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The problem is stated as follows:

With a balance scale and six given weights; 1g, 2g, 3g, 4g, 5g and 6g, is it possible to make sure the labels on the weights are put on correctly only using the scale twice? Assume there is no way to identify weights by apperance, and that only the labels might have gotten in the wrong place (ie there is exactly one of each).

I've come as far as realizing a good way to start would be to balance 6g towards 1+2+3g. If balanced, the 6g weight must be correct, and 1g, 2g, 3g can only have switched labels with each other. It gives no direct info on 4g and 5g, but they can only have switched with each other.

With that in mind, the next step must include exactly 2 out of 1g, 2g, 3g but not both on the same side of the scale (as we still wouldn't know if they have each other's labels in that case), and atleast one of 4g, 5g. Furthermore, if we choose 1g and 2g we might confuse ourself and not realizing that the switch 1g->2g->3g->1g has been made, and same thing with 2g+3g, therefore the weight labeled 1g on one side and and 3g on the other is the best option. But where do I go from here? I've tried so many things but always end up with some sort of ambiguity. That is not a solid proof that it is not possible, however. Do anyone have a way of succeeding in two uses of the scale, or a way of prooving that it is not possible?

There are many ways to figure it out in $3$ weighings. For example, after confirming that $1+2+3=6$ you could see if $2+3>4$, and if that holds, you know that $1,4,5,6$ all are correct, and you just need to confirm that $3>2$ to make sure the last two are correct.

Another pair of interesting weighings is to confirm $4>1+2$ and $6>1+4$. If so, you know $1,2,4,6$ are all correct, so now you just need to confirm $5>3$

I suspect two weighings is not going to be sufficient. One way to try and prove that is to verify that all possible pairs of equations have multiple solutions, where each equation matches the possible outcome of a possible weighing:

$a= b+c tag{**}$

$a= b+c+d tag{***}$

$a+b=c+d tag{bla}$

$a+b= c+d+e tag{*}$

$a>b tag{nope}$

$a> b+c tag{***}$

$a+b>c tag{bla}$

$a+b>c+d tag{bla}$

$a+b>c+d+e tag{**}$

$a+b+c>d tag{superbla}$

$a+b+c>d+e tag{bla}$

$a+b+c>d+e+f tag{bla}$

$a+b>c+d+e+f tag{nope}$

Actually, there are a bunch more possible results from weighings, e.g. $a + b + c + d > e$, but obviously that one doesn't constrain the possible solutions at all, since that is always going to be true when putting 4 objects on one side and 1 on the other side.

Indeed, some equations have fewer solutions than others, and are therefore more 'interesting' equations ... and the more stars, the more constraining, and hence the more interesting. Hence, a combination of (more) starred equations is more likely to have a single solution than non- or less-starred. So those are the kinds of combinations I would explore first if you hope to find one that will lead to one unique solution. The $(bla)$ ones are hardly constraining at all, so I would wait with those. (and the superbla one is always true except for one case ...)

Now, it is not just a matter of picking two equations (you can pick the same type of equation twice, by the way, just like I picked $4 > 1+2$ and $6 > 1+4$ at the start of my post), but you also have lots of options of where the variables in 1 equation get repeated in the second equation. For example, picking:

$a > b + c$ and $b+d=e$

is a different pair from:

$a > b + c$ and $b +d=c$

Same two types of equations, but different distribution of variables ... and so different solution constraints.

... so ... there are lots of possible pairs of types of equations! And if there is no solution for any pair of equations, then you need to to verify that for all possible pairs! Probably far too many to do by hand ... a computer seems to be called for.

Still, there are some ways to cut down the number of possible pairs. For example, notice that for any one equation, a variable can end up in one of three spots: LHS, RHS, and not in the equation at all. (and, if you deal with an equality, then obviously LHS and RHS can be switched). Now, suppose you have two variables that end up in the same location for both equations. Well, then obviously you can swap the values for those variables without affecting the truth of the equations, and hence you have multiple (or zero) solutions for that pair. Hence, there is no need to check any pair of equations with two variables in the same spot. For example:

$a = b + c$ and $a+d>e+f+c$

have $e$ and $f$ in the same spot for both equations (both not in the first equation, and both in the RHS for the second). Hence, you can swap the values of $e$ and $f$ and nothing will change. So, this is not a pair that will result in a unique solution, and so no need to explore this pair.

Notice that this also means that you never want $4$ variables in the same spot for any one equation, since that means that there will always be at least two variables in the same spot for the second equation. This is why I tagged the equation $a+b=c+d+e+f$ with $(nope)$; it seems like an interesting equation (indeed, it forces $a$ and $b$ to be $5$ and $6$), but since $c,d,e,f$ all end up in the same spot, this equation can never be paired up with some other equation resulting in a unique solution. Likewise, $a>b$ is a $(nope)$: $c,d,e,f$ all end up in the same spot (being not in the equation at all)

Now, I already tried a bunch of combinations of 'interesting' equations, and couldn't make anything to work ... so my suspicion is that there is no combination of two weighings that gives a unique solution. But to prove that, yeah, you'll need to go through all possible combinations and variable distributions: I'd use a computer to do that!

I presume that you can distinguish only $3$ cases per weighing: the left pan is lighter, the left pan is heavier, the two pans are the same. If so, there are only $3^2 = 9$ possible outcomes of two weighing. On the other hand, $6$ labels could be rearranged in $6! = 720$ ways so it is not possible. $n$ weighings allow $3^n$ outcomes. The smallest power of $3$ which exceeds $720$ is $6$. This does not prove that it is possible with $6$ weighings, maybe not since $3^6 = 729$ which is only slightly bigger. However, it does prove that fewer than $6$ weighings is not sufficient.