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. ...