July 2015

[Jayder]

Originally Posted by R. Gerbicz

they have agreed on a strategy in advance."

That was not at all obvious to me in the original version.

[tgan]

Originally Posted by R. Gerbicz

There is an update on Ibm (we also discussed on this issue):

"Update (9/7, thanks James):
The last person (first to guess) is helping the other players to win, and that they have agreed on a strategy in advance."

And I was again the first solver of the puzzle.

Hi good work that you are the first. I am trying to solve this and this is what I was able to understand.
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 1-6.
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 (A1-A120) numbers from 1-6 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.

[R. Gerbicz]

Originally Posted by tgan

I am trying to solve this and this is what I was able to understand.

You write good ideas, what you need is a program (or a long paper?)

Originally Posted by tgan

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.

My solution has not that form, for example it has got only sixteen 5's. But there could be such solution (not checked).

[tgan]

thanks

[R. Gerbicz]

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).

[R. Gerbicz]

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.