Actually, many cases of my variant (in the case where we can tell which wires are of each type) can be solved in two trips - using the triangular method. If m is between n triangular and n+1 triangular, connect 1 through n forward cables to each of the reverse cables (I'm adopting the convension that there are n "reverse" cables and m "forward" cables), leaving any remainder unconnected. At the other end, we can identify the reverse cables uniquely by the number of forward cables connected to them, and then a similar linkage will let us identify the forward cables when we get to the other end. There may be embellishments of this to handle other values of m. I don't think any solution is possible in less than log_{n+1}(m) trips, however - the m forward wires can at most be distinguished by which of the n backward wires they are connected to, or by not being connected to anything, at each step. (Connecting to multiple backward wires when m >> n just loses information.) The case where we can't tell which wires are oriented which way can be solved in at most 2 more trips than the other case. Connect all the wires together at one end, go to the other, and you can identify which ones are which at the second end. Now connect all of them together at that end, and you can make the same identification at the other end. This works even if we initially know only the number of wires, not how many go each way, as long as at least 1 goes each way. At least a little more information can be gained in this process: only some of the wires going in one of the two directions need be connected at the far end, and these can then be distinguished back at the near end. This actually provides a complete solution for n+m=3 in two trips. Franklin T. Adams-Watters -----Original Message----- From: franktaw@netscape.net I don't see that this variation _allows_ the triangular numbers approach. I think all you can do is connect some subset of the cables to the shield, then see which ones they are. This will require ceiling(log_2(n)) walks. This does suggest a variant (which I don't know the answer to) - suppose there are n wires with a diode towards one end of the cable, and m oriented the other way (m, n both > 0). There are two variants here: in one, we can tell which wires have the diodes oriented in which direction; in the other, we can't. Now how many trips are required? I suspect that when n+m>3, more than 2 trips will be required. Franklin T. Adams-Watters -----Original Message----- From: Michael D Beeler <mbeeler@csc.com> ... Proposal: Would this variation require the triangular numbers approach, and not the pairs-and-pairs approach?: The n wires each have a diode in them. All diodes point in the same direction, say cathode toward end A of the cable and anode toward end B. This precludes current through loops of paired wires. To make the problem solvable, suppose there is a conductive shield on the cable, which acts as one additional wire with no diode. ...