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Laying a Cable…

Posted by Karl Sharman on September 11, 2010 – 9:04 am

You are an internet technician. You have a job to do where the cabling has already been laid, so you head to the site expecting an easy job. Unfortunately, it seems this network cable runs underwater from one island to another nearby island. The second problem is that the cable consists of ten individually shielded wires, all unmarked.

Your networking kit was lost on the journey to the first island (eaten by sharks). The only equipment you have with you is an indelible marker, a lightbulb and a battery. While the company was thoughtful enough to provide a canoe, you really want to minimize trips between the islands. You’re not lazy, it’s just that you’ve met the local sealife.

In how few trips can you determine which wires at the network’s end correspond to which wires on the deserted island’s end? And of course, explain your answer!

Assume that not only are you incredibly dedicated to risk shark infested waters, but also:
• The individual wires are stripped bare, and may be connected to each other, and or the battery or light bulb.
• The light bulb will light if it completes a circuit with the battery.
• The cabling has negligible resistance.
• No-one is available, willing or able to assist you, you’re on your own.
• You may use the marker to label the wire ends however you like.

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27 Responds so far- Add one»

  1. 1. Nathan Said:

    It can be done in one trip. The solution will wait for more responses.

  2. 2. Eketahuna Said:

    One trip ? I’m impressed, and will have to think about it some more – initial thoughts give me 5 (and assumes you can connect the lightbulb direct to one terminal of the battery):

    Mark two cables (1 & 2) connect one to each side of battery
    Go over with lightbulb (trip 1)
    Find the two that create the connection (one on each side of bulb) and mark A & B
    Go back, disconnect 1 and connect a new one (3) (trip 2)
    Go over (trip 3)
    Disconnect A and find either – another that completes the circuit New one is 3, B is 2, A is 1
    - Or if none complete, then A is 2, B is 1
    Connect 1 to a new one (4), 2 to another (5), 3 to another (6)
    Go back with bulb (trip 4)
    Now connect 1 to battery and bulb to other side of battery (can you do that ?), and find the one that completes the circuit – (4)
    Repeat for 2 / 5, and 3 / 6
    Connect 1 and a new one (7), 2 and another (8), 3 and another (9)
    Go over with bulb (trip 5)
    Connect 1 to battery and bulb to other side – find the one that completes, being 7
    Repeat for 2 / 8, and 3 / 9
    Remaining one is 10

    Extra trips will be needed to disconnect wires at the other end and connect as required for the network, but thats not part of the scope

  3. 3. Ian Said:

    Haul all 10 cables on to your island. Mark both ends of each. Use canoe to take cables to other island.

    Low-tech solution needs only 1 trip.

    May need to work at night. Hence battery and light bulb

  4. 4. Morte Said:

    Probably not the right answer, but I’d pull the cable back to the first island and test and mark the wires there. Then I’d roll the wire up so it would unspool out of the canoe as I rowed my way back to the second island.

  5. 5. Eketahuna Said:

    Ah yes, the one trip ! I’d discounted that, but being a Monday morning, and this being Trick of Mind, I suspect I shouldn’t have :-)

  6. 6. Karl Sharman Said:

    Once you’ve hauled the one cable in, with 10 unmarked, shielded cores, how do you work out which end corresponds to another?

    So, to avoid all the trial and error, and dragging in a couple of hundred yards worth of cable by hand (very difficult!) and then relaying it….

    How else can you do it? There is a solution that leaves the cable in situ.

  7. 7. Rantelvena Said:

    I can see how it could be only one trip but I can’t quite get a hang of the logic circuit that should be used… I’ll need a bit of time, but I’ll get there :)

  8. 8. gadi Said:

    connect cable 1,2,3,4 – group A
    connect cable 5,6,7 – group B
    connect cable 8,9 – group C
    leave cable 10 unconnected. – group D

    go to the other island.
    use the battery and lightbulb to find which group each cable belong.

    connect 1 cable from each group – group A’
    connect 1 cable from groups A,B,C – group B’
    connect 1 cable from groups A,B – group C’
    leave 1 cable from group A unconnected – group D’

    Go back to the first island (the second part of the trip?)

    Now every cable has unique ID: (A,A’),(A,B’),…

  9. 9. 123Tai_mai_shu Said:

    2 trips.

    take all 10 cables on island 1 and connect 5 to positve and 5 to the negative.

    row over to island 2 (trip 1) and using 1 cable (Z), keep it connected to the bulb. there should be 9 cables left. switch up all 9 cables until u find 5 which will match up (y) and 4 which will not (x).

    z = main cable
    y = cables that will match
    x = cables that doesnt match

    keep z and 2y attached to the light bulb. go back to island 1 (trip 2), take the cables off the battery one by one starting with negatives first until the light on island 2 has turned off. Put back the cable that made it turn off onto the battery by itself. lets mark this cable 1b

    take all the positive cables off and try matching up each of the 5 cables to 1a until u get a match. mark that positive cable 1b. continue until u have 2 positive cabls that matches with 1a. if that isn’t posible then take all cables off the battery and connect cable 1b to the postive side on the battery. connect the negative cables excluding 1a, until u find the last cabal that matches with 1b.

    in the end there should only be 2 negatives and 1 positive, OR 2 postives and 1 negative.

    By doing this you’ll find out which ones are positive and neagtives on island 2.

    if needed hop on that canoe go for back to island 2 (trip 3) and finish the job and get paid!

  10. 10. 123tai_mai_shu Said:

    or just 1 trip to island 2, because after u figure out which one match up, you have your answer to which ones correspond to each other.

  11. 11. 123tai_mai_shu Said:

    am i correct?

  12. 12. David Said:

    One trip.
    Island one, twist all the wires together.
    Go to island 2, twist all the wires together. Now you have created one wire.

    Lame, I know, but it’s all I could think of that made any sense.

  13. 13. Karl Sharman Said:

    Gadi starts off well….

  14. 14. Karl Sharman Said:

    123tai_mai_shu… nope not correct :-(
    David – yup, lame ;-)

  15. 15. Eketahuna Said:

    Connect 4 cables – Group A
    Connect another 3 – Group B
    Connect 2 of the others – Group C
    Mark last as 1

    Trip 1 (take battery and bulb)

    Determine each group via lot’s of trial and error
    (take two cables and find if they complete a curcuit, one wont, that’s Cable 1 – repeat with a pair and a single etc, do I need to expand ?)
    Mark them all with group

    Take 2 from group A and add one from group B (mark as 4) to make new group D
    Take 1 from group A (mark it 7) and add to cable 10 – make group E
    Join another from group B (mark as 5) to one from group C (mark as 6) – new group F
    Mark remaining cable from B as 2, and remaining from C as 3

    Trip 2 – take battery and bulb

    As before, determine new grouping – one from old group B wont work, neither will one from group C – label B as 2 and C as 3. Also, one from group C is now in group F, mark as 6, and one from group B is now in group F too, so mark that as 5.
    Another from group B is now with a group of 3 (with 2 A’s), so mark that as 4.
    One from group A is now with cable 10, mark that as 7

    Now just split the 2 unknowns from group A and join one with cable 10 – mark that 8 and the other 9

    Trip 3 – take bulb and battery

    Find the unnumbered cable from group A that works with cable 10 and mark that 8 so the other is 9.

    Hope I didn’t type any of that wrong !

    Down to 3 trips so far…

  16. 16. Paul H Said:

    I think I might have 2 solutions. I’ll post the simplier one first, which is probably the right answer…

    Starting on Island 1:

    We’re given that each wire is individually shielded. Before leaving, we take the 10 wires and lay them out in order numbering them from 1 to 10.

    On wire 1, we twist the conductor to the shielding of wire 2. Wire 2 gets twisted to the shielding of wire 3 and so on. Wire 10 is all by itself.

    Go to Island 2:

    We can use the battery and bulb to figure out which shielding is connected to which conductor. After locating the order of all 10, we know the order. That’s because wire #1 tests against the shielding of #2, whereas wire #10 isn’t attached to anything.

    Done.

    ———————————-
    Solution 2 (much harder and requires going back to island 1)

    Starting on Island 1:
    The second solution has to do with starting with 5 pairs on the first island and label each group of pairs for later.

    Go to Island 2:
    On the 2nd island we determine the pairs, but we don’t care about the order on island 1. Each pair is wire 1&2, 3&4, 5&6, 7&8, 9&10.

    Now untwist and shift everything over and re-twist. Leave pin 1 open. Twist 2&3, 4&5, 6&7, 8&9, 10 to the shield.

    Back to Island 1:

    You now have all the wires in series. So test for which is pin 1 (nothing lights) and pin 10 to its shield. That sets the order. To determine which pair is which, connect the battery and bulb to pin 10 and open and close the remaining pairs in combinations and shorting some pairs together to figure out which pairs are required for the other pairs to complete the circuit.

    -Paul

  17. 17. Paul H Said:

    Funny, I think I just wacked my earlier post as a non-registered when I just now registered. Doh!

    Ok..

    We start on Island 1 and lay out the 10 wires, numbering them from 1 to 10.

    Each wire is individually shielded. We twist wire 1 to the shielding of wire 2. Wire 2 is twisted to the shield of wire 3. Wire 3 is twisted to the shield of wire 4. Repeat for all wires.

    Wire 10 is left alone.

    Go to Island 2.

    Now we use the battery and bulb on each wire and another wire’s shielding to light the bulb. When we find one that lights, move on to the wire whose shield we just used – or to the shield of whose wire we just used.

    We’ll end up with all 10 wires in order with wire #10 being the last one that doesn’t light the bulb when tested against any other shielding.

    Now label 1-10.

  18. 18. Chris Said:

    Label the wires 1 to 10. Join core and shield of wire 1 together and also to the core of wire 2. Join the shield of wire 2 to the core of wire 3, etc. until end with shield 9 joined to core 10. Shield 10 is left disconnected.

    Go to the island. Determine which wire has it’s core coonnected to its shield – that’s wire 1. Now find the other core that’s connected to wire 1 – that’s wire 2. Now find the core that connected to shield 2 – that wire 3. Keep on going until you have identified wire 10’s core.

    That’s one trip to the island. You’re done.

  19. 19. Chris Said:

    … or just connect shield 1 to core 2, shield 2 to core 3 … ahield 9 to core 10. Nip over to the island. Wire 1 is the only one whose core is cpen circuit, blah blah blah.

    It also happens that wire 10 is the only one whose shield is open circuit.

  20. 20. Karl Sharman Said:

    I’m off on holiday for a few days, so I’ll leave you with this answer to argue over…..:-)

    The minimum number of round trips is one – 2 journeys in shark-infested waters – one there and one back. More surprisingly, one round trip could be used to label any number of wires.
    Here’s how:
    Label the wires 0,1,2,3…9.
    Connect the wires into the following groups: {0} {1,2} {3,4,5} {6,7,8,9}. So wire 0 is left alone, and wires 6,7,8,9 are all connected.
    Paddle to the island. Using the light bulb and battery as a circuit detector, begin testing the wires to see which ones can pair into a complete circuit. One of the wires won’t complete a circuit at all; this is wire 0. Label it “A”.
    Two of the wires will complete a circuit with each other, and no other wires; these are {1,2}. Note that you don’t know which is which, but label them “B” and “C”.
    Three wires will complete circuits with two other wires. These are {3,4,5}; label them “D”, “E”, and “F”.
    The last four wires will each complete a circuit with the remaining others. Label them “G”, “H”, “I”, and “J”.
    Now group the wires as follows: {A,C,F,J} {B,E,I} {D,H} {G}. Note that none of the wires are in the same groups that they are at the other end.
    Paddle back, and disconnect the original wires. Using the same method as before, test the wires to determine groupings: For example, if the wire you numbered “8″ now completes a circuit with two other wires, it must be in group {B,E,I}. The only one of that set that was originally connected to the {6,7,8,9} group is I. So 8 connects to I. If, on the other hand, “8″ completes a circuit with just one other wire it must be connected to “D” or “H”. But we know “8″ originally was connected to “G”, “H”, “I”, and “J”. So it must be “H”.
    This system not only allows you to test all ten wires in a single round trip, it works with an unlimited number of wires. Connect the wires in to groups of 1,2,3,4,5… etc. If you have left over wires (for example if you started with 17, then there would be 2 left over wires) group leftovers together. This means you’ll have two groups of the same size. But you can discriminate these two groups by leaving one terminal of the battery connected to one of them when you head for the island, and the other terminal connected to wire “0″ (the single solo wire).
    You think I need the battery for the tests? Nope. Once you’ve figured out which is wire “0″ and which is the same-size group, you can use wire 0 and one of the strays to substitute as your battery.

  21. 21. Chris Said:

    My answer is far better than yours. One crossing only (half a round trip) and far easier to understand.

  22. 22. Paul H Said:

    We said the same thing Chris. Half trip.

    Starting with the wire to the next wire’s shield (mine), or starting with the shield to the next wire’s wire (yours).

    But I’m curious if we’re right.

  23. 23. Chris Said:

    Hi Paul. I managed to miss yours when I posted. Of course we’re right :) However, if we didn’t have the shields, the Karl’s solution looks likely to be the optimum one.

  24. 24. Paul H Said:

    Hi Chris,

    I think I have a second way, without using the shielding at all, but it requires a trip back.

    Island 1:

    You twist the wires into 5 pairs and label each pair of wires. Something like..

    AA BB CC DD EE

    Go to Island 2:

    Use the battery and bulb to find the 5 pairs. We won’t know which pair is which, but that doesn’t matter. Let’s say we call the pairs…

    GG HH II JJ KK

    Twist the wires so the groups are now…

    G GH HI IJ JK K

    Number each wire.

    G GH HI IJ JK K
    1 23 45 67 89 10

    Twist wire 1 to wire 4&5. This will help us to identify wire 1 and 4&5 later.

    Back to Island 1:

    Now we previously marked the pairs of wires from A to E. Untwist these wires and locate the new pairs that were twisted on island 2. It might look like this…

    ACD CA BE BD BE E(open)

    You now know that the last E is wire 10, Wire A is the 1st wire. The other old pair names lie next to each other, forming wires 1-10

    A AC CD DB BE E
    1 23 45 67 89 10

    3 half trips, no shielding harmed in the making of this solution :)

    Paul

  25. 25. Karl Sharman Said:

    There you have it – those who spotted that the wire was shielded got the right answer – Chris & Paul H.
    The answer I gave prior to departing for a few days off was, as Chris says, the optimal solution for insulated cables.

  26. 26. Euclid's Brother Said:

    An internet engineer would know that it only takes 4 wires to create a ethernet link (in an 8 pin RJ45 plug only pins 1,2,3 & 6 are used.

    (Part of this is going to look like Eketahuna’s post.)

    So starting on island 1: Connect 4 wires together and label A. Connect 3 more wires together and label B. Conect 2 more wires together and label C. Lable the last wire D.

    “Sail away, sail away, sail away…” to the other island.

    Use battery and lightbulb to identify the groups of 4, 3, 2 and 1. (for more explination on the testing methods, see below). Once idetified, connect them together and label them A, B, C, D respectivly.

    You know have a 4 conductor cable with labeled ends to use to make your RJ45 ethernet connection between the islands.

    ——-
    Testing Methods.

    Connect battery and lightbulb to an unidentified wire. Test each additional unidentified wire. Use table to below to identify the group.

    No wires cause the light to come on = Group D
    only 1 other wire causes the light to come on = Group C
    2 other wires cause the light to come on = Group B
    3 other wires cause the light to come on = Group A

  27. 27. Turtle Said:

    i come up with 2 solutions . the first is more likely possible ina real world enviroment .

    1.) 3 trips. assuming that the light can be seen from one island to the other . first trip . label wires A-J. connect A to negative terminal and B & C to posative terminal .go to other island . Using the light you an determine which cable is “A” and which two are “B” & “C” . Connect one of the two wires that are possible “B” or “C” to one side of the light and label 2 wires 1 & 2 and connect to the other .On tghe return trip you can determind weither it is “B” or “C” connected. You have now identified “A” “B” & “C” definitively and also Identified 2 new wires (1 & 2) . Attach the cooresponding wire for 1 to the negative side and 2 new unused wires to the posative side . using this method you can identify 3 wires per trip . the 10th wire will be the one unidentified after 3 trips .

    2) cut all the wires to the same length leavign plenty of spare cable. go over to the other island and do the same . one at a time pull the wires each out 1 inch further than the last . label them a-j from shortest to longest . go back to the otehr island and mark them A-j from longest to shjortest

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