Who wants to solve a riddle?

[BeccaKirstyn]

Alright, there seems to be a lot of intelligent people on this forum. Let's see if anyone can solve this riddle (and no cheating!).

 

 

 

[Crypto]

Everyone passes their hat back, the person in back passes it forward. Everyone now should know what the hat is that is on them(since it was in front of them), except for the person in the front.

But I have a feeling this would be considered cheating. I'd work though, and I don't think there was any rule against passing hats. Works for the cooperation part.

Edited November 5, 2015 by Crypto

[BeccaKirstyn]

It might be considered cheating since I don't think they're allowed to take their hats off (although it doesn't specifically say this, but I think it's implied....not 100% sure though), but that was a good idea! Definitely works with the cooperation aspect. 

Edited November 5, 2015 by BeccaKirstyn

[Guest]

42

[Anddenex]

I am unintelligent.  I like Crypto's answer.  If not spoken in the rules of the riddle (or as given by the aliens who ate me), then Crypto's answer is intelligent.  Crypto for the win!

 

[Crypto]

I am unintelligent.  I like Crypto's answer.  If not spoken in the rules of the riddle (or as given by the aliens who ate me), then Crypto's answer is intelligent.  Crypto for the win!

 

It works, but there is certainly a mathematical alternative. I'm better at breaking the system rather than finding creative ways to work within it.

[BeccaKirstyn]

Unless someone comes up with the answer provided in the video (again, without cheating) then I think Crypto wins! 

[Crypto]

I figured out the other method...

but someone around me gave me a one word hint ...  so when someone, likely Vort, shows up with the answer they win.

This was my hint (white text):

,,,,,,,,,,,,,,

|RATIO|
``````

Edited November 5, 2015 by Crypto

[Vort]

No, I already looked at the answer. Clever puzzle. Wish I'd given it a go.

[Crypto]

Teehee!

Edited November 5, 2015 by Crypto

[mordorbund]

I got it. I went down a false path first of having the first person state the majority hat seen, with n-1 being the tie-breaker. Obviously didn't pan out.

 

Crypto, how did your hint help you? My mental prompt is below.

 

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

|               I recognized that the nth person needed to convey information about the group and that took me to the parity bit.      I work in software by the way.                                                                              |

..........................................................................................................................................................................

Edited November 5, 2015 by mordorbund

[Guest]

Clues - only 2 hat colors.

People can see all the hats infront of them.

There is an even number of participants, only 9 of which matters... so 9 divided into 2 sets of colors, one color will always be odd, the other will always be even (all 9 are white, then white is odd, only 8 is white, then black is odd, etc.)

 

This is a perfect problem for Parity checking.  With white or black parity bit being your own hat that you can't see and the guy behind you dictating the parity check.  The first one can be wrong, so all he needs to do is signal the parity check - doesn't really matter if he guesses his hat color right.  They have to agree first that they're gonna use odd parity (or even parity - whatever code they agree on).  So then the guy in the very back who can see all 9 hats calls out the color that matches their agreed upon parity.  So the next guy has to match his color hat to keep the parity out of the colors already called out combined with the colors he can see (everyone only has 1 unknown - their own hat). 

 

Disclaimer:  I'm a programmer... we had a similar problem in one of my exams.  So, I already knew the answer - didn't really have to figure it out.

Edited November 5, 2015 by anatess

[AnnieCarvalho]

I decided that the fellow in the back could just tell the person in front of him the color of his hat, then that person could tell the next one, etc.

 

The only person not knowing the color of his hat would be the guy in the back, and he'd have a 50/50 chance of getting it right. But I think I remember that only 9 out of 10 had to be correct?

 

When I listened to the rest of the tape, it made sense, but it hurt my head. 

[BeccaKirstyn]

 

 

Disclaimer:  I'm a programmer... we had a similar problem in one of my exams.  So, I already knew the answer - didn't really have to figure it out.

I need to go take some programming classes....

[Guest]

Anatess,

 

That's not fair.  I was working on it, and almost had it.

 

I had deduced:

 

1) The first guy did not have to get it correctly.  So, he would forget trying and instead, give some sort of signal to the others through his guess.

2) The others would have to pay attention to the previous participants and keep track.

3) Each response would then have to take into account the previous responses.

4) Then without knowing the exact number of each type, there would have to be some other method of sort of counting.

 

I had not gotten to the odd/even thing.  But it makes sense.

Edited November 5, 2015 by Guest

[Crypto]

I got it. I went down a false path first of having the first person state the majority hat seen, with n-1 being the tie-breaker. Obviously didn't pan out.

 

Crypto, how did your hint help you? My mental prompt is below.

That sounds like the path I went down towards the beginning.

I realized that black or white had to convey some sort of information likely a binary state, true and false, greater than or less than. After hearing the hint I lined up ratios of what every person would see and every ratio had even or odd. It then sorta   clicked after that.

[Guest]

I need to go take some programming classes....

Correction: it's one of my classes in Computer Engineering - Boolean Algebra.