20150709, 17:56  #12 
Dec 2012
2×139 Posts 

20150722, 07:36  #13  
Jul 2015
10_{10} Posts 
Quote:
There are 216 possible number’s from 1 1 1 to 6 6 6 (6^3) Out of them only 120 are legal (Those without duplications) from 1 2 3 to 6 5 4 (6*5*4) So in the answer we will have 96 – 0s (the illegal cases) and 120 numbers from 16. My idea is that we need to find a way that given any 3 numbers (pointers) will always be mapped to a single result. So the challenge is to find the lead numbers (A1A120) numbers from 16 that will satisfy the below For example lead 2nd 3rd 4th A1 1 2 3 A2 1 2 4 A3 1 2 5 A4 1 2 6 A5 1 3 2 A6 1 3 4 A66 4 2 3 A67 4 2 5 A120 6 5 4 A1 is 4 or 5 or 6 A2 is 3 or 5 or 6……. A66#A1 since if they are the same the 2nd will not know if he is 1 or 4 In the same method A3#A67 2nd will not know if he is 1 or 4 A2#A6 in order that the 3rd will know if he is 3 or 2 In a similar way A4#A3#A2#A1 in order that the 4th will know if he is 3 4 5 or 6 I also believe that for symmetry reasons we will have I the solution 20 times 6s and same 20 5s… comes together to 120 numbers. 

20150730, 16:17  #14  
"Robert Gerbicz"
Oct 2005
Hungary
2534_{8} Posts 
Quote:
My solution has not that form, for example it has got only sixteen 5's. But there could be such solution (not checked). 

20150731, 15:28  #15 
Jul 2015
12_{8} Posts 
thanks

20150802, 17:02  #16 
"Robert Gerbicz"
Oct 2005
Hungary
2^{2}·7^{3} Posts 
The official solution has been posted:
https://www.research.ibm.com/haifa/p.../July2015.html My solution used integer programming, pretty straightforward. See my Glpk code and screenoutput. Additional note for my code: I write: "at the last person's guess do not use the same hat what he is seeing" and we don't need to see this for the other people, as they are guessing correctly their hat's number, and those are different numbers. The third person guesses correctly thanks to p3 condition, now see the second person: he is seeing only the first person's hat, but the third person made a correct guess, so he knows that number also, using p2 he can make a correct guess. And this is also true for the first person (see the p1 condition). Last fiddled with by R. Gerbicz on 20150802 at 17:05 Reason: grammar 
20150819, 16:13  #17 
"Robert Gerbicz"
Oct 2005
Hungary
2^{2}·7^{3} Posts 
There was a request to me how to learn the Glpk language.
If you have installed Glpk (the newest has version number 4.55) then you will see a doc folder containing gmpl.pdf that describes how to build a model. In the examples folder you can find many glpk codes, I have already posted some, these files are really useful and glpk independent in that sense that you can learn how to build a lp/ip for a problem.Note that in this folder you can find the glpsol program that solves the glpk programs. 
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