20190501, 12:48  #1 
"Mike"
Aug 2002
3·2,683 Posts 
May 2019

20190508, 05:20  #2 
Sep 2017
3^{2}×11 Posts 
Any ideas on how to solve this?

20190508, 10:34  #3 
Oct 2017
2·5·11 Posts 
Nice and difficult challenge. I have needed some time to program the “breadth first search” (which I didn’t know until this weekend) for verifying the shortest path of length 13 of the example with 8 people and with the restrictions given by 4672616e63657320452e20416c6c656e.
Fortunately the puzzlemaster doesn’t want to know restrictions like “child a isn’t willing to stay with child b” or “the husbands are so jealous, that no woman can be in the presence of another man unless her husband is also present” or so, but he only wants to know the number. So I’ll vary the 32digit hexadecimal number randomly, hoping, that I'll find a number leading to a shortest path length of 73. Last fiddled with by Dieter on 20190508 at 10:36 
20190509, 06:30  #4 
Romulan Interpreter
Jun 2011
Thailand
83×113 Posts 
We solved the bonus part (which is quite easy) and learned few things on the way by googling her name and her connection with groups of 19 people.
For the main task, we don't have much time to try... Real life ^{(TM)} is killing us right now... Last fiddled with by LaurV on 20190509 at 06:30 
20190509, 06:39  #5 
Dec 2012
The Netherlands
3175_{8} Posts 

20190516, 06:47  #6 
Oct 2017
2·5·11 Posts 
Some remarks on the challenge –at the risk of saying only trivial points:
Last fiddled with by Dieter on 20190516 at 06:48 
20190516, 22:07  #7  
Sep 2017
3^{2}×11 Posts 
Quote:


20190517, 07:21  #8  
Oct 2017
156_{8} Posts 
Quote:
I use BFS for calculating the shortest path length of a set of restrictions defined by a 32digit number, because I don't know another procedure. My current number (lenght 45) contains many "f's" and other digits. I try to improve the number by choosing 6 or 7 digits : two digits = f and four/five digits <> f, and then checking all combinations (2^24 respectively 2^28 combinations). Sometimes I find a better number. If I use the four threads of my vostro 3560 from 2012 (i53210M) I am able to check 2^23 combinations in a little bit more than one hour. The puzzlemaster hinted the possibility of a prolongation of the challenge to one more month... 

20190517, 07:37  #9  
Sep 2017
3^{2}×11 Posts 
Quote:


20190520, 17:59  #10  
May 2019
2^{2} Posts 
Quote:


20190521, 08:12  #11  
Oct 2017
2·5·11 Posts 
Quote:


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