Author |
Topic: An interesting paper on coin weighting (Read 1693 times) |
|
wonderful
Full Member
Posts: 203
|
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
Junior Member
Gender:
Posts: 77
|
|
Re: An interesting paper on coin weighting
« Reply #1 on: Apr 14th, 2008, 12:43am » |
Quote Modify
|
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
|
|
IP Logged |
|
|
|
rmsgrey
Uberpuzzler
Gender:
Posts: 2873
|
|
Re: An interesting paper on coin weighting
« Reply #2 on: Apr 19th, 2008, 9:58am » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
wonderful
Full Member
Posts: 203
|
|
Re: An interesting paper on coin weighting
« Reply #3 on: Apr 19th, 2008, 5:25pm » |
Quote Modify
|
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!
|
|
IP Logged |
|
|
|
rmsgrey
Uberpuzzler
Gender:
Posts: 2873
|
|
Re: An interesting paper on coin weighting
« Reply #4 on: Apr 21st, 2008, 4:08am » |
Quote Modify
|
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?
|
|
IP Logged |
|
|
|
wonderful
Full Member
Posts: 203
|
|
Re: An interesting paper on coin weighting
« Reply #5 on: Apr 21st, 2008, 1:59pm » |
Quote Modify
|
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. Have A Great Day!
|
|
IP Logged |
|
|
|
|