Tricky Graph Theory Puzzle I ran into an intriguing puzzle on Reddit that I thought could use some attention.
You start with a 3x3 grid labeled with numbers 1-9 like this:
$$ 1 \text{ }\text{ }\text{ }2\text{ }\text{ }\text{ }4
\\ 5 \text{ }\text{ }\text{ }6\text{ }\text{ }\text{ }8
\\ 9 \text{ }\text{ }\text{ }3\text{ }\text{ }\text{ }7$$
To complete the puzzle, you must connect directed edges between the numbers so that: 


*

*The numbers connect in order from 1-9. There can be intermittent numbers when connecting them. For example, some puzzle may have the solution: 1, 3, 2, 9, 4, 3, 9, 4, 8, 5, 7, 8, 6, 7, 8, 2, 9.

*The edges cannot overlap each other.

*The vertices (the numbers) can have multiple edges connecting to it.

*You cannot go back over edges (no overlaps).


Here's a picture of a completed puzzle:

I have a couple of question relating to this problem:
If there are impossible puzzles, then what criteria determines if a puzzle is unsolvable? If I were to make it 4x4 or 5x5 or NxN, would the criteria extend to those? What proportion of puzzles for an NxN puzzle are solvable.
Here's the link to the Reddit post: https://www.reddit.com/r/math/comments/8b032u/need_help_for_an_eulerian_path_game_i_made/

Here's a possible counterexample: Consider the board
$$\begin{array}{ccc}
3& 5& 7\\
1& 9& 2\\
4& 6& 8
\end{array}$$
Let the first row be set $A$, the second row be set $B$, and final row be set $C$. 
First, notice that there can be at most $7$ edges crossing between $A$ and not-$A$ (you can check the possible cases for yourself), and similarly for $C$. So if you take a path from $1$ to $2$ that enters $A$, you use up $2$ of those crossings. 
Since there has to be at least $6$ crossings going into $A$ that deal with hitting $3$ $5$ and $7$ (you have to go into $A$ before each number, and out of $A$ after each number), if you went into $A$ between $1$ and $2$, you wouldn't have enough edges to accomplish this. Thus, you cannot go into $A$ between $1$ and $2$. Same for $C$. 
Thus, you have to start off your walk as $1\to9\to2$. But by doing this, you cut off the possibility of two of these edges between $A$ and not-$A$, leaving only $5$ possible edges, which isn't enough to hit everything in $A$.
Disclaimer: This answer was adapted from the linked reddit thread, and is not my own work.
