Just that it doesn’t use the Knowlton-Graham partition. If you look at Graham’s original paper http://www.math.ucsd.edu/~ronspubs/66_01_finite_set.pdf, there is one constraint that your solution doesn’t meet: When arriving back on the starting end, he disconnects all wires, does conductivity tests, and from the cardinality of the connected set alone identifies the labeling. In your solution, you have to keep a little memory of which wire the wire #2 was connected to (on the original side), so that you can “reason backwards” etc.

That said, it looks like a perfectly good practical solution.

At the other end, find a pair of connected wires. Label one of them #2 and the other #3. Find the second pair of connected wires. Label one of them #4 and the other #5. The remaining unconnected wire can be labelled #1. Now connect #1 to #2, #3 to #4, and leave #5 unconnected.

Back at the first end, disconnect the previous connections, but keep track of the pairings. Find the wire now connected to #1 at
the other end. Label it #2. The wire that was connected to #2 at this end can be labelled #3. The wire now connected to #3 at the other end is #4, and the wire that was connected to #4 at this end is #5.

We have made two trips and have labelled each wire with the same number at both ends.

Dave

And yes, n/2 splitting (for this modified problem) _should_ be optimal as it transmits maximum information. And I think the _proof_ should be obvious from some information theoretic point-of-view.

So, I think this modified optic-fiber problem turns out to be easy after all; compared to the original copper-wire problem that involved more interesting math. And that deeper problem always (when n != 2,5,9) has a 2-step solution compared to the best Log[2,n]-step solution of the simpler problem!

I think your Ceiling[Log[2,n]]-step assertion is correct, but I’m confused about your method. For eight wires are you saying you should light in the following pattern?
00000001
00000010
00000011
00000100
00000101
00000110
00000111
00001000

Most of those are redundant and you still wouldn’t know four of them. I’m either misunderstanding you and/or you meant something like this:
11110000
11001100
10101010

This is such a simple method, I wonder if there isn’t some more complexity lying within this problem once n gets bigger…

I’m starting to be convinced that there is not. Any initial n-splitting with k “on” wires, no matter how complex, is essentially “isomorphic” to an (n-k)k split, that is, 11001100 ~= 11110000 ~= 10101010. Then it essentially becomes two problems – (n-k) and k. Just judging from our small samples clearly it is better when k = n/2, or as close as possible if n is odd. If k = n/2 is the ideal split, then Ceiling[Log[2,n]] should be correct, no?

For example if instead of splitting 8 into 4-4 we split into 6-2 we’d end up taking four steps instead of three, since we have the initial split and 6 takes three steps.

hmm…

For 5 wires, by working out all possible two-step solutions, it looks clear to me that a three-step solution is the best possible.

In general, a Ceiling[Log[2,n]]-step solution is easy: Number the wires {0,1,…,n-1}. In the kth step, turn on the lights of wires the binary representation of which have kth bit ON. On the other end, keep building the IDs of each wire, a bit at each step.

Now the question is: Could some clever scheme do better than Ceiling[Log[2,n]]-steps?

The two-wire situation is now obvious. Besides that I’ve only looked at the five wore situation which can be solved in three steps as follows:

For wires A,B,C,D, and E, light up A, B, and C and mark that group somehow. Then light up A, and D. This will give you A (lit up both times), D (lit up once), and E (never lit up). Now you have the two wire situation (B and C) and so one more step will do it.

Not sure if it can be done in two steps, nor do I have an inkling towards the general solution.

Is anyone aware if this is a solved problem? Please don’t tell me Erdös solved it :).