4 ms·

Try doing it with just two people. Then extend the concept somehow. All the details of the puzzle are important.

steve_musk 9y ago
Well it’s easy to do with two people because the first person can just guess the color of the second, since we are allowed one wrong guess. I don’t see how this can be extended.
- Too 9y ago
  You can extend to three people if the first "guess" indicates if nr 2 and 3 have the same color or not. But doesn't go further because after the first all guesses must be 100% correct and can't be used for communicating additional information.
  - marcan_42 9y ago
    It does extend. It's just an XOR/parity trick. The last person shouts out the XOR of all the previous colors (if you prefer: whether the number of visible white hats is even or odd). From there, every other subsequent person knows their color by XORing the hats they see together with the guesses made so far. For example, the second to last person knows their hat is black (0) if the last person's guess matches the XOR of the hats they see - if it doesn't, their hat flipped it, and thus is white (1). And so on and so forth. Your hat color is everything you see and everything that was said XORed together.
    - huftis 9y ago
      That’s correct. Or, to explain in (slightly) more non-mathematical terms: The last person (‘person 10’) shouts out (e.g.) ‘white’ if the number of white hats in front of him is an odd number (1, 3, 5, 7, 9) and ‘black’ otherwise. Let’s say he shouted ‘white’. The person in front of him looks at the people in front him. If it’s still an odd number of white hats, his must obviously be black; otherwise it must be white. If his hat is black, his shouts out ‘black’, and the person in front of him knows that the rest of the line (8 people) must still have an odd number of white hats, and he applies the exact same logic. But if the hat of person 9 was white, he would shout out ‘white’, and person 8 would know that the rest of the line (including himself) should now have an even number of white hats. So, basically the rule is: Define ‘the rest of the line’ to mean the people who have not yet shouted out a colour. When the first person (person 10) shouts out ‘white’, this means that ‘the rest of the line’ has an odd number of white hats (parity: odd). Whenever someone shouts out ‘white’, the parity (even/odd) of ‘the rest of the line’ is flipped. Using this, each person only has to count the number of white hats of the people in front of them and observe if it is even or odd. If it matches the parity of ‘the rest of the line’, the person’s hat is black, otherwise it’s white. (This also works for the person in front of the line (person 1). He sees no (or 0) white hats, i.e. he sees an even number of white hats.)
- fyi1183 9y ago
  I assure you that it's possible. To give you a hint without giving everything away: the guess of the first person must be derived from the colour of all other person's hats. You could think of it as a puzzle in coding theory, and it's not implausible. If you sum up the number of hats that each person sees and the number of guesses they hear, you'll note that everybody has 9 bits of information available. And they're supposed to make 9 correct guesses. It adds up. Another hint would be to think of the case of 3 persons first. That can be done with patient case analysis, and it's likely to get you an idea for generalisation.
- OJFord 9y ago
  Think about it as if each guess preceding a person is a bit of information, the goal is to come up with a strategy for using that information (you control encoding as well as decoding) to correctly infer another bit.

Well it’s easy to do with two people because the first person can just guess the color of the second, since we are allowed one wrong guess. I don’t see how this can be extended.

Too 9y ago
You can extend to three people if the first "guess" indicates if nr 2 and 3 have the same color or not. But doesn't go further because after the first all guesses must be 100% correct and can't be used for communicating additional information.
- marcan_42 9y ago
  It does extend. It's just an XOR/parity trick. The last person shouts out the XOR of all the previous colors (if you prefer: whether the number of visible white hats is even or odd). From there, every other subsequent person knows their color by XORing the hats they see together with the guesses made so far. For example, the second to last person knows their hat is black (0) if the last person's guess matches the XOR of the hats they see - if it doesn't, their hat flipped it, and thus is white (1). And so on and so forth. Your hat color is everything you see and everything that was said XORed together.
  - huftis 9y ago
    That’s correct. Or, to explain in (slightly) more non-mathematical terms: The last person (‘person 10’) shouts out (e.g.) ‘white’ if the number of white hats in front of him is an odd number (1, 3, 5, 7, 9) and ‘black’ otherwise. Let’s say he shouted ‘white’. The person in front of him looks at the people in front him. If it’s still an odd number of white hats, his must obviously be black; otherwise it must be white. If his hat is black, his shouts out ‘black’, and the person in front of him knows that the rest of the line (8 people) must still have an odd number of white hats, and he applies the exact same logic. But if the hat of person 9 was white, he would shout out ‘white’, and person 8 would know that the rest of the line (including himself) should now have an even number of white hats. So, basically the rule is: Define ‘the rest of the line’ to mean the people who have not yet shouted out a colour. When the first person (person 10) shouts out ‘white’, this means that ‘the rest of the line’ has an odd number of white hats (parity: odd). Whenever someone shouts out ‘white’, the parity (even/odd) of ‘the rest of the line’ is flipped. Using this, each person only has to count the number of white hats of the people in front of them and observe if it is even or odd. If it matches the parity of ‘the rest of the line’, the person’s hat is black, otherwise it’s white. (This also works for the person in front of the line (person 1). He sees no (or 0) white hats, i.e. he sees an even number of white hats.)
fyi1183 9y ago
I assure you that it's possible. To give you a hint without giving everything away: the guess of the first person must be derived from the colour of all other person's hats. You could think of it as a puzzle in coding theory, and it's not implausible. If you sum up the number of hats that each person sees and the number of guesses they hear, you'll note that everybody has 9 bits of information available. And they're supposed to make 9 correct guesses. It adds up. Another hint would be to think of the case of 3 persons first. That can be done with patient case analysis, and it's likely to get you an idea for generalisation.
OJFord 9y ago
Think about it as if each guess preceding a person is a bit of information, the goal is to come up with a strategy for using that information (you control encoding as well as decoding) to correctly infer another bit.