Pirates of the Caribbean

[Guest]

In the Caribbean Sea, 13 pirates, while plundering an English ship, come upon a chest full of gold. Since the

pirates have found the chest simultaneously, no one can claim the chest as his own.

To protect the chest from the avarices of the pirates, the pirate leader Captain Jack Sparrow suggests a

scheme. “We must put a certain number of locks on this chest and distribute their keys amongst ourselves in

such a way that it can be opened only when 7 or more then 7 pirates dicide to open it. In other words, only

when a majority of the 13 pirates agree should the chest be able to be opened.”

How can this scheme be implemented? What would be the minimum number of locks required and how must

their keys be distributed?

Let us rephrase our problem slightly.

We want that only a group of 7 or more pirates should be able to open the chest. This means that whenever

such a majority group decides to open the chest, they should have amongst themselves the keys to all the

locks on the chest.

Suppose on the other hand, that only 6 pirates decide to open the chest. Then there should be at least one

lock on the chest whose key(s) are not with anyone amongst that group of 6. Thus, that single lock will

prevent the minority group of 6 pirates from opening the chest.

This is the approach we now follow. For every possible group of 6 pirates, we put a lock on the chest and

distribute 7 keys of the lock amongst the remaining 7 pirates. That lock will prevent our group of 6 pirates

from opening the chest.

Such a lock will exist for every group of 6 pirates. Thus, whenever any group of 6 pirates decides to open the

chest, they will be prevented by one lock whose keys are with the other 7 pirates (the case when even less

than 6 pirates decide to open the chest is automatically solved because then there will be more than one lock

to prevent that group from opening the chest)

Also, whenever any group of 7 pirates decides to open the chest, there is no lock whose key is not amongst

one of the 7 members of that group. Thus, any group of 7 pirates (or more) will be able to open the chest.

Thus, what we need to do is put ¹³C₆ (=¹³C₇) locks on the chest and for each lock, we select a group of 7

pirates, make 7 keys of that lock and give one key to each member of this group.

[Guest]

I didn't look at the spoiler in the original post. My solution is 7 locks and 13 keys. All keys are identical. Each pirate gets one key. A majority vote of 7 or more in favor of opening the chest will allow it to be opened.

[Pickett]

I didn't look at the spoiler in the original post. My solution is 7 locks and 13 keys. All keys are identical. Each pirate gets one key. A majority vote of 7 or more in favor of opening the chest will allow it to be opened.

If all of the keys were identical...then assuming that the locks behave like most locks I know and most locks I've seen in the movies on pirate chests, then any one man could just take his key and go unlock all 7 of the locks whenever he wanted. they don't have to unlocked simultaneously...

[Guest]

If all of the keys were identical...then assuming that the locks behave like most locks I know and most locks I've seen in the movies on pirate chests, then any one man could just take his key and go unlock all 7 of the locks whenever he wanted. they don't have to unlocked simultaneously...

Not the answer intended, but he could be right if he speficified that they were locks that only stayed unlocked while the key was in. They do exist (and are quite common if you look at things like auto ignitions)

[Guest]

If all of the keys were identical...then assuming that the locks behave like most locks I know and most locks I've seen in the movies on pirate chests, then any one man could just take his key and go unlock all 7 of the locks whenever he wanted. they don't have to unlocked simultaneously...

Very true if the lock is designed to remain open after it is removed. I was thinking of the type of lock such as my fire safe, which will not allow the key to be removed when it is open. Nothing in the puzzle states the design requirements of the lock.

[Pickett]

Not the answer intended, but he could be right if he speficified that they were locks that only stayed unlocked while the key was in. They do exist (and are quite common if you look at things like auto ignitions)

Very true if the lock is designed to remain open after it is removed. I was thinking of the type of lock such as my fire safe, which will not allow the key to be removed when it is open. Nothing in the puzzle states the design requirements of the lock.

Very true...I had forgotten about them (and my fire safe is also like that). Since the OP didn't suggest the type of lock, I agree, this is a good answer

[Guest]

Riddle Update: The Locks are the old school type which remain open even after the key is withdrawn. (or else the puzzle would have been very simple)

[Guest]

Well one obvious solution would be

Since you didn't say ANY 7 pirates decide, you could just have 7 locks with 7 keys. This would mean those 7 pirates all had to be present. If somebody else wanted it open, they'd have to consult with one of the special 7.

[Guest]

I never did advanced mathematics, but I don't see how this is a solvable situation...

Here's where I'm stuck. We'll name the pirates A-M. A,B,C,D,E,F represent a general minority of six pirates. We need one more pirate from G-M to get minority. Any one of them will give us the solution. If we add G, we get seven. Locks are openable.

But say A get's cold feet, and bows out. Back to six. But by your rules, any of the pirates from H-M can represent A, and before they also brought everything to the party G, did. So if H had replaced A from the beginning, wouldn't it then be openable with just 6 pirates? I see no way that the lock can be openable with any seven random pirates.

Another approach... The pirates split the island down the middle. A-F on one side, H-M on the other. G is in the middle, can't figure out which group to help open. No matter how many pirates from one side switch spots with the other, they wouldn't open it. G must have a unique key.

I don't know, maybe I'm just too daft for this one.

[Guest]

Wait, doesn't the dog have the keys?

[Guest]

This riddle doesn't make sense. Pirates plunder to AQUIRE WEALTH. There is NO reason why ANY of the pirates would decide NOT to open the chest. They would simply divide it's contents (gold, jewels, womens clothing) among themselves. There is no reason for them to complicate something so simple as plundering.

Not to mention, the riddle is only about OPENING the chest (a minor detail), and has nothing to do with the CONTENTS of the chest (the ONLY thing they care about).

And what's wrong with a verbal vote? "How many of us want to open it? Everybody? GREAT!!" End of discussion.

No pirate would even consider a ridiculous key scheme when a friggin' TREASURE CHEST is sitting there. And try explaining the multi-key-mathmatical-lock-sequence concept to an alcoholic murderer that never went to school. He would kill you.

--No wonder they left Sparrow on an island to die.

--And then chained him to the boat to let the Krakken eat him.

[Guest]

This riddle doesn't make sense... No pirate would even consider a ridiculous key scheme

Not surprisingly, Mr Logic seems to make perfect sense. Split it up, get drunk, carouse, etc. It's all part of the code.

[plasmid]

Just put one lock on it and give me the key. I'll make sure the treasure is nice and safe.

Oh, wait, this isn't one of my puzzles, is it.

[Guest]

Well one obvious solution would be

Since you didn't say ANY 7 pirates decide, you could just have 7 locks with 7 keys. This would mean those 7 pirates all had to be present. If somebody else wanted it open, they'd have to consult with one of the special 7.

The Pirates were democratic and there was no such special 7 group

[Guest]

I never did advanced mathematics, but I don't see how this is a solvable situation...

Here's where I'm stuck. We'll name the pirates A-M. A,B,C,D,E,F represent a general minority of six pirates. We need one more pirate from G-M to get minority. Any one of them will give us the solution. If we add G, we get seven. Locks are openable.

But say A get's cold feet, and bows out. Back to six. But by your rules, any of the pirates from H-M can represent A, and before they also brought everything to the party G, did. So if H had replaced A from the beginning, wouldn't it then be openable with just 6 pirates? I see no way that the lock can be openable with any seven random pirates.

Another approach... The pirates split the island down the middle. A-F on one side, H-M on the other. G is in the middle, can't figure out which group to help open. No matter how many pirates from one side switch spots with the other, they wouldn't open it. G must have a unique key.

I don't know, maybe I'm just too daft for this one.

there is more than 1 lock, the pirates are all unique and different, next time I wont copy a riddle from my school book.

[Guest]

They need one lock and one key. The captain keeps it, and should any majority decide to open the box, they can ask/request/demand/steal the key from him. a Minority would know better than to mess with a majority of pirates.

[Guest]

Not the answer intended, but he could be right if he speficified that they were locks that only stayed unlocked while the key was in. They do exist (and are quite common if you look at things like auto ignitions)

Those are security locks. We use them at work. They are quite more secure than the common lock one buys at the grocery store.

[Guest]

They need one lock and one key. The captain keeps it, and should any majority decide to open the box, they can ask/request/demand/steal the key from him. a Minority would know better than to mess with a majority of pirates.

Perfect answer..... I don't think Jack could do advanced mathematics anyway.

[Guest]

groups. Group A has 6, group B has 6, and group C has the last person by himself. In each team there is persons 1-6. give 1 on each team the same key, 2 on each team the same key, etc. etc. Then give the poor 13th person his own key and noone can open it because he has the last key, and only the last key. Thus, you need 7 locks.

EDIT: o dang Dio de Goat gave pretty much my same answer...

Edited April 12, 2009 by skirider

[Guest]

Well, I certainly hope The Black Pearl is heavily loaded with locks...

they will be needing 1.716 locks and 12.012 keys.

If you have a group of an uneven number, and you make a sub-group containing more than half of the full group, this sub-group will have at least one element in common with all possible groups of the same size.

Give each member in such a group the key for a unique lock. Do this for all other possible combinations (sorry, permutations) of equally sized groups, and lock the chest with all these locks (1 lock for each group).

In other words...

For a group of X participants (X is uneven and at least 3) we can make P(x , x/2) groups, and give each group a unique lock, and each member of that group a key for that lock. And finally lock the chest with the 1716 locks. (x/2 is an integer division - rounds down to nearest integer... I don't know the proper notation for this)

So for this particular task we need:

P(13,6) = 13! / (6! * 7!) = 8*9*10*11*12*13 / 1*2*3*4*5*6 = 1716 unique locks with 7 keys for each

Edited April 12, 2009 by uhre

[Guest]

Well, I certainly hope The Black Pearl is heavily loaded with locks...

they will be needing 1.716 locks and 12.012 keys.

If you have a group of an uneven number, and you make a sub-group containing more than half of the full group, this sub-group will have at least one element in common with all possible groups of the same size.

Give each member in such a group the key for a unique lock. Do this for all other possible combinations (sorry, permutations) of equally sized groups, and lock the chest with all these locks (1 lock for each group).

In other words...

For a group of X participants (X is uneven and at least 3) we can make P(x , x/2) groups, and give each group a unique lock, and each member of that group a key for that lock. And finally lock the chest with the 1716 locks. (x/2 is an integer division - rounds down to nearest integer... I don't know the proper notation for this)

So for this particular task we need:

P(13,6) = 13! / (6! * 7!) = 8*9*10*11*12*13 / 1*2*3*4*5*6 = 1716 unique locks with 7 keys for each

Uhm... these are combinations, not permutations. However, everything should be correct... just replace P( ) with C( )

[Guest]

Take two boxes, one fits snug over the other. Drill a hole in the top and fix a pole to the base of the other. The pole extends throught the hole of the upper box and has a end stop at its top. Now add 13 padlocks round the pole and give a key to each pirate. The hieght of the pole is arranged so that all 13 locks fit snug between box and end stop. When each lock is removed the upper box is free to rise a little each time. The Height of the upper box is arranged so that when 7 padlock are removed it can swing free of the base.

Now any 7 can open the box it doesn't matter which ones.

Edited April 24, 2009 by Tinners

[Guest12345]

For every group of 6 people put on a lock that none of them have the key to. Then give the remaining 7 people each a key to the lock. This ensures no minority can open it and ensures that any group larger than 6 can open the chest. In total there are 13 choose 7 locks. And 7 * (13 choose 6) keys. The keys are distributed equally so each pirate (7 * (13 choose 6))/13 keys