How many steps in a cycle of pouring water from one jug to another? If one jug has M units  and another has N units, (M even, N odd)
and the only operation is doubling the contents of the
lesser by pouring from the greater, how many operations
does it take to return to the original configuration?
(M+N-1)/2 will do it since there are only (M+N-1) pairs with total M+N,
but shorter cycles are also possible.
Given M, and N  how can one predict the number of pourings needed
to return,  without actually doing all the steps empirically?
Also, What is the MINIMUM number of units a jug has in the course of a cycle?
 A: Set $G:=M+N$ (which we know is odd). The operation here - on both jugs - is doubling $\bmod G$.  Looking just at the $M$ jug, you either have
$M\to 2M \quad$ or
$\begin{align}
M\to\  &G-2N \\ 
&= G-2(G-M) \\ 
&= 2M-G
\end{align}$
So the cycle length will divide the order of $2\bmod G$ (the smallest value $k$ such that $2^k\equiv 1 \bmod G$), which itself will divide the Carmichael function $\lambda(G)$.
The minimum cycle occurs when $M=2N$ for example $(M,N)=(6,3)$
If $G$ is prime and  $2$ is a primitive root $\bmod G$, and assuming the jugs are labelled, you will have a full cycle of all $G{-}1$ possibilities for every starting configuration.
In the case $G=7$, we have that $2$ is not a primitive root, giving cycles of length $3$:

(here the blue squares are valid starting possibilities as defined, again taking the jugs as labelled)

For your example $M=104, N=47$, we have $G=151$ which is prime. The order of $2 \bmod 151$ will divide $151{-}1=150$ by Euler's theorem. We'd need to check the values of $2^i \bmod 151$ to find the cycle length; depending on what tools you have available, it's probably just as easy to run the calculation on $M$.
$104 \to 57 \to 114 \to 77 \to 3 \to 6 \to 12 \to 24 \to 48 \to 96 \to 41 \to 82 \to 13 \to 26 \to 52 \to 104$
So when $M\le 75$ - meaning $M<N$ - the next value is $2M$, otherwise the next value is $2M-151$ as I explained before.
I don't know of any particular shortcut that will give you the minimum values, unless you have a full cycle of $G-1$ (in which case the minimum is $1$).

For $G=185$, we have $185=5\times 37,$ then $\lambda(185) = {\rm lcm}( \lambda(37),\lambda(5)) = {\rm lcm}(36,4) = 36$ So we expect a cycle length of (or dividing) $36$ (marked jugs). Also, for initial states with multiples of $37$, we'd expect a cycle length of $4$. 
Checking the cycle on $1$, we indeed get a cycle length of $36$ (the second half just being the same as the first but with jug quantities reversed). So there should be $5$ such cycles, since each cycle runs the the full set of congruences $\bmod 37$ (coprime to $37$) and there are $5$ of each such value in the range. And then the $4$-cycle on multiples of $37$ will complete the set, $5\times 36+4=184$.
The minimum value in the cycle will obviously not be a member of any other cycle. Also in the case of $G=185$ you know that $5$ and $37$ will be the smallest member of whichever cycles they are in, since they are prime divisors of $185$ and all quantities in that cycles will be divisible by those starter primes. But in general finding the minimum of every cycle is not quick. I believe it's always prime (or $1$).
