, Karo´nskiKaro´nski and Pfender [8] 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-sum-distinguishing {?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 ?, Slight modifications of the proof of Kalkowski
, ? 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
, Karo´nskiKaro´nski and Pfender [8] that every nice graph G admits a neighboursum-distinguishing 5-edge-weighting can be modified to prove that every nice graph admits a neighboursum-distinguishing {1, 2, 3, 4, 6}-edge-weighting. We voluntarily do not give the full proof of this claim in details, as the proof would be identical to the original one, The proof of Kalkowski
, At the beginning of the algorithm, all edges v i v j are assigned weight f (v i v j ) = 4
, the possible valid moves are the following: -If f (v j v i ) = 4, then doing either ?2 (changing the weight to 2) or +2 (changing the weight to 6) is a valid move
, the first forward edge (i < j) incident to v i in order to define the two allowed sums for v i , doing either ?1 (changing the weight to 3) or ?3 (changing the weight to 1) to f (v i v j ) is a valid move. Furthermore, we can choose the two sums allowed for v i and perform valid moves on edges incident to v i so that: -If f (v i v j ) is changed to 3, then the current sum of v i is the biggest of the two allowed ones, ? Whenever it is needed to modify the weight f
, Let ? be a neighbour-sumdistinguishing {1, 2, 3, 4, 6}-edge-weighting of G. We deduce a weak (2, 3)-colouring of G by 2-colouring and (possibly) altering the weights assigned by ?
, Note that if the red sums of u and v are equal, then their blue sums cannot be equal too: in such a situation, we would get ? ? (u) = ? ? (v), a contradiction
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