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 2016-11-01, 13:35 #1 Xyzzy     Aug 2002 37×229 Posts November 2016
 2016-12-06, 16:41 #2 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 112·13 Posts The official solution is at: https://www.research.ibm.com/haifa/p...ember2016.html That is Motty Porat's math solution (this earned a star for him, the only star solution!). My sent solution was: "The minimal is N=42, and one possible solution: 0 13 29 11 12 19 10 11 21 9 10 23 8 9 25 7 8 27 6 7 29 5 18 19 4 17 21 3 16 23 2 15 25 1 14 27 found this in 18 minutes with a backtracking code: for each month we store the possible triplets: if we fix the i-th month's triplet, then we store those triplets in the further months (j=i+1,..,12) for that we don't get a violation for the (i,j) month dual. In this way we ensure that (k,j) month dual will be valid for all ki). If we reach i=12, then obviously we found a solution." [...] ps. After I have sent this observed that we can use symmetry: we can assume that N1

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