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Open By Majority (Posted on 2004-03-03) Difficulty: 3 of 5
A group of five people want to put a set of locks on a chest and distribute keys to the locks amongst themselves in such a way that all the locks on the chest could be opened only when at least three of them were present to open it.

How many locks would be needed, and how many keys?

See The Solution Submitted by Brian Smith    
Rating: 4.1429 (7 votes)

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re(3): Solution (Penny's) | Comment 16 of 45 |
(In reply to re(2): Solution (Penny's) by Tristan)

Tristan wrote: "I don't think that you could have five keys that each open a different set of locks."
 
Here is one simple scenario. There are a billion others you could think of.
 
Let's say that Mr. A's key has a tiny magnetic strip on it with the letter "a", B's key has the letter "b", C's key has the letter "c", D's key has the letter "d", and E's key has the letter "e". 
 
Lets say each lock has a sensor that can detect these letters on a key. Each lock has an "unlock rule" that states that "if any of three specific letters are detected on a key, I will unlock."
 
If these unlock rules are:  
 
Lock #1: any key with an "a", "b" or "c" unlocks me
Lock #2: a,b,d
Lock #3: c,d,e
Lock #4: b,d,e
Lock #5: b,c,e
Lock #6: b,c,d
Lock #7: a,d,e
Lock #8: a,c,e
Lock #9: a,c,d
Lock #10: a,b,e  
 
Then A's key will unlock [1 2 7 8 9 10]
B's will unlock [1 2 4 5 6 10]
C's will unlock [1 3 5 6 8 9]  
D's will unlock [2 3 4 6 7 9] 
E's will unlock [3 4 5 7 8 10]
 
Then any three of A,B,C,D,E will be able to unlock all the locks, but no two of them alone can. (E.g. A and E will not have a key that unlocks lock #6)
  
 

 

 

 

 

 

 

Edited on March 4, 2004, 8:02 am
  Posted by Penny on 2004-03-03 22:18:47

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