 2) ≥ d(v) − col + col + 4 ≥ col + 6. |L′u1 |, |L′u2 |, |L′u3 | ≥ ∆ + 3
3
3
Associate variables x1 , x3 , x5 and x0 , x2 , x4 , x6 with v u1 , v u2 , v u3 and v, u1 , u2 , u3 , respectively. Let s0,1 , s0,2 , . . . , s0,p denote the sums of all the neighbours of v except for u1 , u2 and u3 , hence p = d(v) − 3 ≤ ⌈ 43 col⌉ − 3. For every i ∈ {1, 2, 3}, let also si,1 , si,2 , . . . , si,qi be the sums of all the backward neighbours of ui preceding v , thus qi ≤ col − 2. Denote additionally by s0 , s1 , s2 , s3 the sums of v, u1 , u2 , u3 , resp., and consider the following polynomial:
P (x0 , x1 , x2 , x3 , x4 , x5 , x6 ) = (x0 − x1 )(x0 − x2 )(x0 − x3 )(x0 − x4 )(x0 − x5 )(x0 − x6 ) × (x1 − x2 )(x1 − x3 )(x1 − x5 )(x3 − x4 )(x3 − x5 )(x5 − x6 ) × (s0 + x0 + x1 + x3 + x5 − s1 − x1 − x2 ) × (s0 + x0 + x1 + x3 + x5 − s2 − x3 − x4 ) × ( s0 + x0 + x1 + x3 + x5 − s3 − x5 − x6 ) q1 q2 × (s1 + x1 + x2 − s1,i ) (s2 + x3 + x4 − s2,i ) ×
i =1
i =1
q3
p
i =1
(s3 + x5 + x6 − s3,i )
(s0 + x0 + x1 + x3 + x5 − s0,i ).
i =1
Analogously as in the corresponding application of Lemma 7, if we find a nonzero substitution for P from the lists defined above, then the respective colours of v, v u1 , u1 , v u2 , u2 , v u3 and u3 shall be consistent with all our requirements. Note that for 1 col , t := 3 t + 3 < |L′v |, 2t + 1 < 2t + 2 < 2t + 3 ≤ ⌈ 23 col⌉ + 4 < |L′v u1 |, |L′v u2 |, |L′v u3 | and analogously 2t + 2 ≤ ⌈ 23 col⌉ + 3 < |L′u1 |, |L′u2 |, |L′u3 |. On the other hand, p ≤ ⌈ 34 col⌉ − 3 ≤ 4t and qi ≤ col − 2 ≤ 3t for i = 1, 2, 3. Let us consider the polynomial:
P (x0 , x1 , x2 , x3 , x4 , x5 , x6 ) × (x1 + x2 )3t −q1 Q (x0 , x1 , x2 , x3 , x4 , x5 , x6 ) := × (x3 + x4 )3t −q2 (x5 + x6 )3t −q3 (x0 + x1 + x3 + x5 )4t −p .
J. Przybyło / Discrete Applied Mathematics 202 (2016) 163–173
169
+1 2t +2 2t +2 2t +2 2t +3 2t +2 Note that deg Q = 13t + 15, hence xt0+3 x2t x2 x3 x4 x5 x6 is a maximal monomial of Q . Moreover, its coefficient 1 in the expansion of Q is the same as in Qt , given by (4), and thus equal to bt , which is nonnegative by Lemma 8. By Combinatorial Nullstellensatz we thus obtain that there is a nonzero substitution for Q (and hence also for P) from the lists L′v , L′v u1 , L′u1 , L′v u2 , L′u2 , L′v u3 , L′u3 . To conclude this case it thus remains to prove Lemma 8.
Proof of Lemma 8. Let us define a matrix W = [wij ] by setting:
1 2 3 4 5 6 2 W = [wij ] = 3 5 4 5 6 2 4 6
∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅
0 0 0 0 0 0 1 1 1 3 3 5 5 5 3
3 1 1
∅ ∅ ∅ ∅ ∅ ∅ ∅ ∅ . ∅ ∅ ∅ ∅ 0 0 0
(Note that the consecutive rows of W correspond to the first 15 linear factors of Qt from (4) and include the variables’ indicators arranged in the reversed order, so that the indicator of a unique variable preceded by a minus sign from a given bracket is always in the first column; the entries ‘∅’ shall be insignificant for the forthcoming formulas, and thus could as well be replaced by any other symbol.) Also define a function O : Z → Z such that
O(k) =
1 0
if k = 0, if k = ̸ 0.
Analogously as in the proof of Lemma 7 we may thus write that
15 2 4 4 4 3t 3t O(ji −1) 15 15 × bt = ... (−1)i=1 2t + 2 − O(wiji − 2) 2t + 2 − O(wiji − 4) j 1 =1 j 2 =1 j12 =1 j13 =1 j14 =1 j15 =1 2 2
i=1
3t
×
2t + 2 −
15
4t
O(wiji − 6)
t +3−
i =1
4t −
t +3−
O(wiji )
15
O(wiji )
15 15 2t + 1 − O(wiji − 1) − 3t − 2t + 2 − O(wiji − 2) i =1
i =1
15 i =1
×
i=1
i=1
15 15 O (w ) − 2t + 1 − O (w − 1 ) − 3t − 2t + 2 − O (w − 2 ) 4t − t + 3 − iji iji iji i =1 i =1 i=1 × . 15 15 2t + 2 − O(wiji − 3) − 3t − 2t + 2 − O(wiji − 4)
15
i =1
i =1
The summation above runs through all ordered 15-tuples within every of which each subsequent entry corresponds to 15 the index of one of the variables appearing in the appropriate consecutive bracket. The sum i=1 O(ji − 1) within the formula above counts the number of variables corresponding to a given 15-tuple which are preceded by a minus symbol in the respective brackets. Further, the first binomial coefficient counts the number of ways we may complement a given (obtained) monomial of degree 15 to obtain (altogether) 2t + 2 ‘copies’ of x2 , where these remaining copies (i.e, except for 15 the ones appearing already in the monomial of degree of 15, and counted by the formula i=1 O(wiji − 2)) must be chosen from the factor (x1 + x2 )3t . The following three binomial coefficients count the analogous numbers of ways we may obtain all together 2t + 2, 2t + 2 and t + 3 copies of x4 , x6 and x0 by complementing those from a given monomial of degree 15, while +1 choosing from (x3 + x4 )3t , (x5 + x6 )3t and (x0 + x1 + x3 + x5 )4t , respectively. Then we count in how many ways x2t might 1
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be achieved by choosing the necessary number of copies of x1 from the remaining (unused) factors of (x0 + x1 + x3 + x5 )4t , +2 and afterwards — in how many ways x2t might be achieved by choosing the necessary number of copies of x3 from the 3 4t remaining factors of (x0 + x1 + x3 + x5 ) (x5 shall be automatically chosen from the remaining factors). We thus obtain:
bt =
2 2
2 4 4 4
...
j1 =1 j2 =1
15
(−1)i=1
j12 =1 j13 =1 j14 =1 j15 =1
O(ji −1)
3t
3t
×
2t + 2 −
15
O(wiji − 2)
2t + 2 −
4t
3t
×
15
2t + 2 −
O(wiji − 6)
t +3−
3t − 3 +
× O(wiji )
i=1
i=1
15
O(wiji − 4)
i =1
i =1
15
15
O(wiji )
i =1
t +3−
15
(O(wiji − 1) + O(wiji − 2))
i=1
15
2t − 6 + (O(wiji ) + O(wiji − 1) + O(wiji − 2)) i =1 × 15 t + 4 − (O(wiji − 3) + O(wiji − 4)) i=1
=
2 2
2 4 4 4
...
j 1 =1 j 2 =1
15
(−1)
O(ji −1)
i=1
j12 =1 j13 =1 j14 =1 j15 =1
(3t )!
× (2t + 2 −
15
O(wiji − 2))!(t − 2 +
i=1
O(wiji − 2))!
i=1
(3t )!
× (2t + 2 −
15
O(wiji − 4))!(t − 2 +
i=1
15
O(wiji − 4))!
i=1
(3t )!
× (2t + 2 −
15
O(wiji − 6))!(t − 2 +
i=1
×
15
15
O(wiji − 6))!
i=1
(4t )! 15 (t + 3 − O(wiji ))!(t + 3 − (O(wiji − 1) + O(wiji − 2)))! 15
i =1
i=1
1
× (t + 4 −
15
(O(wiji − 3) + O(wiji − 4)))!
i =1
1
× (t − 10 +
15
,
(O(wiji ) + O(wiji − 1) + O(wiji − 2) + O(wiji − 3) + O(wiji − 4)))!
i=1
where we assume every ingredient containing a factor a! with a < 0 (in a denominator) to be equal to 0 in the sum above. Similarly as in the proof of Lemma 7 we may thus now extract the common factors of all the elements under the sum. In order to do it possibly most efficiently, we analyse the first 15 brackets of the polynomial Qt (or the rows of matrix W ) and note that x2 appears in just 3 of these, and so does x4 and x6 . Moreover, (at least) one bracket is formed exclusively of x1 and x2 , and one of x3 and x4 . Similarly, there is (at least) one bracket which contains none of x0 , x1 , x2 , x3 , x4 . We thus finally a−1 have (assuming that k=a f (k) = 1 for any function f and integer a): bt =
(3t )! (3t )! (3t )! (4t )! · · · (2t + 2)!(t + 1)! (2t + 2)!(t + 1)! (2t + 2)!(t + 1)! (t + 3)!(t + 2)!(t + 3)!(t + 4)! 2
×
2
j1 =1 j2 =1
2
...
4
4
4
15
j12 =1 j13 =1 j14 =1 j15 =1
(−1)
i=1
15
O(ji −1)
i=1
×
O(wij −2)−1 i
3
(2t + 2 − k)
k=0 k=
15 i=1
(t − 2 + k)
O(wij −2)+1 i
J. Przybyło / Discrete Applied Mathematics 202 (2016) 163–173 15
15
O(wij −4)−1 i
i=1
×
3
(2t + 2 − k)
k=0 k=
15 i=1
15
15
O(wij )−1 i
i=1
15 i=1
15 i=1
15 i=1
(t − 2 + k)
O(wij −6)+1 i
(O(wiji −3)+O(wiji −4))−1
(t + 3 − k) ×
14
k=
k=
i
(t + 4 − k)
k =1
k=1
×
3
(2t + 2 − k)
k=0
O(wij −4)+1
k=0
=
(t − 2 + k) ×
(O(wiji −1)+O(wiji −2))−1
(t + 3 − k)
i
i=1
×
O(wij −6)−1
i=1
171
(t − 10 + k)
(O(wiji )+O(wiji −1)+O(wiji −2)+O(wiji −3)+O(wiji −4))+1
(3t )! (2t + 2)!(t + 1)!
3
(4t )! · 12 · (1152 + 18336t + 127472t 2 (t + 2)!((t + 3)!)2 (t + 4)!
+513920t 3 + 1347344t 4 + 2444534t 5 + 3188541t 6 + 3062765t 7 + 2197996t 8 +1186064t 9 + 479902t 10 + 143110t 11 + 30108t 12 + 4066t 13 + 285t 14 + 5t 15 ), where the last equality above is accessible only via application of a computer program.2 By the Rational Root Theorem, every possible integer-valued root of the polynomial of degree 15 in t above must divide 1152 = 27 · 32 . By substituting all (altogether 24) positive divisors of 1152, we then obtain that this polynomial has no roots which are positive integers, and thus bt ̸= 0 for every integer t ≥ 1. 3.4. Case 4 Assume eventually that d(v) ≥ ⌈ 43 col⌉ + 1 (hence d+ (v) ≥ 3 and ∆ ≥ ⌈ 43 col⌉ + 1). We then first fix a colour of v chosen from Lv so that it is distinct from the colours of (at most ∆) neighbours of v and from the coloured (at most col − 1) edges incident with v . Each forward edge v ui of v has then at most d− (v)+(col − 2) colours blocked by its coloured incident edges, and at most one by v (we disregard the colour of ui , which shall be changed later if necessary). Thus every v ui is left with the list L′v ui of at least ∆ +⌈ 35 col⌉− d− (v)−(col − 2)− 1 ≥ (d− (v)+ d+ (v))+⌈ 32 col⌉− d− (v)+ 1 ≥ d+ (v)+ 3 available colours. We shall simultaneously choose colours from these lists for all forward edges of v so that all these edges’ colours are pairwise distinct and the obtained sum of v is different from the sums of its backward neighbours. This might be achieved using Combinatorial Nullstellensatz via analysis of a properly defined polynomial. Such approach would however require a quite complex argument, see the last section for further comments. We thus use a simpler method inspired by [8], applicable in this particular case, which might be viewed at as a generalisation of Lemma 6. For this purpose, we first dynamically modify the lists L′v u1 , . . . , L′v u + . Subsequently, for i = 1, 2, . . . , d+ (v) − 1, we take min L′v ui (where L′v ui always refers to the up-tod
(v)
date remainder of this list on a given stage of our modifying procedure) and remove it from all current lists L′v uj with j > i.
Then, subsequently, for i = d+ (v), d+ (v) − 1, . . . , 2, we find max L′v ui and remove it from all up-to-date lists L′v uj with j < i,
and denote the finally constructed respective lists by L′′v u1 , . . . , L′′v u
d+ (v)
. This way at most d+ (v) − 1 element were removed
from every list L′v ui and for every i < j < k, L′′v uj contains neither min L′′v ui nor max L′′v uk . Thus if we denote the consecutive (increasing) elements of each of the obtained lists L′′v ui by ci,1 , ci,2 , . . . , ci,li (hence ci,1 = min L′′v ui and ci,li = max L′′v ui ), then
d+ (v) each of the following at least i=1 |L′′v ui | − d+ (v) + 1 ≥ d+ (v)((d+ (v) + 3) − (d+ (v) − 1)) − d+ (v) + 1 = 3d+ (v) + 1 distinct sums consists of d+ (v) pairwise distinct integers: c1,1 + c2,1 + · · · + cd+ (v)−2,1 + cd+ (v)−1,1 + cd+ (v),1
< c1,1 + c2,1 + · · · + cd+ (v)−2,1 + cd+ (v)−1,1 + cd+ (v),2 ... < c1,1 + c2,1 + · · · + cd+ (v)−2,1 + cd+ (v)−1,1 + cd+ (v),ld+ (v)
< c1,1 + c2,1 + · · · + cd+ (v)−2,1 + cd+ (v)−1,2 + cd+ (v),ld+ (v) ... < c1,l1 + c2,l2 + · · · + cd+ (v)−2,ld+ (v)−2 + cd+ (v)−1,ld+ (v)−1 + cd+ (v),ld+ (v) .
2 E.g., Mathematica.
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J. Przybyło / Discrete Applied Mathematics 202 (2016) 163–173
Since v has at most col − 1 ≤ 3(( 43 col + 1) − (col − 1)) ≤ 3(d(v) − d− (v)) = 3d+ (v) backward neighbours, at least one of these sums must yield s(v) distinct from the sums of all of these neighbours. We then choose colours of the forward edges of v corresponding to this sum. At the end we subsequently adjust the colour of every vertex ui , with i = 1, 2, . . . , d+ (v), so that its colour is different from the colours of its backward edges and neighbours, and so that the obtained sum of ui is distinct from the sums of its backward neighbours. Altogether at most 3(col − 1) colours might thus be forbidden for ui . However, in the analysed case, |Lui | ≥ ∆ + ⌈ 35 col⌉ ≥ (⌈ 43 col⌉ + 1) + ⌈ 35 col⌉ > 3(col − 1), thus we may always choose an appropriate colour from Lui for ui , i = 1, . . . , d+ (v). The proof of Theorem 4 is thus completed. 4. Remarks Within the general approach presented in this paper we could not obtain a better bound than roughly, i.e. up to a small additive constant, ∆ + 53 col, even with more detailed algebraic analysis applied. Our main result thus could not be significantly improved. It is due to the second case analysed in the proof above, which we could not omit. To see that this case prevents a significant reduction of the number of colours necessary for our algorithm, assume that e.g. ∆ is roughly equal to col, and a vertex v , which we are about to analyse, has only one forward neighbour u1 , where v and u1 have both roughly col neighbours preceding v . Suppose that we strive at proving a somewhat smaller upper bound than ∆ + 53 col for ′′ ch′′ (G) (or χ (G)). Then the variables x0 , x1 , x2 , representing colours sought for v, v u1 , u1 , resp., each have at most roughly
∆ + 53 col − 2col ≈ 23 col colours available (i.e., not blocked by the colours of other edges or vertices), say 1, 2, . . . , 23 col for all of these. Now within these possibilities, e.g. col smallest achievable values for x0 + x1 might be blocked by the backward neighbours of v (hence x1 would have to be larger than 31 col), while col largest achievable values for x1 + x2 might be blocked
by the backward neighbours of u1 (and thus x1 should be at most 31 col, a contradiction). Using the same approach, we might however be able to improve our main result for graphs with maximum degree (much) larger than their colouring number, i.e. decrease for them the multiplicative factor 35 preceding col(G) in the bound from Theorem 4. The main obstacle on the way towards such goal is Case 3 from the proof above. Within this, instead of just one polynomial (4), we would have to analyse a whole (possibly infinite) family of its correspondents of increasing complexity. Moreover, as already occurred while investigating polynomial (4), a monomial with a non-zero coefficient in the expansion of a given representative of such a family might presumably lack symmetry (in its variables’ powers), what might cause additional complications for our reasonings. (Our experimental analysis showed e.g. that Lemma 8 was false +1 2t +2 2t +2 2t +2 2t +3 2t +2 for ‘more symmetrical’ correspondents of the monomial xt0+3 x2t x2 x3 x4 x5 x6 .) 1 To complement our aforementioned comment, let us also note that a polynomial whose analysis would be equivalent to our reasoning in Case 4, has the following form:
(xi − xj )
1≤i
3d+ (v)
(s0 + x1 + x2 + · · · + xd+ (v) − si ),
(5)
i =1
where the investigated vertex v has a (temporary) sum s0 , variables x1 , . . . , xd+ (v) correspond to colours of consecutive forward edges of v , while we estimated in Case 4 that v may have at most 3d+ (v) backward neighbours, whose sums are denoted by s1 , . . . , s3d+ (v) above. The reason why in this case we could come up with a simpler approach than the one based on Combinatorial Nullstellensatz was the fact that, unlike in the previous cases, we could concentrate on just one sum of variables (corresponding to the sum of v ), namely x1 + x2 + · · · + xd+ (v) , in order to evade a number of potential conflicts (while in previous cases we had to control two or even four corresponding sums associated with respective vertices simultaneously). It is also worth noting that the construction used for handling Case 4 is optimal (and might also be easily generalised to ‘less symmetrical’ cases). To see that, suppose accordingly to our assumptions from that case, that every variable xi is associated with the list of d+ (v) + 3 available values, say 1, 2, . . . , d+ (v) + 3 for each of them. Then, given that the first factor of the polynomial (5) corresponds to the required feature that distinct variables take different values (colours), there might be at most (4 + 5 + · · · + (d+ (v) + 3)) − (1 + 2 + · · · + d+ (v)) + 1 = 3d+ (v) + 1 distinct values available for the sum x1 + x2 + · · · + xd+ (v) (disregarding the second factor of (5)), and exactly this many were guaranteed by our construction in Case 4. Acknowledgements The research contained in this paper were supported by the National Science Centre, Poland, grant no. 2014/13/ B/ST1/01855, and partially supported by the Polish Ministry of Science and Higher Education. References [1] L. Addario-Berry, K. Dalal, C. McDiarmid, B.A. Reed, A. Thomason, Vertex-colouring edge-weightings, Combinatorica 27 (1) (2007) 1–12. [2] L. Addario-Berry, K. Dalal, B.A. Reed, Degree constrained subgraphs, Discrete Appl. Math. 156 (7) (2008) 1168–1174. [3] M. Aigner, E. Triesch, Irregular assignments of trees and forests, SIAM J. Discrete Math. 3 (4) (1990) 439–449.
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