Riddle

Well, I think the key is, that the farmers wish isn't going to be exactly fullfilled any way, so it's really subjective.

Of course, one could simply kill one of the children, or split it up according to whichever was born first.
 
never mind
 
Prison Riddle:

There is a prison with 23 inmates. In the prision there is a room with two switches that don't do anythin. One is on the left and the other is on the right. These swiches may be set to either on or off.

The gaurds make a deal with the prisoners. They that they will send send prisoners into this room one at a time in no particular order, non-sequencially. The prisoners must then flip 1 switch before returning to their cell. At any point, a prisoner may guess whether or not all of the prisioners have visited the room at least once. If he guesses correctly, the gaurds will let all the prisoners go. If he guesses incorrectly, the gaurds will shoot all the prisoners. The prisoners are given time to discuss any stratagy they may want to use to make sure everybody has gone to the switch room, before one of them makes a guess.

The prisoners are to be placed in solitary confinement for the rest of there stay at the prison, so they will not have any way of knowing when or who is visiting the switch room. The gaurds do not touch the switches. A prisoner may be sent to the room multiple times, even if his fellow inmates have only been sent a once or not at all; there is no forclosed order to who is sent to the switch room and when. The initial position of the switches is unknown.

How should the prisoners ensure that the all of them have visited the switch room before one of them makes the "guess"?
 
I'm guessing the solution has nothing to do with the switches, i.e., the prisoners write their names on the wall or something like that.
 
There are actually a number of solutions to this problem. It's a very fascinating problem and finding the solution that take the least amount of time is very difficult.

Weasel Op said:
I've heard one similar to this, but this one is harder.

In the version I heard, there was 1 switch, and it operated a light. One randomly chosen prisoner was sent into the room once a day.

The answers are different, because the answer to this one wouldn't work for the other one.
Click to expand...
The answers are exactly the same, the systems are equivalent
 
Souron said:
At any point, a prisoner may guess whether or not all of the prisioners have visited the room at least once.
Click to expand...
Can't one guess "Not all of the prisoners have visited the room" at the start?
 
The answer does involve switches. The prisioners are not allowed to do anything in the switch room exept flip a switch.

The first time I read this I took it to mean that the initial settings were unknown to the prisoners. If it means they can set the switches to on or off, then the solution(s) would be the same.
Click to expand...
The initial settings are indeed unknown. The statement is ment to underscore that you cannot put a swich half way up.

A guess must be that "all the prisoners have visited the switch room", no other guess is acceptable.

I only know of one solution, and I am certain that it is the shortest solution. I think it is the only solution.
 
I'm not sure of the anwser, however I believe some math is involved, the number 23 is quite conspicuious. anyway, they should lick there finger touch the switch, the next prisoner would then wipe it away and so on. since 23 is odd the first condition of the switch would be the last. the first state of the switch is dry, the first prisoner into the room must but salivia on the switch. Since they should touch the switch before guessing they can determine if all have entered the room by feeling it. Dry not everyone has, wet possibly. to prevent a miss guess on wet, a specific position of the wall must be wet at all times weted progressivly. (north, south, east, west, NW, NE, SW, SE, going higher up each time the process gets repeated, as long as the wall is brick or tile this works, if it isn't you make sure a space is between each wet section, the first as long as no one messes up this will work) a third section of SW and SE won't be wet so when all section are wet except for SE, that person will know he is last.
 
Souron said:
I only know of one solution, and I am certain that it is the shortest solution. I think it is the only solution.
Click to expand...
Lemme guess what you think the "shortest solution" is:

First thing you do is only use one switch for storing info, the other is just to switch up and down when you don't have to store info. Hereafter I will only be concerned about the info switch.

You predetermine one prisoner to be the collecter. The first prisoner to enter puts the switch into the up position (provided he isn't the collector, if the collector enters first he puts it down). When a non-collector enters the switch room and the switch is down he puts it up unless he has already put it up, otherwise he does nothing. Every time the collector comes to the room and the switch is up he pulls it down. When the collector has pulled it down 22 times he knows all the other prisoners are accounted for and can safely declare that all prisoners have been in the room.
 
vikingruler said:
I'm not sure of the anwser, however I believe some math is involved, the number 23 is quite conspicuious. anyway, they should lick there finger touch the switch, the next prisoner would then wipe it away and so on. since 23 is odd the first condition of the switch would be the last. the first state of the switch is dry, the first prisoner into the room must but salivia on the switch. Since they should touch the switch before guessing they can determine if all have entered the room by feeling it. Dry not everyone has, wet possibly. to prevent a miss guess on wet, a specific position of the wall must be wet at all times weted progressivly. (north, south, east, west, NW, NE, SW, SE, going higher up each time the process gets repeated, as long as the wall is brick or tile this works, if it isn't you make sure a space is between each wet section, the first as long as no one messes up this will work) a third section of SW and SE won't be wet so when all section are wet except for SE, that person will know he is last.
Click to expand...
But putting saliva on the wall is not flipping a switch, and thus not allowed.

Also, a person may be sent to the switch room multiple times, while another may not be sent at all, untill perhalps a year later.
 
Perfection said:
Lemme guess what you think the "shortest solution" is:

First thing you do is only use one switch for storing info, the other is just to switch up and down when you don't have to store info. Hereafter I will only be concerned about the info switch.

You predetermine one prisoner to be the collecter. The first prisoner to enter puts the switch into the up position (provided he isn't the collector, if the collector enters first he puts it down). When a non-collector enters the switch room and the switch is down he puts it up unless he has already put it up, otherwise he does nothing. Every time the collector comes to the room and the switch is up he pulls it down. When the collector has pulled it down 22 times he knows all the other prisoners are accounted for and can safely declare that all prisoners have been in the room.
Click to expand...
Nope. You forgot something. Your very close though. Look over the instructions again.
 
If a person gets in the room twice or more, he'll only flip the main (the one for storing info) switch once. If he gets in for the second time, he'll just flip the other switch.
 
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