General Discussions

**OmuCuSucu** · 09:48 AM

since the other thread seems to have success i'm hoping this will be a long one too. i'm going to start it with this one. some mathematic and algorithm knowledge needed for it.

Quote:

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled A and B, each of which can be in either the 'on' or the 'off' position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must move one, but only one of the switches. He can't move both but he can't move none either. Then he'll be led back to his cell.

"No one else will enter the switch room until I lead the next prisoner there, and he'll be instructed to do the same thing. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back.

"But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time anyone of you may declare to me, 'We have all visited the switch room.' and be 100% sure.

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators."

What is the strategy they come up with so that they can be free?

EDIT: answered by Insensus

**Insensus** · 03-06-2007, 03:18 PM

Will I get talkupation if I solve it, just like the other riddle?

I won't look on the site! *honsety*

*Solving*

**Insensus** · 03-06-2007, 03:21 PM

1. Select the youngest person and make all the other persons declare everyone has visited. They'll all die and then the one remaining will visit the room and can leave.

**Learning Newbie** · 03-06-2007, 03:39 PM

I don't know, but I certainly hope this isn't the prison Scooter Libby winds up in. He should at least have to wait for the Republicans to regain the white house before he's pardoned and set free.

**travelagent** · 03-06-2007, 04:02 PM

Quote:

Originally Posted by Learning Newbie

I don't know, but I certainly hope this isn't the prison Scooter Libby winds up in. He should at least have to wait for the Republicans to regain the white house before he's pardoned and set free.

LMAO -- you do crack me up alot of times!

**memberpro** · 03-06-2007, 04:05 PM

The riddle solution is there is no solution.

The warden says that he will select "a prisioner" to move a switch... the riddle also says that "23 new prisioners" arrived. so we have to assume two things...

1) there are more prisioners than the 23 new prisioners and
2) there will always be more coming in, and some that leave

Therefore, everyone should shut their mouths and never say anything, because there will never be a time when any of them are 100% certain.

They should just be happy working out, and watching cable tv on my tax dollars.

**OmuCuSucu** · 03-06-2007, 04:54 PM

Quote:

Originally Posted by memberpro

The riddle solution is there is no solution.

The warden says that he will select "a prisioner" to move a switch... the riddle also says that "23 new prisioners" arrived. so we have to assume two things...

1) there are more prisioners than the 23 new prisioners and
2) there will always be more coming in, and some that leave

Therefore, everyone should shut their mouths and never say anything, because there will never be a time when any of them are 100% certain.

They should just be happy working out, and watching cable tv on my tax dollars.

1. wrong.
2. wrong.

none matter. and there is a time there will be 100%, just as stated

**Learning Newbie** · 03-06-2007, 06:12 PM

Quote:

Originally Posted by memberpro

They should just be happy working out, and watching cable tv on my tax dollars.

You do realize most of the TVs in US prisons are donated by churches, right? You know, private sector charity, faith-based initiative, The Big J.C., I'm the decider, all that great stuff. But as long as we're talking about riddles, why don't we have a look at the two things you implied, but probably don't really believe?

Prisoners have a better life than Memberpro. They get to do fun things all day like watch TV and lift weights, but you have to pay for it like a slave. Those prisoners sure out smarted you!
You're the only one on the planet who pays taxes. Man, how did you ever wind up in such an unlucky situation as that? Furthermore, you must be richer than Bill Gates if your tax withholdings cover the expenses at every prison in America. That must be why the rest of us don't pay taxes, huh?

Seriously, people, think before you speak! That is all.

**OmuCuSucu** · 03-06-2007, 07:16 PM

could you two pretty please stop it there?

i hate messy threads

divert to the riddle at hand please

**ForrestCroce** · 03-06-2007, 08:33 PM

We know the prisoners have an opportunity to organize, and that starting tomorrow the only way they'll be able to communicate is through the switches.

I'm assuming there's a solution to this riddle, one that would have all of the prisoners set free, and none fed to the lions. I suspect there might not be, unless the prisoners can all watch the comings and goings.

There are four states possible: a on / b on, a on / b off, a off / be on, a off / b off. The system doesn't have a "memory," like when you pull a card out of a deck, it's not available anymore. So the prisoners can't use it to count, and somehow "carry the 1."

So I'm not sure. They could have one switch set to the on position if they think all the prisoners haven't had their turn, and to the off position if they think all have, then when someone sees two off switches he could assume all have had their chance. But that's a far cry from 100 % certainty.

**vangogh** · 12:29 AM

Ok I'm going to think out loud here and see what I come up with.

Forrest I think you're on the right track. Here again are the different states.

A B
on on
on off
off on
off off

We have 23 prisoners and at some point they will all have visited the room and flipped a switch the exact same number of times so we're dealing with multiples of 23. I think we can for the sake of the riddle assume they each only go in once and the logic will still carry over if they all end up going 34 times.

I'm also thinking there are 4 states for the switches and since 4 x 6 = 24 or one more than the number of prisoners it will be that 24th flip where the prisoner knows.

I'm thinking one possible strategy is going to have something to do with only one or some specific number of prisoners flipping one of the two switches while all the other prisoners flip the other one.

After a couple minutes in thought another strategy might be something along the lines of the first time any prisoner flips a switch they move one to either the on or off positions and the second time they always do either the opposite or the same. Of course since the prisoners don't know how the switches are initially set they can't automatically have decided going in which way to flip the switch.

So the strategies can involve specific prisoners only flipping a certain switch or prisoners always flipping one of the switches either the same direction they did the last time or the opposite direction as the last time or maybe even some combination of both.

Well that's my out loud thinking for now. I'm too tired at the moment to try to work something out, but hopefully some of the above will help someone further this along or solve it.

**OmuCuSucu** · 03-07-2007, 04:23 AM

some of that can help vangogh ... the problem is just a bit more complicated and there are many more aspects to be taken into consideration

**Insensus** · 03-07-2007, 11:47 AM

I've got one question:
If someone is escorted to the switches room, will the other prisoners now it?

**vangogh** · 03-07-2007, 11:57 AM

I would think they can't know otherwise they could just keep track of who has and hasn't gone, which would make it easy to know when they'd all gone.

Dragos I assumed it was more complicated. I figured I would toss out a few variable that could possibly be used in the hopes that it would stir some thought in everyone. I could have kept it all to myself, but I'm thinking this one is tough and if we all help each other a little one of us may solve this.

I'll have to start thinking of some of the other aspects that can be controlled by the prisoners and I'm thinking the solution will need a combination of these aspects.

This one is going to be both difficult and time consuming, but it's also going to be fun.

**Learning Newbie** · 03-07-2007, 01:54 PM

Quote:

Originally Posted by vangogh

I figured I would toss out a few variable that could possibly be used in the hopes that it would stir some thought in everyone. I could have kept it all to myself, but I'm thinking this one is tough and if we all help each other a little one of us may solve this.

Only by working together can we go free.

**memberpro** · 03:46 PM

I started to solve this problem empirically using MiniTab to generate random prisioners walking into the room and using only one switch rather than two. I then figured out the exact instance when all #edit Crap!! I should have use 23 prisioners# 32 prisioners had visited the room and then I looked for any kind of pattern in the data..

I saw that over the course of 10 simulations that when all 32 prisioners visited the switch room (and the simulation alerted me that the criteria was met) that one prisioner noted that they saw the switch position off (1) four times and on (2) five times (or visa versa).

But statisically speaking, a false positive could occur somewhere beyond 10 simulations and that is why there are two switches, not just one( to increase to resolution of the "measurement")

Given the added switch and the addition of two possible combinations, I should be able to determine when all 32 prisioners have visited the switch room in half the number of combinations seen (2 combinations each). I haven't run this simulation, so feel free in doing so.

Here is my proposed solutions:

Have each person who goes into the room note the original configuration of the switches when they visit. Each person will change the switches in a predictive order depending on the number of times they visit the room.

For example: 1st visit - 1st switch reverse
2nd visit - 2nd switch reverse
3rd visit - 1st switch reverse
4th visit - 2nd switch reverse

the order shouldn't matter as long as everyone is doing it the same.

Each person will keep track of when they see the original configuration and the inverse configuration. When any person sees the original + the inverse configuration #edit most likely once #twice that means that all the prisioners have visited the room.

Like I said... I haven't run this senerio out... but the senerio using one switch would work based upon my findings. #same thought process will work to determine correct solution#

**Insensus** · 03-07-2007, 05:14 PM

I think I've come very close now.
I'm only having problems with the starting position.
Here's my solution:

a | b
0 | 0 ||| 1
0 | 1 ||| 2
1 | 0 ||| 3
1 | 1 ||| 4

Positions 1 and 2 are 'false', positions 3 and 4 are 'true'.
One prisoner is made to 'counter'.
Each time he sees the position 3 or 4, he adds 1 to his imaginary variable 'i'.
He then sets the switches to position 1 or 2.

Each time a normal prisoner comes in and he sees position 3 or 4, he switches it to position 4 or 3.
Each time a normal prisoner comes in and he sees position 1 or 2, if he hasn't yet visited he switches to position 3 or 4, if he has already visited he switches to 2 or 1.

This way, if a prisoner comes in that hasn't visited, the positions 3 or 4 carry on to the counter, which adds to 'i' and then resets the positions to 1 or 2 which then carry on to a new prisoner.
The problem is now that that if the counter counts to 22 (number of prisoners need to visit) but he is the first one in the room and the positions are 3 or 4, he would actually have to count one further.
If he counts to 23 but he isn't the first and the starting positions is 1 or 2, he'll get to 22 but then there will be no more new prisoners left.

**OmuCuSucu** · 03-07-2007, 05:26 PM

the problem was meant for programmers ... so this might help you a tiny bit

please be fair and try to solve it here and not search for an answer elsewhere

yes vangogh it's gonna be both time consuming and fun

**vangogh** · 03-07-2007, 05:58 PM

I did consider searching for an answer, but that would kind of ruin the fun of trying to figure it out.

So I'm going to have to break out my programming skills to solve this. Lets hope they're up to par. I haven't had time to give this my concentration yet. I just thought I'd pop in and see what others had come up with.

memberpro I had been thinking along the same lines of what you attempted and just haven't had the time to work something out. I was thinking that by giving different prisoners some kind of instruction on which switch to change in succession at some point after one sees the same configuration of switches they'd know everyone had visited at least once.

Insensus I had been thinking along your lines too, but I think not being able to know the initial position throws things off. I haven't tried to develop anything along those lines though so maybe it's possible.

**memberpro** · 03-07-2007, 06:23 PM

I thought about the bit, byte, Nybble anology... can't see how that would work as the starting position would be relevant. Or can it???