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# It's Math-Puzzle Time

Doesn't look like anything to meRegistered User regular
edited April 2008
It's that time of year again (Ok, so I made that part up) which means that it's time to play the "let's post amusing mathematical puzzles, oddities or anomalies game". Why? Because we can, and who doesn't love maths?

And to begin, here is the first mathematical puzzle of the thread.

---

There are three train stations, A, B and C, B sitting equidistant between A and C. Next to stations A and C there are Klimpy's chain restaurants. Given that stations A and C are in commercial districts of the city and B sits in the only nearby residential district the restaurants at A and C are almost entirely reliant upon business from people who have caught the train from B.

Customers, knowing that the distance between their station and the stations either side do not plan which Klimpy's they will eat at, instead they arrive at the train station and simply catch the next train, it's it's going to station A, they eat at that restaurant, if it goes to B, they eat at that restaurant.

However, Klimpy's central management discovers that the Klimpy's restaurant at station A does about three times as much business as the restaurant at station C.

Assume that the same number of trains travel to each station each day, that the gap between any particular two trains traveling to the same station is 20 minutes and that the times at which the customers show up to the train station is random/evenly distributed throughout the day.

Explain.

Tide goes in. Tide goes out.
Es-annon NEVA 4GET
Apothe0sis on

## Posts

• Registered User, ClubPA regular
edited April 2008
Apothe0sis wrote: »
It's that time of year again (Ok, so I made that part up) which means that it's time to play the "let's post amusing mathematical puzzles, oddities or anomalies game". Why? Because we can, and who doesn't love maths?

And to begin, here is the first mathematical puzzle of the thread.

---

There are three train stations, A, B and C, B sitting equidistant between A and C. Next to stations A and C there are Klimpy's chain restaurants. Given that stations A and C are in commercial districts of the city and B sits in the only nearby residential district the restaurants at A and C are almost entirely reliant upon business from people who have caught the train from B.

Customers, knowing that the distance between their station and the stations either side do not plan which Klimpy's they will eat at, instead they arrive at the train station and simply catch the next train, it's it's going to station A, they eat at that restaurant, if it goes to B, they eat at that restaurant.

However, Klimpy's central management discovers that the Klimpy's restaurant at station A does about three times as much business as the restaurant at station C.

Assume that the same number of trains travel to each station each day, that the gap between any particular two trains traveling to the same station is 20 minutes and that the times at which the customers show up to the train station is random/evenly distributed throughout the day.

Explain.
If the trains going to A leave at :00, :20, and :40 and the trains going to C leave at :05, :25, and :45, then there is a 15 minute period where people will be arriving and then boarding train A, but only a 5 minute period where people will be arriving and then boarding train C.

Doc on
• Doesn't look like anything to me Registered User regular
edited April 2008
Hooray for Doc!

Correct.

And now I have no idea what the next puzzle should be.

Apothe0sis on
Tide goes in. Tide goes out.
Es-annon NEVA 4GET
• Registered User, ClubPA regular
edited April 2008
Apothe0sis wrote: »
Hooray for Doc!

Correct.

And now I have no idea what the next puzzle should be.

How about as the winner of the previous puzzle, I get to post one? Others are free to post problems as well, but this way there's a (sort of) prize.

Doc on
• Registered User, ClubPA regular
edited April 2008
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

Doc on
• Doesn't look like anything to me Registered User regular
edited April 2008
Doc wrote: »
Apothe0sis wrote: »
Hooray for Doc!

Correct.

And now I have no idea what the next puzzle should be.

How about as the winner of the previous puzzle, I get to post one? Others are free to post problems as well, but this way there's a (sort of) prize.

Sounds like a plan.

Apothe0sis on
Tide goes in. Tide goes out.
Es-annon NEVA 4GET
• Doesn't look like anything to me Registered User regular
edited April 2008
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?
How many prisoners? Does it even matter?

Apothe0sis on
Tide goes in. Tide goes out.
Es-annon NEVA 4GET
• Registered User, ClubPA regular
edited April 2008
Apothe0sis wrote: »
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?
How many prisoners? Does it even matter?

Let's say 50.

Doc on
• Registered User regular
edited April 2008
Are the switches known to be stable across days? Do they start in known positions? Can the number of days passed be counted by the prisoners?

Linden on
What if this weren't a rhetorical question?
• Pokemon Champion (retired) Ann ArborRegistered User regular
edited April 2008
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

Can prisoners be selected more than once?

enlightenedbum on
Never use the phrase "alt right." Call them what they are: white supremacists.
• Registered User, ClubPA regular
edited April 2008
Linden wrote: »
Are the switches known to be stable across days? Do they start in known positions? Can the number of days passed be counted by the prisoners?

Let's say they start both down/off. The only people that mess with the switches are the prisoners, and the prisoners are free to count days passed. But a prisoner can be selected multiple times, possibly even twice in a row (as it's random every day).

Doc on
• Registered User
edited April 2008
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I've seen this puzzle before. For it to be solvable, the prisoners must be given enough information beforehand to be able to describe the different possible positions of the switches. That is, they must be able to make a list of instructions for themselves that refers to switches as either in position A or position B. On or Off, Left or Right, Up or Down - it doesn't matter what the two states are, but the prisoners must be able to walk into the room and know whether each switch is in position A or B. Example: what if the switches are not labeled, and do not turn anything "on" or "off" that the prisoners can see? What the first prisoner walks into the room on day one only to discover that switch 1 flips Left/Right while switch 2 flips Up/Down? What if the switches are buttons that get pressed and change colors from Red to Green with each push? etc.

This is a great puzzle though. Number of prisoners doesn't matter. Solution is easier if no prisoner can be called more than once, but it can be solved even if any prisoner can be called any random number of times. It's also easier if you know the initial state of the switches (both down/off, as Doc posted here) but it can even be done without knowing the initial state. All in all, rockin hard puzzle!

I'll leave the solution to somebody else and post one of my favorite tricky puzzles.

Absurdist on
[SIGPIC][/SIGPIC]
• Registered User, ClubPA regular
edited April 2008
Absurdist wrote: »
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I've seen this puzzle before. For it to be solvable, the prisoners must be given enough information beforehand to be able to describe the different possible positions of the switches. That is, they must be able to make a list of instructions for themselves that refers to switches as either in position A or position B. On or Off, Left or Right, Up or Down - it doesn't matter what the two states are, but the prisoners must be able to walk into the room and know whether each switch is in position A or B. Example: what if the switches are not labeled, and do not turn anything "on" or "off" that the prisoners can see? What the first prisoner walks into the room on day one only to discover that switch 1 flips Left/Right while switch 2 flips Up/Down? What if the switches are buttons that get pressed and change colors from Red to Green with each push? etc.

I'll leave the solution to somebody else and post one of my favorite tricky puzzles.

Right, they are physical switches with an up and a down. This can be known ahead of time.

Doc on
• Registered User regular
edited April 2008
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I think I have a solution for this, although it may not be the fastest one.
They actually only need one switch for this. The inmate pick one person as the control. He's the guy that keeps track of things and eventually makes the announcement to the warden. Whenever one of the other people enters the room and sees the switch in the "off" position, if they have never toggled the switch before, they set it to "on". Otherwise they leave the switch alone( if they HAVE to toggle a switch, they use the second one, which is ignored by the control person ). Every time the control person enters the room and sees the switch is "on", he turns it "off". Once he's turned the switch off X-1 times, in this case 49, he knows that all the other inmates have been to the room at least once.

Cantide on
• Registered User regular
edited April 2008
That won't work if someone is called more than once.

TheLawinator on
My SteamID Gamertag and PSN: TheLawinator
• Registered User regular
edited April 2008
Yes it will, because
the only time their appearance "counts" is the first time that they show up and find the switch off. If the switch is already on, or if they've toggled the switch before, they do nothing.

Cantide on
• Pokemon Champion (retired) Ann ArborRegistered User regular
edited April 2008
That won't work if someone is called more than once.

No, it's fine because if they have already switched the important switch they switch the other one and the control dude ignores it. Of course, the expected length of time for this whole method to work is some enormous times, surely longer than the shortest sentence any of the prisoners is serving, but I think it would work.

enlightenedbum on
Never use the phrase "alt right." Call them what they are: white supremacists.
• Registered User regular
edited April 2008
That won't work if someone is called more than once.
yes it does, after you flip the first one once you flip the second one on any other trips into the room.

Proto on
and her knees up on the glove compartment
took out her barrettes and her hair spilled out like rootbeer
• Registered User
edited April 2008
The solution to like every math puzzle ever is on the internet at this point, so no cheating if you want a good challenge! Some people get this right away, but it was really hard for me. Like, keep-me-up-all-night-and-then-eureka-while-showering-the-next-morning kind of hard. PM me for a hint.

What is the next number in this sequence?

1, 4, 7, 12, 15, 18, 21, 24, 27, ?

Absurdist on
[SIGPIC][/SIGPIC]
• Registered User regular
edited April 2008
Actually, I think that will work, but it will take a really long time.

Corlis on
But I don't mind, as long as there's a bed beneath the stars that shine,
I'll be fine, just give me a minute, a man's got a limit, I can't get a life if my heart's not in it.
• Registered User regular
edited April 2008
Corlis wrote: »
Actually, I think that will work, but it will take a really long time.

That's why I said it's probably not the fastest, although that's partly due to the relatively large example value of X. If we were talking about 10 prisoners, my method would probably take only a few months.

And if the prisoners don't have to flip any switches, there's an easy way to cut down the time:
Instead of flipping just the one switch, they could do things binary style, with the first guy setting the switch positions to 01, the second to 10, and the third to 11, with the control guy resetting things back to 00 each time. That way, he could potentially write off 3 people on each of his visits instead of just 1.

EDIT: Actually, that doesn't work because it would require flipping two switches on the same visit. Never mind.

Cantide on
• Registered User regular
edited April 2008
Absurdist wrote: »
The solution to like every math puzzle ever is on the internet at this point, so no cheating if you want a good challenge! Some people get this right away, but it was really hard for me. Like, keep-me-up-all-night-and-then-eureka-while-showering-the-next-morning kind of hard. PM me for a hint.

What is the next number in this sequence?

1, 4, 7, 12, 15, 18, 21, 24, 27, ?

I love ambiguous sequences!
32. Just choose the integers ending in 1, 4, and 7 for the first 10 natural numbers. Then choose the integers ending in 2, 5, and 8 for the next set of 10. Repeat, alternating between the two different sets of three numbers. I'm sure there's a different "correct" answer, but poorly defined sequences irk me.

cyphr on
• Registered User
edited April 2008
cyphr wrote: »
Absurdist wrote: »
The solution to like every math puzzle ever is on the internet at this point, so no cheating if you want a good challenge! Some people get this right away, but it was really hard for me. Like, keep-me-up-all-night-and-then-eureka-while-showering-the-next-morning kind of hard. PM me for a hint.

What is the next number in this sequence?

1, 4, 7, 12, 15, 18, 21, 24, 27, ?

I love ambiguous sequences!
32. Just choose the integers ending in 1, 4, and 7 for the first 10 natural numbers. Then choose the integers ending in 2, 5, and 8 for the next set of 10. Repeat, alternating between the two different sets of three numbers. I'm sure there's a different "correct" answer, but poorly defined sequences irk me.

Jeez, some people.

Ok, to clarify, what is the next number in this sequence that does not require the use of multiple rules or "if...then" statements. Since, you know, you can obviously define any sequence via an infinite number of functions if you allow the use of compound functions.

1, 4, 7, 12, 15, 18, 21, 24, 27, ?

Absurdist on
[SIGPIC][/SIGPIC]
• Registered User regular
edited April 2008
Alright alright, point taken. But now I'm also going to be up for a while thinking about it. Bastard. :-p

cyphr on
• Registered User
edited April 2008
Here is another tricky one!

What's the next number in this sequence?

10, 0, 3, 20, 16, 51, 32, 67, 74, ?

Absurdist on
[SIGPIC][/SIGPIC]
• Registered User regular
edited April 2008
I'll post my own while I ponder yours.

1, 11, 21, 1211, 111221, 312211, 13112221, ?

cyphr on
• Registered User
edited April 2008
Cantide wrote: »
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I think I have a solution for this, although it may not be the fastest one.
They actually only need one switch for this. The inmate pick one person as the control. He's the guy that keeps track of things and eventually makes the announcement to the warden. Whenever one of the other people enters the room and sees the switch in the "off" position, if they have never toggled the switch before, they set it to "on". Otherwise they leave the switch alone( if they HAVE to toggle a switch, they use the second one, which is ignored by the control person ). Every time the control person enters the room and sees the switch is "on", he turns it "off". Once he's turned the switch off X-1 times, in this case 49, he knows that all the other inmates have been to the room at least once.

Technically that's not a solution, as technically the problem isn't solvableâ€” every prisoner is not necessarily going to be called into the room after any amount of time.

That is a "good strategy", though

• Registered User
edited April 2008
cyphr wrote: »
I'll post my own while I ponder yours.

1, 11, 21, 1211, 111221, 312211, 13112221, ?

A classic! This sequence was the first of it's type that I had ever come across, and I never did get it. I had to ask for the answer. Since then, of course, I look for solutions of a similar type in any stumper puzzle.
Each number is a description of the digit sequence, left-to-right, in the previous number. So 1 is "one", 11 is "one one", 21 is "two ones", 1211 is "one two and one one", 111221 is "one one, one two, and two ones", etc.

Absurdist on
[SIGPIC][/SIGPIC]
• Registered User
edited April 2008
Technically that's not a solution, as technically the problem isn't solvableâ€” every prisoner is not necessarily going to be called into the room after any amount of time.

That is a "good strategy", though

He's got you there, Doc. You need to stipulate that the prisoners and the universe they live in are immortal.

hehe

Absurdist on
[SIGPIC][/SIGPIC]
• Registered User, ClubPA regular
edited April 2008
Cantide wrote: »
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I think I have a solution for this, although it may not be the fastest one.
They actually only need one switch for this. The inmate pick one person as the control. He's the guy that keeps track of things and eventually makes the announcement to the warden. Whenever one of the other people enters the room and sees the switch in the "off" position, if they have never toggled the switch before, they set it to "on". Otherwise they leave the switch alone( if they HAVE to toggle a switch, they use the second one, which is ignored by the control person ). Every time the control person enters the room and sees the switch is "on", he turns it "off". Once he's turned the switch off X-1 times, in this case 49, he knows that all the other inmates have been to the room at least once.

That is a solution. It does take a really long time.

Doc on
• Registered User
edited April 2008
Doc wrote: »
Cantide wrote: »
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I think I have a solution for this, although it may not be the fastest one.
They actually only need one switch for this. The inmate pick one person as the control. He's the guy that keeps track of things and eventually makes the announcement to the warden. Whenever one of the other people enters the room and sees the switch in the "off" position, if they have never toggled the switch before, they set it to "on". Otherwise they leave the switch alone( if they HAVE to toggle a switch, they use the second one, which is ignored by the control person ). Every time the control person enters the room and sees the switch is "on", he turns it "off". Once he's turned the switch off X-1 times, in this case 49, he knows that all the other inmates have been to the room at least once.

That is a solution. It does take a really long time.
Does that mean there's a better solution? I must know.

Rufus_Shinra on
• Registered User regular
edited April 2008
Cantide wrote: »
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I think I have a solution for this, although it may not be the fastest one.
They actually only need one switch for this. The inmate pick one person as the control. He's the guy that keeps track of things and eventually makes the announcement to the warden. Whenever one of the other people enters the room and sees the switch in the "off" position, if they have never toggled the switch before, they set it to "on". Otherwise they leave the switch alone( if they HAVE to toggle a switch, they use the second one, which is ignored by the control person ). Every time the control person enters the room and sees the switch is "on", he turns it "off". Once he's turned the switch off X-1 times, in this case 49, he knows that all the other inmates have been to the room at least once.

Technically that's not a solution, as technically the problem isn't solvableâ€” every prisoner is not necessarily going to be called into the room after any amount of time.

That is a "good strategy", though

If every prisoner isn't called in at some point, it's unwinnable anyway.

jothki on
• Registered User
edited April 2008
The game as described is not necessarily winnable, yes.

• Registered User regular
edited April 2008
Doc wrote: »
Cantide wrote: »
Doc wrote: »
The solution to this is available on the 'net, so no cheating!

A prison warden decides to play a game with the inmates. If they can win, they all go free. If they lose, they all get stuck in there for life. The prisoners can all talk to each other to come up with a strategy before the game starts, but not after. The game is as follows: a prisoner will be randomly selected daily. He will enter a room where he sees two switches. He must toggle exactly one of the two switches and then exit the room. To "win," a prisoner must go to the warden and say "All prisoners have visited the room." If he is correct, they win. If not, they lose.

Again, after the game starts, the prisoners cannot communicate with each other. What's a good strategy for them to come up with before the game starts?

I think I have a solution for this, although it may not be the fastest one.
They actually only need one switch for this. The inmate pick one person as the control. He's the guy that keeps track of things and eventually makes the announcement to the warden. Whenever one of the other people enters the room and sees the switch in the "off" position, if they have never toggled the switch before, they set it to "on". Otherwise they leave the switch alone( if they HAVE to toggle a switch, they use the second one, which is ignored by the control person ). Every time the control person enters the room and sees the switch is "on", he turns it "off". Once he's turned the switch off X-1 times, in this case 49, he knows that all the other inmates have been to the room at least once.

That is a solution. It does take a really long time.
Does that mean there's a better solution? I must know.

I've come up with a solution, but it seems a little slow (but not as slow as the previous one). This assumes that they HAVE to flick a switch every day.
Assign every prisoner a number from 1 to X. Designate a "prisoner month" to be X days, and keep track of what day of the "prisoner month" it is. Basically, we are going to have everyone accumulate a list of prisoners they've known to have entered the room.

A prisoner enters the room on day of the "month" Y. If the first switch is on, he notes that prisoner Y has been in the room at some point. Otherwise he notes nothing and always ignores the previous setting of the second switch. Then, if he is prisoner Y+1 (modulo X) or he has noted that prisoner Y+1 has been in the room, he toggles the switches so the first one is on and the second is whatever. Otherwise, he toggles the switches so the first one is off and the second is whatever.

For example, let's say there are 7 prisoners, assigning each one a day of the week and they start by marking off their day of the week. Some prisoner enters on a certain day of the week, say Monday. If he sees the switch on, he marks off Monday. Then if he has Tuesday marked off, he sets the first switch to on and leaves, otherwise he sets it to off. The first prisoner with all the days marked off goes to the warden.

This means that some prisoner will have to enter the room on all of the other assigned days at least once, and that every other prisoner will have to enter the room on his assigned day. That could take awhile.

Savant on
• Registered User regular
edited April 2008
The traditional prisoner puzzle only has one switch, and a lot of work has been done on it:

http://www.segerman.org/prisoners.pdf
http://www.ocf.berkeley.edu/~wwu/papers/100prisonersLightBulb.pdf

Edit: Obviously, these links can be considered massive spoilers for the prisoner problem. You've been warned.

Most methods can be extended to two lightbulbs (switches) in some respect; the first paper has a small section on two switches, as well as a bunch of other variants.

Interestingly, to my knowledge, no-one has been able to prove what the best solution is.

area on
• Registered User
edited April 2008
Nobody can solve my puzzlers. Do I win something?

10, 0, 3, 20, 16, 51, 32, 67, 74, ?

1, 4, 7, 12, 15, 18, 21, 24, 27, ?

Absurdist on
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• Registered User regular
edited April 2008
The jump from 7 to 12 is obviously the sticking point in that second series. A hint on that one would be appreciated. I haven't given as much thought to the first one. They're both very good.

cyphr on
• Doesn't look like anything to me Registered User regular
edited April 2008
This is most assuredly not my series, and math geeks will probably see why this is funny.

2, 5, 877, 27644437, 35742549198872617291353508656626642567, 359334085968622831041960188598043661065388726959079837 ... ?

Apothe0sis on
Tide goes in. Tide goes out.
Es-annon NEVA 4GET
• Registered User
edited April 2008
Apothe0sis wrote: »
This is most assuredly not my series, and math geeks will probably see why this is funny.

2, 5, 877, 27644437, 35742549198872617291353508656626642567, 359334085968622831041960188598043661065388726959079837 ... ?

Usually when they jump right from small numbers into enormous numbers it means that the series is a list of numbers with an unusual property.
The first three are prime, and I will bet that the rest are too. But if I am right about the "unusual property" thing (i.e. if the series is not derived from performing operations on its members) then it's just a case of either recognizing these numbers or not recognizing them. I don't recognize them. Will you tell me if the series is operationally derived or not? Because if it is, then I will try to solve it. If it isn't, then I give up

EDIT:
My guess is that these are primes that also have some other property, or primes that are also some other kind of number like being Markov numbers (they aren't) or something like that.

Absurdist on
[SIGPIC][/SIGPIC]
• Doesn't look like anything to me Registered User regular
edited April 2008
Absurdist wrote: »
Apothe0sis wrote: »
This is most assuredly not my series, and math geeks will probably see why this is funny.

2, 5, 877, 27644437, 35742549198872617291353508656626642567, 359334085968622831041960188598043661065388726959079837 ... ?

Usually when they jump right from small numbers into enormous numbers it means that the series is a list of numbers with an unusual property.
The first three are prime, and I will bet that the rest are too. But if I am right about the "unusual property" thing (i.e. if the series is not derived from performing operations on its members) then it's just a case of either recognizing these numbers or not recognizing them. I don't recognize them. Will you tell me if the series is operationally derived or not? Because if it is, then I will try to solve it. If it isn't, then I give up
You probably have to recognise the series. It's concievably solvable, but you'd have to have an Erdos number of 0 or something to do so.

REAL SPOILER
You do not want to solve it anyway, I gave it as a joke because the next number has 6539 numbers in it

Apothe0sis on
Tide goes in. Tide goes out.
Es-annon NEVA 4GET
• Registered User
edited April 2008
Apothe0sis wrote: »
Absurdist wrote: »
Apothe0sis wrote: »
This is most assuredly not my series, and math geeks will probably see why this is funny.

2, 5, 877, 27644437, 35742549198872617291353508656626642567, 359334085968622831041960188598043661065388726959079837 ... ?

Usually when they jump right from small numbers into enormous numbers it means that the series is a list of numbers with an unusual property.
The first three are prime, and I will bet that the rest are too. But if I am right about the "unusual property" thing (i.e. if the series is not derived from performing operations on its members) then it's just a case of either recognizing these numbers or not recognizing them. I don't recognize them. Will you tell me if the series is operationally derived or not? Because if it is, then I will try to solve it. If it isn't, then I give up
You probably have to recognise the series. It's concievably solvable, but you'd have to have an Erdos number of 0 or something to do so.

REAL SPOILER
You do not want to solve it anyway, I gave it as a joke because the next number has 6539 numbers in it

I'm just happy that I know what an Erdos number is.

Absurdist on
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