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Topic: Please help me with the following question
Replies: 78   Last Post: Mar 25, 2013 1:38 AM

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Robert Hansen

Posts: 7,617
From: Florida
Registered: 6/22/09
Re: Please help me with the following question
Posted: Mar 9, 2013 8:23 AM
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On Mar 8, 2013, at 5:45 PM, Joe Niederberger <niederberger@comcast.net> wrote:

> Then you just have to figure out what 3 weighings will do that (leave the trace) for you.

That was my point. I don't see that problem as being any easier than the original problem.


> How do you think Mr. Coles came up with this doozy?

Frank Cole's Solution

We can start with identifying the sequences that can result in the numbers 1 through 12. We assume the coin is heavier, if it is lighter then all that does is switch all of the signs.

1 then 1, 0, 0 or -1, 0, 0
2 then -1, 3, 0 or 1, -3, 0
3 then 0, 3, 0 or 0, -3, 0
4 then 1, 3, 0 or -1, -3, 0
5 then -1, -3, 9 or 1, 3, -9
6 then 0, -3, 9 or 0, 3, -9
7 then 1, -3, 9 or -1, 3, -9
8 then -1, 0, 9 or 1, 0, -9
9 then 0, 0, 9 or 0, 0, -9
10 then 1, 0, 9 or -1, 0, -9
11 then -1, 3, 9 or 1, -3, -9
12 then 0, 3, 9 or 0, -3, -9

Whether the coin is heavier of lighter doesn't matter since that will only reverse all of the signs and swap the groups.

Note (from the zeros above) that in each weighing 4 coins must NOT be weighed...

3, 6, 8, 9 must not be weighed in the first round.
1, 8, 9, 10 must not be weighed in the second round.
1, 2, 3, 4 must not be weighed in the third round.

After that, you can then place the coins in the left or right (weighed groups) using the encodings. If at some point, you find that the left or right side is filled, simply use the alternate encoding. The encodings do dictate which weighings the coins must participate in but for each coin there are two possible sequences.

I came up with (the numbers in the parenthesis are not weighed)...

First Weighing: 1, 4, 7, 11 / 2, 5, 8, 10 (3, 6, 9, 12)
Second Weighing: 2, 3, 4, 12 / 5, 6, 7, 11 (1, 8, 9, 10)
Third Weighing: 5, 6, 7, 8 / 9, 10, 11, 12 (1, 2, 3, 4)

There are number of possibilities. If you sort the encodings as follows...

1, 3, 9
1, 3, 0
1, 3, -9
0, 3, 9
0, 3, 0
0, 3, -9
-1, 3, 9
-1, 3, 0
-1, 3, -9
...

Then I think you are more inclined to get Frank's particular selection, although there is still chance involved.

Bob Hansen


>
> (from: http://www.iwriteiam.nl/Ha12coins.html )
>
> Another alternative solution by Frank Cole
> Probably in the early 60's, we enjoyed a pre-determined solution which enabled mental calculation of the result; pre-determined in the sense that the 3 set-ups are in writing before any weighings, thus eliminating adjustments at weighings II and III. In addition, once the 3 set-ups are written and the 3 weighings are done, one can orally and immediately announce the number of the odd ball and whether it is light or heavy. The derivation is readily extendable to any number of weighings (>= 2) and the appropriate number (>= 3) of balls.
>
> Left Pan Right Pan
> Weighing I 4 8 10 11 1 2 5 7
> Weighing II 2 4 7 12 3 5 6 11
> Weighing III 5 6 10 12 7 8 9 11
>
> Start with a Sum = 0 in the mind. Perform the 3 weighings, observing the tilt, if any.
>
> Weighing I: Add -1 if tilt is left pan down, +1 if tilt is left pan up.
> Weighing II: Add -3 if tilt is left pan down, +3 if tilt is left pan up.
> Weighing III: Add -9 if tilt is left pan down, +9 if tilt is left pan up.




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2/24/13
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2/24/13
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3/7/13
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3/7/13
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3/7/13
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3/7/13
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3/7/13
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3/7/13
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3/8/13
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Robert Hansen
3/8/13
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3/8/13
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3/8/13
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3/8/13
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3/8/13
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3/8/13
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3/9/13
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3/8/13
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3/8/13
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3/9/13
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3/9/13
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3/8/13
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3/9/13
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3/8/13
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3/9/13
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Joe Niederberger
3/9/13
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3/10/13
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3/10/13
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3/10/13
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3/10/13
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3/10/13
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3/11/13
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3/11/13
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3/11/13
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3/11/13
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3/11/13
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3/11/13
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3/11/13
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3/11/13
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3/11/13
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3/12/13
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3/12/13
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3/13/13
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3/13/13
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3/13/13
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3/13/13
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3/13/13
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3/14/13
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3/14/13
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3/15/13
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3/14/13
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3/14/13
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3/14/13
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3/15/13
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3/25/13
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