Island Networks

[araya]

Label the wires on whichever island you start on 0-9. Connect 1 to 2. Twist 3,4 and 5 all together. Twist 6,7,8 and 9 all together.

Paddle to the other island. Run to the cables, avoiding contact with the locals. Checking different combinations of wires, find out which one doesn't complete a circuit with any other wire. This one is wire 0. (checking for complete circuit is easy, just touch one wire to the lightbulb which is connected to the battery, and touch the other wire to the other battery terminal). Then find 4 wires which complete a circuit no matter which 2 you touch together. These are 6 through 9 (you don't know which is which). Then find 3 wires, any 2 of which complete a circuit. These are 3 through 5. The other two are 1 and 2, and they should only complete a circuit with each other. Label them, say, A is the single, B and C are doubles. D,E and F are the triple and G,H,I and J are the quad. Paddle back.

Now disconnect them all. Twist 0,1,3 and 6 together. Twist 2,4 and 7 together. Twist 5 and 8 together. Leave 9 unconnected.

Paddle back to the island. Find 4 wires, any 2 of which complete a circuit. One of these will be B or C, which you know is 1. One of them will be D, E or F, which you know is 3. One of them will be G,H,I or J, which you know is 6. The last is A, which you knew is 0 already.

The rest follows..find the 3-pack. Whichever one is B or C is 2. Whichever one is D,E or F is 4. The last is G,H,I or J and is 7.

Of the double, whichever is D,E or F is 5. The other is G,H,I or J and is 8.

The single, unconnected wire is 9. You're done.

[Ghost Post]

Conjure up a demon and make him go one way in 30 seconds, and back in 15, all the while noting down 1 connection. This way you don't have to travel once.

[Ghost Post]

I think you can do it in one round trip.

Label the wires 0 to 9. Connect one terminal of the battery to wire 0. Connect the other terminal to wires 1 and 2. Connect wires 3, 4, and 5. Connect 6 and 7. Leave 8 and 9 unconnected.

Paddle across.

Test the lightbulb across every pair of wires. Find that two pairings cause the bulb to light, and both pairings have a particular wire in common. Label the common wire A. Label the other two wires B and C. At this point we know that 0<->A and {1,2}<->{B,C}.

Next create a “continuity tester” by connecting one terminal of the bulb to A. We will test continuity by connecting to the other terminal of the bulb and to wire B. Examine the remaining wires. Look for a set of three which are all connected to each other. Label these D, E, F in any order. Then look for a set of two which are connected to each other but to no others. Label these G, H in any order. Label the last two I, J in any order.

Now we know that 0<->A, {1,2}<->{B,C}, {3,4,5}<->{D,E,F}, {6,7}<->{G,H}, {8,9}<->{I,J}.

Connect C to D. Connect F to G. Connect H to I.

Paddle back.

Now we’ll start relabeling the numbered wires with letters. Relabel 0 as A.

Disconnect the battery and all the wires from each other. Then use the battery+lightbulb as a “continuity tester”. Test the six combinations of connecting 1 or 2 with 3 or 4 or 5. Only one of these will cause the bulb to light. Whichever of 1 or 2 caused the bulb to light, label it C and label the other one B. Whichever of 3, 4, or 5 caused the bulb to light, label it D.

Likewise test the four combinations of 3 or 4 or 5 (excluding the one relabeled D) with 6 or 7. Only one of these will cause the bulb to light. Whichever of 3, 4, or 5 caused the bulb to light, label it F. The other of 3, 4, or 5 (not already labeled D or F), label it E. Whichever of 6 or 7 caused the bulb to light, label it G. Label the other one H.

Finally, determine whether H is connected to 8 or 9. The one that is connected, label I. The one that is not connected, label J.

Now all the wires are correctly labelled A through J at both ends.

[Tom]

I'm not adding anything new here, I don't think, but it seems you can explain it a bit more clearly.

So - you can do it in one round trip.

On island 1, label the wires 0-9, and group into sets of 1,2,3 and 4. So we have {0}, {1,2}, {3,4,5} and {6,7,8,9}.

Row over to the other island, test the wires to see how they are grouped together (by araya's method). Label them {A}, {B,C}, {D,E,F} and {G,H,I,J} corresponding to the 4 sets. Now, attatch them together in sets of 1,2,3,4 as follows:
{G}
{H,D}
{I,E,B}
{J,F,C,A}
If you read downwards, you can see no wires from the original sets are now in the same set. Row back to the first island, unlink the wires, and test to see which groups they are in. Relabel the wires A-J accordingly, and you are done.

Full credit has to go to araya though. JRR's solution may well work, but this is simpler.

[This message has been edited by Tom (edited 05-31-2000).]

[HyToFry]

Man, i gotta get to work earlier.

One trip is my calculation to

{0} {1,2} {3,4,5} {6,7,8,9}

get to the other side

lable them

conect {0,9} {1,3} {2,6} {4,7} {5} and {8}

You guys post to early

[HyToFry]

Man, my solution is different, maybe i should explain it?

[HyToFry]

Okay here goes my explanation.

group the wires in sets of 1,2,3, and 4.

go to the other side, lable them 0-9 like this

{0} {1,2} {3,4,5} {6,7,8,9}

now group them like this
{0,9} {1,3} {2,6} {4,7} {5} and {8}

Now back to the island,
the wire that was by itself is 0, it will have continuity with one other wire from group 4, that wire is nine

one wire in group 2 will have continuity with one wire from group 3, they are 1 and 3 respectively.

one wire in group 2 will have continuity with one wire from group 4, they are 2 and 6 respectively.

one wire in group 3 will have continuity with one wire from group 4, they are 4 and 7 respectively.

one wire from group 3 will not have continuity with any wires, thats 5.

and finally one wire from group 4 will not have continuity with any wires, thats 8.

there is two answers to this apparently.

[ZenBeam]

[Ghost Post]

ZenBeam, I looked at your comment and araya/HyToFry/Tom's solutions and now I think the problem can be solved with a single round trip for an arbitrarily large number of wires (or at least an arbitrary TRIANGULAR number such as 21=1+2+3+4+5+6).

Same sort of trick as before. At the starting point, connect up the wires in bunches of 6, 5, 4, 3, 2, and 1. Hook the battery up to any two bunches.

Paddle across.

Use the light to figure out the corresponding bunches of 6, 5, 4, 3, 2, and 1 on the other side.

Label the six wires in the bunch of 6 as follows: 6.1, 6.2, 6.3, 6.4, 6.5, 6.6
Label the bunch of 5 as follows: 5.2, 5.3, 5.4, 5.5, 5.6
Label the bunch of 4: 4.3, 4.4, 4.5, 4.6;
the bunch of 3: 3.4, 3.5, 3.6;
the bunch of 2: 2.5, 2.6;
the singleton: 1.6

Then reconnect the wires:

Leave 6.6 unconnected
Connect 6.5 to 5.6
Connect 6.4 to 4.6
Connect 6.3 to 3.6
Connect 6.2 to 2.6
Connect 6.1 to 1.6
Leave 5.5 unconnected
Connect 5.4 to 4.5
etc.

Paddle back to the starting side.

On the starting side, try all pairs of wires, with one drawn from the original bunch of 6 and the second drawn from the original bunch of 5. One of these pairs will connect. Label these wires 6.5 and 5.6

Repeat for every pair of bunches, and you've got it.

[This message has been edited by JRR (edited 05-31-2000).]

[This message has been edited by JRR (edited 05-31-2000).]

[ZenBeam]

JRR, I had come to the same thought, that with one exception*, any number of lines could be correctly labeled in two trips.

I think you can start with your triangular solution, then add unconnected lnes in the first step, which you can connect to groups in the second step. You might have to have the battery connected to your singleton line.

* What's the exception?

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It is so clear, and so it is hard to see.

[Ghost Post]

ZenBeam, I’m pretty sure you can do it in a single round trip for an arbitrary number of wires. I’d use a slightly different trick to handle the leftover wires…you can distinguish the two groups of equal size when you're on the far side if you've left one group connected to the battery and the other not connected. I’ll try to think about your exception…meanwhile, here’s the modified technique I’d propose.

I’ll illustrate with the example of 18 wires, but it should work for any number.

Pick a wire, label it 1.X. Pick two more wires, label each of them 2.X. Pick three more wires, label each of them 3.X. Continue until there are only three wires left; label them 3A.X. Now the wires are labeled as follows:
1.X,
2.X, 2.X,
3.X, 3.X, 3.X,
4.X, 4.X, 4.X, 4.X,
5.X, 5.X, 5.X, 5.X, 5.X
3A.X, 3A.X, 3A.X

Connect all the 2.X’s to each other, all the 3.X’s to each other, etc. Connect one terminal of the battery to the 3A.X’s; connect the other terminal to 1.X. (In the general case, the trick is to connect one terminal to the “leftover” bunch and the other terminal to a bunch of a different size).

Paddle across.

On the far end, connect the lamp across each of the (18!/2) pairs of wires. The lamp will light in three cases; each case will have one wire in common (the opposite end of wire 1.X). Label the common wire 1.5. Label the other three wires 3A.3, 3A.4, and 3A.5.

Now create a continuity tester by connecting one lead of the lamp to 1.5. The continuity tester consists of the other lead of the lamp plus the wire 3A.1. Use the continuity tester to test every pair of wires, aside from 1.X, 3A.3, 3A.4, and 3A.5. Discover a unique group of two wires which are connected to each other and no others; label these 2.4 and 2.5 in no particular order. Likewise, discover a unique group of three wires which are connected to each other and no others; label these 3.3, 3.4, and 3.5 in no particular order. Likewise for a group of 4 and a group of 5.

Now the wires are labeled as follows:
1.5,
2.5, 2.4,
3.5, 3.4, 3.3,
4.5, 4.4, 4.3, 4.2,
5.5, 5.4, 5.3, 5.2, 5.1,
3A.5, 3A.4, 3A.3

Reconnect the wires as follows: 1.5 to 5.1, 2.5 to 5.2, 2.4 to 4.2, etc. Connect 3A.5 to 3.5 (it will also be connected then to 5.3). Connect 3A.4 to 3.4 (and 4.3). Leave 5.5, 4.4, 3.3, and 3A.3 disconnected.

Paddle back.

Disconnect the battery and the wires. Use the battery and the lamp as a continuity tester.

Relabel 1.X as 1.5. Use the continuity tester to find out which of the 5.X’s it is connected to. Relabel this one as 5.1.

Use the continuity tester to identify the unique 2.X which is connected to a 5.X. Relabel these as 2.5 and 5.2.

Continue for every wire which hasn’t yet been relabeled, saving the 3A’s for last. Then use the continuity tester to determine which 3A.X is connected to 3.5, and relabel this one 3A.5. Repeat for the remaining 3A’s.

Done.

[HyToFry]

JRR, one word man.. NICE!!!!!

I like

[HyToFry]

So basically, after all of our ideas come together, you can determan which wires are wich in one trip for any number of wires less than infinity.

I wonder if this is what minotaur was looking for?

[Griffin]

ZenBeam - 2!

[ZenBeam]

Griffin: Yep!

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It is so clear, and so it is hard to see.

[HyToFry]

I don't get it, griffen says zenbeam 2 and i thought he meant also, but then zenbeam says yep, did i miss something.

Oh Btw, I think everyone did great, JRR just did a wonderfull job of summarizing it.

[ZenBeam]

HyToFry: Yep!

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It is so clear, and so it is hard to see.

[HyToFry]

LOL! <-- sorry i know thats rather lame, but it felt appropriat

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I'm just working here until a good fast-food jop opens up.

[HyToFry]

There's got to be a way, come on ppl.

[HyToFry]

Ya know ZenBeam, when i first read that, i didn't know you were asking, i thought maybe JRR had said "with one exception" and then deleted it out.. LOL ...I didn't know you were asking

[daniel801]

why don't you just canoe out and hook all of the wires up to the lightbulb, but have none of the wires touching each other. then paddle back and try each wire, noting which wire is which according to when the bulb lights up.
be sure to do this at night so you can see it.

[HyToFry]

This wouldn't work Daniel, in order to hook up more than one wire to the light bulb, you would have to in escence touch the two wires together, its not possible to conect more than one set of wires without them haveing the same effect as if they were connected to each other.

I've been doing some thinking about the two wires, and if you have two batteries, you can discover which is which in one trip across the water, but no less of coarse you would have to travel back across to disconect the first battery

[Ghost Post]

Why don't you just pick up the one end of wires, canoe them over to the other island, and figure it all out there. This way you're done in one trip. Of course the network might want you to replace the wire afterwards, so it still ends up being two trips.

[HyToFry]

HaHa great answer.
It did say the wires have "negligable" resistance

[ZenBeam]

I can do it in one crossing:

Using you wire, some coconuts, crushed clamshells and some palm fronds, construct a crude set of headphones. Hook up the multiple output signal generator, which washed up on the lagoon, to nine of the wires, conecting the ground to the tenth. Set the output frequencies to nine increasing notes taken from the Gilligan's Island theme song. Paddle to the island, and label the wires in order of increasing frequency. Reward yourself for a job well-done with a big slice of coconut cream pie. Ask Mary Ann what she's doing later that evening.

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It is so clear, and so it is hard to see.

[Borodog]

Yeah, but you still can't fix the hole in the Minnow.



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Insert humorous sig here.

[Alfie]

What if you labled the first side in order:
0-1-2-3-4-5-6-7-8-9
Then, you used some resistant material, palm fronds, coconut juice, human flesh, whatever, to connect them in order 1-9. You then connect 0 to the negative end of the battery and connect 1 to the positive end. When you get to the other side, finding zero is easy. Then just leave 0 on the bulb and test the brightness of the bulb on each of the other wires. The brightest wire will be 1 and the dimmest 9.

[IrishJoe]

Trying to create a semiconductor of any consistancy in resistance would be ludicrous, but if you add the time factor you can dosomething similar.

Strip the wires in incrementally shorter lengths, say 1/8-inch difference in lengths from one wire to the next, leaving the outer cable insulation intact

Then prop the cable up so the open end of the cable insulation is up. Fill the cable insulation with water, connecting all the wires electrically.

Paddle across once, and start testing the connections with the battery and light bulb while the water is evaporating. The first wire to lose continuity is the longest, the second is second-longest, etc.

One trip.

[Ghost Post]

I'm with Mathieu

Canoe over with one end of the cable. Figure it all out and canoe back. One round trip.

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I could be wrong.

[Bicho the Inhaler]

You don't even need the light bulb. By attaching any 2 wires to opposite terminals of the battery, you can tell if the wires are connected on the other side just by touching one of them with your feet on the ground. If the battery is strong enough, and the wires are connected, you'll get shocked. If you're still alive by the end, you can return the light bulb brand new, get a full refund, save a fortune and retire early.

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[John Pye]

Here is my way to do it in one trip:

First, connect one lead of the light bulb directly to one end of the battery. The other ends of the light bulb and battery form a short circuit sensor. No big deal.

Prior to going out to the island, separate the 10 leads of the cable into 5 pairs. For 4 of the pairs connect each lead to its partner, leaving the fifth pair unconnected.

Paddle out to the icky island with the sensor.

Using the sensor it is simple to match 4 pairs of connected leads.
Label these A1, A2, B1, B2, C1, C2, D1, D2. Label the open pair E1, E2.
Now connect the following leads:
A2-B1, B2-C1, C2-D1, D2-E1

Paddle back from the island (with sensor).

Back on the main land, first note which leads were paired with which, and then disconnect them.
Take the pair that hadn’t been connected and label it E.
Now one of the E leads will be floating and one will form a short circuit with one of the other leads.
Label the open lead E2 and the connected one E1.
Label the lead E1 is connected to D2.
Get D2’s original partner and label it D1.
Repeating this process, you can then get C2, C1, B2, B1, A2, and finally A1.

Of course, one more trip out to the island may be required to disconnect the leads and hook up the system.

[John Pye]

Hmmm... it just occurred to me that this method could be extended to any number of even numbers... cool.

I haven't thought about the case of odd numbers.

[Ghost Post]

John Pye's method sounds good. I was trying to figure a way to do it with pairs, but couldn't figure out how to avoid the ambiguity of which wire was which in a pair.

Bicho's idea of not using the light bulb can be done less painfully, although it will be rather tedious. When short circuited with a low resistance wire a battery will start to heat up. Hook up the battery to a pair of wires and wait five or ten minutes to see if it gets warm.




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"I hate relativism more than I hate anything else, excepting, maybe, fiberglass powerboats."
Jerry Fodor

[Ghost Post]

Bravo, John Pye!

Your technique indeed works just as well for an odd number of wires. Just leave a single wire unconnected before you paddle across.

For example, with 11 wires, connect up 5 pairs, then paddle across, then use the continuity tester to label them A1, A2, B1, B2, C1, C2, D1, D2, E1, E2, and F.

Reconnect F to E2, E1 to D2, D1 to C2, etc.

Paddle back. You already know which one is F (it's the only wire that didn't get paired up originally). Then you can identify the rest using your technique.

[John Pye]

Thanks.

Now I am wondering if it is solvable if there is not ten, but rather, TWO wires. I can’t see a good way to tell which is which.

I am assuming the light bulb is not a diode.

[IrishJoe]

Hrmf!
I've been away for a while, but ever since the day I posted what i thought to be a good, if mechanical, solution, I was driving home and realized that a trip back would be required for mine, as it still didn't allow me to differentiate between the last two wires.
That part is easy with a return trip, however - just link up one of the last wires to to the first wire to lose continuity, and find out which it is when you get back.
The thing that gets me is that nobody else caught that in the intervening time.

OK, so having read posts since by John Pye and JRR, I have to conclude that there are at least three ways of doing it in a single round trip.

It will be interesting to see which is the 'official' solution, but I suspect it won't be mine.

Kudos to JRR, HyToFry, ZenBeam, John Pye, et.al.

Determining the difference between two wires would only be possible, I think, if the light bulb had polarity. Since nobody mentioned a diode, I think we can safely assume that it's not.

[Random Idiot]

After all your construtive discussion, I still don't know what do you mean by connecting. Is it like there are 10 socket on both sides and 10 wires that are unconnected, or is it all connected on one side but unconnected to the other (like ten electrical appliances to ten sockets)?

[IrishJoe]

Per the puzzle, it can be assumed that "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."

Connected, in this case, seems to mean attaching in such a way as to establish electrical continuity for a period sufficient for you to complete a (circuit or circuits)in order to determine which wires are which.

Hope that helps

[HyToFry]

HAHAHA Minotaur is slipping.. J/K

"The minimum number of round trips is one. More surprisingly, one round trip could be used to label any number of wires."

Almost any number of wires .

Until one of us (and i know we will) discovers a way to tell two wires apart, that statement will be false.

[Ghost Post]

Well, as long as we're on an island...

Dip one wire end in the ocean. Go to the other island, hook up one terminal of the battery to the light bulb and put the other end in the ocean. See which wire completes the circuit, with the salt water of the ocean providing the second path between the islands. We are able to detect miniscule amounts of current with this set up, aren't we?

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"I hate relativism more than I hate anything else, excepting, maybe, fiberglass powerboats."
Jerry Fodor