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There are $n=4$ coins and we know exactly $2$ of them are counterfeit. In $2$ weighings, how can we find which of the four coins are the counterfeit ones?

Assumptions: the counterfeit coins look exactly the same but weigh less. Both counterfeit coins are of equal weight.

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  • $\begingroup$ I assume that the scale can weigh only two masses at a time (at least that's what you intended), but I have a solution with a 4-mass balance/scale. $\endgroup$ – Brandon_J Feb 12 at 18:09
  • $\begingroup$ By weighings do you mean comparing the masses of two objects, or weighing as in placing some objects and obtaining as a result their total mass? $\endgroup$ – Daniel Duque Feb 14 at 1:40
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Step 1:

Weigh two coins (we'll call them C1 and C2, the other two coins are denoted C3 and C4).

Step 2:

  • If they weigh the same, take C1 and weigh it with another coin (C3). If C1 weighs more than C3, C3 and C4 are counter-fit, otherwise, C1 and C2 are counter-fit.

  • If they have different weights, take the one that weighs less (we'll call it C# where C# is whichever coin, C1 or C2, weighs less) and weigh it against another coin, say C3. If C3 weighs less than C#, then C3 and C4 are counter-fit. If C3 weighs more than C#, C# and C4 are counter-fit. If they weigh the same, C# and C3 are counter-fit.

Flow chart (CFC mean Counter-fit coin):

enter image description here

Note:

I believe this works if the two real coins have the same or different weights, just assumes that the two counter-fits coins weigh less than both the real coins.

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  • $\begingroup$ I slightly modified my answer which I think will give you a better understanding. Let me know if you still have questions. $\endgroup$ – eye_am_groot Feb 12 at 18:23
  • $\begingroup$ Solution makes a lot of sense! $\endgroup$ – user57107 Feb 12 at 18:30
  • $\begingroup$ I don't quite see how all those possibilities exist in the tree. E.g. if C1<C2, then C1 is fake and C2 is real. So the solution can't be C3 and C4 are fake. $\endgroup$ – Dr Xorile Feb 12 at 21:48
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    $\begingroup$ That assumes that there are two weights, a real weight and a counter-fit weight. My solution allows for 3 weights, real weight one (RW1), real weight two (RW2), and the counter-fit weight (CFW), where RW1 > RW2 > CFW. So if C1 < C2, C2 has RW2. C1 could have RW1 with C3 and C4 having CFW. As I mentioned on another answer, assuming RW2 = RW1 is probably a safe assumption, though, technically not one that OP made (and it appears he deactivated his account, so we'll never truly know). $\endgroup$ – eye_am_groot Feb 12 at 23:31
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One weighing. 100% of the time.

More of an engineering solution than a math one but whatever. Also assuming the good coins are identical.

Take an equilateral triangle and suspend it in a bowl of water with one coin taped to each corner. Materials used to create the triangle will need to be light but three equal pieces of very thin plastic taped together will work for most coins.

You’ll be able to tell from which sides/points are more depressed which two coins are equal in mass to each other AND whether the third coin is heavier or lighter than those two.

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  • $\begingroup$ Oh yeah...duh. I thought of that, but I kept excluding stuff like that in my head because in similar problems you don't know whether or not the counterfeits are heavier. Also, I guess you could just use a three-pan balance instead of the water. $\endgroup$ – Brandon_J Feb 12 at 22:55
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At the risk of flogging a dead horse, it can always be done by:

  1. measuring $C_1$ v $C_2$, and then
  2. $C_1$ v $C_3$.

Then just look up the results on this table:

\begin{array}{lcc} & C_1-C_2 & C_1-C_3 \\ RRFF & - & / \\ FFRR & - & \backslash \\ RFRF & / & - \\ RFFR & / & / \\ FRRF & \backslash & \backslash \\ FRFR & \backslash & - \end{array}

This is a simpler process because you don't need any if... then...

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  • $\begingroup$ The table certainly helps to see the logic of the answer. $\endgroup$ – Overmind Feb 26 at 12:25
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@Greg came up with a perfectly good solution, but I've thought of an alternative method that will work within one weighing 50% of the time and two weighings 50% of the time - a tad more efficient, but not much.

  1. Get a four-object balance. Put one coin on each balance. If the balance does not balance, then you are done, and the 2 higher coins are counterfeit. If it balances,

  2. then switch two adjacent coins and try again. The two higher coins are counterfeit.

There is a small technical problem with this solution - we aren't told for sure that

the real coins are of the same type/weight. It's likely, but we aren't guaranteed this in the question.

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  • $\begingroup$ This does assume that the real coins are equal weights (which I'd think is a fair assumption, but technically not a given assumption in the question) $\endgroup$ – eye_am_groot Feb 12 at 18:17
  • $\begingroup$ Good point, will edit. @Greg $\endgroup$ – Brandon_J Feb 12 at 18:18
  • $\begingroup$ But, working with that assumption, is, as you said, the more efficient method :-) $\endgroup$ – eye_am_groot Feb 12 at 18:19
  • $\begingroup$ It works the first time 66% of the time. The odds of a coin being across from it's equal mass partner are only one in three. $\endgroup$ – Dark Matter Feb 12 at 21:54

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