Considering a general fake coin problem. There are $n$ coins in total and $k$ of them are fake. Fake coins are lighter than the normal ones. You only have a balance to compare two groups of coins (no amount limit). How many times you will compare at least?

Note: All the fake coins have the same weight $w_f$; all the good coins have the same weight $w_g$; $w_f < w_g$; The target is to find all the fake coins using minimum time complexity (expected $O(\log n)$).

  • $\begingroup$ Is the value of $k$ known beforehand ? $\endgroup$
    – Evargalo
    Oct 5, 2022 at 8:43
  • 1
    $\begingroup$ @Evargalo Yes, the $k$ is a known constant here. We can suppose $k <= \frac{n}{2}$ $\endgroup$
    – picker
    Oct 5, 2022 at 15:39

1 Answer 1


For $n$ coins of which $k$ are fake, there are $\binom{n}{k}$ potential assignments. Each weighing has three possible results (left heavier, right heavier, equal). So at least $\left\lceil \log_3\binom{n}{k }\right\rceil$ weighings are required to distinguish between the possibilities.

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    $\begingroup$ I think the number of weighing rounds is correct theoretically, but can you please provide some details of how to weigh them in each round pratically? $\endgroup$
    – picker
    Oct 4, 2022 at 14:56
  • $\begingroup$ This is a theoretical minimum, I don't think it can always be achieved. $\endgroup$
    – Florian F
    Oct 5, 2022 at 13:27
  • $\begingroup$ @FlorianF hm, interesting, I first read this answer as "the minimum number of weighings is at least this much". But I can see how it can also mean "the minimum is this much". $\endgroup$
    – justhalf
    Oct 6, 2022 at 2:03

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