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Friday Puzzler -- Cable

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Friday Puzzler -- Cable

#1

Friday Puzzler -- Cable

Alex Y

You have a cable with 66 unmarked wires running from the basement to the penthouse of a building. You need to match the ends of the wires, and the only tools you have are pliers that you can use to twist wires together, a continuity tester to determine which wires are connected together, and tape to mark the ends.

For instance, you may twist three of the wires in the basement together, and label them A, B, and C. You could go to the penthouse and find which wires are connected, and label them "ABC" (since when you find two that are connected, you will not know if they are AB, AC, or BC) You can then twist any wires together in the penthouse and go to the basement and test wires there.

How many trips must you make to be able to correctly label both ends of each wire?

Re: Friday Puzzler -- Cable

#2

Re: Friday Puzzler -- Cable

Larry Barrett

I think Bill Earl would say two trips. Where is Bill when we need him?

Re: Friday Puzzler -- Cable

#3

Re: Friday Puzzler -- Cable

Alex Y

I think Bill would be right -- one round trip.

How does he do that in absentia? :-)

I take it this was a rerun? I didn't remember.

Re: Friday Puzzler -- Cable

#4

Re: Friday Puzzler -- Cable *LINK*

Larry Barrett

Not a rerun, but seemed familiar to me, and I remembered that Bill Earl had responded to the earlier puzzle. Google search found that one. It was more complicated, but I think the same approach would work here.

Here is a link to the "Earl"ier puzzle:

http://www.woodcentral.com/woodworking/forum/trivia.pl/md/read/id/23570/sbj/friday-puzzle-unlabeled-cable/


http://www.woodcentral.com/woodworking/forum/trivia.pl/md/read/id/23570/sbj/friday-puzzle-unlabeled-cable/

Re: Friday Puzzler -- Cable

#5

Re: Friday Puzzler -- Cable

Alex Y

Obviously, my memory is shorter than 11 years :-)

I was thinking of Bill's first solution, which has the advantage that it works for any number of wires (even or odd), tying the initial group of wires into a bundle of n, where n*(n+1)/2 >= total number of wires. Each other bundle is one smaller than the previous, until the last bundle is whatever is left over when you can't go one smaller.

Guess I better find some new material!

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