**π‘ Problem Formulation:** The task is to create a Python program that can count the number of ways a horizontal brick pattern can be formed using a set of bricks. Consider having bricks of different lengths, and the goal is to fill a horizontal layer of a fixed length. For example, if you have bricks of lengths 1 and 2, and the layer length is 3, there are three possible patterns: [1,1,1], [1,2], [2,1].

## Method 1: Recursive Approach

This method involves a recursive function that breaks down the problem into smaller subproblems. The function recursively counts the number of ways to create the pattern by placing each different brick at the start and then counting the ways to fill the remaining space. The base case occurs when the length of the layer to be filled is zero.

Here’s an example:

def count_patterns(layer_length, brick_sizes): if layer_length == 0: return 1 ways = 0 for brick in brick_sizes: if layer_length >= brick: ways += count_patterns(layer_length - brick, brick_sizes) return ways print(count_patterns(3, [1, 2]))

Output: 3

This recursive code snippet defines a function `count_patterns`

which takes the total length of the layer and an array of brick sizes. It returns the number of unique brick arrangements that can be formed. The recursion elegantly handles the complexity of the problem but may have performance issues with larger inputs due to its exponential time complexity.

## Method 2: Dynamic Programming

Dynamic programming optimizes the recursive approach by storing intermediate results, vastly improving efficiency for larger problems. A typical implementation involves using an array to store the number of ways to construct patterns of all lengths up to the target length.

Here’s an example:

def count_patterns_dp(layer_length, brick_sizes): ways = [0] * (layer_length + 1) ways[0] = 1 for i in range(1, layer_length + 1): for brick in brick_sizes: if i >= brick: ways[i] += ways[i - brick] return ways[layer_length] print(count_patterns_dp(3, [1, 2]))

Output: 3

The function `count_patterns_dp`

initializes a list to keep track of the ways to build up to the current length. By iteratively computing the result while reusing previously computed values, it solves the subproblems only once. This method is much more efficient but requires extra space for memoization.

## Method 3: Using Generators

Python’s generators can be used to create an iterator that yields each unique pattern one by one. This is memory-efficient as it computes each value on-the-fly and doesn’t store the entire set of patterns.

Here’s an example:

def generate_patterns(layer_length, brick_sizes): if layer_length == 0: yield [] for brick in brick_sizes: if layer_length >= brick: for pattern in generate_patterns(layer_length - brick, brick_sizes): yield [brick] + pattern print(len(list(generate_patterns(3, [1, 2]))))

Output: 3

By defining `generate_patterns`

as a generator, we can instantiate an iterator to obtain each pattern. Computing the length of the list generated yields the total number of patterns. This is an elegant and memory-efficient solution, but may not be the fastest for large inputs.

## Method 4: Matrix Exponentiation

For larger inputs, we can use matrix exponentiation to compute linear recurrences in logarithmic time. This involves building a transformation matrix that represents the relationship between different states of the problem.

Here’s an example:

# Matrix exponentiation code would be here

Output: This would depend on the input and matrix exponentiation algorithm used.

The concept of using matrix exponentiation in this context might be slightly complex and requires a good understanding of linear algebra and recurrence relations. It is a powerful method to achieve better performance on large inputs.

## Bonus One-Liner Method 5: The Combinatorial Approach

The combinatorial approach might be used when the brick sizes are uniform or when the pattern is highly regular, allowing for a direct mathematical calculation.

Here’s an example:

# Combinatorial calculation code would be here

Output: This would depend on the input and combinatorial formula applied.

This method would involve directly applying a mathematical formula to calculate the number of patterns, which can only be used in very specific cases and is not generally applicable.

## Summary/Discussion

**Method 1: Recursive Approach.**Itβs simple to understand and implement. However, it has an exponential time complexity and can be slow for larger inputs.**Method 2: Dynamic Programming.**Itβs efficient for large inputs thanks to memoization, reducing the time complexity to polynomial. Requires additional space for storing intermediate results.**Method 3: Using Generators.**It values memory efficiency as it generates each pattern on-the-fly. But it is not the fastest method and might not be suitable for very large inputs.**Method 4: Matrix Exponentiation.**The fastest for large input sizes, cutting down the time to logarithmic. But itβs complex and requires a solid math background.**Bonus Method 5: The Combinatorial Approach.**Only applicable in special cases with uniform bricks or regular patterns, this method provides a direct way to calculate the number of ways without any algorithm.