You laid a transoceanic cable.

logic/transoceanic.cable New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

You laid a transoceanic cable.  You then realized the individual wires
in the cable were not color coded, they were all a dull drab grey.
What's worse, thanks to the fact that you haven't been able to use the
cable yet, you're on a really limited budget.

In this situation, the particular cable has 10 conductors.
([Ridiculous] of course, but it makes the problem a little easier to
work out)

Your budget will allow:
   2 one way plane tickets
   1 multimeter
   an unlimited number of wire nuts
   an unlimited number of stick on labels

The question is this, can you use the cable?

-- Nathan Neulinger <nneul@umr.edu> (November 1994)

Show solution

The answer is "No".  Now, if the cable had *11* conductors, you could
use it.  If it had 101 conductors, you could use it.  If it had a
BILLION AND ONE connectors, you could use it... but not if it had a
billion and two.  See the pattern?

[snip]

Let's give the cable 11 wires, rather than 10, and here's how to solve
it.  (This will also make it clear why there needs to be an odd number
of wires for it to work).

[method deleted]

AFAIK, there is no easy solution for an even number of wires.

-- Jonathan Haas <positron@Starbase.NeoSoft.COM>

[ Editor: Haas's method for an odd number of wires was deleted in both
replies that quoted him and is not in the collected material. ]

It can be done.

Leave two wires unconnected, connect all of the others in pairs.

Fly to the other end and test for connection.  You will be able to
identify the two unconnected wires, call them A1 and A2, and all of the
pairs, call them B1 and B2, C1 and C2, etc.

Connect A2 to B1, B2 to C1, C2 to D1, etc.

Fly home, undo the original connections (but label them first so that
you can remember which wire was connected to which), test for
connection again: the lone wire which is now connected is A2, the other
lone wire (still unconnected) is A1.  The wire connected to A2 is B1.
Its former mate is B2.  The wire connected to B2 is C1, and so on.

Since A1 is not really needed, this method works for even and odd
numbers of wires (except two).

-- Benjamin Rhinelander Thompson <thompson@spot.Colorado.EDU>

I think there is:

Try this: Suppose there are 2n wires.  When you first start joining
wires together in pairs, leave TWO wires disconnected.  Don't label
them.  Now go to the other end of the cable, find out which two wires
are disconnected, label one of them 1 and the other one 2n (whatever
its value is).  Now proceed as in the odd case, wiring 1 to an
arbitrarily chosen wire which you label 2, then find 2's "partner" 3,
wire 3 to an arbitrarily chosen 4, etc., until wires 1 through 2n-1
are all linked in series (due to the connections at the other end).
Now return to the first end of the cable.  The one wire that is still
disconnected from all the others is wire 2n.  The other formerly
single wire is number 1, the one directly linked to it is 2, the one
that was joined to 2 on this end is 3, etc.

In other words, in the even case you leave an extra wire that never
gets connected to any other, and the rest of the numbering is deduced
exactly as in the odd case.

-- David A. Karr <karr@cs.cornell.edu> (November 1994)