Ugh, I decant even

Question

You know those decanting-problems, where you have two jugs of different sizes, and a water fountain, and you need to get a specific amount of water into one of the jugs using the smallest number of transfers? I came up with a decanting problem like that, but then it turned out that there were two solutions tied for "smallest number of transfers".

In those solutions, how full did the larger jug end up?

CLARIFICATION: In problems of this sort, it doesn't matter which jug ends up containing the target volume.

Answer 1

For two solutions to be equally fast

a kind of symmetry is required. More precisely the empty part of the big jug must be symmetric to the filled part. In other words, the big jug will end up half full.

Explanation:

The big jug (volume $V$) can start either with $0$ or $V$ filling, which is already symmetric around $\frac{V}{2}$. In the first step one can transfer the volume of the small jug $v$ in or out of the big one, so its filling after the first transfer is either $v$ or $V-v$, which again has the same distance to the final filling.
In the next step one could either revert the previous one (which doesn't bring you closer to the required filling) or again withdraw or add a volume of $v$ to the big jug – unless the volume would exceed $[0, V]$. In the second case the big jug has now a filling of $V$ or $0$ and the small one of $2v - V$ or $V - v$, respectively. Again, one can see that the symmetry between around the fillings $\frac{V}{2}$ and $\frac{v}{2}$ are maintained.
Also in all further steps this symmetry can not be broken, because for every transfer there is an opposite operation on the "mirror image", that instead of pouring $n$ liters from jug A to jug B pours it the other way round.

Two symmetric paths leading to the same solution in the same number of transfers requires the two paths to meet. This is exactly the case in the symmetry axis, i.e. when the big jug is half full.

Example:

One can clearly see the symmetry showing as the amount of air in the left path being equal to the amount of water in the right path.

Answer 2

My second thought is that:

The larger jug will always be half full.

While I don't have a formal proof, I think that:

If there are two optimal solutions, then the goal must be to get half the value of the larger jug. One solution will be to fill the small jug and move it to the large jug until it's half full, and the second will be to fill the large jug once and empty it into the smaller N times, until half of it had been emptied.

For example, consider when:

The jugs are A at 4L and B at 1L and the target is 2L
Two optimal solutions are:
Fill A, move A to B, dump B, move A to B
And,
Fill B, move B to A, Fill B, move B to A**

If I have time I'll try to work on a more formal proof for why this would have to be the case.

Answer 3

My guess is

In one case it's empty, in another it's full

Because

You have two jugs, a smaller one, denoted X and a larger one Y (Y > X)
there are two solutions you can get to in two different 4-step ways:
1) x:0 - 0:x - x:x - 2x-y:y
2) 0:y - x:y-x - 0:y-x - y-x:0
for both solutions to have the same amount in the small jug, we need y-x=2x-y, e.g. y=1.5x

So I'm guessing the riddle is, take two jugs size 2,3 L and out 1L in the smaller one

I can't prove (yet?) that it's the only solution, but it's one anyway :)