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What's the least amount of weighings you'd need to be able to successfully determine the weights of 4 balls weighing 1, 2, 3 and 4 kilograms, using a balance scale (counting the worst scenario)?

Any amount of balls can be placed on a single side of the scale.

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    $\begingroup$ Looking at the 2 answers here so far, can we please have a clarification: Are we able to put more than one ball on each side of the scale? $\endgroup$
    – Chronocidal
    Apr 27, 2020 at 15:11

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A simple / basic answer, to provide an Upper Bound, is to use the scales

4 times

First, label our unknown balls "A", "B", "C", and "D". Next

Weigh them in pairs, and write down the winning pairs:
A&B v C&D
A&C v B&D
A&D v B&C

If, for demonstrative purposes, we assume that A=1, B=2, C=3 and D=4, then you would get this table:

1&2 = 3 v 3&4 = 7
1&3 = 4 v 2&4 = 6
1&4 = 5 v 2&3 = 5 : If we get the other 2 first, then we can skip this one!

We can then tally up each ball by "Heavy" versus "Light"

1: Light / Light / Draw
2: Heavy / Light / Draw
3: Heavy / Light / Draw
4: Heavy / Heavy / Draw

This means that our final step is

Compare the 2 balls with 1 "Heavy" and 1 "Light", to work out which is 2 and which is 3

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  • $\begingroup$ With the update I have deleted my answer as it is no longer correct. This seems to be the optimal stratergy with the rules now being clarified :) $\endgroup$
    – Beastly Gerbil
    Apr 27, 2020 at 15:39
  • $\begingroup$ @BeastlyGerbil It's a pity that Puzzling isn't suitable for having a "sandbox" like Worldbuilding does to spot those queries slightly earlier ^_^' And, I certainly can't find a way to knock a number off at the moment $\endgroup$
    – Chronocidal
    Apr 27, 2020 at 15:54
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    $\begingroup$ There is no approach that guarantees success in fewer tests. Simple brute force on the three possible non-trivial first tests shows this. $\endgroup$
    – user3294068
    Apr 27, 2020 at 16:58

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