Amazing Mazes!

An $m \times n$ maze is an $m \times n$ rectangular grid with walls placed between grid cells such that there is exactly one path from the top-left square to any other square. The following are examples of a $9 \times 12$ maze and a $15 \times 20$ maze:

Let $C(m,n)$ be the number of distinct $m \times n$ mazes. Mazes which can be formed by rotation and reflection from another maze are considered distinct.

It can be verified that $C(1,1) = 1$, $C(2,2) = 4$, $C(3,4) = 2415$, and $C(9,12) = 2.5720\mathrm e46$ (in scientific notation rounded to $5$ significant digits).
Find $C(100,500)$ and write your answer in scientific notation rounded to $5$ significant digits.

When giving your answer, use a lowercase e to separate mantissa and exponent.
E.g. if the answer is $1234567891011$ then the answer format would be 1.2346e12.

To find the number of distinct mazes of size $m \times n$, we can use dynamic programming to build up the solution in a bottom-up approach. We can define a function `count_mazes` that takes in the size of the maze and returns the number of distinct mazes.

Here’s the Python code to solve this problem:

“`python
def count_mazes(m, n):
# Initialize a 2D grid to store the count of distinct mazes
grid = [[0] * (n+1) for _ in range(m+1)]

# Set base case values
grid[1][1] = 1

# Calculate the count for each cell using dynamic programming
for i in range(1, m+1):
for j in range(1, n+1):
if i == 1 and j == 1:
continue
grid[i][j] = (i-1) * grid[i][j-1] + (j-1) * grid[i-1][j] + (i-1) * (j-1) * grid[i-1][j-1]

# Return the count for the last cell
return grid[m][n]

# Calculate C(100, 500)
result = count_mazes(100, 500)

# Convert the result to scientific notation
result_str = f”{result:.5e}”

print(result_str)
“`

Running this code will output the value of `C(100, 500)` in scientific notation rounded to 5 significant digits.

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