Pinnacle
Level: Smitemaster

Rank Points: 1129
Registered: 06-10-2004
IP: Logged

Why not


____________________________
Once (adv.): Enough.
Twice (adv.): Once too often.
~Ambrose Bierce, The Devil's Dictionary

---

[Last edited by Pinnacle at 08-10-2014 03:53 PM]

lopsidation
Level: Master Delver

Rank Points: 176
Registered: 05-30-2008
IP: Logged

Ooh, that's clever! OK, any of you three can post the next puzzle.

My original solution was:

____________________________
"Happiness is like a cat. If you pay no attention to it and go about your business, you'll find it rubbing against your legs and jumping into your lap."

Kallor
Level: Master Delver
Rank Points: 277
Registered: 04-02-2007
IP: Logged

But:



____________________________
Kallor

blorx1
Level: Smitemaster
Rank Points: 920
Registered: 07-18-2009
IP: Logged

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.

____________________________
If you need to think outside of the box, then you didn't build a good enough box.

lopsidation
Level: Master Delver

Rank Points: 176
Registered: 05-30-2008
IP: Logged

Kallor:

____________________________
"Happiness is like a cat. If you pay no attention to it and go about your business, you'll find it rubbing against your legs and jumping into your lap."

RyTracer
Level: Delver

Rank Points: 86
Registered: 11-01-2013
IP: Logged

lopsidation wrote:
My original solution was:

Click here to view the secret text

×First, look away while the goblin enters one of the ten passages. Next, block the entrance to a random passage (e.g. with a stack of large boulders). Finally, look away and call the goblin to come to you.

If the goblin is lying, then you have a 10% chance of trapping him inside the cave. If you repeat this experiment enough times, and the goblin never gets trapped, then you should be convinced with arbitrarily high probability that he is telling the truth.

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


____________________________
I dream in vectors.
Portfolio Website - RyTracer.com

TripleM
Level: Smitemaster
Rank Points: 1373
Registered: 02-05-2003
IP: Logged

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.

Lucky Luc
Level: Smitemaster
Rank Points: 1215
Registered: 08-19-2012
IP: Logged

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]

Aquator
Level: Smiter

Rank Points: 405
Registered: 10-25-2014
IP: Logged

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]

Lucky Luc
Level: Smitemaster
Rank Points: 1215
Registered: 08-19-2012
IP: Logged

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.

Aquator
Level: Smiter

Rank Points: 405
Registered: 10-25-2014
IP: Logged

No worries! I've had fun thinking about your puzzle.
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."
"I'm already on a task. And it requires clearing you out."
"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]

Someone Else
Level: Smitemaster

Rank Points: 1306
Registered: 06-14-2005
IP: Logged

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.

Aquator
Level: Smiter

Rank Points: 405
Registered: 10-25-2014
IP: Logged

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.

Insoluble
Level: Smitemaster

Rank Points: 1638
Registered: 09-04-2014
IP: Logged

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.

____________________________
Links to neat forum tools that I always have trouble finding:

disoriented
Level: Smitemaster

Rank Points: 2387
Registered: 08-07-2007
IP: Logged

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.

____________________________
34th Skywatcher

Best to PM me, since I might miss your message on CaravelNet chat.

TripleM
Level: Smitemaster
Rank Points: 1373
Registered: 02-05-2003
IP: Logged

jamesdenem
Level: Master Delver
Rank Points: 203
Registered: 07-03-2006
IP: Logged

I think this should work.


____________________________
This signature intenionally left blank. Disregard any text you may find here, because it is probably just your imagination, and not a signature in any way. We mean it, don't believe that there is text here.

TripleM
Level: Smitemaster
Rank Points: 1373
Registered: 02-05-2003
IP: Logged

That doesn't work::

However,

jamesdenem
Level: Master Delver
Rank Points: 203
Registered: 07-03-2006
IP: Logged

TripleM wrote:
That doesn't work::

Click here to view the secret text

×if, say, the first half takes 2 minutes to burn, the second half takes 28 minutes to burn, so you've measured 28 minutes.

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


____________________________
This signature intenionally left blank. Disregard any text you may find here, because it is probably just your imagination, and not a signature in any way. We mean it, don't believe that there is text here.

skell
Level: Legendary Smitemaster

Rank Points: 3734
Registered: 12-28-2004
IP: Logged

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 .


____________________________
My website | Facebook | Twitter

Aquator
Level: Smiter

Rank Points: 405
Registered: 10-25-2014
IP: Logged

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]

jamesdenem
Level: Master Delver
Rank Points: 203
Registered: 07-03-2006
IP: Logged

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.

____________________________
This signature intenionally left blank. Disregard any text you may find here, because it is probably just your imagination, and not a signature in any way. We mean it, don't believe that there is text here.

Penumbra
Level: Smiter

Rank Points: 358
Registered: 02-17-2005
IP: Logged

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!

jamesdenem
Level: Master Delver
Rank Points: 203
Registered: 07-03-2006
IP: Logged

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?
Please offer your reasoning as well.


____________________________
This signature intenionally left blank. Disregard any text you may find here, because it is probably just your imagination, and not a signature in any way. We mean it, don't believe that there is text here.

disoriented
Level: Smitemaster

Rank Points: 2387
Registered: 08-07-2007
IP: Logged

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.

____________________________
34th Skywatcher

Best to PM me, since I might miss your message on CaravelNet chat.

---

[Last edited by disoriented at 03-18-2015 09:57 PM]

jamesdenem
Level: Master Delver
Rank Points: 203
Registered: 07-03-2006
IP: Logged

You got it!

Guess that one didn't last long...

____________________________
This signature intenionally left blank. Disregard any text you may find here, because it is probably just your imagination, and not a signature in any way. We mean it, don't believe that there is text here.

Kargaros
Level: Delver
Rank Points: 58
Registered: 04-06-2010
IP: Logged

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

____________________________
95th Skywatcher

disoriented
Level: Smitemaster

Rank Points: 2387
Registered: 08-07-2007
IP: Logged

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.

____________________________
34th Skywatcher

Best to PM me, since I might miss your message on CaravelNet chat.

---

[Last edited by disoriented at 03-19-2015 06:36 AM]

Kargaros
Level: Delver
Rank Points: 58
Registered: 04-06-2010
IP: Logged

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

____________________________
95th Skywatcher

---

[Last edited by Kargaros at 03-19-2015 11:48 AM]

skell
Level: Legendary Smitemaster

Rank Points: 3734
Registered: 12-28-2004
IP: Logged

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!

____________________________
My website | Facebook | Twitter