, Thus, c(u 1 ) is different from the 23 used colors. Therefore, we obtain a contradiction

, = c(v 1 ) = c(x 1 ) it can be remarked that every vertex of B ? {(0)} is at distance at most 3 of (at least) one vertex among {u 2 , v 1 , x 1 } and, as previously, we obtain a contradiction since

B. ?{, Suppose that c(u 1 ) = c(v 2 ) = c(y 2 ) It can be easily verified that every vertex of (B \ {(1.5, 2.5, 1)}) ? {(0)} is at distance at most 3 of (at least) one vertex among {u 1 , v 2 } and that every vertex of (B ? \ {(?1.5, 2.5, 1)}) ? {(0)} is at distance at most 3 of one vertex among {u 1 , y 1 }) and (3, 1, 2) has a color not given to the vertices of B and one vertex among (0, 1, 2) and (?3, 1, 2) has a color not given to the vertices of B ? . Since both B?{, )} is at distance at most 3 of one vertex among {u 1 , v 2 }. By Lemma 5.7, one vertex among?3, 1, 2)} contain vertices of 23 different colors, we have c(u 1 ) = c((1.5, 2.5, 1)) and c(u 1 ) = c((?1.5, 2.5, 1)). However, since d F0, 2001.

B. , since every vertex of (B \ {(1.5, 2.5, 1)}) ? {(0)} is at distance at most 3 of one vertex among {u 2 , v 1 } and every vertex of)} is at distance at most 3 of one vertex among {u 2 , y 1 }, we have also a contradiction since, as previously, ?3, 1, 2)} contain vertices of 23 different colors)) and c(u 2 ) = c((?1.5, 2.5, 1)), 2002.

, = c(v 1 ) holds. First, suppose c(u 1 ) = c(v 2 ) By Case 1, we have c(u 1 ) = c(x 2 ) and by Case 2, we have c(u 1 ) = c(y 2 ). Consequently, by Lemma 5.5, we have c(u 2 ) = c(x 1 ) = c(y 1 ) and by symmetry and Lemma 5, we can not have both c(u 1 ) = c(y 2 ) and c(u 2 ) = c(y 1 ) and a contradiction with Case 1. Second suppose c(u 2 ) = c

, = c(x 1 ) and by Case 2, we have c(u 2 ) = c(y 1 ). Consequently, by Lemma 5.5, we have c(u 1 ) = c(x 2 ) = c

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