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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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Solution Really Simple | Comment 34 of 45 |

I'm sure this has been covered already, but not nearly as clearly as it should. It's been proposed that we should only need 10 locks and 30 keys, but maybe figuring out how to distribute them is a problem. Everyone I've seen so far has given the arrangement, and then shown how their arrangement solves the problem.

The simplest approach is to simply say, if the lock will be opened if and only if at least three people are present, then any given pair should be unable to unlock at least one lock. That's where the 10 locks comes from, the 10 combinations of 2 people out of a group of 5. Make those assignments first:

AB 1
AC 2
AD 3
AE 4
BC 5
BD 6
BE 7
CD 8
CE 9
DE 10

Read this as A and B alone should be unable to open lock 1. Thus, C, D, and E should all have a key to that lock, and the same reasoning follows for the other 9 locks.

The assignment of keys follows naturally from there:

A:         5 6 7 8 9 10
B:   2 3 4       8 9 10
C: 1   3 4   6 7     10
D: 1 2   4 5   7   9
E: 1 2 3   5 6   8

I realize this is redundant of what other people have posted, but without the right method this can get way too messy ..

  Posted by DJ on 2004-03-04 12:19:42
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