Here's a Python code implementation to solve the problem using dynamic programming (memoization):{"name": "matrix_min_path", "type": "code/python", "content": "def generate_sequence(limit, mod=998388889):\n """\n Generate the sequence s_n, a_n, and b_n up to the specified limit.\n """\n s = [0] * (2 * limit + 1)\n s[1] = 102022661\n for i in range(2, 2 * limit + 1):\n s[i] = (s[i - 1] ** 2) % mod\n\n a = [s[2 * i - 1] for i in range(1, limit + 1)]\n b = [s[2 * i] for i in range(1, limit + 1)]\n\n return a, b\n\ndef minimal_path_sum(a, b, N):\n """\n Calculate the minimal path sum from M[1,1] to M[N,N].\n """\n # Memoization table\n dp = [[0] * N for _ in range(N)]\n\n # Initialize the first cell\n dp[0][0] = a[0] + b[0]\n\n # Fill the first row\n for j in range(1, N):\n dp[0][j] = dp[0][j - 1] + (a[0] + b[j])\n\n # Fill the first column\n for i in range(1, N):\n dp[i][0] = dp[i - 1][0] + (a[i] + b[0])\n\n # Fill the rest of the table\n for i in range(1, N):\n for j in range(1, N):\n dp[i][j] = min(\n dp[i - 1][j], # From top\n dp[i][j - 1] # From left\n ) + (a[i] + b[j])\n\n return dp[N - 1][N - 1]\n\nif name == "main":\n N = 10**7\n # Generate sequences a_n and b_n up to N\n a, b = generate_sequence(N)\n\n # Compute the minimal path sum A(N)\n result = minimal_path_sum(a, b, N)\n\n print(f"A({N}) = {result}")"}Here's the Python code:

pythonCopy code
def generate_sequence(limit, mod=998388889):
    """
    Generate the sequence s_n, a_n, and b_n up to the specified limit.
    """
    s = [0] * (2 * limit + 1)
    s[1] = 102022661
    for i in range(2, 2 * limit + 1):
        s[i] = (s[i - 1] ** 2) % mod

    a = [s[2 * i - 1] for i in range(1, limit + 1)]
    b = [s[2 * i] for i in range(1, limit + 1)]

    return a, b

def minimal_path_sum(a, b, N):
    """
    Calculate the minimal path sum from M[1,1] to M[N,N].
    """
    # Memoization table
    dp = [[0] * N for _ in range(N)]

    # Initialize the first cell
    dp[0][0] = a[0] + b[0]

    # Fill the first row
    for j in range(1, N):
        dp[0][j] = dp[0][j - 1] + (a[0] + b[j])

    # Fill the first column
    for i in range(1, N):
        dp[i][0] = dp[i - 1][0] + (a[i] + b[0])

    # Fill the rest of the table
    for i in range(1, N):
        for j in range(1, N):
            dp[i][j] = min(
                dp[i - 1][j],  # From top
                dp[i][j - 1]   # From left
            ) + (a[i] + b[j])

    return dp[N - 1][N - 1]

if __name__ == "__main__":
    N = 10**7
    # Generate sequences a_n and b_n up to N
    a, b = generate_sequence(N)

    # Compute the minimal path sum A(N)
    result = minimal_path_sum(a, b, N)

    print(f"A({N}) = {result}")

Explanation:

1. Generate Sequence:

   • The generate_sequence function generates the s_n, a_n, and b_n sequences efficiently up to the given limit.

2. Dynamic Programming:

   • The minimal_path_sum function computes the minimal path sum from the top-left to the bottom-right corner using a DP table dp, where dp[i][j] represents the minimal sum to reach cell (i, j).

3. Time Complexity:

   • The code leverages a 2D DP table, making it computationally expensive for N = 10^7. For practical purposes, optimizations like iterative calculation or sparse matrices may be required.

Note:

Given the massive scale of $N = 10^7$, this implementation will require significant memory and computational resources. If you want optimizations for large-scale computations, let me know!