Riddles!

[TheGreatErenan]

I'm assuming you can consider the paint to be just a dot on your forehead, so you can't see it. The point is: This is a logic puzzle. It's not a trick question, like "identify the color of the paint by detecting the presence or absence of the odor of blue pigment in the paint." I'll let awpertunity confirm this, though.

This one's a real headscratcher. I'm going to think about this more...

[TheGreatErenan]

Okay, I think I have it. I was trying to figure out an optimum RLE method where the first few people try to communicate how many of each color there are or something like that, but then I realized that if you treat the blue and red paint like a binary value, then you can use the parity of the colors you can see to communicate those colors to subsequent prisoners.

So the first prisoner counts all the red prisoners he can see. If there are an even number of reds, he guesses red. If odd, then he guesses blue.

The second prisoner counts all the red prisoners that he can see. From the number of reds he sees and the guess given by the first prisoner, he can deduce his own color.

Similarly, the third prisoner can do the same, using the guesses given by the prisoners behind him.

Using this strategy, 39 prisoners are guaranteed survival, and the first has a 50/50 chance.

[awpertunity]

(05-10-2013 02:04 AM)TheGreatErenan Wrote:  Okay, I think I have it. I was trying to figure out an optimum RLE method where the first few people try to communicate how many of each color there are or something like that, but then I realized that if you treat the blue and red paint like a binary value, then you can use the parity of the colors you can see to communicate those colors to subsequent prisoners.

So the first prisoner counts all the red prisoners he can see. If there are an even number of reds, he guesses red. If odd, then he guesses blue.

The second prisoner counts all the red prisoners that he can see. From the number of reds he sees and the guess given by the first prisoner, he can deduce his own color.

Similarly, the third prisoner can do the same, using the guesses given by the prisoners behind him.

Using this strategy, 39 prisoners are guaranteed survival, and the first has a 50/50 chance.

Nailed it!

This problem is called the hat guessing game in math. I just slightly changed the details to make it un-googleable

[TheGoldenGriffin]

How does person 2 know what person 1 guessed if person 2 isn't allowed to know?

[TheGreatErenan]

Everyone can hear the guesses. They just aren't informed of whether the guesses were correct or not.

[TheGreatAnt]

(05-10-2013 05:23 AM)awpertunity Wrote:  I just slightly changed the details to make it un-googleable

Sneaky, sneaky! Good riddle, though! =]

[awpertunity]

Four people need to go through a narrow tunnel filled with monsters. Only two people can go through at a time and the group only has 1 sword to ward off the monsters. It takes the people different amounts of time to get through the tunnel:

Person A takes 1 minute
Person B takes 3 minutes
Person C takes 7 minutes
Person D takes 10 minutes

So if two people need to go through together, say A and C, it will take them 7 minutes because it takes C 7 minutes to get through.

What is the fastest time they can all get through the tunnel (and how eventually but let's see what people can do!)?

[TheQwertiest]

22 mins?

[joelduque]

13 mins. Combine 2 of the slowest, 10 mins. Then the remaining 2 slowest, 3 mins.

[awpertunity]

(05-10-2013 12:41 PM)joelduque Wrote:  13 mins. Combine 2 of the slowest, 10 mins. Then the remaining 2 slowest, 3 mins.

But once C and D go through the tunnel, the sword is now on the wrong side. So A and B cannot get through the tunnel without the sword!


22 minutes is a valid solution, but I can do better...