while reading the first post in the thread here, various logic riddles should be added as most exposed. some logic riddles are very deceptive. one of my favorites is:


You are one of 20 prisoners on death row with the execution date set for tomorrow.

Your king is a ruthless man who likes to toy with his people's miseries. He comes to your cell today and tells you:

“I’m gonna give you prisoners a chance to go free tomorrow. You will all stand in a row (queue) before the executioner and we will put a hat on your head, either a red or a black one. Of course you will not be able to see the color of your own hat; you will only be able to see the prisoners in front of you with their hats on; you will not be allowed to look back or communicate together in any way (talking, touching.....)

(The prisoner in the back will be able to see the 19 prisoners in front of him
The one in front of him will be able to see 18…)

Starting with the last person in the row, the one who can see everybody in front of him, he will be asked a simple question: WHAT IS THE COLOR OF YOUR HAT?

He will be only allowed to answer “BLACK” or “RED”. If he says anything else you will ALL be executed immediately.

If he guesses the right color of the hat on his head he is set free, otherwise he is put to death. And we move on to the one in front of him and ask him the same question and so on…

Well, good luck tomorrow, HA HA HA HA HA HA!”

Now since you all can communicate freely during the night, can you find a way to guarantee the freedom of some prisoners tomorrow? How many?


this solution to this riddle is absolutely amazing!!! it took about 30 posts for a forum filled with experienced logicians to solve it.

i found this puzzle on this forum: http://brainden.com/forum/

It has 402 replies now.
http://brainden.com/forum/index.php?/topic/825-hats-on-a-death-row-one-of-my-favorites-puzzles/

Saving a minimum of 10 is trivial. It's amazing how high the number can go though.

Back to the original post, since Japanese trick boxes are popular, of course we should be open minded. But that's only in this topic. In the rest of the forum our minds should be twisted.

That riddle and the supposed solution (using a secret parity code) are ridiculous, because no matter how you solve it you must ultimately break the no-communication rule of the riddle itself: the prisoners MUST communicate with each other (at least tacitly by some pre-agreed code) as they say their own hat colour.

A much simpler solution than the one given would be to say your own hat colour in a high tone if the next hat is red, or a low tone if the next hat is black, and just listen to the tone of the guy behind you, as he says his hat colour, to determine your own.

Both solutions involve tacit communication, but my solution is far less complicated, easier to explain to a bunch of prisoners (who presumably aren't in the mood to listen to complex abstract logic concepts as they argue over their impending execution!), less susceptible to mistakes, and easier to put right when mistakes are made (since it's not dependent on any complex chain of logic).

Sometimes people try to be too clever for their own good...

I am moving this to the Non-Twisty Puzzles forum. They are not mechanical puzzles,
but they are surely ON-topic!

I will also start a more generalised thread.




Pantazis

a very good solution but, not in the spirit of the problem.




you do not have to explain all the logic to the prisoners, tell them:

the person in back says red if the number of red hats are even, black if odd.

before 2nd to last says his answer, count the number of reds ahead of you. if it matches the parity of what the person in the back say you think black. if it doesn't match you think red.

next, starting at the second to last, if someone says red you change the color you are thinking of.

They'd probably just stare at me blankly ... and then beat me up.

one person stands still, the next stand next to him. The third looks and see if the others match, if they do match, then he stands next to them. the next does the same, if they dont match then he places himself inbetween the two colors.
repeat!

I like this one:

"The man with the hat"
There are four man standing in front of a firing-squad. Two of them (nr.1 & 3) wear a black hat and two of them (nr.2 & 4) wear a white hat. They are all facing the same direction and between nr.3 and nr.4 stands a brick wall. So nr.1 can see nr.2 & 3, nr.2 sees nr.3, nr.3 sees only the wall and nr.4 doesn't see a thing. The men know that there are two white and two black hats.
The commander of the firing-squad is willing to let the men go if one of them can say what color hat he is wearing. The men are not allowed to talk. The only thing they may say is "I'm wearing a white/black hat". If one of the men knows which hat he is wearing he must tell it and all men will be free.
Which man knows 100% sure what color hat he's wearing?

how to draw a simple house in one line!

one house is easy! two in a row is HARD! please help me!

The second one is impossible because there are now four vertices with an odd number of lines, rather than just two as required to draw with one continuous line: you have to start and end at two different odd vertices (or at the same even vertex), but all other vertices must be even.

Exactly.




Pantazis

damn...

how about 3 houses in a row then?

Nope, for the same reason.

I know a puzzle similar to the first one but a lot simpler. There are three people sitting down in a line and they can only look forward. There are three red hats and two blue. The hats are placed on the people at random. The person in the back says " I can't deduce what color my hat is." The person sitting in the middle says "I can't deduce the color of my hat either." The person in front says " I know what color my hat is. " How does he know what color his hat is?

This is the same as the Island of blue/brown eyed people http://www.subzeroblue.com/archives/2007/09/blue_eyes_riddle.html and the solution is recursive. You can extend the solution to any number of people. It's pretty easy to spot a solution with only 3 people but trying to start at 100 is quite difficult to spot the recursion.

I love that one! It was actually that one I was thinking of. But I can never remember it (tell it correctly) so I found that other riddle I posted instead (I found it on the internet) I have never heard the original name before.
The way I was introduced to this one, was back when I was about 14 or 15 (I am 24 today) And we were playing D&D, our brilliant dungeon master used it as a riddle we had to solve to be allowed to pass some sort of gate! So great! I will never forget..

There are two possibilities.

If man #1 sees two hats of the same colour, he figures out that his hat is the opposite colour, and man #1 speaks up immediately.

If man #1 doesn't speak up immediately, then man #2 can figure out that his hat is the opposite of the one he sees, and man #2 speaks up.

The problem is how fast is "immediately". If man #1 needs 5 seconds to figure it out and man #2 only waits 3 seconds, they die.