Is the solution to this holiday puzzle unique? I read the following question on internet. 

Start at 2011. By moving through the maze and doing any arithmetic operations you encounter, exit the maze with a result of 2012. You may pass through an operation several times, but not twice in a row.


Out of curiosity, I'd like to ask the following two questions:


*

*One solution is on the site. Is the solution to this problem unique?

*How can we find a solution?
 A: Note that if a fractional value is ever reached (by applying $[\div 2]$ to an odd number), no integer is subsequently reachable.  Therefore, the first move must be $[+7]$, giving $2018$.  Similarly, since $2012$ is not divisible by three, the last move must be $[-5]$, starting from $2017$.  The walk from $2018$ to $2017$ must consist of any number of the following "loop" operations, taken in any order:


*

*$A=[\times 3][-5],\qquad A(x)=3x-5$

*$B=[-5][\times 3],\qquad B(x)=3x-15$

*$C=[\div 2][+7], \qquad C(x)=\frac{1}{2}x+7$; apply to even $x$ only

*$D=[+7][\div 2], \qquad D(x)=\frac{1}{2}x+\frac{7}{2}$; apply to odd $x$ only.


The shortest walk from $2018$ to $2017$ takes 13 such operations: $DAC^3A^4D^2C^2$, corresponding to a solution with 28 moves:
$$
\begin{eqnarray}
2011&\xrightarrow{[+7]}&2018\xrightarrow{C}1016\xrightarrow{C}515\xrightarrow{D}261\xrightarrow{D}134\xrightarrow{A}397 \\
&\xrightarrow{A}&1186\xrightarrow{A}3553\xrightarrow{A}10654\xrightarrow{C}5334\xrightarrow{C}2674 \\
&\xrightarrow{C}&1344\xrightarrow{A}4027\xrightarrow{D}2017\xrightarrow{[-5]}2012.
\end{eqnarray}
$$
The solution is not unique; the next-shortest walk is of length 16: $C^2D^2C^3B^2ABDACAD$.  Not surprisingly, there are also families of solutions of arbitrary length.  For instance, $C(14)=14$, and so any solution passing through $14$ can be padded with any number of copies of $C$. The shortest such family is $C^2A^2B^2A^2C^{k}D^3CDACD^2C^2$.
