, This weighting is not equitable but is indeed good and neighbour-sum-distinguishing

,. .. and ;. .. , We claim that G is a (1, {2k + 1})-gadget. In any neighbour-sum-2-distinguishing 22, respectively. In the second case, we get sum conflicts between u 2 and its neighbours in G 1

. Proof, Since the problem is obviously in NP, we proceed with the proof of its NP-hardness. is: ? if F has a clause (x i 1 ? x i 1 ? x i 1 ), then F is not satisfiable in a 1

, ? if F has a clause (x i 1 ? x i 1 ? x i 2 ), then x i 2 and x i 1 are forced to true and false, respectively, by any truth assignment making F

.. .. Variables-of, F. , and ,. .. , For each variable x i of F , we add to G a star V i with root v i and 2k i leaves u i,1 ,. .. , u i,2k i , where 2k i ? max{10, n i } is any even integer, and n i is the number of clauses of F that contain x i. Next we add (1, {4k i + 1})-, (2, {4k i + 3})-, (2, {4k i + 5})-,. .. , (2, {6k i ? 3})-and (2, {6k i ? 3})-gadgets G 1, We denote by x 1

. Proof, For any fixed k ? {1, 2, 3, 4}, we introduce below an algorithm that checks in polynomial time whether T admits a neighbour-sum-2

.. .. , .. .. Being-t-u-1-+-vu-1, ,. .. Root-;, .. .. , .. ;. et al., For every shrub, we call the edge incident to the root the root edge. The non-root end of the root edge is called the subroot. We are now ready to describe our algorithm for deciding whether T admits a neighbour-sum-2-distinguishing k-edge-weighting. The rough ideas are the following. The tree T can be seen as a union of d := d(r) shrubs S 1 ,. .. , S d whose roots were identified, resulting in r. A neighbour-sum-2-distinguishing k-edge-weighting of T is thus essentially the union of (relaxed, see below) neighbour-sum-2-distinguishing k-edge-weightings of the d shrubs attached to r, with the additional property that the resulting ?(r) does not create any sum conflict. Therefore, in order to construct a neighbour-sum-2-distinguishing kedge-weighting of T , it suffices to find convenient neighbour-sum-2-distinguishing k-edgeweightings of S 1, Designate a node r of T as being its root. This defines a root-to-leaf orientation of T in the usual way, where every non-root node v has a parent, and every non-leaf node v has children. By the descendants of v, we refer to the nodes of T for which we find v when iterating the parent relationship. The subtree T v of T rooted at v is the subtree whose nodes are v and all its descendants. 1

, implying that this subgraph is not locally irregular, thus that the 4-edge-colouring is not locally irregular, a contradiction. So, necessarily, one of the four colours does not appear around x and v in G , and the previous case applies. Assume now that d(u) = d(v) = d(w) = d(x) = 3. We denote by u , v , w , x , respectively, the neighbour of u, v, w, x, respectively, necessarily one of wx and wv is isolated in the subgraph induced by its assigned colour

. Assume, Note first that we cannot have ?(vw) = 1 or ?(wx) = 1. Indeed, in such a situation (say ?(wx) = 1), so that all four colours appear in the neighbourhood of u, v, x, one would need, without loss of generality, ?(xx ) = 2, ?(vw) = 3 and ?(vv ) = 4. But then either wx is an isolated edge in the 1-subgraph 1 , or vw is an isolated edge in the 3-subgraph, contradicting the fact that ? is locally irregular. So we may assume that 1 ? {?(vw), ?(wx)}. We consider two cases depending on whether ?(vw) and ?(wx) are equal or not

, Because the 2subgraph is locally irregular, we necessarily have ?(ww ) = 2, which implies, because the 3-subgraph is locally irregular, ?(vv ) = 3. Therefore, if u is a 2-vertex in the 1subgraph, then we can extend ? to G by setting ?(ux) = ?(uv) = 1. So assume u is a 3-vertex in the 1-subgraph. Analogously, if v is not a 3-vertex in the 3-subgraph, then we can extend ? to G by setting ?(uv) = 3 and ?(ux) = 1. So assume v is a 3-vertex in the 3-subgraph. Now, if w is not a 3-vertex in the 2-subgraph, then we can extend ? to G by setting ?(wv) = ?(wx) = 2, and ?(ux) = 1 and ?(uv) = 3, Without loss of generality, assume that ?(vw) = 3 while ?(wx) = 2, and also that ?(xx ) = 4 (since colour 4 appears in the neighbourhood of u, v, x)

, ? Case 2: ?(vw) = ?(wx)

, We may assume that ?(vw) = ?(wx) = 2, and that ?(vv ) = 4 and ?(xx ) = 3

, Recall that every result on strong (2, 2)-colourings directly transfers to standard and weak (2, 2)colourings. We start off with complete graphs

, So we may assume that G is an odd multicactus, and thus that G is not strongly (1, 2)colourable

, Standard (, k)-colouring

, Note that a standard (, k)-colouring is nothing but a decomposition into graphs admitting neighbour-sum-distinguishing k-edge-weightings. From that perspective, it could be interesting to wonder whether graphs, general, decompose

. Conjecture, Towards the Standard Conjecture, we thus also raise the following related conjecture, which is

. .. , Recall that every locally irregular graph H verifies ? e ? (H) = 1. Furthermore, all nice bipartite graphs verify the 1-2-3 Conjecture. From these arguments, using a set of 40 coloured weights 1, 2, 3 to independently weight the edges of each of the H i 's and the H i,j 's, i ? {a, b, c}, j ? {1, vol.36

, Together with Theorem 7.5.2, this yields the following: then we call G cubic. Furthermore, if G is connected and not cubic, i.e., G has vertices with degree 1 or 2, then we say that G is strictly subcubic. We first prove the Standard Conjecture for 2-degenerate graphs

, 2)-colourings), each of F r and F b , independently, admits a standard (1, 2)-colouring; let ? r and ? b be any such standard (1, 2)-colourings of F r and F b , respectively, Our proof of Theorem 7.5.4 relies on the following lemma, which is proved later in this i.e., admits standard

. Proof, Throughout the proof, which is by induction on |V (G)|+|Enow consider the general case |V (G)| ? 5. If G is strictly subcubic

G. By-v-1 and .. , We have p, q ? 1. If p ? 2, then we recolour uw 1 , the first edge of the red path P , blue. From the point = 1, and by symmetry q = 1, then we can proceed as above except when the red let x 1 ,. .. , x n denote the vertices of G. We 2-edge-colour G with colours red and blue, yielding two subgraphs G R and G B , respectively, as follows. An edge x i x j is coloured red if and only if i = j mod 3, G R , different from w 1 , w l , respectively

G. and G. , By Lemma 7.5.9, it can be decomposed into two 3-colourable graphs: an r-colourable graph G R and a bcolourable graph G B with r, b ? 3. Since G is at least 4-chromatic we have also r, b ? 2. We distinguish two cases: ? If G has no isolated triangles, then, by Lemma 7.5.8, it can be decomposed into two nice graphs G R and G B with ?(G R ) ? r and ?(G B ) ? b with r, b ? 3. So, both of them verify the 1-2-3 Conjecture, and thus admit standardly (1, 3)-colourings, G is standardly (1, 3)-colourable

, 3)-colouring of what is left, and then give the same color to all the edges of the isolated triangles. Finally, by weighting the edges of each isolated triangle with weights 1, ? If G has isolated triangles, then we remove those from G, apply the previous point to get a, vol.2

. Conjecture,

, Towards the Weak Conjecture, we here first prove that all nice graphs are weakly (3, 2)and (2, 4)-colourable. Both proofs are based on the fact that every nice graph admits a neighbour-sum-distinguishing 5-edge-weighing

, Weak (3, 2)-and (2, 4)-colourability

, Karo?ski and Pfender [KKP10] allow to show that every nice graph even admits a neighbour-sum-distinguishing {s ? 2, s ? 1, s, s + 1, s + 2}-edge-weighting, for any integer s. Let thus ? be a neighbour-sumdistinguishing {?2, ?1, 0, 1, 2}-edge-weighting of G. We deduce a weak (3, 2)-colouring of G by modifying and colouring the weights of ?

, ? we colour blue every edge with value in {?2, ?1}, and multiply its value by ?1

, ? we colour green every edge with value 0, and change its value to 1

, The key point is that, through ?, every two adjacent vertices u and v are only distinguished via their incident edges with weight in {?2, ?1, 1, 2}. Said differently the edges with weight 0 are useless for distinguishing u and v. This implies that, in the obtained (3, 2)colouring, it is not possible that both the red and blue sums of u and v are equal

, Since G is nice, it admits a neighbour-sum-distinguishing 5-edge-weighting ? according to the result of Kalkowski, Karo?ski and Pfender

, and change its value to 1. i (v), the weight of each edge might be modified only once

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