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Caravel Forum : Other Boards : Forum Games : Puzzle "tag" (Don't forget to read the rules in the first post!)
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Pinnacle
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Re: Puzzle "tag" (0)
Why not

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[Last edited by Pinnacle at 08-10-2014 03:53 PM]
08-10-2014 at 03:52 PM
lopsidation
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Re: Puzzle "tag" (0)
Ooh, that's clever! OK, any of you three can post the next puzzle.

My original solution was:

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08-10-2014 at 08:16 PM
Kallor
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Re: Puzzle "tag" (0)
But:

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08-11-2014 at 08:10 AM
blorx1
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Re: Puzzle "tag" (0)
Other point on the string solution. You can just as easily ask the goblin to bring it back out (while still not looking), which only requires that he can't duplicate your string.

Also, I don't have any clever puzzles, so another person can feel free to go.

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08-11-2014 at 08:51 AM
lopsidation
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Re: Puzzle "tag" (0)
Kallor:

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08-11-2014 at 03:14 PM
RyTracer
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Re: Puzzle "tag" (0)
quote:
lopsidation wrote:
My original solution was:

Okay, this doesn't make sense to me. I think you missed a key point.

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09-24-2014 at 09:17 AM
TripleM
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Re: Puzzle "tag" (0)
I think the point was that the goblin wants to prove to you there is a passage, thus chooses to enter the connected caves each time.
09-24-2014 at 10:19 AM
Lucky Luc
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Re: Puzzle "tag" (+1)
So, is this still a thing? Cause I've learned a pretty cool puzzle just the other week.

Suppose 100 (immortal) mathematicians (let's say 501st to 600th) are taking part in an experiment. Each of them is held in solitary confinement. Every now and then, a guard picks one of them and leads him into an interrogation room. The only things inside this room are a lamppost and an orb that toggles said lamppost.

The order in which the guard picks the mathematicians is almost completely random, so it might be that one mathematician gets picked twice in a row or not in a million times. The only condition is that every mathematician will visit the interrogation room at least once at some point of time infinitely often over the course of time.

The mathematicians may have discussed the problem beforehand, but once they are locked up, they have no other means of communicating but the lamppost in that interrogation room (also no cheap tricks like "That lightbulb is still hot!"). They have no clue how many - if at all - other mathematicians were in the interrogation room between their own turns.

The goal for the mathematicians is that one of them tells the guard as soon as he's absolutely sure that every mathematician has been to the interrogation room at least once. Is this possible if:
a) They know that the lamppost is initially turned off?
b) They don't know the initial state of the lamppost?

When I heard it, someone in the round solved the puzzle almost immediately, so if you're an electronic engineer or a mathematician or something, you might find this easy. If you're just a chemist (or a chemistry student like me), it might hopefully take a bit longer.

[Last edited by Lucky Luc at 03-16-2015 03:06 PM]
03-16-2015 at 09:59 AM
Aquator
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Re: Puzzle "tag" (+2)
Turns out the guard tricked all of the mathematicians, because they won't be able to actually toggle the lamppost unless there's some kind of weapon or a power token in the interrogation room with which to activate the orb.

I have to admit, I think I may remember a similar problem to a) from ages ago, but I did have to think a bit to solve b).

[Last edited by Aquator at 03-16-2015 01:01 PM]
03-16-2015 at 12:55 PM
Lucky Luc
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Re: Puzzle "tag" (0)
Yup, that's correct for both cases Sorry, I got that information wrong (not the one with the power token, though that's a valid point, too - the one in your spoiler). Corrected it in the original text.
03-16-2015 at 03:10 PM
Aquator
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Re: Puzzle "tag" (+1)
Well, I guess I'll have to come up with something now. Uhhh...

An inexperienced Smitemaster finds himself in a rather peculiar situation. During one of his contracts he ends up trapped in a room with several metal beings, able to talk and move in what can only be described as a clumsy imitation of the Eighth's usual inhabitants. More importantly however, no matter how often he strikes them down with his Rather Large Blade they seem to get up completely unfazed, not even giving him enough time to catch his breath. Despite his attempts at smiting the iron beasts they do not seem particularly hostile.
The Smitemaster finally ceases his attacks and one of the constructs approaches him.

"WE REQUIRE YOU TO COMPLETE A TASK. THEN WE WILL LET YOU LEAVE."
"WE HAVE A ROPE. YOU MUST BURN IT. THEN WE WILL GO AND YOU CAN CONTINUE."
"Well, that sounds easy enough", the Smitemaster replies, slowly giving up hope of finishing up this dungeon in time for dinner.
"THE ROPE MUST BE BURNED TO MEASURE 15 MINUTES. LIGHTING IT ON ONE END WILL CAUSE IT TO BURN UP IN EXACTLY ONE HOUR."
"Oh. I know this puzzle... But are you sure about the 15 minutes? I don't think that's how it went..."
"ONE ROPE. 15 MINUTES. WE WILL COUNT ONLY THE TIME THE ROPE IS BEING BURNED."
"Well then, this is no problem. I've got a sword. You and your buddies here just sit there and watch me cut off a fourth and burn it."
"THE ROPE DOES NOT BURN HOMOGENOUSLY. DIFFERENT PARTS OF THE ROPE TAKE DIFFERENT PARTS OF TIME TO BURN. WE CAN ASSIST YOU. WE ARE EQUIPPED TO START FIRE. THE LAST SUBJECT BURNED ITSELF ALIVE. WE ARE NOW ALSO EQUIPPED TO EXTINGUISH FIRE. PLEASE DO NOT SET YOURSELF ON FIRE."
"I wasn't about to just yet."

Can the Unnamed Smitemaster escape this situation?!
Sorry for the caps.

[Last edited by Aquator at 03-17-2015 02:17 AM]
03-16-2015 at 04:56 PM
Someone Else
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Re: Puzzle "tag" (0)
I'm thinking the answer isn't "Light it on one end, wait 14 minutes, then light the whole thing on fire." But that's the best I've got so far.
03-17-2015 at 01:55 AM
Aquator
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Re: Puzzle "tag" (0)
You're right, the constructs won't accept that as even if the whole process of burning it ends up taking just about 15 minutes, claiming that you actually used the rope to measure out the 15 minutes is kind of a stretch.

For all intents and purposes you can just assume the Smitemaster has no means of keeping time other than the rope. Only then would the constructs consider the task of using the rope to measure the time frame as fulfilled.
03-17-2015 at 02:12 AM
Insoluble
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Re: Puzzle "tag" (0)
Okay, so measuring time is not allowed (which makes sense since that would render the problem trivial). What kind of measurements are allowed? Length & angle measure?

Would this work:

Arrange the rope in a perfect circle, then light both ends. Once the shape becomes a perfect semicircle, extinguish the flames. Arrange the remaining rope in a perfect circle and repeat the process.

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03-17-2015 at 03:34 AM
disoriented
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Re: Puzzle "tag" (0)
quote:
Insoluble wrote:
Okay, so measuring time is not allowed (which makes sense since that would render the problem trivial). What kind of measurements are allowed? Length & angle measure?

Would this work:

Arrange the rope in a perfect circle, then light both ends. Once the shape becomes a perfect semicircle, extinguish the flames. Arrange the remaining rope in a perfect circle and repeat the process.

The rope is non-homogeneous so there's no guarantee that any amount of time has elapsed when you reach a semicircle.

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03-17-2015 at 04:20 AM
TripleM
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Re: Puzzle "tag" (+2)

03-17-2015 at 04:40 AM
jamesdenem
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Re: Puzzle "tag" (0)
I think this should work.

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03-17-2015 at 04:46 AM
TripleM
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Re: Puzzle "tag" (+2)
That doesn't work::

However,

03-17-2015 at 04:50 AM
jamesdenem
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Re: Puzzle "tag" (+1)
quote:
TripleM wrote:
That doesn't work::

There seems to be a bit of a flaw in your reasoning there (How do you burn it?)
Here is maths that shows how my solution works

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03-17-2015 at 11:52 AM
skell
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Re: Puzzle "tag" (0)
It's even simpler to check for specific cases:

Case A:
- Half Alpha burns for 30 minutes
- Half Beta burns for 30 minutes

So burning Alpha from both sites will take 15 minutes and Beta will take 15 minutes. If you lit them at the same time it will take exactly 15 minutes for both of them to burn down.

Case b:
- Half Alpha burns for 50 minutes
- Half Beta burns for 10 minutes

So burning Alpha from both sites will take 25 minutes and Beta will take 5 minutes. If you lit them at the same time, Beta will burn after 5 minutes and Alpha will burn after 25 minutes. Then you are killed because Constructs didn't like your math .

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03-17-2015 at 12:16 PM
Aquator
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Re: Puzzle "tag" (0)
The original intended solution is indeed the one TripleM proposed in that the Smitemaster applies a simple recurring method to get arbitrarily close to the 15 minutes until the constructs are satisfied.

Jamesdenem's math accurately explains why this recurring method will result in 15 minutes. However, simply burning both halfs on all of their ends does not ensure that the time needed for both of them to burn up is anywhere close to the average time.

Edit: Insoluble's geometric approach does not work in this case, because the rope is non-homogenous and the length of a part of the rope does not determine how long it takes to burn, but it's certainly a good reminder of why you don't want to use mathematicians as construction workers. I assume there would be more time spent on perfect circles than on actual construction.

[Last edited by Aquator at 03-17-2015 01:16 PM]
03-17-2015 at 01:03 PM
jamesdenem
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Re: Puzzle "tag" (0)
Yeah, I agree that I was thinking about averaging the burning time, and not how long the rope would actually burn.
So congrats to TripleM for getting the solution.

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03-17-2015 at 01:13 PM
Penumbra
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Re: Puzzle "tag" (+2)
I have been hearing puzzles about these "non-homogeneous fuses" for years. Somewhere, there is a factory that can create perfectly precise fuses, yet is completely unable to determine how long any portion of their fuses will burn.

How are they still in business!

03-17-2015 at 08:06 PM
jamesdenem
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Re: Puzzle "tag" (+1)
Since it seems that TripleM doesn't have a puzzle ready, I would like to submit one.

You are running late for your flight and find yourself at the very back of the line to board the plane. This particular plane has 50 passenger seats, and every seat has been booked.
The attendants have a peculiar policy of only letting one passenger board and take a seat at a time. The first to board loses his ticket in the wind as he is about to board, and in the confusion, forgets his seat number.
When he enters the plane, he chooses a random seat and sits there. Another strange policy of the plane is that if your assigned seat is already being sat in, you must sit in another seat.
Therefore, as the other 48 people ahead of you board the plane, they will either sit in their assigned seat or some random seat.

Being the last person to board the plane, what are the chances that your assigned seat is empty?

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03-18-2015 at 08:44 PM
disoriented
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Re: Puzzle "tag" (+2)
My somewhat-educated hunch is 50%.

Consider a small airplane with two seats. In this case, it's obviously 50% whether or not the first passenger gets his assigned seat, and thus it's the same for the second passenger.

Now to extend it to the 3-case...
For three seats, the first passenger A has an equal chance of sitting in A's seat, B's seat, or C's seat (1/3 each). In the first case, C will always get their assigned seat. In the third case, C will never get their assigned seat. In the second case, there's a 1/2 chance that C gets their seat depending on where B randomly goes. So I still see it as 50% overall.

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[Last edited by disoriented at 03-18-2015 09:57 PM]
03-18-2015 at 09:47 PM
jamesdenem
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Re: Puzzle "tag" (0)
You got it!

Guess that one didn't last long...

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03-18-2015 at 10:01 PM
Kargaros
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Re: Puzzle "tag" (0)
I've seen jamesdenem's puzzle several times throughout the years, but the answer 50% always seemed mysterious to me. I kept feeling that there should be an intuitive explanation.

Today, I finally managed to convince myself with the following argument (though at the time I should perhaps rather have followed the conference talk that I was actually attending). It is a generalization of disoriented's argument to n people:
Say for simplicity that Person k is supposed to sit in Seat k, for 1 <= k <= n.
Person 1 takes a random seat as stated in the problem. If Person 1 takes his own seat (Seat 1), then the question is resolved and Person n gets Seat n. If Person 1 sits in Seat n, the question is also resolved, and Person n does not get Seat n. In the remaining possibilities the question remains unresolved after Person 1 sits down.
So if the question is resolved by Person 1, then there is an equal chance that Person n gets Seat n or not.
If the question was not resolved by Person 1 (say Person 1 sits in Seat k), then at a later point another person will be choosing a seat at random (e.g. Person k). Then the question gets resolved if that person sits in either Seat 1 (then Person n gets Seat n) or Seat n (then Person n does not get Seat n), and in all other cases the question remains unresolved.

At any point, the question is resolved precisely when a Person k picking at random (among the remaining seats) sits in either Seat 1 (all remaining passangers, including Person n, gets their seats) or Seat n (Person n does not get his seat), and remains unresolved otherwise. Whenever the question gets resolved, the chances of picking Seat 1 and Seat n were equal, so the changes of positive and negative outcomes must be equal as well, i.e. a 50/50 chance.

Hopefully, my thoughts are not too convoluted, and at least I finally managed to convince myself

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03-19-2015 at 02:12 AM
disoriented
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Re: Puzzle "tag" (0)
Yup, that's how I convinced myself too! It's an inductive argument.

I thought of a puzzle to post but I see TripleM already posted it... I guess I'll have to think of another one.

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34th Skywatcher

[Last edited by disoriented at 03-19-2015 06:36 AM]
03-19-2015 at 05:58 AM
Kargaros
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Re: Puzzle "tag" (+3)
quote:
disoriented wrote:
Yup, that's how I convinced myself too! It's an inductive argument.

I will claim that the argument is not inductive, which actually surprised me while writing my previous post. Here is a proper proof, with some mathematical pseudo-LaTeX for good measure:
The question is only resolved when a person sits down in Seat 1 (all remaining people get their seats) or Seat n (Person n doesn't get Seat n).
Let
p_k = P(Person k is the decision point)
be the propability that the question is resolved by Person k -- this is essentially a sum of propabilities of all the histories where Person k decides the outcome, but the beauty is that we don't care about the actual probability.
The question is decided at some point, by the unique person who sits down in Seat 1 or Seat n. Therefore
\sum_{k from 1 to n-1} p_k = 1.
Suppose that the question is decided by a particular Person k. We then know that Person k has to be choosing at random, and that Seat 1 and Seat n are both still available -- otherwise the question would already be decided previously. Given that Person k decides the problem, we know that he sits in either Seat 1 or Seat n -- otherwise he would not be the decision point. The conditional probability for a positive outcome, under the assumption that Person k is the decision point, is therefore
P(Person n gets Seat n | Person k is the decision point) = 1/2
Using all the conditional propabilities we can calculate the total probability:
P(Person n gets Seat n) = \sum_{k from 1 to n-1} p_k * P(Person n gets Seat n | Person k is the decision point)
= \sum_{k from 1 to n-1} p_k * 1/2
= 1/2 * \sum_{k from 1 to n-1} p_k
= 1/2 * 1 = 1/2
So we don't actually need induction (though induction is great), we just need that the "local" probability of a positive outcome is always the same as the local propability of a negative outcome.

Sorry for derailing the puzzle thread with math

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[Last edited by Kargaros at 03-19-2015 11:48 AM]
03-19-2015 at 11:45 AM
skell
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Re: Puzzle "tag" (0)
I just wrote down all possible combinations of seatings and counted the ratio and the result is...

...will be ready in a few years, you just wait!

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03-19-2015 at 12:19 PM