Comments on: The Wire Identification Problem http://ashutoshmehra.net/blog/2009/09/wire-identification-problem/ Notes on Math, Computer Science & Programming Thu, 29 Jul 2010 04:20:10 +0000 http://wordpress.org/?v=2.9.2 hourly 1 By: Ashutosh http://ashutoshmehra.net/blog/2009/09/wire-identification-problem/comment-page-1/#comment-1527 Ashutosh Sat, 27 Feb 2010 14:23:21 +0000 http://ashutoshmehra.net/blog/?p=598#comment-1527 @v: Never tried ;-) @v: Never tried ;-)

]]> By: v http://ashutoshmehra.net/blog/2009/09/wire-identification-problem/comment-page-1/#comment-1526 v Mon, 25 Jan 2010 06:40:39 +0000 http://ashutoshmehra.net/blog/?p=598#comment-1526 hi, you have got great credentials.. ever tried to work for google ? hi,
you have got great credentials.. ever tried to work for google ?

]]> By: Ashutosh http://ashutoshmehra.net/blog/2009/09/wire-identification-problem/comment-page-1/#comment-1525 Ashutosh Fri, 08 Jan 2010 22:45:33 +0000 http://ashutoshmehra.net/blog/?p=598#comment-1525 Ceiling[Log[2,n]]-step method is for your modified problem description (where each optical fiber can be ON/OFF). For eight wires labeled {000,001,...,111} in binary, you'll need 3 steps: 1. Turn on the just wires 1,3,5,7 on end A. On end B, label the wires -- the ON and OFF groups will be {xx0, xx1}, each with 4 wires. 2. Next turn on just the wires 2,3,6,7. On end B, the two group classifications will be re-classified as {x00, x01, x10, x11}. 3. Next turn on just 4,5,6,7. End B will now have 8 "groups": {000, 001, ..., 111}. 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! Ceiling[Log[2,n]]-step method is for your modified problem description (where each optical fiber can be ON/OFF). For eight wires labeled {000,001,…,111} in binary, you’ll need 3 steps:
1. Turn on the just wires 1,3,5,7 on end A. On end B, label the wires — the ON and OFF groups will be {xx0, xx1}, each with 4 wires.
2. Next turn on just the wires 2,3,6,7. On end B, the two group classifications will be re-classified as {x00, x01, x10, x11}.
3. Next turn on just 4,5,6,7. End B will now have 8 “groups”: {000, 001, …, 111}.

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!

]]> By: Nick http://ashutoshmehra.net/blog/2009/09/wire-identification-problem/comment-page-1/#comment-1524 Nick Fri, 08 Jan 2010 21:41:40 +0000 http://ashutoshmehra.net/blog/?p=598#comment-1524 Ashutosh - 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... Ashutosh -

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…

]]> By: Ashutosh http://ashutoshmehra.net/blog/2009/09/wire-identification-problem/comment-page-1/#comment-1523 Ashutosh Fri, 08 Jan 2010 20:14:52 +0000 http://ashutoshmehra.net/blog/?p=598#comment-1523 That's an interesting problem, Nick! 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? That’s an interesting problem, Nick!

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?

]]> By: Nick http://ashutoshmehra.net/blog/2009/09/wire-identification-problem/comment-page-1/#comment-1522 Nick Fri, 08 Jan 2010 19:46:18 +0000 http://ashutoshmehra.net/blog/?p=598#comment-1522 Another interesting related, but different (I think) problem arises if you let the wires be optical fibers capable of representing two states (on and off or whathaveyou). 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 :). Another interesting related, but different (I think) problem arises if you let the wires be optical fibers capable of representing two states (on and off or whathaveyou).

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 :).

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