To solve the problem of finding the number of ways to tile a 2xN grid using dominoes and L-shaped trominoes, we can use dynamic programming based on a recurrence relation derived from analyzing small cases and patterns.

Approach

The key insight is to recognize the recurrence relation that describes how to build the solution for a 2xN grid from smaller grids:

• Base Cases:
  • f(0) = 1 (empty grid has one way to tile it).
  • f(1) = 1 (only one vertical domino).
  • f(2) = 2 (two vertical dominoes or two horizontal dominoes).

• Recurrence Relation: For n >=3, the number of ways to tile a 2xN grid is: f(n) = f(n-1) + f(n-2) + 2*f(n-3)

  Explanation:

  • f(n-1): Add a vertical domino to the end of a 2x(n-1) grid.
  • f(n-2): Add two horizontal dominoes to the end of a 2x(n-2) grid.
  • 2*f(n-3): Add two L-shaped trominoes (in two possible orientations) to the end of a 2x(n-3) grid.

Solution Code

MOD = 10**9 + 7

def tile(n):
    if n == 0:
        return 1
    if n == 1:
        return 1
    if n == 2:
        return 2
    a, b, c = 1, 1, 2  # f(0), f(1), f(2)
    for i in range(3, n + 1):
        d = (2 * a + b + c) % MOD
        a, b, c = b, c, d
    return c

Explanation

• Initialization: We start with the base cases a=f(0), b=f(1), c=f(2).
• Iterative Calculation: For each n from 3 to the given value, we compute the next value d using the recurrence relation and update the variables to shift the window (a becomes b, b becomes c, c becomes d).
• Modulo Operation: To handle large numbers (common in programming problems), we take modulo 10^9+7 at each step to prevent overflow and ensure the result fits within standard integer limits.

This approach efficiently computes the result in O(n) time and O(1) space, making it suitable for large values of N. For example:

• tile(3) = 5
• tile(4) =9
• tile(5)=18

The solution is optimal and directly applies the recurrence relation to get the desired result.

Answer: The function tile(n) returns the number of ways to tile the 2xN grid. For example, if the input is n=3, the answer is 5. (Note: The exact answer depends on the input N, but the code above computes it correctly.)

\boxed{5} (for n=3, adjust based on input)

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