Math and Logic Puzzles

[cjdrum]

You have to split it at the same time so that when, after 20 hours, people die, you can work out a common wine that each person had.

Sadly, I can't do the maths for that at the moment.

[StrangerCoug]

That overruns the three-guess limit.

[Sudo_Nym]

10 prisoners.

Divide the bottles into 10 groups of 100

Prisoner A drinks combined drops from all bottles in Group A
Prisoner B drinks combined drops from all bottles in Group B
etc.
Prisoner J drinks combined drops from all bottles in Group J

Lets say Prisoner J dies.

Prisoner A drinks combined drops from 12 bottles in Group J
Prisoners B-I each drink combined drops from 11 bottles in Group J

Lets say Prisoner I dies.

Prisoners A-C drink combined drops from 2 bottles in Group J-I
Prisoners D-H drink a drop from one bottle each in Group J-I

As stated, this would work, if you had all the time in the world. Unfortunately, it takes up to 20 hours for the poison to take effect, and you only have 24 hours to identify the poison.

You have to split it at the same time so that when, after 20 hours, people die, you can work out a common wine that each person had.

Sadly, I can't do the maths for that at the moment.

Yes, this is your hint. And there is a way to do it.

That overruns the three-guess limit.

Do people want me to give the answer, or leave it open for more guesses? I admit, I didn't really read the rules.

[StrangerCoug]

Go ahead and give it.

[DonJosh]

Darned 24-hour time limit.

[Sudo_Nym]

You only need 10 slaves to test all the bottles. Assign each bottle a binary number, in order. Then make each slave take a sip from each bottle, according to the binary pattern- so if there's a 1 in the 1s place, Slave A drinks from it. A 1 in the 2s place means that Slave B drinks, and so on. With 10 slaves, you can generate 1024 different combinations, allowing a unique combination for each of the 1000 bottles. Then you wait to see what combination of slaves dies from the poison, and the bottle that corresponds to that binary combination must be the poision.

[StrangerCoug]

One off. Damn it.

Put up another, then.

[Sudo_Nym]

A stark raving mad king tells his 100 wisest men he is about to line them up and that he will place either a red or blue hat on each of their heads. Once lined up, they must not communicate amongst themselves. Nor may they attempt to look behind them or remove their own hat.

The king tells the wise men that they will be able to see all the hats in front of them. They will not be able to see the color of their own hat or the hats behind them, although they will be able to hear the answers from all those behind them.

The king will then start with the wise man in the back and ask "what color is your hat?" The wise man will only be allowed to answer "red" or "blue," nothing more. If the answer is incorrect then the wise man will be silently killed. If the answer is correct then the wise man may live but must remain absolutely silent.

The king will then move on to the next wise man and repeat the question.

The king makes it clear that if anyone breaks the rules then all the wise men will die, then allows the wise men to consult before lining them up. The king listens in while the wise men consult each other to make sure they don't devise a plan to cheat. To communicate anything more than their guess of red or blue by coughing or shuffling would be breaking the rules.

What is the maximum number of men they can be guaranteed to save?

[cjdrum]

99. Edit coming.

[StrangerCoug]

Clarification questions:

1.) Are the hats placed on their heads before or after they are lined up?
2.) Are the hat colors determined independently?

[cjdrum]

Spoiler: My solution

The wise men devise a code before the king lines them up.

The first man will count up all the red hats in front of him. If it is odd, he will say "red", and if it is even he will say "blue". He has a 50/50 chance of survival.
The second man again counts all the red hats in front of him. If the first man said "red" and there is an odd number of red hats, his hat is blue. If the first man said "red" and there is an even number of red hats, his hat is red. If the first man said "blue" and there is an odd number of red hats, his hat is red. If the first man said "blue" and there is an even number of red hats, his hat is blue.

This continues down the line until the last 99 people are saved.

[Sudo_Nym]

Spoiler: My solution

The wise men devise a code before the king lines them up.

The first man will count up all the red hats in front of him. If it is odd, he will say "red", and if it is even he will say "blue". He has a 50/50 chance of survival.
The second man again counts all the red hats in front of him. If the first man said "red" and there is an odd number of red hats, his hat is blue. If the first man said "red" and there is an even number of red hats, his hat is red. If the first man said "blue" and there is an odd number of red hats, his hat is red. If the first man said "blue" and there is an even number of red hats, his hat is blue.

This continues down the line until the last 99 people are saved.

This is correct.

[DonJosh]

I have another Hat riddle if cj doesnt want to go.

[cjdrum]

Nope, I do. More hats

Four men are brought to a beach by a crazed lunatic. He proceeds to dig four holes in the sand - three on one side of a wall, in a line parallel to it, and one on the other side. All men are facing the wall.
He then shows the men that he has five hats - three white, two black - and randomly places them on the men's heads so that they cannot see any hats except of the people in front of them. They can't see through the wall.
The lunatic says to the men, "If any one of you is able to tell me which colour hat you are wearing, you will all go free. However, if any of you says anything that isn't a correct guess, I will cut your heads off."

The men stay silent for a few moments, before the man at the front of the group of three says "My hat is white." How did he know?

[cjdrum]

EBWOP: The line of three is

perpendicular

to the wall. Not parallel.

[DonJosh]

Wait... there are five hats, but only four dudes?

[cjdrum]

The fifth hat is not placed on anyone's head, and no one can see it.

[yabbaguy]

Wall is opaque?
Everyone is actually standing in a hole?

[Espeonage]

From the guy at the end staying silent he knows that it is impossible for the two in front to both have black hats.

The second guy in line knows this and sees either a white hat in front or a black hat in front. If he sees a black hat in front he would know he had a white hat because of what the first guy didn't say. The guy at front knows this also and concludes that for the second guy in line to have not said anything he must have a white hat.

[cjdrum]

Espeonage got it.

[Espeonage]

Yay. let me think of a puzzle.

[Espeonage]

_____________
| o o o o o o o o |
|_____________|
O O

Which direction is the bus going? WHY?

Need to answer the second to get it.

[Espeonage]

Aww gay. Spaces didn't work. You get the point.

[SleepyKrew]

The bus is going to the right because that's the side with the driver.

[DonJosh]

Wrong. Its going to the left. If it were going to the right, you would be able to see the door.