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Old 2016-11-01, 13:35   #1
Xyzzy
 
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"Mike"
Aug 2002

11110000100112 Posts
Default November 2016

https://www.research.ibm.com/haifa/p...ember2016.html
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Old 2016-12-06, 16:41   #2
R. Gerbicz
 
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"Robert Gerbicz"
Oct 2005
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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 k<i (where j>i). 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<N2<N3, with this the running time is only 2 seconds... Btw it is the lex. smallest solution.
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