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Pirates of the Caribbean - the sequel


plasmid
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As I was pondering Pirates of the Carribean, I thought of some approaches that were probably off topic but nonetheless interesting. Suppose you have four pirates and want to make a treasure chest that will open when any three decide to open it. You only have four locks, and one key for each lock. How would you do it?

If you have four master keys, but each is marked with a different, specific lock's number, then each lock will still have it's own key.

Then the chest can be opened when any three want to, because any of them can open the fourth lock also.

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If you have four master keys, but each is marked with a different, specific lock's number, then each lock will still have it's own key.

Then the chest can be opened when any three want to, because any of them can open the fourth lock also.

yaeh i agree sounds about right to me. unless we're missing a variable.

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As I was pondering Pirates of the Carribean, I thought of some approaches that were probably off topic but nonetheless interesting. Suppose you have four pirates and want to make a treasure chest that will open when any three decide to open it. You only have four locks, and one key for each lock. How would you do it?

One key for each lock, could it also means one key opens all locks? Then the three who decide to open the chest will be majority, and can demand that the one with the key open the locks.

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Maybe the chest is just the top, placed on the bottom part. Put 1 lock at each side and when 3 people unlock it, the fourth lock with just act as sort of like a pivot point to open it. There doesn't have to be any "master key" or anything.

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Maybe the chest is just the top, placed on the bottom part. Put 1 lock at each side and when 3 people unlock it, the fourth lock with just act as sort of like a pivot point to open it. There doesn't have to be any "master key" or anything.

Got it, Avalanche!

I was also trying to think of a way to use all 12 edges of a cube in a problem like this, but that just got too complicated for me.

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The first lock can be opened by any of the 4 keys and the key can be removed. (perhaps with an outer box). Once that lock is open (exposing an inner chest?) you must have 3 keys (any 3) turning the lock mechanism open WITH THE KEY IN THE LOCK for the chest to be open.

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if we are assuming that each key uniquely corresponds to a lock, that the all locks are "closed," that the keys are distributed to the pirates before attempting to open the chest, and that the chest must be open in one attempt given this initial distribution of keys, then the problem (as it is posed) does not have a solution:

suppose that it can be done. then there is some way you can hand out the keys to the four pirates so that any combination of three of them will be able to open the chest. in particular, take any group G of three pirates (from the original four) and name the pirate that's left alone P. obviously, the chest can't be opened without having all four keys. so the group G has all four keys among them and P has 0. notice that no one of the pirates from G can be holding all the keys. otherwise, the remaining three from G would have no chance to open the chest (let alone any lock), which would contradict our assumption that any group of three pirates can open the chest. on the other hand, since the three pirates in G have all four keys, one of them MUST have at least two keys. this means that the other two pirates (from G) has AT MOST two keys. call these pirates K1 and K2. here's the contradiction: the group P, K1, K2 have only two keys among them. so they can't open the chest.

Edited by five0to180
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you give each pirate a key then you make a teasure chest with a removable top and have a lock hinge on each side. so when locked each side has a lock . hence if thre pirate used there keys it will hinge on the four lock to open.

As I was pondering Pirates of the Carribean, I thought of some approaches that were probably off topic but nonetheless interesting. Suppose you have four pirates and want to make a treasure chest that will open when any three decide to open it. You only have four locks, and one key for each lock. How would you do it?
Edited by Venslo
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This "solution" is a little more elaborate but I think it still works. I hope my crummy diagram helps you visualise. :)

Consider a chest that is held closed by 4 rods. The placement of the 4 rods is such that two run from one side of the chest to the other (parallel to one another) and the other two rods are positioned diagonally between the end points of the first 2 rods. The end of each rod is secured by one of the locks and each lock secures the end of 2 rods.


(1) (2)
|\ /|
| \ / |
| \ / |
| \/ |
| /\ |
| / \ |
| / \ |
|/ \|
(3) (4)

Each rod is secured in place by 2 locks. Unlocking any 3 locks enables the pirates to remove all of the rods and thus open the chest. For example. Opening locks 1 and 3 enables the lefthand rod to be removed. Opening lock 2 allows the diagonal to be removed. Lock 4 is still locked to the remaining rods but since these rods have one free end (since the other locks have been opened) these rods can simply be slipped out, still attached to lock 4. This works regardless of which pirate is missing.

I may need to clarify this some more. :)

Edited by HibsMax
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This "solution" is a little more elaborate but I think it still works. I hope my crummy diagram helps you visualise. :)

Consider a chest that is held closed by 4 rods. The placement of the 4 rods is such that two run from one side of the chest to the other (parallel to one another) and the other two rods are positioned diagonally between the end points of the first 2 rods. The end of each rod is secured by one of the locks and each lock secures the end of 2 rods.


(1) (2)
|\ /|
| \ / |
| \ / |
| \/ |
| /\ |
| / \ |
| / \ |
|/ \|
(3) (4)

Each rod is secured in place by 2 locks. Unlocking any 3 locks enables the pirates to remove all of the rods and thus open the chest. For example. Opening locks 1 and 3 enables the lefthand rod to be removed. Opening lock 2 allows the diagonal to be removed. Lock 4 is still locked to the remaining rods but since these rods have one free end (since the other locks have been opened) these rods can simply be slipped out, still attached to lock 4. This works regardless of which pirate is missing.

I may need to clarify this some more. :)

Under these circumstances, there would be certain combinations of 2 that could open the chest.

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This "solution" is a little more elaborate but I think it still works. I hope my crummy diagram helps you visualise. :)

Consider a chest that is held closed by 4 rods. The placement of the 4 rods is such that two run from one side of the chest to the other (parallel to one another) and the other two rods are positioned diagonally between the end points of the first 2 rods. The end of each rod is secured by one of the locks and each lock secures the end of 2 rods.


(1) (2)
|\ /|
| \ / |
| \ / |
| \/ |
| /\ |
| / \ |
| / \ |
|/ \|
(3) (4)

Each rod is secured in place by 2 locks. Unlocking any 3 locks enables the pirates to remove all of the rods and thus open the chest. For example. Opening locks 1 and 3 enables the lefthand rod to be removed. Opening lock 2 allows the diagonal to be removed. Lock 4 is still locked to the remaining rods but since these rods have one free end (since the other locks have been opened) these rods can simply be slipped out, still attached to lock 4. This works regardless of which pirate is missing.

I may need to clarify this some more. :)

I agree with Romulus' assessment, but that's an interesting approach that inspired another solution. Such creative people here at brainden!

The treasure chest would need to have a "lip" sticking out from the bottom of the lid and the top of the base. If you drill holes in the lip and thread a chain through, you could put the locks on the chain's links as shown in this picture, and the chain could only be removed and the chest opened when 3/4 pirates remove their locks.

post-15489-1239500695.jpg

BOW TO MY MAD PAINT SKILLZ

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