so you're in one of those common puzzle situations with a bunch of other people in a room, waiting to be executed in a playful manner, unless, of course, you can collectively figure out a good strategy to solve the puzzle. so, here it is:
All twenty of you will be placed in a solitary cell with no way of communicating or seeing each other. Then, one at a time, you will be brought to a room where there are two switches on a wall that, as far as you know, are not connected to anything. Once in the room, you have to flip just one of the switches, and only flip it once. Then you are returned to your cell. You can be selected in any order, and there is replacement - in other words, the same person can be chosen 15 times and someone else may not have been picked at all yet. Once someone thinks that everyone has been into the room at least once, he can say so, and if he's right then you all go free. If he's wrong ... a cup of instant DEATH for everyone! (mwahahaha)
You have time before being divided up to work out a plan. What do you do?
Originally posted by kyngjIf your description of the puzzle is complete, then it could be the case that there are some who never are taken from their cell to the switch-room.
so you're in one of those common puzzle situations with a bunch of other people in a room, waiting to be executed in a playful manner, unless, of course, you can collectively figure out a good strategy to solve the puzzle. so, here it is:
All twenty of you will be placed in a solitary cell with no way of communicating or seeing each other. Then, one at a ...[text shortened]... yone! (mwahahaha)
You have time before being divided up to work out a plan. What do you do?
Originally posted by kyngjEveryone must leave their boxers on the floor. When there are 19 boxers on the floor the last person can come back and say "We did it".
so you're in one of those common puzzle situations with a bunch of other people in a room, waiting to be executed in a playful manner, unless, of course, you can collectively figure out a good strategy to solve the puzzle. so, here it is:
All twenty of you will be placed in a solitary cell with no way of communicating or seeing each other. Then, one at a ...[text shortened]... yone! (mwahahaha)
You have time before being divided up to work out a plan. What do you do?
But if any player does not have boxers on they must leave a sock or a shoe or something else to show they were there.
Phla-
Originally posted by bbarrYou were correct in assuming that it is theoretically possible that some are never taken from the box.. of course, if that is the case, then execution would be inevitable...
This was fun to work through. If nobody posts the solution in the next day or so I'll post mine.
I'm looking forward to the solution, been thinking through this one for a little while, and aside from trial and error, I have come across no good way to approach it so far
Originally posted by kyngjI gave the solution, there is no way know if 20 people have come unless they leave a mark just once of some kind... each will count the marks left behind, leaving only one on their first visit. When there are 19 marks, and you have never been in the room you are number 20.
You were correct in assuming that it is theoretically possible that some are never taken from the box.. of course, if that is the case, then execution would be inevitable...
I'm looking forward to the solution, been thinking through this one for a little while, and aside from trial and error, I have come across no good way to approach it so far
If there is a solution with just 2 switches to show the visit of all 20 people I'll eat my chess pieces.
Originally posted by PhlabibitI ike your solution, although I don't think it's allowed within the narrow and heavily contrived rules of the puzzle... I get the feeling there is going to be a solution with just the two light switches, better get the salt and pepper ready ;-)
I gave the solution, there is no way know if 20 people have come unless they leave a mark just once of some kind... each will count the marks left behind, leaving only one on their first visit. When there are 19 marks, and you have never been in the room you are number 20.
If there is a solution with just 2 switches to show the visit of all 20 people I'll eat my chess pieces.
joe
Originally posted by kyngjExcept we don't know that they are light switches, (they might be power socket switches etc...) So how can we decide what our strategy is going to be? 😕
I ike your solution, although I don't think it's allowed within the narrow and heavily contrived rules of the puzzle... I get the feeling there is going to be a solution with just the two light switches, better get the salt and pepper ready ;-)
joe
Unless...............
We all agree to become Jehovah's witnesses then each promise only to say something after having flipped the switch on the 1000 time. Thus statistically greatly increasing our chances that everybody should have been in the room at least once. However if we receive the cup of instance death it does not matter, for on the day of judgement Jehovah will grant the rightous everlasting life in paradise.... 😉
Originally posted by Jay PeateaHere are a few hints (good ones, I hope).
Except we don't know that they are light switches, (they might be power socket switches etc...) So how can we decide what our strategy is going to be? 😕
Unless...............
We all agree to become Jehovah's witnesses then each promise only to say something after having flipped the switch on the 1000 time. Thus statistically greatly increasing our ...[text shortened]... for on the day of judgement Jehovah will grant the rightous everlasting life in paradise.... 😉
1. One prisoner must be selected as the person who will eventually inform the guards that everyone has been in the room.
2. The two switches serve different fuctions within the context of this puzzle, and everyone must be aware of these different functions.
3. The selected prisoner will not be able to confidently tell the guards that all the prisoners have been in the room until they've all been in there twice.
Originally posted by bbarrDoes someone need to flip a switch when they go in every time, or can they enter the room and leave the switches alone????
Here are a few hints (good ones, I hope).
1. One prisoner must be selected as the person who will eventually inform the guards that everyone has been in the room.
2. The two switches serve different fuctions within the context of this puzzle, and everyone must be aware of these different functions.
3. The selected prisoner will not be able to confi ...[text shortened]... the guards that all the prisoners have been in the room until they've all been in there twice.
One switch is light on and off.
One switch is buzzer on and off.
both start in off position... light on the left, buzzer on the right.
Lets hear an answer so I can discredit it.... message me if you don't want to give the answer here in the thread.......
I think the first question I asked is valuable information that needs to be known.
I have a very small set of pieces should I be wrong, and I think they would be quit good with salad dressing on them.... I'll post a picture should I need to eat them.
🙂
I believe that I have a solution, but it may be a little difficult to explain so bear with me.
There are four possible switch combinations (both off, left on, right on, both on). The key I believe is to plan out a pattern and assign each person a certain point in the pattern. The puzzle does not state how many people there are, so certain numbers vary upon how many times the pattern must be repeated. For example, of the four possible combinations only three can be used by the majority of the people. The fourth must be reserved by one individual alone to do, that person also being the designated person to say when everyone has been into the room. Each person is to execute one action in the pattern when they enter the room, and to do so only once (EXCEPT the "odd man", who does his assigned part EVERY time he is capable). By doing this, the designated "odd man" can safely tell the gaurds when everyone has been there once he has entered the room to find the pattern complete as many times as there are groups of people to execute the pattern. I'll give an example:
Let's say there are 13 prisoners. Let us also say that the agreed upon pattern will be: both switches off, left switch on, right switch on, both switches on. The "odd man out" will be the only one to turn both swithes off once he enters the room and see both of them on. The other 12 prisoners are divided up into groups of four. Four people turn the left switch on once they enter the room and find them both off. Each person only does so ONCE. No matter how many times they actually enter the room, any individual only does their assigned action once and then never touches the switches again. This ensures that if anyone hasn't been to the room, the pattern will never be completed for the "odd man" to finish it the fourth time untill they do.
As I said before, some things depend on the number of people. In my example above the "odd man" would be the one to finish the sequence. If that were not to be the case, then the "odd man" would have to return to the room again untill he finds the switches in the position that they should be in at the ende of the chain.
I have found this really hard to explain in a simple way (because there is little simple about it), but if I need to explain further just let me know. Likewise, if anyone can find fault in my reasoning just let me know and I will hit the drawing board again.
Originally posted by Omnislashthere are twenty people who will eventually visit the room.
I believe that I have a solution, but it may be a little difficult to explain so bear with me.
There are four possible switch combinations (both off, left on, right on, both on). The key I believe is to plan out a pattern and assign each person a certain point in the pattern. The puzzle does not state how many people there are, so certain numbers vary ...[text shortened]... nyone can find fault in my reasoning just let me know and I will hit the drawing board again.
Each person who enters the room must flip a switch, and flip it only once.
explain the answer so I can eat chess pieces.
😉
Originally posted by bbarrWe're waiting...
This was fun to work through. If nobody posts the solution in the next day or so I'll post mine.
I haven't gone through this rigorously, but I'm curious as to how the two switches (which seem to be equivalent to two binary bits) can preserve a sufficient amount of information.
Originally posted by richjohnsonSorry for the delay. First, I designate myself as the only one with the authority to inform the guards that every other prisoner has visited the room. I'll do this by counting the number of times the left switch is flipped. I pay no attention to the right switch. These are the instructions I give to the prisoners:
We're waiting...
I haven't gone through this rigorously, but I'm curious as to how the two switches (which seem to be equivalent to two binary bits) can preserve a sufficient amount of information.
1) When you are taken to the room, flip the left switch up. If the left switch is already up, leave it alone and flip the switch on the right.
2) After you have flipped the left switch up a total of two times, you are no longer allowed to touch the left switch. On any visit subsequent to the visit where you flipped the left switch up for the second time, you must flip the right switch, regardless of the left switch's position.
3) Never, under any circumstances, flip the left switch down.
So, when I enter the room and find the left switch up, I'll know a prisoner has been there. I'll then 'reset' that switch by flipping it down. If I come in and the left switch is down, then it is ready to record the visit of the next prisioner, and thus I flip the switch on the right. Once I've counted that the switch on the left has been flipped to the up position a total of 38 times, then I'll know each prisoner has been there at least once.
Excellent solution. I am not sure I agree that two visits are required. Would it not work equally well if each person flips the left switch on their first visit and only on their first visit (which would vastly reduce the number of visits required to conclude that all persons had visited the room)?