Jaunt Through the Garden (Problem J, CS3233 Midterm Contest 2025, 2025)
Time limit: 1 sec
Memory limit: 1024 MB

# Problem Statement

**TL;DR:** Let $T = \{(r, c) \mid 1 \le r, c \le N\}$. Count the number of distinct functions  
$f : T \rightarrow \{0, 1, 2\}$ such that $\exists X \subseteq T,\ T \cap f^{-1}(1) = f^{-1}(1),\ T \cap f^{-1}(2) = \emptyset,$  
$(r, c) \in X\ \Longrightarrow\ (r = N \lor (r + 1, c) \in X) \land (c = 1 \lor (r, c - 1) \in X)$.

---

There is a square garden consisting of $N \times N$ tiles in the School of Computing. Each tile is either empty, has a blue flower, or has a red flower. A cat can walk on the border of the garden or between the sides of two adjacent tiles, but not on the tiles themselves.

SoCCat (the cat) is a perfectionist, and considers a path through the garden starting from the top-left corner and ending at the bottom-right corner to be *good* if and only if:

- SoCCat always moves optimally towards the bottom-right corner, i.e., the path only moves right or down.
- Whenever SoCCat looks to the left, all tiles **(from SoCCat’s direct line of sight)** are either empty or have a blue flower, and whenever SoCCat looks to the right, all tiles **(from SoCCat’s direct line of sight)** are either empty or have a red flower. In other words, if we consider the path as splitting the garden into two parts, the top-right part must contain only empty or blue flower tiles, and the bottom-left part must contain only empty or red flower tiles.


For example,  
![IMAGE](CS3233_2025_Midterm-jaunt_through_the_garden-1.png)

The orange path and the green path in the garden on the left are examples of a *good* path. On the other hand, the pink path is not a good path, since there is a tile with a red flower when SoCCat looks to the left when traversing the path (in the garden on the left), or SoCCat doesn’t move optimally (in the garden on the right). Note that there are no good paths in the garden on the right.

SoCCat hasn’t seen the garden yet, and so wonders: among the $3^{N \times N}$ possible gardens, how many possible gardens are there such that there exists a good path through the garden? Since the answer can be very large, output the answer modulo $10^9 + 3233$.

# Input

The input contains a single integer $N$ ($2 \le N \le 2500$) denoting the size of the garden.

# Output

Output a single integer denoting the number of possible gardens such that there exists a good path through the garden, modulo $10^9 + 3233$.

# Sample Input #1
```
1
```
# Sample Output #1
```
3
```
# Sample Input #2
```
2
```
# Sample Output #2
```
52
```
# Sample Input #3
```
3
```
# Sample Output #3
```
4032
```
# Sample Input #4
```
4
```
# Sample Output #4
```
1314816
```

Sample inputs:
1

2

3

4

Sample outputs:
3

52

4032

1314816

# Solution Analysis

For an arbitrary garden configuration of red and blue flowers, there can be many possible good paths. To be able to count the number of good garden configurations, we first build a bijection between a garden configuration and a unique good path.

We define the _canonical path_ as a good path where we always prioritize going down, then going right. In the first example given in the statement, The green and orange paths are good paths, but both are not canonical. The canonical path is the path that goes in the following direction sequentially: down, right, down, down, right, down, down, right, right, right.

If you observe canonical paths of several grids, you can observe several properties:

- whenever a canonical path goes right followed by going down, there must be a red flower on this corner. Otherwise, if there is no red flower, then the canonical path will still go down instead of going right for this particular step.
- aside from the constraint above, all flowers on the bottom-left split must be red flowers, and all flowers on the top-right split must be blue flowers.

Since we have a unique canonical path for every good configuration, we can now try to count the number of good configurations by counting canonical paths. We use DP to count this: let $f(i, j, k)$ be the number of gardens with $i$ rows and $j$ columns, and the last movement in the canonical path is $k \in \{\text{RIGHT}, \text{DOWN}\}$.

The transition for the DP is to extend the canonical path DOWN (we multiply by $2^j$ or $2^{j - 1}$, since there are $j$ new cells on the bottom-left split, and depending on $k$ the rightmost cell must be a red flower or can be an empty cell) or RIGHT (the transition is similar). This yields $\mathcal{O}(1)$ transition, thus a $\mathcal{O}(N^2)$ dynamic programming solution.

There is also an explicit formula that you can compute in $\mathcal{O}(N)$ which is left as an exercise to the reader.


Special judge (C++ scorer):