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   An interesting paper on coin weighting
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wonderful
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An interesting paper on coin weighting   coin_2.pdf
« on: Apr 12th, 2008, 5:19pm »
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Attached is a nice paper on coin weighting with two defected coins. I post it here since it provides a solution to the question "Coin weighting with two defected coins" posted in this forum.
 
The paper is short (10 pg) and easy to read. I am particularly interseted in page 562 in the paper since it discusses a scheme to detect 2 defected coins among 13 coins. However, I am not sure if I understand the authors completely.
 
Specifically, the authors propose a scheme in which the first weighting includes (A, B, C) ? (D, E, F). They then say that if this weighting is in balance then the two defected coins must be in the remaining set e.g., the 7 remaining coins (G,H,I,J, L, M, N).
 
This seems questionable. For instance, if the two defected coins D and E weight w-2 and w+ 2 respectively, then (A, B, C) = (D, E, F).
 
Am I missing something here? It would be great if we can discuss a bit here.
 
Have A Great Day!
« Last Edit: Apr 12th, 2008, 5:21pm by wonderful » IP Logged
pscoe2
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Re: An interesting paper on coin weighting  
« Reply #1 on: Apr 14th, 2008, 12:43am »
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i think they meant tht the 2 questions both had either higher weight or both had lower and the point the one has higher and other lower (is quite possible) but may hv assumed not possible. and also many of such qn.s assume tht the other 2 i.e the defective coins have the same weight
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rmsgrey
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Re: An interesting paper on coin weighting  
« Reply #2 on: Apr 19th, 2008, 9:58am »
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For (ABC)=(DEF) on the first weighing to tell you that none of those 6 are defective, the two defective coins must differ from a normal coin's weight by different amounts - if they are both over or both under by the same amount, then having one in each pan would cause them to balance; if one's over and the other under by the same amount, then having them both in the same pan would cause them to balance.
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wonderful
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Re: An interesting paper on coin weighting  
« Reply #3 on: Apr 19th, 2008, 5:25pm »
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Yes, that's exactly the point I mentioned. I looked at the paper again but didn't find where the authors explicitly excluded the case in which the two defective coins weht w+ a and w - a respectively.
 
Have A Great Day!
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Re: An interesting paper on coin weighting  
« Reply #4 on: Apr 21st, 2008, 4:08am »
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on Apr 19th, 2008, 5:25pm, wonderful wrote:
Yes, that's exactly the point I mentioned. I looked at the paper again but didn't find where the authors explicitly excluded the case in which the two defective coins weht w+ a and w - a respectively.
 
Have A Great Day!

What about the case with w+a and w+a?
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wonderful
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Re: An interesting paper on coin weighting  
« Reply #5 on: Apr 21st, 2008, 1:59pm »
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That should work. My question is that if  the proof given in the paper will work in every case including the two interested coins weight w-a and w+a.
 
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