<s>[INST] Minimum number of steps in water measure problem I've been struggling with this problem for a while and gone through questions about the "Water Jug Problem/Puzzle". 
A person wants to have $2$ separate $1$ L measures of water at the same time. However the only measures she has are for $6, 10$ and $15$ L. Show how this could be done with the minimum number of steps without marking the measures or using any container other than the original large beaker of water. The only steps allowed are filling or emptying a measure or transferring water from one measure to another.
I've read a question here which is pretty similar (also involving minimum operations) to this but only used $2$ measures/jugs instead of $3$. I didn't understand how I could adapt it into my problem. It would be a great help if someone could explain in simpler terms...
 [/INST] Solutions in $9$ steps (without proof of optimality yet): 
denote jugs as $a$($6$L), $b$($10$L), $c$($15$L); then: 
\begin{array}{|c|c|c|c|c|}
\hline
\# & move & a & b & c \\
\hline
1) & \bullet \rightarrow a & 6 & - & - \\
2) & \bullet \rightarrow c & 6 & - & 15 \\
3) & c \rightarrow b & 6 & 10 & 5 \\
4) & b \rightarrow \bullet & 6 & - & 5 \\
5) & c \rightarrow b & 6 & 5 & - \\
6) & a \rightarrow c & - & 5 & 6 \\
7) & \bullet \rightarrow a & 6 & 5 & 6 \\
8) & a \rightarrow b & 1 & 10 & 6 \\
9) & b \rightarrow c & 1 & 1 & 15 \\
\hline
\end{array}
\begin{array}{|c|c|c|c|c|}
\hline
\# & move & a & b & c \\
\hline
1) & \bullet \rightarrow a & 6 & - & - \\
2) & \bullet\rightarrow b & 6 & 10 & - \\
3) & b \rightarrow c & 6 & - & 10 \\
4) & a \rightarrow b & - & 6 & 10 \\
5) & \bullet \rightarrow a & 6 & 6 & 10 \\
6) & a \rightarrow c & 1 & 6 & 15 \\
7) & c \rightarrow b & 1 & 10 & 11 \\
8) & b \rightarrow \bullet & 1 & 0 & 11 \\
9) & c \rightarrow b & 1 & 10 & 1 \\
\hline
\end{array}

Actually there are up to $12$ valid moves at each stage:
 $\bullet \rightarrow a$, $\;\bullet \rightarrow b$, $\;\bullet \rightarrow c$;
 $a \rightarrow b$, $\;a \rightarrow c$, $\;a \rightarrow \bullet$;
 $b \rightarrow a$, $\;b \rightarrow c$, $\;b \rightarrow \bullet$;
 $c \rightarrow a$, $\;c \rightarrow b$, $\;c \rightarrow \bullet$;
so there are not more than $12^{8}\approx 430\:000\:000$ chains of $8$-step moves (not too much as for computer check).
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