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Old 2009-09-11, 03:04   #1
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Default Counterfeit Coin2

You have 120 coins, one of which differs in weight from the other 119. Yo have the typical two pan balance for identifying the odd weight coin in just five (5) weighings.
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Old 2009-09-11, 12:09   #2
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This is obviously possible since 120 < (3⁵-1)/2. As an added bonus, you can even say whether the counterfeit coin weighs more or less than the genuine ones. I'll leave the actual methodology as an exercise to the interested readers who never designed a Hamming code before.
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Old 2009-09-11, 19:15   #3
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Trade the scale at the flea market for a package of athletic socks, duct tape and a chair.

Find the counterfeiter and duct tape him to the chair.

Fill a sock with the 120 coins and beat the counterfeiter with it until he tells you which one is counterfeit.

0 weighings.
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Old 2009-09-11, 21:00   #4
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Quote:
Originally Posted by Orgasmic Troll View Post
Trade the scale at the flea market for a package of athletic socks, duct tape and a chair.
.......
0 weighings.
Reminds one of this thread:
http://www.mersenneforum.org/showthr...160#post180160
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Old 2009-10-19, 06:09   #5
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Quote:
Originally Posted by Orgasmic Troll View Post
Fill a sock with the 120 coins and beat the counterfeiter with it until he tells you which one is counterfeit.
how could he tell ?
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Old 2009-10-19, 06:55   #6
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Quote:
Originally Posted by jvwert View Post
You have 120 coins, one of which differs in weight from the other 119. Yo have the typical two pan balance for identifying the odd weight coin in just five (5) weighings.
This puzzle becomes slightly more interesting (but still possible) when we are not told the results of the five weighings until after the fifth weighing is completed.

i.e.

Weigh:
A vs B
C vs D
E vs F
G vs H
I vs J

then ...

Results given:
Left down, even, left, right, even.

then ...

Answer is:
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Old 2009-10-19, 13:35   #7
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Quote:
Originally Posted by m_f_h View Post
how could he tell ?
Any decent counterfeiter would be able to distinguish their product.

If they aren't a decent counterfeiter, look for the coin that has a shoddy design.
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Old 2009-10-23, 02:17   #8
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Quote:
Originally Posted by Orgasmic Troll View Post
Trade the scale at the flea market for a package of athletic socks, duct tape and a chair.

Find the counterfeiter and duct tape him to the chair.

Fill a sock with the 120 coins and beat the counterfeiter with it until he tells you which one is counterfeit.

0 weighings.


I remember a scale used in Chemistry class in high-school. It was in a wooden and glass case. It was all mechanical and very sensitive. I remember the instructor asking one of the guys to write his name on a little slip of paper, but leave the t's uncrossed in his name and weigh it. Then cross the t's and weigh it again. There was a difference. I don't know what type of scale it was. It had to be protected from dust, moisture, and so on.
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Old 2009-10-23, 05:40   #9
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Quote:
Originally Posted by storm5510 View Post
I remember a scale used in Chemistry class in high-school. It was in a wooden and glass case.
It was likely a balance. I used to use 1/10th milligram balances on a daily basis. We use balance tables that weighed ~300lbs and had rubber isolators between the deck and the legs. It was made of solid stone (marble is memory serves.)

You find scales in a bathroom and at the checkout counter, but balances in the lab.
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Old 2009-10-23, 09:21   #10
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Counterfeit Coin

You are given 120 coins, one of which differs in weight (and in weight only) from the other 119. You are given a pan balance to identify the odd coin in just 5 trials.

To distinguish from among 120 equally likely objects requires log2 120 (= 6.907 bits) of information.

You will receive log2 3 (= 1.585 bits) of information on each trial so long as each of the 3 possible outcomes (left pan up, down, or balanced) is equally likely. Since 5 log2 3 (= 7.925) is greater than log2 120, a solution is feasible in 5 trials.

The need to specify whether the counterfeit coin is heavy or light requires one additional bit of information. Since log2 120 + 1 (= 7.907 bits) is less than 5 log2 3 (= 7.925 bits), a solution in 5 trials is feasible.

Notice that this does not guarantee a solution unless the requirement for equally likely outcomes on each trial is met. With equally likely outcomes, the maximum number of coins that may be distinguished in 5 trials is 2 raised to the power 5 log2 3 (= 243).

If the counterfeit is heavy or light, the maximum number that may be distinguished will be less than 2 raised to the power 5 log2 3 - 1 (e.g. less than 121.5).

Note that it is meaningful to carry out the math and say less than 121.5 (as opposed to less than 122). The reason for this is that, for some fraction of the time, it is possible to get the answer for 122 coins.

This will happen whenever the pans do not balance in the first trial. (If they are unequal this reduces to the solution for 121 coins.) If the pans do balance in the first trial, you will be able to discover which coin is counterfeit, or whether it is light or heavy, but not both.
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Old 2009-10-24, 17:43   #11
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Quote:
Originally Posted by Uncwilly View Post
It was likely a balance. I used to use 1/10th milligram balances on a daily basis. We use balance tables that weighed ~300lbs and had rubber isolators between the deck and the legs. It was made of solid stone (marble is memory serves.)

You find scales in a bathroom and at the checkout counter, but balances in the lab.
Actually, I do believe you are correct. I don't remember a blank piece of paper being used, but perhaps there was. It was a very long time ago.

The case this device rested in was maybe five inches tall and six inches front to back, but well over a foot in length. Possibly as much as 18 inches.
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