[Combinatorics]
Problem: Derek starts at the point $(0,0)$, facing the point $(0,1)$, and he wants to get to the point $(1,1)$. He takes unit steps parallel to the coordinate axes. A move consists of either a step forward, or a $90^{\circ}$ right (clockwise) turn followed by a step forward, so that his path does not contain any left turns. His path is restricted to the square region defined by $0 \leq x \leq 17$ and $0 \leq y \leq 17$. Compute the number of ways he can get to $(1,1)$ without returning to any previously visited point.

Solution: Divide into cases according to the number of right turns Derek makes.

- There is one route involving only one turn: move first to $(0,1)$ and then to $(1,1)$.
- If he makes two turns, he could move up to $(0, a)$ then to $(1, a)$ and then down to $(1,1)$. In order to do this, $a$ must satisfy $1<a \leq 17$, leading to 16 options.



- If Derek makes three turns, his path is entirely determined by the point at which he turns for the second time. If the coordinates of this second turn point are $(a, b)$, then both $a$ and $b$ are between 2 and 17 inclusive, yielding $(17-1)^{2}$ possibilities.
- If Derek makes four turns, his last turn must be from facing in the $-x$-direction to the $+y$-direction. For this to be his last turn, it must occur at $(1,0)$. Then his next-to-last turn could be at any $(a, 0)$, with $1<a \leq 17$, depending on the location of his second turn as in the previous case. This adds another $(17-1)^{2}$ possibilities.
- It is impossible for Derek to make more than four turns and get to $(1,1)$ without crossing or overlapping his path.

Summing up the possibilities gives $1+16+16^{2}+16^{2}=\mathbf{5 2 9}$ possibilities.

Final answer: 529