“Imagine a 64-square chessboard with a coin on each square.
Each of the coins has either head or tails facing up, chosen at random.
I check the board and decide which coin will be my favorite one.
One of you (say A) will be with me, see the chessboard and I will reveal to him (only to him) which coin is my favorite. He then must flip over exactly one of the coins on the chessboard, while the other mathematician (B) is in another room not looking.

Once the coin is flipped over, the uninformed mathematician (B) is summoned into the room and must deduce which coin is my favorite only by examining the coins on the chessboard.
To secure absence of any other hints A is escorted out of the room.

Clearly, prior to the procedure, you are free to discuss the problem between the two of you and establish its solving strategy. You have no time limit, you are free to use any kind of calculator, but any communication between you two is strictly prohibited”

What strategy can the two mathematicians devise to ensure that my favorite coin can always be correctly identified?

Thanks,  Charlie

There was no serious attempt to solve this beautiful puzzle,other than non-related remarks and general bla-bla.
Although there are several versions of this problem (jailer, devil etc) on the web and some provide quite obscure solutions, IMHO none of the solvers bothered to explore the googling option.

I will try to provide 100% sure strategy, illustrate it with an example, assuming  that you are familiar with binary strings and the basic axioms of the XOR operation.  If not ,  just read the definitions at https://en.wikipedia.org/wiki/Exclusive_or

I will publish the official solution as soon as I  receive a feedback confirming that my presentation was coherent and “doable”.

Or, what changes are needed  to make it totally clear
So ��" here we go:

1. A will agree with his friend B that they map the board from top-left to bottom-right from 0 to 63.
2.Heads  get the 6-digits strings corresponding to their locations.

3. A will  XoR between all the Heads "addresses" and get a 6 bit binary number as a result, say string RES

4. After being told by the challenger a non ��"related number, say NRN  A will evaluate RES XOR NRN , get a string NEW , and flip the coin on a corresponding square.

5. A exits, B enters and evaluates the total XOR of heads  ( i.e. their strings )

6. The result is  NRN -  non related number

why?    Because if X  XOR Y= Z    then  X XOR Z =Y

Example:
Let’s say that there are tails only on 3,7,20,61 and the "chosen" square is 8: binary XoR on HEADS{3,7,20,61} will be:

000011
000111
010100
111101
101101    (DEC 45)  

101101    (DEC 45)  XOR  001000 (DEC 8) =100101( DEC 37)

FLIPPING THE COIN ON THE 37 SQUARE CREATES NEW RESULT FOR B      when he scores the heads  -  he gets the secret number

IF  45 XOR 7 WAS 37 THEN 45 XOR 37 GETS 8 (SEE THE BOLDED THEOREM)

Btw, Hamming was mentioned in the context  of parity bits evaluation, isomorphic  to XOR operation on strings and not implying any error-correction technique.

Edited on February 3, 2016, 6:25 pm

Posted by Ady TZIDON on 2016-02-03 12:25:18